Lab Power Supply - The Lost Current - Page 1
Forum section: electronics / projects

author: radoczi94
author_rank: Contributor
timestamp: January 30, 2018, 06:17:37 pm
content: Hi.I built the power supply shown on the schematic.It looks like working as a treat, but not really.The CV mode is ok, the CC is what causing the nightmare. I started the testing with a massive short circuit on the output, slowly turning the current pot up. The comparator doesn't wanted to switch over. I probed around for a while, found out, there isn't enough voltage drop across the shunt (R35-R25) resistors.I started to question the quality of those, modified the voltage divider on the comparator input, then came the idea: measuring the current thru the shunts.Okay, now, here comes the funny part.I measured 1A on the shunt, while on the output was 3A. I lost track of 2amps!!! I measured over and over again with 4 different DMMs and analog meters, the results were the same. It turned out, the lost 2A is not a fixed value, it depends on the load on the output, it's about 2/3 of the output current. I removed the shunts and measured resistance to see, if there is anything weird, but only measured something in the 200MOhms range, which is basically nothing. Anyone has any idea, what should I try, or measure? Because I'm out of ideas.Thanks.

author: C
author_rank: Super Contributor
timestamp: January 30, 2018, 07:14:03 pm
content: just a quick lookIf you have a power supply that has a CV mode & a CC mode, you have a OR.Output = the lesser of (set voltage or set current)

author: Kilo Tango
author_rank: Regular Contributor
timestamp: January 30, 2018, 07:17:43 pm
content: Hi,most power circuits I have come across measure current in the +ve rail, otherwise it means your -ve output is different to your circuit common terminal, the thick blue line, so which one is earth ?.You can't loose current (easily !) , it has to flow somewhere. if it is flowing out of X6-2 and back into X6-1, it has to make its way back to the input along the thick blue line, which means either thro your shunt resistors or by another path. By the way you don't need to remove the shunt resistors to check them, there isn't anything significant in the circuit to disturb their measurement easily, also if they now measure 200M then they are blown.I'm still trying to get my head round the Q4 emitter follower pulling up the bases of the 2 2N3055's and both op amps are running without any feedback ?. is this a published circuit ?Cheers Ken

author: radoczi94
author_rank: Contributor
timestamp: January 30, 2018, 08:51:43 pm
content: I know, the schematic is a bit annoying, working on a version for better readability. You can find many versions of this schematic on the web, even an ebay kit is based on this PSU.The IC2 gets feedback directly from the output, IC2 has no feedback, since it works as a comparator.

author: Damianos
author_rank: Frequent Contributor
timestamp: January 30, 2018, 10:39:46 pm
content: How you measured 1A on 200MOhm resistance? Is there a 200MV source?Also how you measured the output current?On the circuit:- What is the role of T1?- What about C16?- Missing capacitors on D4 and D5 (bases of respective transistors)...- The reference voltage (VREF 8V) is relative to the other side of the shunt, so the current limit adjustment will vary with load current...- The 100nF capacitors, here and there, are like the designer had a bag of them, without knowing where to use them! I'm wondering if any calculation of time-constants / filter-poles had been made.- ...

author: C
author_rank: Super Contributor
timestamp: January 30, 2018, 10:49:58 pm
content: Quote from: radoczi94 on January 30, 2018, 08:51:43 pmIf IC2 is a comparator then it would be a switcher supply.IC2 feedback includes Q4,Q5 & Q6If you look at T1, as current drop changes across current sense resistors it's resistance changes effecting output.As R17 voltage changes due to output voltage change, IC2's output changes.This is not an OR, it is a voltage modulated by current.With this happing then remaining parts of circuit has to compensate for this error or you have a voltage modulated by current output.

author: radoczi94
author_rank: Contributor
timestamp: January 30, 2018, 11:36:25 pm
content: Quote from: Damianos on January 30, 2018, 10:39:46 pmI did not measured 1A on 200MOhm resistance, I removed the shunts and tried to measure resistance on the pcb to see, if there is some sort of short circuit or something.T1 transistor shorts out the output of IC2, when the negative supply dies, which happens almost instantly, when the mains goes off.Whitout that, after switching off, the output would jump up to the maximum voltage.Honestly, I have no idea, what C16 supposed to do.I placed caps there on the PCB, just missing on the schematic.Quote from: C on January 30, 2018, 10:49:58 pmSorry, I meant IC1 has no feedback.Which is actually not true, because it has a cap, so its an integrator.My bad.I do not know if this thing oscillates or not, I do not have access to a scope to see what is happening.One thing came into my mind: if I tear off IC1 from the circuit, and put a reasonable load on the output, I would be able to disqualify a lot of theories. If the missing 2A comes out, then something crazy happens with IC1, if not, then it's an another kind of problem.Anyway, made a new schematic, I hope, it's better.

author: C
author_rank: Super Contributor
timestamp: January 31, 2018, 03:47:20 am
content: Most people think of Ground as something that does not move.In latest where you labeled GND changes based on current and needs to float.missed Q7 acting like a switch.If you see problems on power off you could also have problems at power up.I would say best cure is to control power up and down of power supplies and circuit.Think of it this way on power down. Circuit is losing power, output is dropping and circuit reacts by turning on harder using even more power and kills output in process.If you replace D3 with a switch controlled by -5v supply getting to say -4v then when power loss happens, V+ is no longer one diode drop from X1-1.All V+ current demands then drops V+ faster. The harder IC2 output goes positive the faster output dies.Controlling Vref is another way to control output on power up/down.Many ways to do something.I just do not like that circuit.

author: radoczi94
author_rank: Contributor
timestamp: January 31, 2018, 08:07:45 am
content: You may not like that solution, but it works perfectly since the first power up. If you look, D1-D2 and C1-C2 makes a Villard voltage duobler. At the power up, the negative supply needs some time to stabilise,it takes a few sine cycle. When powering off it goes away rapidly, because there is no large capacity like on the positive side. That part of the circuit works just fine.

author: xavier60
author_rank: Super Contributor
timestamp: January 31, 2018, 08:29:22 am
content: It's odd that the Emitter resistors are 0.2 ohms and the shunt resistors are 0.22 ohms. I would run the output into an ammeter or a shunt resistor to know the output current then measure the voltage drop across all of the low value resistors in the current path to make sense of it.Maybe this is what you have been doing?

author: radoczi94
author_rank: Contributor
timestamp: January 31, 2018, 08:36:29 am
content: Quote from: xavier60 on January 31, 2018, 08:29:22 amI used the same kind of resistors for the shunt and the emitter resistors. Measured the voltage drop across the emitter resistors too, they had around 1,5 A flowing thru each. Nothing odd there.Tried the same kind of resistors as a load on the output, they had 0,65V voltage drop at 3A, so the resistors are correct value.

author: xavier60
author_rank: Super Contributor
timestamp: January 31, 2018, 08:51:54 am
content: Quote from: radoczi94 on January 31, 2018, 08:36:29 amAnd the voltage drop across R25 or R23 indicates only 1 amp?

author: radoczi94
author_rank: Contributor
timestamp: January 31, 2018, 08:54:58 am
content: Quote from: xavier60 on January 31, 2018, 08:51:54 amYes. And the series connected current meter shows that too.

author: xavier60
author_rank: Super Contributor
timestamp: January 31, 2018, 09:00:15 am
content: Although you have checked, there must be a short somewhere bypassing current around the shunt resistors. What supplies input power?Can you take some photos?

author: radoczi94
author_rank: Contributor
timestamp: January 31, 2018, 09:19:14 am
content: Quote from: xavier60 on January 31, 2018, 09:00:15 amThe power is supplyed by a 250VA transformer, a cooled rectifier bridge and 10kuF 100V caps.That board ind the front-middle is just only a reverse polarity protection board, 2 diodes and a fuse. It's not attached to the circuit right now. The small board on the left is a small supply for the relays and the fan control.And yes, the non marked wires have a reversed polarity on the rectifier board output,I corrected that before powering on.https://drive.google.com/drive/folders/1BI_EM4ImJGpWINuNeS3gs_Wgnjbidrzf?usp=sharing

author: xavier60
author_rank: Super Contributor
timestamp: January 31, 2018, 09:32:41 am
content: Wow! that is one serious power supply.All I can suggest, if you have not done so already, is to remove one of the shunt resistors and solder in PCB pins so that you can experiment and take measurements while the PCB is mounted.

author: radoczi94
author_rank: Contributor
timestamp: January 31, 2018, 11:11:02 am
content: Yeah, next time will make some wire loops for test points. Always learning.

author: Damianos
author_rank: Frequent Contributor
timestamp: January 31, 2018, 05:54:41 pm
content: Quote from: radoczi94 on January 30, 2018, 11:36:25 pmSo, the 1A is imaginary or real current?How you lost the 2A? How you know that the total was 3A?T1(Q7) and IC2: See the notes of the maximum ratings section in the Op-Amp datasheet.C16: see what does C14 and you will see that it, at least, cancels the compensation.Consider the current reference, noted above...QuoteThe IC1 is neither a comparator nor an integrator, it is intended to operate linearly by reducing the voltage setting when the current tends to be greater than the set limit...

author: radoczi94
author_rank: Contributor
timestamp: January 31, 2018, 09:50:19 pm
content: No, I measured that 1A series with the shunt resistors, 3A was present AT THE SAME TIME on the output, parallel(Intentionally).If IC1 oscillates, it may have tricked the dmm.I do not know which data should I look in the table, but it has unlimited short circuit protection in the temperature range, if you think abot that.Yes, it moves a little bit, but this was not meant to be a high precision thing. It can be solved by whacking in an another reference chip, tied to the fixed GND. I was aware of that, but probably it will be good enough for the girls I go out with.

author: xavier60
author_rank: Super Contributor
timestamp: February 01, 2018, 02:41:28 am
content: It wouldn't be too surprising if it is oscillating in CC mode because both op-amps are in the loop making the response rather complex.I see in other designs that either the CV op-amp or the CC op-amp can take sole control of the power transistors. The response of both loops can be individually optimized.

author: Damianos
author_rank: Frequent Contributor
timestamp: February 01, 2018, 11:16:56 am
content: I lost the communication!If someone gives me an algorithm, to decode the messages, it may be reestablished!

author: TimNJ
author_rank: Super Contributor
timestamp: February 02, 2018, 05:16:56 am
content: Try a "diode-OR" to switch between the CV and CC control loops. This is a common and effective way to implement constant current limiting on a "normally CV" power supply.Check out the attached schematic of an HP6214A lab power supply.Also check out this Keysight article:http://powersupply.blogs.keysight.com/2012/07/how-does-power-supply-regulate-its.html

author: xavier60
author_rank: Super Contributor
timestamp: February 02, 2018, 12:01:13 pm
content: There is a 5W 0.68 ohm resistor on the regulator PCB. I can't see the marking of the resistor next to it? Are these the shunt resistors?

author: radoczi94
author_rank: Contributor
timestamp: February 02, 2018, 09:51:56 pm
content: Quote from: TimNJ on February 02, 2018, 05:16:56 amThanks, will try it.Edit: One thing came into my mind. I am not shure, that I can put a diode in series with the CV opamp. What I could do is to move the diode on the CV opamp output (D7) and tie it's anode to the driver transistors' (Q4) base.Quote from: xavier60 on February 02, 2018, 12:01:13 pmYes, it was in there, when I took the picture. When I put those 0,68R resistors,then I realised, that the "missing" 2 amps is more or less linear with the output load current.

author: xavier60
author_rank: Super Contributor
timestamp: February 03, 2018, 02:52:04 am
content: Don't modify it until you have found a definite reason for the problems. Have you checked for oscillations?When you checked the regulator PCB for a possible short bypassing the shunt, did you have all of the wires connected?Quote from: radoczi94 on February 02, 2018, 09:51:56 pm

author: xavier60
author_rank: Super Contributor
timestamp: February 03, 2018, 06:25:49 am
content: Also, does the discrepancy in current readings exist in CV mode? The loop should be stable in CV mode.

author: radoczi94
author_rank: Contributor
timestamp: February 03, 2018, 08:36:44 am
content: Quote from: xavier60 on February 03, 2018, 02:52:04 amThe wires were connected, just removed the shunts, when I was checking for shorts. I'm not sure, how to check for oscillations without a scope. My idea is to give this thing a try without th CC opamp installed.Quote from: xavier60 on February 03, 2018, 06:25:49 amYes, it is stable in CV mode.The CC mode output current is pretty stable too. There are some smaller deviations, like 200mA or so, typically downwards. But it is stable, the meter does not moving at all, the current value just deviating a little bit every time the CC kicks in.

author: xavier60
author_rank: Super Contributor
timestamp: February 03, 2018, 09:32:29 am
content: I was manly curious to know if there is a difference between output current and shunt current in CV mode also.

author: radoczi94
author_rank: Contributor
timestamp: February 03, 2018, 10:08:33 am
content: That is one thing I did not measured, at least I don't remember. Will do that too, if I have time.You may probably noticed, my english is not the best, sorry for that.I'm trying.

author: C
author_rank: Super Contributor
timestamp: February 03, 2018, 04:21:45 pm
content: Quote from: radoczi94 on February 03, 2018, 10:08:33 amAn doing good, Very simple, If you have problems to or from english just say so.oscillations without a scopeWith no scope the best you can do is trying to measure AC with meter.On DC mode with meter, only last digit value should change.Keep in mind that a Lab Power Supply is to help you when building new circuits & protect the circuit.Think of what happens when you are building a PWM dimmed LED circuit. At some point in this design, the power supply will see a load changing from almost no current to very high current very quickly. With volt meter, you can not see total effect on lab power supply.If you had a scope, a good test of power supply is a test load is one that changes a lot.Could help you a lot if you created a spreadsheet to record your measurements in. You should measure circuit many times with different loads.All these measurements will help you understand how the circuit is working and spot where it is not working as you think it should..You are having problems with CC mode, Would be good to have all the measurements so you can compare before change to after a change.It might help you a lot if you make your schematic a lot better.You havePower paths to output.Power to run control circuit.Reference block.Output voltage sense.Output current sense.Voltage control circuit.Current control circuit.Output control circuit.Some quick cut & paste moves could make circuit easer to under stand.One way is to make top of schematic output & control with bottom power for control & reference..Make a box around circuit that supplies power to control circuit.Box top is between D3 & X-1Box right is between x2-2 & LED1Move this box down below where X1-3 goes across schematic.Move Reference block. down also where X1-3 goes across schematic.Box left is between IC1 & X3-3Box right is between X4-2 & R19What I am calling Output Control is area above R26.Voltage SenseR21 & R27 could be vertical next to C17Think these changes and a few more will change the schematic from cram it in to a more logical schematic.A more logical schematic can let you see problem areas.===

author: radoczi94
author_rank: Contributor
timestamp: February 05, 2018, 09:43:49 am
content: QuoteThanks for the idea, will try that. My meter has freq counter function, if the amplitude is high enough, it is able to estimate the osc. freq. Will try that too. If I have time.About the schematic.I think I understand the schematic as it is, but yeah, it could be a lot easier to read, definitely needs some optimizing. I put in those regulating circuits and the reference.http://www.electronics-lab.com/project/0-30-vdc-stabilized-power-supply-with-current-control-0-002-3-a/Here is the original schematic.I want to redesign the PCB, it had some problems, the circuit needs some fusing and a handful of protection diodes. It went trough a lot of soldering and desoldering, now it looks like it's made by some blind one handed apprentice.

author: C
author_rank: Super Contributor
timestamp: February 05, 2018, 12:36:13 pm
content: Before you start changing thinks, would be good to understand the circuit.Most times each circuit has problem areas. There are normally many ways to work around a problem, some better then others. Some fixes help in one area while making other areas worse.In a lot of ways, the Electronics-lab link of last post is much better.Look at something simple.Last schematicTransistor labeled Q1With Q1 acting like a switch, why short out the output of U2? Not good for U2 especially when there is another way to do same thing.Q1 could pull down on junction of R13 and Q2!Quote from: radoczi94 on February 05, 2018, 09:43:49 amSo if you really understand current schematic and compare it to this schematic some of the bad changes should really stand out.These days it is not uncommon to have a load that is doing high current PWM. Not uncommon to have many different PWM loads. During testing/design a Lab Power Supply is the power supply for this new circuit. Until you know exactly what value of series resistor to use for that PWM dimmed LED the law power supply CC mode is the protector of that high dollar LED, would be a good idea for it to function properly.With this being a simple cheap supply, you may not have the best but still want the best it can be.So think about a load that changesInitially with Max current of CC mode the output of this supply will be rapidly changing from VC mode at some voltage to some current value in CC mode. How does the lab power supply handle this? What changes in the lab power supply circuit.To make it easy the negative output of supply is connected to ground referenceWith respect to ground referencewhat is point labeled 7 on last schematic doing?What is cathode of D5 doing?What is anode of D7 doing?What is voltage across C5 doing?Is C5 a good idea or is it causing problems?What is base of Q2 doing? How does it change with low/high output voltage, low/high output current?Each little change can effect output of the supply.A change that makes one part of circuit better that at same time makes supply output worse is not a great change. Some changes that could effect output can be hidden by other parts of circuit, but there is a limit to how much.Try to understand last post schematic and then look at your first posted schematic. What does each change do?What is changes to circuit when load changes such that same CC mode current is two different output voltages?Quote from: radoczi94 on February 05, 2018, 09:43:49 amIf you are going to redesign the PCB, I would strongly suggest a redesign of circuit first to fix the many problems both versions have.

author: radoczi94
author_rank: Contributor
timestamp: February 05, 2018, 02:37:45 pm
content: QuoteYou meant R15? Because if I pull down the R13-Q2 juntion, Q1 shorts C1 to ground, causing rather large currents and it would not be so good for the cap and the transistor too.The opamp should not care about that short, it has "infinite" output short circuit protection to each power rail and ground. But still, it's better to do that trough a resistor (if you meant to pull down the base of Q2).QuoteOn the last, linked schematic:All with the respect of the output negative-Point 7It is the "high current" power rail. With a varying load it is also varying, depending on the main transformator and the capacity of the filtering cap.-Cathode of D5The cathode of D5 is on the ground.If we measure that point in the respect of the negative output, there should be the voltage drop across the shunt resistors. A square wave if the varying load is on the output.(If the load is pwm)-Anode of D7That's the negative voltage rail, there sould be the zener voltage+the shunt voltage drop.A square vawe with a dc offset.-C5should be same as the point 7. I think it is a deoupling cap, but not shure. Therefore, I do not really know if it is necessary or it is bad.-Base of Q2There should be the output voltage + Q4 and Q2 B-E voltage drop. At maximum voltage, there should be 31-32, at minimum voltage there should be the between 1-2 volts, depending on the transistors. At low output current there should be output voltage+ the B-E voltage drops, at max. current, there should be 0 volts. Or even negative voltages.I wanted to make a stabilised supply for the IC-s, because my transformer is 28Vac and capable of 5amps. The rectified voltage would be much higher, than the opamps would tolerate, that is why tose parts were put in. Changed the voltage reference also, because I did not liked that circuit at all, never understood, actually. Why are these changes so critical?QuoteI was intended to do so, but I can not really found the problems with this circuit on my own. If you guys could help me find those probelms, explaining, why are they a problem and how can it be solved, I will correct them. I'm just not at the level of do this on my own. The more I learn about electronics, the more dumb I feel myself.

author: C
author_rank: Super Contributor
timestamp: February 05, 2018, 09:22:57 pm
content: Yes R15Q1 controls base of Q2Think this through a step at a time.What happens if I short Q2 base to the positive output?U2 supplies current to other side of R15. Assume for a moment that U2 output is shorted to point labeled circle 7, In practice U2 will be limited to less then this. What is max current possible.After you think you have some answers look at end of this post.QuoteIf you think about it, Base of Q2 is output control. It will change based on voltage in VC mode and try to change by current in CC mode.So all of the circuit is changing based on current drop across R7 except for what is directly connected to circle 4 ( negative output.C5C5 is trying to prevent voltage changes between circle 7 & circle 4.You stated that circle 7 changes based on current demand of output. So C5 is also trying to prevent fast CC mode response to a current change.My initial thoughts is that C5 is bad, makes supply worse in both CC & VC modes.Quotelook at last schematic at U1Think of how an op amp works, Output is like a variable resistor to +V and a variable resistor to -V.Effects of Changes to +V are reduced by power supply rejection ratio of op ampYour first posted schematic has R13 changing by load current & filtered some by C10You have power supply rejection ratio of U1 vs rejection of TL431.But by creating the stabilised supply for IC's you now have all IC's having to fight against Load current changes on their supply lines vs IC output.You have a bunch of capacitors connected to what is noise for what you want on output. Examples are C3, C4 C5,C7,C9,C6.So you are starting from last schematic that has a poor CC Mode and then increasing source voltage creating more problems.Might be a good idea to fix CC Mode in such a way that more source voltage becomes less of a problem, instead of hack the hack the hack.There are many ways to do this.With no scope and limited testing capability you will get a "I think it is good lab power supply". With changes you have made from last schematic, your limited testing is saying "I know it is broke/bad"A simple circuit for a power supply that has a CV mode and a CC mode is as follows.A pull up resistor that turns on the output to MAX.Your Voltage reg & Current reg then limit the max.Your current regulator's diode is setup to do this. Needs to control output directly not via voltage regulator as this adds a second control loop that is talked about.Output of Voltage regulator needs to connect to diode with anode of two diodes common and controlling output.So you have one diode for CC Mode & one diode for VC mode & you could add a third diode or more diodes for shutdown or other limit conditions.With low side current sense you have a problem that any control circuit current changes connected to negative output also effects your current sense.If this is a 0V to 30V supply, You could think of it as a 5V to 35V supply.The negative output regulates the -5V supply you have.With -5v Controlled by output negative, you could then use this to control +V to op amps. You have just removed supply bounce by current.This leads to control circuit power around negative output.You could think of floating control circuit power with positive output as the reference. This can make it easy to have higher voltages and make high side current sense easer. The down side is that today most think of negative ground.What I am trying to show is how you think about a circuit can change what you see. A small change can make a huge difference.So step one is understanding both circuits better and comparing what the differences do.Any change needs you to understand the effects of the change.AnswerQ1 controls base of Q2 question.You missed that when Q1 turns off Q2, collector of Q2 can rise and will not short out C1D10 keeps Q2 base from going below circle 3 (positive output ) by diode drop. At a Diode drop below Q2 is off, This in tern means that Q4 is off. The only thing left trying to hold base of Q2 high is R15.Max current is R15 with voltage of circle 7 - circle 4Now Think of what U2 will do. Say output was set at 1 V.Pulling Q2 base low will cause output voltage drop which will cause U2 to turn on harder trying to make U2 output more positive.So at start very little current through R15, but will increase to U2 max output voltage current with Q1 having to conduct this current.

author: radoczi94
author_rank: Contributor
timestamp: February 05, 2018, 11:17:29 pm
content: Huh, that is a lot of information in one go. Now I see, why this circuit is a total failure. Maybe with extra, stabilised supply voltages for the opamps,with a decent voltage reference and with the OR diode gate, this thing will be able to work like as it should.Or maybe I should just look for another schematic. Designing for another 4-5 days, taking away from my precious time, building it, for 2 days, experimenting for weeks, because why would it work for the first time. Or just getting an old, well proven circuit from somewhere, understanding how it works and building it as it is. I need to think this trough.I really wanted to build an indestructible, foolproof power supply, that is why I used that massive transformer and heatsink. With the case I built, it's weight is around 20 kgs, and not even used the full 5 amps of the transformer.Anyway, thanks for all help, it's really appreciated!

author: C
author_rank: Super Contributor
timestamp: February 06, 2018, 07:05:35 pm
content: Quote from: radoczi94 on February 05, 2018, 11:17:29 pmNo such thing exists. All have limits and is up to user to keep connected load in limits.Example a 30v power supply and idiot connects a 48v battery to output.For power supply to protect against this is very hard to do, best to think of limiting the damage.Example 1000F cap bank connected to output. The power supply would be at current limit for days. How long can power supply survive with output at 0V & Max current.Sure you can add to this list.So the best you can do is the limit damage and user has to supply remaining protection.One thing I would not pass on is using what you have to learn. A lot could be learned.Back in the 80's HP was doing a lot of internal research. A lot of things HP sold were first designed,built & used to cure a research problem. Often a power supply consisted of many modules or boards. This makes upgrades or new designs easer.So before you look for something better, would be good to know what to look for.The separate parts have to work together to make the whole thing good.You can have very good blocks in schematic, but they have to be connected together in a way that makes total good.You also have to keep in mind that nothing is perfect.You have a cap across output terminals.For CC mode you want no cap or small cap. The reason is that cap charge gets dumped into a short on output.For CC mode any cap on output slows rate of current change to hold desired current.For VC mode to hold constant output you want that cap huge.For VC mode to detect a load change on output, smaller lets circuit see same change sooner.The right cap or caps can stop supply from oscillations.There are more thingsSo output cap is war of sizes with an option of making CC or VC better.Then it could be that expected loads will survive the dump of output cap on short such that a little larger cap is ok.Or load is not effected badly with a voltage change.For a lab power supply you need to think of all the nasty loads that could be connected.Step one for lab power supply is to protect connected load.If connected load is real bad then step two is for lab power supply to try to protect it's self.Then you have option of big load power supply & small load power supply where you pick best for loadSo at cost of some of your time you could learn a lot from what you have.Keep in mind the many ways to do things.One change from last schematic to first is power and reference.Last schematic used op amp's power supply rejection to lower noise.First schematic I see creating problems by coupling noise into control power.Again bad schematic hides details.Take last schematic and redraw it.Power sectionon left is D1-D4, R1 & C1on Right iscurrent sensor R7Output control Q2,Q4 & R16? C3Some protection by D10, D11Output filter C7What is left is Control circuit.Now if you do not have ability to easily do this, use your versions.Now the Control circuit needs inputs to function.Voltage and Current inputs should be differential.Voltage sense is connection to Circle 3 and Circle 4Current sense is connection to each side of R7For control circuit to control output you could need connections to circle 7, circle 3, circle 4, left side of R7Note that control and sense should be separate wires or paths, AKelvin connection.Time to learnDoes current PC board do this separation?Does your first do this?It's not shown on any schematic posted, yet is simple and cheap to do.If you look for new design, something to look for.Note also this separates the high heat parts, parts that fail on heat.With the schematic redone, You should be able to see some problem areas and have some ideas on how to improve circuitSo a great schematic is very important.Makes it easy to understand the circuit quickly. Can show problem areas & and many more things.Now for control section separate out it's parts.Reference sectionU1, D8, R4,R5,f6it;s outputs are connection to R18 & circle 6Should note a few things herePower is via U1 V+ & V-Part of differential sense is by R4 & R6 being connected to circle 4.Should note that V+ and current set pot is main source current changes.Should note that last schematic makes great use of op amp's power supply rejection to lower noise. A change to less is not an improvement.The V+ connection of U1,U2 & U3 is connection to circle 7 for output control.Adding a second line across to circle 7 could show this.So I see a good first step is to learn from what you have now & fix schematic to show details and make understanding easer.Take a huge number of measurements with your meter with different settings & loads. Save this to look back to to see a change is good or bad.Each change a new spreadsheet of measurements.Unknown is how well the last schematic functioned before changes.Could be original has problems or you created problem for CC mode. Only measuring will tell.One common thing is to have separate power for control & load supply.So you have a massive transformer, would be a shame not to use it if you see a need for the power supply it can be a part of.The way I see a single PCB is a way to mass produce something.If you want something good that can upgrade then many PCB's or circuits is better. A good circuit design also makes this easy.If you look at old HP supplies you often see it is made of many modules. A module is often used in many different supplies.Think of future some, today you build best power supply you can. Next year you could think of it as junk. Easy to change junk could become new supply better then first.

author: radoczi94
author_rank: Contributor
timestamp: February 06, 2018, 09:21:47 pm
content: QuoteYes, I know that, i did not really meant it as it is, just basic protection and really good cooling. I have some desperate ideas, like having a huge parallel diode reverse biased and a fuse on the output in series (may be fast blow), it could prevent a reverse polarity battery for example. A huge truck battery will kill the whole thing anyway, so not much protection, but better than nothing. A series diode with proper cooling could help also, but the voltage drop can be an issue.I see now, I can not really build an ok-ish psu without external voltages. So I have to go to the store, and buy a pair of small transformers, build a good secondary supply for the control circuit. Even if I hate wiring work.I was not aware of the Kelvin-connection thing.I did not knew, that such thing exist. Saw in a bunch of EEVBlog vids, but never knew, if it is intended or not. Therefore my design has absolutely noting like that. Thank you for that!Otherwise my pcb design is a bit beta-ish, tried to maintain wide traces for the 3A current paths, so the contol gone into a tiny little space between huge polygons.Never again.I think I will do some further investigation on what is happening around the circuit, to see if I can make it work, but I don't really feel the motivation to stick to this schematic and make it an acceptable psu.My buddy sent me an old schematic he built two of this. It has separate power supplies, opamps controlling trough an OR gate etc. A lot of things I seen in older Tektronix schematics. The schematic itself is a perfect example of how not to draw a schematic. My only concern is the old parts used. If it works, I can replace the 741s with better opamps, like OP07, which has excellent DC characteristics ( the BW is a bit low, so certainly not the fastest opamp on the market). The other thing is the 723. It can be used as a switching regulator so it's not hard to make it oscillate.https://drive.google.com/file/d/0B7yeYY6jyIVVaHZpTlhTQVl2b0wzVHpEUXAtS1VtU1E0b2o0/view?usp=sharingHere it is.I kinda like it, have not found any better yet. The description is pretty good too, explains the principle.

author: C
author_rank: Super Contributor
timestamp: February 06, 2018, 11:58:21 pm
content: In my past I do not remember any Tektronix supplies.Used a lot of HP and a few other brands.If you spend some time looking at old, look at HP power suppliesProtectionAs a last resort protection the simple thing is have control circuit blow a fuse.On input side you have many choices.Blow line fuse, blow a fuse on secondaryOn output side you want to get disconnected from load.Think the best device is the SCR or triac to do this.QuoteYou can use any type of supply to supply control circuit power. If you have any electronics junk you have something you could probably use.The important part is the separation.Think of this, you have a possible large current draw on input bridge diodes. Current = voltage drop, changing current = noiseA simple change is just using a second diode bridge and filter. You gain some isolation from the noise. Separate source is more noise isolation.Actual power voltages is not that critical. Some designs are easer if control power exists below or above output terminals of supply.Important thing here is op anps have a sweet spot in center between supply with problems at close to supply rails.Kelvin-connectionYou may have seen some current sense resistors with a Kelvin-connection. For a lab power supply you want a Kelvin-connection to output terminals for voltage sense. Better supplies add remote voltage sense terminals to extend Kelvin-connection to load.Can be important to use on a PCB as well. Think of When this changes where do the currents go/flow. Think of this, a voltage based temp sensor and a current based temp sensor. On end of a 1000 feet of wire one is crap with other still great.QuoteGuess why you see separate boards for high current or one small area for high current.QuoteInvestigate to learn. If you chop up the schematic like I suggest, then future is replacing blocks on schematic.The 723 is ok for a cheap simple power supply but is not great.The link you post from my quick read is not great.Keep in mind that you want the lesser value of Voltage or Current.Link Instead of simple you have a mess.In link voltage control A1 if not modified wants to turn on output more in current limit mode while current control needs to turn it off more.Any change in A1 output messes up good current control!Add a pull up resistor to output transistor causing output to turn on full blast.A2 is all ready, Too much current and it pulls down lowering current to set point.A1 needs a diode and then will work same way and reduce to set point.Just want you want for a supply that has CC & VCLower output level win.Now you have two separate circuits controlling output. You do not have to modify one for other to work properly.Now look at range for the change.A diode has a voltage vs current curve. The curve is fairly flat but not flat. Change over goes from one output control, a current to the two outputs equal , 50% of one controlling. If it takes 20 ma to hold down output then you would look at two spots on diode curve, 20 ma & 10 ma. and get some voltage value.Now due to feed back this becomes mostly hidden and change from CC to VC is fast with few if any problems.Want something great just read a HP manual for a power supply. Every thing is in it, how it works, schematic, parts list and often PC Board layout.Gave you a hint above but you may have not caught it.Current goes the distance, You can connect low side circuit to high side circuit with current. A current change then can control output.

author: xavier60
author_rank: Super Contributor
timestamp: February 07, 2018, 12:12:18 am
content: Because of the way the voltage adjust potentiometer is configured as a VR in the Elektor design, it is not fail safe. If the potentiometer becomes scratchy, the PSU output could spike to full voltage. I think it is better to use the potentiometer as a potentiometer to directly vary the reference voltage. Including a wiper bleed resistor would make it fail safe. The reference voltage will need to be negative.I don't like the remote sensing. It should be omitted.

author: radoczi94
author_rank: Contributor
timestamp: February 07, 2018, 09:31:44 am
content: One thing I do not really understand about this old schematic. 723 makes the reference. As far as I know emitter followers have a voltage gain of 1. How is 0-30 volts on the output, if the Vref is 7,15V? I'm shure, that some resistor makes this magic. Maybe R17 is talking back? Why don't just pull it up to the collectors? I've seen in some HP and Tektronix supplies, the base were just pulled up.What I do not like about this, is the 723. Maybe a little bit too crusty. I'm shure there are much better solutions for a good reference.Quote from: xavier60 on February 07, 2018, 12:12:18 amYou mean, like the current set pot? Dividing the vref with the pot itself?

author: xavier60
author_rank: Super Contributor
timestamp: February 07, 2018, 12:10:49 pm
content: In your old design, The output Darlington functioned as an Emitter follower which buffered the output of the CV op-amp which was able to swing by the whole range required for 0-30 volts output.In the Elektor design and many others, everything is referenced to the positive output. This includes the Darlington Emitter, 0 volt reference for the low voltage split rails(+/- 12 volts) and the voltage reference source, a TL431 possibly.A good way to properly drive the Darlington's Base is to source current from a voltage source that is always a fixed amount higher than the Emitter so that there is enough available Base drive current even when rising Emitter voltage approaches Collector voltage. This can only be from the +12 volt rail.This makes it possible to fully turn on the Darlington. This is desirable when the PSU is operating at high output voltage and current so that as much as possible of the unregulated voltage on the Collector can be transfered to the output. The regulator will be said to have a low dropout voltage, a good thing.In the Elektor design, the Darlington can no longer be thought of as an Emitter Follower. It is a series pass element conducting current from the unregulated rail to the output under control of either the CV op-amp or CC op-amp.Google "Darlington pair"Do you understand what op-amps do?Quote from: radoczi94 on February 07, 2018, 09:31:44 amYes

author: radoczi94
author_rank: Contributor
timestamp: February 07, 2018, 12:53:19 pm
content: So the floating supply's ground is referenced to the output.This means, that the floating positive rail and the Vref is on higher potential than the output, and the floating negative rail is lower. This is because the floating supply'ss ground tied up to the output by R23 47R. Right?This way the Vref is able to source current to the Darlington bases, because the Vref's potential is higher than the emitter(s).As I understood, the opamps are sucking away current from the Darlington base, or the CV oamp tries to push it, if the output voltage is not high enough, but it can not be happen, because the diode is reverse biased,so the Vref opens the Darlington more.If I'm right, the Vref needs to be capable of providing enough current to drive the transistors, and the voltage dividers at the opamps.In the same time, it needs to be stable with low drift. Maybe a bare TL431 is not enough for this, a lot depends on the beta of the Darlington. Edit: The total beta of the darlingtons in worst case 400, so 8-10mA is enough for 3-4A output.Quote from: xavier60 on February 07, 2018, 12:10:49 pmThere are some violations to this, because there is some voltage drop across the shunt. But we want this, so the voltage drop is compensated by the control circuit. Right?Quote from: xavier60 on February 07, 2018, 12:12:18 amWhy would be practical to have a negative reference voltage? Just in case of pots getting scratchy?

author: xavier60
author_rank: Super Contributor
timestamp: February 07, 2018, 04:34:30 pm
content: Quote from: radoczi94 on February 07, 2018, 12:53:19 pmThat's right. That area of the circuit can be left the same but with the sense wire tied directly to the + output terminal.Quote from: radoczi94 on February 07, 2018, 12:53:19 pmBoth low voltage rails should be regulated for best performance and the drive current for the Darlington should come from the +12v rail. You won't have a positive reference if you decide to go for a negative reference anyway.The low voltage rails don't strictly have to be +/-12v. A bit lower would be ok if your small transformer doesn't output enough to regulate 12v.Quote from: xavier60 on February 07, 2018, 12:10:49 pmPots going scratchy is all too common. It is actually the wiper losing contact with the track. Have a think about what happens next if the pot is configured as a VR as in the Elektor design?As far as I can see, the reference needs to be negative if the control is to be wired as a potentiometer rather than a VR.Time to start sketching some circuits.

author: C
author_rank: Super Contributor
timestamp: February 07, 2018, 04:43:01 pm
content: Simple questionThe Elektor linkHow much trust should you give to an article in a mag where the operating principles does not match the final design?The article describes how it works in figure 2.A classic design where CC mode is hacked on to a VC mode regulator.Some of this hack shows in final circuit.The final circuit figure 3 does not match the theory of operation in figure 2.Here you have the 723 suppling power to R9 & have the analog OR for CC & CV ( D4, D5)So Author could not describe a simple analog OR to make sense of circuit.Then you have if it was truly a VC & CC regulator, why only one indicator of mode? How much effect does driving that indicator have on mode switch?Note that it is possible that a load is CC & VC at same time. With load noise a good supply can be changing fast between the three states.With out a scope, trying to update this very old design I would say is a fail.It is what it is an attempt at a cheap precision power supply.some thing to keep in mind, If you are going to be building analog circuits a Positive & Negative supply is nice to have during design.

author: radoczi94
author_rank: Contributor
timestamp: February 08, 2018, 01:41:48 am
content: QuoteI think I will go with +-15V. I have a few salvaged 2*18V*0,15A PCB mains transformers, I want to use them. Then put around some 78-7915 regulators some filtercaps, and a preload resistor or whatnot.QuoteIf the pot works as just a voltage and the contact is failing in the wiper, the output goes low, because the voltage difference between the inputs is positive. If the VR fails, the goes to full chooch.Oops.Quote from: C on February 07, 2018, 04:43:01 pmHad a feeling about that, but only concentrated on that voltage setting pot. Which was basically the 90% of the theoretical part.As far as i know, this thing is able to work, my buddy made two of these and both worked first-second try.Spent half a day googling around to see in search of a usable schematic, nothing. I will probably end up poking parts into breadboard. With only a multimeter, for now. I wanted to buy a scope since ages, so I guess, the time has come.

author: C
author_rank: Super Contributor
timestamp: February 08, 2018, 04:06:14 am
content: Googlefind a model number for an HP supply and find the manual using google.As I said most are fantastic with everything. Was common in 80's to have everything in manual so user could repair it if necessary.One thing you will find is that controlled power up and down is built in to design not a tack on.Little things are included in a great design. You might design the voltage control pot to change from 0 to Vref. A pot failure becomes 0 a normal setting.The great design can have a 0 output of supply equal some % not 0 of reference. A pot failure becomes something not normal and electronics can detect problem and shut down power supply with an error. That analog OR lets you have many circuits that can shutdown supply.So google and study a HP supply and see what is missing in the simple hacks. Manual not schematic as it's in manual.The problem with a supply is often not the DC it's the load caused AC changes causing the power supply to fail. It really takes some fancy test equipment to check the AC part.One thing some forget is what a power supply is. Most are crippled power op amps. Think it through If you had an op amp that could do X amps with a range of Y volts you have a power supply. Removing half of op amp's power stage is a cost & heat savings. The better lab supplies still have some sink capability.Think of what a direct coupled audio amp is. Could be thought of as a bidirectional power supply.So You have a transformer, what are the spec's, Just one winding?Old school is using pot's to set things, New is using DAC's. Would be good to design so that ether can be used. Some were controlled by resistor decades.Keep your eyes open. In first schematic you posted, you had Q4 as an NPN. A change to a PNP would have let you have a supply to limits of the 2N3055 while having lower voltage control circuit.6227B Power Supplyhttps://literature.cdn.keysight.com/litweb/pdf/06227-90001.pdf?id=734411DC Power Supplies - Discontinued and Obsolete Productshttps://www.keysight.com/en/pc-1000002054%3Aepsg%3Apgr/dc-power-supplies-discontinued-and-obsolete-products?nid=-536902299.0.00&cc=US&lc=eng.

author: xavier60
author_rank: Super Contributor
timestamp: February 08, 2018, 07:36:58 am
content: Most of us would agree that for this type of PSU project that the topology that has everything referenced to the +out is a good starting point. The rest is details, not all will be trivial ones.radoczi94, at the heart of understanding what is going on, it is essential that you understand op-amps. Do you understand how op-amps function and how they are applied?I'm planning on doing some related experiments shortly.

author: radoczi94
author_rank: Contributor
timestamp: February 08, 2018, 06:58:52 pm
content: The various protection circuits are neccessary for shure. One thing I already did, is a switch-on delay with a 555 and a relay switching the output. It works, but yeah, switch contacts doesn't really like switching DC. There could be a switch-off circuit too, like in the Electronics-Lab design, monitoring the AC, even on the mains side trough an optocoupler and then connecting to the OR gate. Or even a thermal shutdown (I don't think that this is absolutely necessary in my case.)But I can throw all the protection to the trash can if the 2 main function doesn't work. Since I do not have a scope (yet), what I can do now is just sit down and learn as much as I can. And maybe search for more suitable parts, do some sketches, collect ideas.That massive transformer is 250VA 2X(0-3-12-24V)*5A. For this thing, 4A output current would be comfortable at 25V output. Could have an automated tap selector circuit. At least for the 12V tap. I think the heatsink could handle it either way, but why piss away 100W?What i know about opamps...well, basics. Understood the DC operation well, and some of the AC. I found a lot of informative stuff on the Analog Devices website, there is a lot of useful stuff. I'm reading those, also found some Linear stuff about undesirable oscillations.

author: xavier60
author_rank: Super Contributor
timestamp: February 09, 2018, 12:59:34 am
content: Just because a PSU has CC, doesn't really mean that it's short circuit protected. Even my Agilent U8002A will supply over 20 amps into a sub ohm load for 100us before it begins to sluggishly current limit.I mocked up a CV/CC regulator to experiment with the CC response. The cause of the delayed CC response was because the output of the CC op-amp normally sits at close to its full + rail voltage. When the PSU is suddenly overloaded, it then takes a long time for the CC op-amp's output to slew down to the point where it takes control of the Base.One of the reasons is that the loop compensation capacitor is usually connected directly between the op-amp's output and inverting input. I have connected the capacitor to the other side of the ORing diode so that the op-amp's output can slew at its maximum rate until the ORing diode conducts.Because the LM358 that I have used doesn't slew that fast anyway, 0.3V/us, I have put a diode and LED in its feedback path that keeps its output at 2.2v so that it doesn't have far to swing down to take control of the Base. I am ordering some faster NE5532 op-amps to see if I can omit this extra complication.Extra: I have used an LED for the CC ORing diode to give CC indication.

author: xavier60
author_rank: Super Contributor
timestamp: February 09, 2018, 09:53:50 am
content: The 13V rails in my regulator experiment are supplied from 2 regulated plug packs. I noticed that the regulator's output would pulse to full voltage when I powered down the plug packs only. I have added protection to prevent this from happening. Q3 cuts Base drive current when the +/-13v rails drop by a few volts.It was also an ideal place to add an on/off switch.

author: John Heath
author_rank: Frequent Contributor
timestamp: February 10, 2018, 02:46:08 am
content: I am sensing unknown variables as part of the problem. Measuring the voltage drop across a resistor relies on that resistor being what it says it is to know the current. 200 m ohms is not easy to measure. A hall effect DC current probe will measure the current in no uncertain terms and it would be a nice addition to your test equipment. My DC current probe cost 35 bucks. Cheap. Make sure it is a DC current probe not the other type that is limited to AC only.Failing this you can use your cell phone as a current probe to know how much current. It requires a little calibration as cell phones were not intended to be used this way. Google magnetic app for android and you will find lots of apps for free. Find a known to be true 1 amp source that can be trusted. Run 1 amp through a wire and find the sweet spot on the back of your cell phone where the magnetic reading is the highest. Mark that spot on your phone and mark what the magnetic strength is for 1 amp. That's it , you now own a free hall effect current probe calibrated to 1 amp and as a side benefit you can make a phone call.

author: radoczi94
author_rank: Contributor
timestamp: February 10, 2018, 01:12:39 pm
content: Quote from: John Heath on February 10, 2018, 02:46:08 amThe resistor was pretty close to the nominal value. It could not have vary that much, it was +-5% 300ppm and it was just a little bit above the ambient temperature. The problem was: I used inadequate equipment. The best theory is that there were some serious oscillation and it tricked the Wun-Hung-Lo multimeters,so they just showed an average.I did not even knew, that such a thing exist, will do some research on it, thanks.Quote from: xavier60 on February 09, 2018, 12:59:34 amThat design is almost identical to the one formed in my head. My concern is, that I want to use fast transistors, around 30MHz.The opamps should be order of magnitude slower to prevent loop oscillation without compensation. But the slew rate of the opamp should be high enough to be able to handle the switch on-off events, with an acceptable waveform on the output. Guess, I should learn to use a simulator software before I start soldering.

author: John Heath
author_rank: Frequent Contributor
timestamp: February 10, 2018, 03:31:13 pm
content: Speaking of oscillations I noted a 10 uf condenser on the output of your power supply. Who ordered that? 10 uf in CC mode means much higher current than the setting for the CC mode in the first 1 m second. It defeats the purpose of current limiting for sudden current demands in the 1 m second range. That is fishy. Did they put it there as an after thought to compensate for an oscillation problem in CC mode? Try a 100 uf quality condenser and see if things improve. It's a long shot but you never know. If it does improve then the current limiting is only for average not sudden current demands. This is not good as one could burn out an IC with a sudden short even if the current limiter is set to 100 m amps. A raw 100 uf quality condenser will be more than happy to provide the destructive current to destroy this IC .

author: xavier60
author_rank: Super Contributor
timestamp: February 11, 2018, 02:23:01 am
content: Quote from: radoczi94 on February 10, 2018, 01:12:39 pmOscillation is a big concern. My experiment did oscillate for various reasons. One time it seemed to be Darlington that was the cause. That's why I put the 220 ohm in series with the Base.One of my suppliers has some 150V 73A MOSFETS(PSMN020-150W) selling cheap, so I bought some to experiment with in my regulator.

author: C
author_rank: Super Contributor
timestamp: February 11, 2018, 04:22:44 am
content: Many ways to stop or limit Oscillation.Rate of change positive could be different from negative.Multi-stage feedback.Areas of circuit where frequency response is different.Positive feedback to get large change with negative feedback for small change.For an analog or you could have two. A min OR to control output & a MAX OR to prevent swing to rail.Many ways other then just slow.

author: jaycee
author_rank: Regular Contributor
timestamp: February 12, 2018, 01:35:36 am
content: Seen this circuit before. To be honest it's not a very good one, and those opamps are probably seeing way too much voltage across them!Take a look at the schematic for the ELV 22532 power supply. A good approach used by many of the big manufacturers is to have a seperate "bias" supply for the opamps, which floats around the output. I've used a tiny little 9-0-9 transformer from an old bedside digital clock along with 7805/7905 regulators to create such a supply before. Things are vastly simpler and performance is much better with this sort of topology

author: xavier60
author_rank: Super Contributor
timestamp: February 19, 2018, 02:12:46 pm
content: My idea of connecting the integrator capacitors to the other side of the ORing diodes was too unstable. I want the current limit fast acting yet stable. I'm experimenting with an idea that enables the feedback through the integrator capacitor only after the CC op-amp has taken control.

author: radoczi94
author_rank: Contributor
timestamp: February 19, 2018, 02:49:16 pm
content: Hi Everyone!Sorry for the late reply, I had a critical exam last friday, studied my brain out. It was succesful, so rewarded myself with a scope,ordered a Rigol 1054Z. It should be here tomorrow.On the weekend I was playing around with the PSU. I removed the CC opamp from the socket, and put on a 12V 10W chinese LED as a load. With this config, I measured 0,5A on the shunts and 1 amps on the output. So the CV opamp was oscillating (or something nasty like that). Then I replaced it (NE5534) with a 741, turned on, the result was 1 (or both) shorted 2N3055 and a smoking chinese LED. Doh. I just lost my temper, whacked out the industrial heatgun, desoldered everything from the pcb and threw it in the scrapbin.Quote from: jaycee on February 12, 2018, 01:35:36 amI decided to build that supply on breadboard and play around with it,try to change small things and measure everything. What I want to do is replace the TIP14x transistors with 1 driver and 3 or 4 end transistors, use a more adequate reference than the 7805, and replace the opamp with a more skookum one. I will put together a transformer tap switch circuit too.Quote from: C on February 11, 2018, 04:22:44 amI will make some research on things like these, thanks. Will need to do some experiments to see what are the exact effects of these methods.

author: C
author_rank: Super Contributor
timestamp: February 20, 2018, 05:50:23 pm
content: So you should now have a scope.Think of the old time test equipment.Back then it was very poor, but the old timers built some fantastic stuff using poor. Very stable circuits that had power supplies in the ball park with a lot of noise.A voltmeter or current meter was treated as a poor slow speed analog display that was not very accurate. A scope was treated as a higher speed video display, again not very accurate.With every thing poor first high res voltmeters where voltage comparators. They built these by comparing resistors to build a voltage divider. They compared the voltage reference to a better voltage reference.The result was a bunch of ratios. This resistor is 2x that resistor, this resistor matches that resistor.. The 2x was hard until you had two matching resistors and then very easy. As the comparator got better the degree of match got better and the resistor divider got better, but it still was all ratios.Assuming that the D'Arsonval galvanomete was repeatable, and by using a Wheatstone bridge you could compare(voltage, current, resistance).If you use your brain some you can work with uncalibrated test equipment using it more as a comparator. Good numbers are then just a way of passing information.Old timers also used positive and negative feedback. The trick for this is balance. At static stage you want positive feedback countered by negative feedback.Think of a switcher power supply, it is just on or off. What makes it work is Time( the rate of change). Good use of analog can act like switcher for large changes and analog for small changes.If you think of an op amp output as a pot, then if you have a pot input you get a scaled pot output. But here you are forgetting that a op amp output is really two pot outputs. You only get a pot output if both output pots change correctly and you are still forgetting the output load.Going a step further the op amp feed back is like a . seesaw( teeter-totter) and output is more like a set of springs.In physical world a seesaw can bend and bounce. Load changes changes spring deflection.A min parts circuit has to work harder or be not as good as a many parts circuit. Many parts also has problems of it's own.The total result is what you want to be as good as possible.A lot of old time transistor circuits would often use differential circuits. This signal goes positive while a second signal goes negative. This can be cheap while opening up options for control.Really think about your control circuit.When you have a big output current change, you have a lot of current change on 2N3055 base. Do you want the op amp handle all this change or would it be better if the circuit around the 2N3055 handled some of this change directly.Do you want source change all handled by op amp or some handled by circuit around 2N3055.Your output stage is based on current, think current.

author: chris_leyson
author_rank: Super Contributor
timestamp: February 20, 2018, 07:20:07 pm
content: Stumbled across an interesting blog from Peter Oakes about building a bench supplyhttps://www.element14.com/community/groups/test-and-measurement/blog/2014/09/15/the-modular-bench-power-system-the-essential-diy-build-for-every-ee-student-and-old-timer-alike

author: C
author_rank: Super Contributor
timestamp: February 22, 2018, 02:38:03 pm
content: First, how much have you tested this with different loads? As a lab power supply I would think you would want to connect any load to this and know how well it will try to protect the load while trying to maintain it's settings.Think of using a second copy of this as a load(shunt regulator) to test the first. The +30volt supply & 1000uf cap is your supply. Q2 & R5 is the regulator.By keeping both the same, you could use one to test the other. By making the same changes to both you improve the Power Supply use & Load use.Might also want to think of some nice future additions and make it easy to add them.One thing I have seen is using an internal known load to get more information about connected load.An additional think to think about. Do you want only the load pulling output to 0 Volts & 0 Current output.Really think about this, Say you are building/testing a circuit connected to this supply. You short something out in the circuit. Do you want all the cap's in test circuit dumping in to short or would you like the Lab Power Supply able to pull excess from test circuit. Right now you can source current, To make faster response to a lower output you would need a shunt across the output, Pull down in addition to existing pull up..If you are going to trouble of creating a PCB, might be nice to be able to use the PCB as part of more test equipment of your test bench.If you plan ahead you can make ground loops less of a problem.

author: xavier60
author_rank: Super Contributor
timestamp: February 22, 2018, 07:00:34 pm
content: There is the concern that some combination of load reactance will make it unstable. I use another large MOSFET driven by 15Hz to cycle various loads. Besides the function of Q1, the op-amps have been applied in a rather standard way, so I'm not expecting any problem that tweaking the compensation won't fix.I had no plans to make it sink current except for a bleed resistor across the output unless I find too much overshoot when it unloads.Designs like that of Peter Oakes have the option of being controlled and monitored by a micro-controller. Every time I think about this, it adds another 4 op-amps, so I won't bother for now.

author: radoczi94
author_rank: Contributor
timestamp: February 23, 2018, 10:31:53 am
content: Did you tried to tie diodes between the fet's DS? That way the CC opamp could pull down the voltage from the capacity on the output, possibly from the connected circuit too. I don't know if it's a working idea or not, just a brainfart.

author: xavier60
author_rank: Super Contributor
timestamp: February 23, 2018, 11:17:31 am
content: Quote from: radoczi94 on February 23, 2018, 10:31:53 amIf I understand properly, that would need another power MOSFET to force down the output. I'm trying to keep it as simple as practical.Even though there looks like a lot of current overshoot, it is actually much better than other PSUs. An even faster op-amp I'm expecting to reduce the overshoot further.

author: radoczi94
author_rank: Contributor
timestamp: February 23, 2018, 12:31:31 pm
content: Quote from: xavier60 on February 23, 2018, 11:17:31 amNo, I just tought about diodes, 2 or 3, they only open if there is a radical intervention by the CC opamp. Something like the on the schematic below. The 2 diodes on the bottom are the analog OR gate.

author: xavier60
author_rank: Super Contributor
timestamp: February 23, 2018, 12:50:45 pm
content: The 2 diodes would clamp the Gate at -1.4 with respect to the Source. The Gate needs to be reduced quickly to below about 4v for the MOSFET I'm using.

author: C
author_rank: Super Contributor
timestamp: February 23, 2018, 03:32:48 pm
content: You have to think of a load as being dynamic. At some point in time you will try to use your lab power supply to power almost anythingYou should have no problem seeing a load switch from no current to max current and back.Think of what this is doing to your current op amp.The higher freq op amp change hides some of this problem.Think of the states of your power supply.You are designing for two states CC & VC. You have a third state where output is not at set voltage or set current due to a load change. The time it takes for output to get to a set state is this third state. Most circuits handle less then set a lot better then greater then set.You need to think about when you connect a circuit you are designing in the future. That circuit could have over voltage protect and/or over current protect circuit that is faster then this power supply. It could be more sensitive.What will happen when the load makes a massive change on just a little overshoot? Protection will most likely go open circuit or dead short. Remember that this lab power supply has to protect it's self & the load.So the higher speed op amp makes circuit respond faster. This also increases freq response range of power supply output. This adds greater chance for circuit to oscillate.Try testing you circuit with smaller value for C6. C6 is there to prevent the fast current changes from causing your power supply to oscillate. THere smaller is better( less un-controlled dump into load).Find where circuit starts oscillating and try to make osc freq higher in rest of circuit first. This will reduce output overshoot some. Then make circuit stable with min parts change.While testing, if you are not using a lab power supply to supply your 30 Volt supply you might want to add some protection to Q2 to prevent magic smoke. A temporary resistor between Q2 & 1000uf could do this.R1 should be the lowest value possible.Need to remember that active circuits take time to change.A good test point is to look at anode of D! & D3 and look at what is happening on cathodes.

author: xavier60
author_rank: Super Contributor
timestamp: February 23, 2018, 07:07:24 pm
content: Quote from: radoczi94 on February 23, 2018, 12:31:31 pmI understand your idea now. After the MOSFET is turned off, the CC op-amp will help pull the output down, It is only 30ma though.Actually the op-amps can't pull the PSU output down directly because they are referenced to the the output rail. Current needs to be sunk to the negative output rail.

author: xavier60
author_rank: Super Contributor
timestamp: February 24, 2018, 04:44:30 am
content: It did oscillate when I reduced C6 to 10uF. I found a choice of 2 possible fixes.One is by simply reducing the Proportional response of the CV op-amp by reducing R3 to 33k and increasing C2.The other is by increasing the frequency response of the MOSFET by reducing the Gate resistor to 100and also making the response of the CV op-amp only Integral by putting only a 100pF capacitor in its feedback path.With C6 changed back to 100uF, the voltage dips by 0.25V with a 3A load and recovers in 5uS.

author: C
author_rank: Super Contributor
timestamp: February 24, 2018, 06:55:46 am
content: The lowest possible gate resistor is best here. Remember that connecting a scope to gate changes things. It is only here to prevent ringing on the gate. You are trying to charge/discharge a gate cap and need a lot of peek current. Digital gate drivers peek current is in the amp's to get fast change.Keep in mind that C6 is like a CV hack and CC harm. It is there to handle very high freq CV & CC changes. Smaller is better for CC while larger is better for CV. C6 also hides change.Your CV circuit has two modes.One is changing between two currents in CV mode.Second is changing to/from CV to CCYour CC circuit also has two modes.You op amp datasheet should list specifications for large and small output changes.Also of note is your op amp has a limited amount of output current.In your testing look for differences.In CV mode1. lower current to higher current.2. higher current to lower current.Here you are looking at voltage change, rate of change and overshoot.3. CV to CC where CC mode is small or large currents.4. CC to CV where CC mode is small or large currents.again for CC modeand many more.Good testing is a must.DC up to freq's hidden by C6 and all possible dynamic loadchanges.Think of each part and what would happen if you larger or smaller and all the effects on the circuit.For example it C6 was huge output changes are slower & at same time change is slowed so sense is slower..At same time it's different for larger or smaller currents.Your circuit can not compensate for what it can not see.Huge also turns a short in to a spot welder.And you have no perfect(matched) parts for a copy.Think of using changed values or added parts to assist in making other parts of circuit function better.For example changing R5 makes current sense larger or smaller & also effects CV mode with more or less change needed.Some cap's are speed up and some are slow down, use with a lot of care.Change I would thing about is using a quad comparator to drive CC & CV leds. This should leave more current for driving Q2 and provide output test points that reduce effect on control circuits.One problem area is mode shift.In CV mode, CC wants more current. You could have one op amp at rail before change and require a huge output change for transition.

author: C
author_rank: Super Contributor
timestamp: February 26, 2018, 12:54:23 pm
content: Quote from: xavier60 on February 26, 2018, 10:37:41 amThis is why I suggested you look at the anode & cathodes of D1 & D3You have a limited rate of change on the outputs of the op amps. In addition more time is usually required for op amp's output to come off it's rails.You need to test both CC and CV mode before making a change. What happens to one should happen to the other.CV mode to current op amp.CC mode to voltage op amp.When you find a problem for one make same change for other.With your current connections, C6 is hiding current changes from R5. In addition R5 voltage drop changes is effecting Q2 directly.moving C6 to Q2 source would let R5 see more of the load current change & let CV control see need to change.Think of what circuit is doing.A negative change of Q2 gate is output increase.A negative change on R20 is output increase.Swapping current sense makes a negative change of R5 and output increase.With all working in same direction simpler control.You have D1 & D3 setting min and controlling Q2Think of another analog OR that sets ready to take control max.Set voltage or D1 anodeSet current or D3 anodeRemember to keep testing with lower values of C6 as this will let you see problem areas better.

author: xavier60
author_rank: Super Contributor
timestamp: February 26, 2018, 01:10:18 pm
content: actually it is stable with 47uF for C6, I neglected to update the schematic.There is a delay of the op-amp's output initially dropping from sitting at full voltage. It is responsible for the remaining current overshoot.I have never seen any designs with the shunt exposed directly to the load. Ill see what happens.

author: xavier60
author_rank: Super Contributor
timestamp: February 26, 2018, 01:58:17 pm
content: Moving C6 to the other side of the shunt allows the shunt to see the capacitor's discharge current but nothing can be done about it anyway.I mainly set out to solve the problem that many PSUs must have, where the short circuit current is unlimited for some time. My Agilent U8002A will supply the full set voltage into a sub ohm load, tens of amps, for 100us before the CC loop begins to respond.

author: C
author_rank: Super Contributor
timestamp: February 26, 2018, 03:00:50 pm
content: Quote from: xavier60 on February 26, 2018, 01:58:17 pmSomething can be done, decrease size of C6 to decrease current dump.Increase control speed to reduce 100us.The more you reduce R1 the faster you can change.The more current you can pump in/out Q2 gate the faster you can change.If you have control then you can add functions.The faster you can make circuit respond the less time output is uncontrolled.Faster also reduces overshoot.Think about it. you can drive a hot rod on the street with proper control. A bad driver will end up in a tree, while a good driver has power to get out of way.The difference is how good is the control.One of the big problems for a power supply is the unknown load.I have seen power supplies that add a scaled internal voltage load & a scaled internal current load not effected by output load to get better control.Need to keep in mind total result & how each part effects local area & total.for exampleC3 has two effects, removes noise & slows changes.A cap across R20 does same for output.A cap across R21 adds output noise & speeds changes.There will be matching places for CC side.Using both an AC divider.You can use separate dividers and mix two at input to op amp.Remember that when you add a cap that the effect is different at DC then at high frequency.Try to have few caps and really think about it before adding one.Right now you are running op amps open loop with total circuit controlling gain. You have option of some local resistor feedback which would slow all changes.Note that you can use many feed back paths. Op amps are commonly used to combine two audio sources for example.Might think of three inputs to op amp, scaled output, reference input and local feedback that gets canceled or over ridden to keep output off rails.Old timers often used positive & negative feedback.If you have 1v positive feedback and 1v negative feedback. A set of matching resistors in series = 0 feedback.Remember that mode change is by conduction curve of diodes. Diodes have capacitance.Diodes only pull one way. Watch for difference in over shoot vs under shoot.Your LED's are adding error to output that is different between low and high. I would think of using a quad comparator for leds.When thinking of a change, really look at the many ways to accomplish same thing.

author: C
author_rank: Super Contributor
timestamp: February 26, 2018, 05:31:15 pm
content: You are usingNE5532phttp://www.ti.com/lit/ds/symlink/ne5532.pdfQ2 = PSMN020-150Whttps://media.digikey.com/pdf/Data%20Sheets/NXP%20PDFs/PSMN020-150W_2.pdfD1 & D3 =_____Here is my thinkingQ2 gate is a voltage control device, current is rate of change.The Q2 gate-source slows rate of change. And with out proper control can lead to ringing.D1 & D3 have non linear conduction curves but only effect positive output change.NE5532p is a voltage output device.From data sheet output short circuit current is 10-60 ma.The current limit of NE5532p limits positive output changeA part of this current goes to over coming diode pull up resistor.diode pull up is source of negative output change output & Q2 gate.To get fast you need to change fast but must remember time delays of Q2 gate-source cap & NE5532p delay.Now look atCC op amp in CV mode or CV op amp in CC mode.Both are swinging full range which takes time.If you can build a window around these, then you could shorten the swing from a little above current output needs to control of output.You can connect two more diodes below D1 & D3and build a max analog ORD1 & D3 are a min analog OR of CV/CC to control output.New Max analog OR can be used to limit the swings of the op amp'sReally try to keep both CV and CC op amps the same circuit with only difference being difference of output & reference.Also note that you can regulate down to 0.With R15 connected as shown in last the rate of change will get slower the closer to 0 you get.Think of dropping the output and catching it with the op amps.You have option of using a constant current or a better third op amp to do this function.You could add a resistor between op amp output and diode to be able to sense op amp output current.You can window a control signal and slide the window.You can also window the window and gain even more control.

author: xavier60
author_rank: Super Contributor
timestamp: February 27, 2018, 10:14:28 am
content: It did become unstable when I reduced C6 to 1uFWithout changing the speed of the MOSFET, I was able to make CV mode stable by tweaking the CV op-amp's feedback components.So I hope this means that I can now more trust the stability when I make C6 larger again.I gave it some Proportional gain also which has helped the CV op-amp regain control much sooner from voltage overshoot after a load dump.This period after an overshoot can be a problem because the CV op-amp drives the Gate to a lower than usual voltage. If load is applied during this period, it catches the CV op-amp off guard and not able to turn the MOSFET on in time causing a very deep dip in output voltage.I really must make the PCB now.

author: C
author_rank: Super Contributor
timestamp: February 27, 2018, 12:49:20 pm
content: I could be wrong but just looking at it, it's still not very good.Guess I am not explaining very good.If the two op amp circuits so not look the same you have a problem.The difference between CV & CC is nothing but gain.Really think it through.R5 uses current to create a Voltage.R20 & R21 a voltage divider output is a voltage.Looking like you just want to hack a poor CC hack on a voltage regulator..Adding cap's is very last step. Should have very good stability with out cap's.An osculation is trying to tell you it's bad.You have to find the reason it's bad and correct the real problem source not hack a patch.To see the true problem requires looking, testing and some times adding or modifying to find the actual problem.Remove feedback from CV amp and just use a resistor to gain some control.What will help is some non linear gain in path. Remember that this is one big mess of interactions and need to get all correct.If is osculates work to make it osculate at a higher frequency.First get you connections to Q2 area correct and good.With your 12V common connected to + output you are adding control noise into what you are trying to sense. This can cause problems.Moving 12V common to drain of Q2 is proper connection to prevent control currents effecting both sense.Start testing CC mode, just set CV to not harm load.Start with a load of the biggest cap you have as a load.The big cap will give you more time to use scope to look at circuit.With cap discharged you should see a quick ramp up to set current.Now think about CC op amp. It's acting like a comparator more then an an amp. This is a cause of osculationRemove C1 and replace it with a resistor.This resistor value is like current gain.As you are not changing to CV mode until load cap is charged, CV circuit should not be effecting circuit.Replace R1 with two resistors in series. Parallel one with back to back diodes. This should be the higher value resistor of the two.What this does is change your rate of change to be non-linear.You are using the conduction curve of the two diodes.With cap as load. when turned on, you will transition from diodes not conducting to diodes conducting hard.As current gets closer to set current, The rate of change will decrease due to diode curve.If this osculates decrease feedback resistor on CC op amp.Change values of two R!'sResistor paralleled by diodes is center control.Other resistor is high rate of change control.Feedback resistor is how hard to go at high rate.All interact.Once you have best action. lesser value load cap should just have voltage across cap changing at a faster rate.If you have osculation, look for cause.bypass caps should be large between op amps V+ & V-A cap to V common is a source of noise.A huge cap on output has no high frequency, as caps are for High frequency No caps, Find the problem.When you have this make CV op amp match. only feedback resistor is different.Test it with out going into CC mode.When you have this, next step is working on mode change.Note when testing in CC mode with large cap you can switch in/out a resistor to discharge the cap.Q2 will be changing from open with resistor doing allmost all the discharge back to some on resistance when resistor is switched out.If you have problems, think of a good change, think you will save time posting the change first.Think along the lines of non-linear with smooth changes.To be clear an osculation keep going.Ringing that decreases is not osculation.For example op amps have no output down control, Only veering up control. A resistor with one bypass diode changes feedback of up vs down.

author: xavier60
author_rank: Super Contributor
timestamp: February 27, 2018, 01:34:34 pm
content: Quote from: C on February 27, 2018, 12:49:20 pmNo, not at all. In my previous post I explained that I had eliminated instability when C6 is 1uF. I can even get it to be stable with no output capacitor. But I will be putting a 47uF on the output when I finish it.What part of the circuit looks like a hack? Because the CV and CC are working very well.Extra: Ill post an updated schematic after I have done the PCB and done more testing.

author: C
author_rank: Super Contributor
timestamp: February 27, 2018, 02:10:04 pm
content: If you still have 12v common connected to positive output.It is not as good as it could be and that connection adds control noise to what you are trying to sense.Proper connection for 12V common is between Q2 and current sense resistor.

author: xavier60
author_rank: Super Contributor
timestamp: February 27, 2018, 02:46:10 pm
content: Quote from: C on February 27, 2018, 02:10:04 pmThat would reduce noise but makes voltage sensing a bit more complicated, needing a balanced input amplifier.This is the schematic for the Agilent U8002A. They seem to have tied the 12V common to the + output. Difficult to be certain with the way they use ground symbols.And I can't find the output capacitor to see what size it is. It appears to be about 100uF when I do an external measurement at the output terminals of my U8002A while it's totally powered down.http://d1.amobbs.com/bbs_upload782111/files_39/ourdev_639010D487UM.pdf

author: C
author_rank: Super Contributor
timestamp: February 27, 2018, 03:02:10 pm
content: If you look at that HP supply you will see a whole bunch of differential connected op amps.And you can create a dual differential mixer using an op ampI call this a mixer as used in audio but it is actually a summer.Think differential output - differential reference.In addition to working it also lowers noise some.

author: Kevin.D
author_rank: Frequent Contributor
timestamp: February 27, 2018, 06:49:28 pm
content: Quote from: C on February 27, 2018, 12:49:20 pmThat's the challenge really :Designing (not copying or combining a few integrated regulator's) a DECENT bench power supply really takes a fair amount of knowledge and skill, many try but few succeed in creating anything decent, the web is full of them.It's far more challenging than designing a C.V/C.C source (linear or switcher) for a 'known' fixed load.Quote from: xavier60 on February 26, 2018, 01:58:17 pmTo be sure a design is up to the job you have to test the stability and frequency response under the most 'unfavorable' loads you can expect.knowing what the most unfavorable conditions possible to test under then is crucial in a design/test process.Here's a few suggestions to try :-For C.V Mode:- test at a set voltage that gives lowest voltage across Mosfet (which gives max Cgd) and near the max current rating of the supply using a current sink as the load (so high Z load) and load the output with ceramics, say 2uF worth to emulate a load which may have lot's of ceramic decoupling cap's (so this test has created the lowest possible frequency gate and load pole's using very low esr caps).Then also test with just a low ohmic load and no load capacitance and at high current and at max/min voltages (now there is no load pole and a high current (= higher FET Gfs) so the Vcntrl loop should be at max gain/speed).For CC mode test. Try load stepping at high current with various pure inductive loads (10uH,1mH,100mH), but the large capacitor you generally find on the output of supplies though makes C.C mode very stable and ruins it's transient load response .The large cap on the output of almost all bench supplies is a bit of a fudge really but it gives a very good load transient response in CV mode and makes for easy stability (rely's on large size and ESR) so the trade off is sacrificing CC mode performance in favor of CV mode performance.Best Regards . Kevin

author: xavier60
author_rank: Super Contributor
timestamp: February 28, 2018, 11:12:18 pm
content: Quote from: Kevin.D on February 27, 2018, 06:49:28 pmThat is a problem, when the regulator goes into drop out. The Gate gets over charged causing a longer than usual delay when either op-amp tries to turn off the MOSFET later. Ill try some higher current op-amps later like TLC072. Reducing the control supply rail voltages should have some advantages also.

author: C
author_rank: Super Contributor
timestamp: February 28, 2018, 11:48:48 pm
content: Most op amps have a limited current output.Here the op amps only have one direction control with the other by resistor.With limited current output of op amps in mind, think a two transistor current booster would be a better choice.This could be placed next to the gate of Q2 and give a better increase/decrease of output.If you have a signal source need to check the booster for speed.Here you can do a non linear voltage to current curve of booster with the center more action of op amp and current increasing for larger changes.have booster powered from 12v rails.Make up side match the down side of booster.Note Booster not just a high current driver.You need to check the range needed to adjust booster properly.Take measurements of Q2 gate to base.Make an xy chart of the values. One axis output voltage and other current.Then test that the booster will function over the range need with a little extra margin.Speed is important as you can easily slow the speed but not boost it.

author: xavier60
author_rank: Super Contributor
timestamp: March 01, 2018, 12:35:49 am
content: I have experimented with an Emitter follower buffer and have found little advantage in turning the MOSFET on faster than the 2ma from the pull up resistor. While Vds is above 2V, the Gate is fairly easy to drive.The TLC072 is suppose to output about 50ma which will turn the MOSFET off fast enough and I can increase the pull up current if needed also.

author: C
author_rank: Super Contributor
timestamp: March 01, 2018, 01:03:36 am
content: Think of a load like a esp32, sleep, running & transmit.very fast change and too much drop and big problems.Todays loads change very fast, and you need to test the dynamics of the supply. Lack of fast change leads to larger cap across output which is bad for CC mode

author: xavier60
author_rank: Super Contributor
timestamp: March 01, 2018, 01:09:02 am
content: This is a load transient test I did at 3A while the output capacitor was 1uF. The voltage dip on the top trace is upside down because I am measuring everything with respect to the + output.This response is comparable to that of my U8002A, and I suspect that it has a 100uf output capacitor. I wish I could be certain about this.

author: C
author_rank: Super Contributor
timestamp: March 01, 2018, 01:57:50 am
content: A transient lets things settle.To know what you have, you need to test with repeated changes.So what I see in picture is a slow response.

author: xavier60
author_rank: Super Contributor
timestamp: March 01, 2018, 02:20:08 am
content: Quote from: C on March 01, 2018, 01:57:50 amKeep in mind that I tested my regulator with only 1uF on the output.This is the same test done on my Agilent U8002A, a well respected model.I think it would be very difficult to get it much faster than this.Maybe the new op-amps will help.

author: xavier60
author_rank: Super Contributor
timestamp: March 01, 2018, 06:30:41 am
content: Quote from: C on February 27, 2018, 03:02:10 pmThe ORing is done in a clever way in the U8002A. The CC op-amp,IC9, can take control of the CV op-amp,IC8, via D42 to pin 8. Pin 8 is one of the op-amp's compensation pins that provides a connection to the point where the transconductance amplifier's output connects to the input of the output stage.So in both CV and CV modes, IC8's output has full source/sink control of the Gate.http://d1.amobbs.com/bbs_upload782111/files_39/ourdev_638271E5Q049.pdf

author: C
author_rank: Super Contributor
timestamp: March 01, 2018, 07:58:12 am
content: And this is the U8002A that has slow response to over-current.

author: xavier60
author_rank: Super Contributor
timestamp: March 01, 2018, 08:49:47 am
content: Quote from: C on March 01, 2018, 07:58:12 amYes, This test was done by connecting a 0.22 ohm resistor to its output with the U8002A set to 15V 1A. There is about 0.06 ohms of lead resistance.The DSO shows over 10V being sustained across the resistor, about 45 amps.I have little doubt that this is caused by too much delay in going from CV to CC mode, likely due to the time taken for IC9's output to slew down to where it can take control of the Gate via IC8.

author: radoczi94
author_rank: Contributor
timestamp: March 01, 2018, 09:01:51 am
content: It's interesting to see an actual supply during development. I didn't even knew, that high-er current opamps exist.

author: xavier60
author_rank: Super Contributor
timestamp: March 02, 2018, 12:44:30 pm
content: I have removed my posts that contained schematics because they had a serious mistake that has found its way to my PCB design. Ill post a corrected one after I get it right.

author: C
author_rank: Super Contributor
timestamp: March 02, 2018, 02:17:14 pm
content: I hope that you have found that CV circuit should match CC circuit.With a CC mode & a CV mode control is Min (current OR Voltage).With 0.1 R current sense you have 0 to 3V current sense for a load range of 0 to 3 amps.With a 0 to 3 volt voltage sense for an output of 0 to 30 volts output you have two matching control circuits.If the two circuits do not match then you start getting a voltage caused change not equal to a current caused change.With matched circuits, testing becomes a little easer.

author: xavier60
author_rank: Super Contributor
timestamp: March 02, 2018, 02:51:05 pm
content: Quote from: C on March 02, 2018, 02:17:14 pmDo you mean they should have the same gain and response? Can you tell me where I can find an example? The Current Sense resistor is now close to 0.05 ohms. I'll eventually get some power resistors and mount them on the heat sink.Anyways, I have corrected my massive blunder with the input circuitry on the CV op-amp.The PCB is working as well as the breadboard mock up.I'm still not totally satisfied with the CV to CC transition. Under certain conditions, there is brief dip in current after the initial spike.Oh yes, also, I ended up adding the extra op-amps to make it digital ready. There are outputs for monitoring voltage and current. And the set inputs are now positive voltage.

author: C
author_rank: Super Contributor
timestamp: March 02, 2018, 03:23:55 pm
content: Total system is what counts and is what will power your load.As I have stated many times you actually have foru states.CV mode where current is not close to your current setting.CC mode where voltage is not close to your voltage setting.CV = CCThe fourth mode actually has two sub modes.One where you are changing from CV to CC.One where you are changing from CC to CVQuote from: xavier60 on March 02, 2018, 02:51:05 pmWhat happened to using common sense for answer?Think of a load change.If the load increases then you have two things happing, Voltage drops & current increases.If you just look at the voltage drop then control will try to increase the Voltage.If you just look at the current change then control will try to decrease current.Only one is correct response.If times do not match, then you can easily have a increase the voltage followed by a delayed decrease the current.For a short this could be increase the voltage and output changes to max voltage the 30V+ followed by the decrease the current to x amps at 0V.I would think that this is common sense.

author: xavier60
author_rank: Super Contributor
timestamp: March 02, 2018, 04:13:40 pm
content: Quote from: C on March 02, 2018, 03:23:55 pmWhen the output is overloaded while in CV mode, first the voltage drops, the CV op-amp immediately starts allowing the Gate voltage to rise. The CV op-amp has to react first because it is in control at the time of the overload.The CV op-amp keeps allowing the Gate voltage to increase for some time after the current has exceeded the set point because the CC op-amp can't react quickly enough. I have reduced this delay to an acceptable level.Slowing down the response of the CV op-amp would allow extra time for the CC op-amp to react to the overload before the current rinses too high. I don't want to slow down the response of the CV op-amp.

author: C
author_rank: Super Contributor
timestamp: March 02, 2018, 04:43:45 pm
content: From you post it is clear to see you are not thinking balanced or matched.Don't slow the voltage side, speed up the current side.You would only slow the voltage side as a last choice.Think about what the test facts are trying to tell you.IF you have a problem going from CV to CC then you also have a problem going fro CC to CV.

author: xavier60
author_rank: Super Contributor
timestamp: March 03, 2018, 12:09:32 am
content: This is the corrected schematic without the extra op-amps, in case anyone wants to experiment with it.The circuitry at the output of the CC op-amp allows it to swing low at its full slew rate to take control of the Gate quickly during an overload. Q1 connects the compensation capacitor, C1, after the CC op-amp is in full control of the Gate.This quick current control response might not be important to every one.My Agilent U8002A would output about 40A for 100us in the same test.The bottom trace shows the current response when the output is short circuited with the regulator set to 10V 3A.The top trace shows the output voltage upside down because I'm measuring with respect to the + output.

author: C
author_rank: Super Contributor
timestamp: March 03, 2018, 12:39:55 am
content: https://www.ti.com/ww/en/bobpease/assets/AN-31.pdfhttps://en.wikipedia.org/wiki/Wheatstone_bridgeWould do your self some good if you studied these.

author: xavier60
author_rank: Super Contributor
timestamp: March 05, 2018, 11:30:39 am
content: Some NE5532 op-amps I had bought on ebay arrived today. Although I have some TLC072 op-amps on order from Mouser, I decided to try the newly arrived NE5532 ICs in my regulator circuit and noticed a drop in performance.I'm not the first to have found this possible problem.https://www.eevblog.com/forum/reviews/chinese-fake-op-amps/msg1443733/#msg1443733

author: radoczi94
author_rank: Contributor
timestamp: March 05, 2018, 11:54:39 am
content: My overall experience with semicons from ebay are not so good. The ST branded semiconductors are generally cheap and low quality, but the chinese ebay thingies are worse than ST parts. I'd rather buy parts from a local distributor for 2x-3x price, than wasting my not-too-much time with some fake garbage. I guess you get what you pay for

author: xavier60
author_rank: Super Contributor
timestamp: July 30, 2018, 11:29:43 am
content: My project has evolved into a useful bench supply.The linear post regulator is supplied by an offline SMPS which does pre-regulation, maintaining 3V across the series pass MOSFET.During early testing I found that the temperature of the MOSFET became much higher than that of the heat sink at high load currents.I soon figured out that the cause was the thermal resistance of the insulating washer used to mount the MOSFET to the heat sink, regardless of the type of material used.The MOSFET's temperature is now much lower. I have mounted the MOSFET directly to a 6mm thick aluminum spreader plate which is insulated from the heat sink using silicone thermal sheet. The much larger surface area results in lower thermal resistance.Ill post an updated schematic of the linear regulator section.There is a messy aspect to my quick CC response idea. Ill try to explain this also.

author: radoczi94
author_rank: Contributor
timestamp: August 02, 2018, 04:07:09 pm
content: Wow, that is a nice looking power supply! Good job, mate!A lot happened around me, since I last posted here. Long story short, I did not had time and mood to work on this project of mine. I dig up some smaller projects and finished them, to have some fun. Learnt to use the new scope I bought on a basic level, helps a lot in troubleshooting. I did a lot of research too. I will design a range switcher circuit to switch between the secondary coils of the trasformer and a dual power supply for the control circuit. Then I will start experimenting with the schematic in the german ELV magazine, swapping out the opamps and try to use a more decent reference. It will be a long run, but I learnt again, that there are no shortcuts. Keep us updated about your PSU, I really like it!

author: xavier60
author_rank: Super Contributor
timestamp: August 03, 2018, 01:28:07 pm
content: This is the schematic with changes that will allow it to work stand alone without the micro-controller.Potentiometers for the settings need to be supplied from a negative reference which can be the -8v rail.And it is possible to use a TL431 in a negative series pass regulator.Btw, those two nice blue ten turn Pots that I was using in my earlier mock up, both developed bad wiper to track contact. Drilling a small hole and injecting some PAO oil got them working again. It is important that both the CC and CV modes are fail safe, they must go low if a Pot goes open circuit.

author: radoczi94
author_rank: Contributor
timestamp: August 04, 2018, 06:23:18 pm
content: Interesting. A few questions.What is that part with the ptc kinda thing doing on the right? The ptc protects the protection diode from reversed voltage tied on the output? That fet helps to kill the output faster?As I see, the opamp opens q2 and shunting down the gate of the fet, right?Why is it better to use negative voltage as reference?

author: xavier60
author_rank: Super Contributor
timestamp: August 04, 2018, 08:38:52 pm
content: Quote from: radoczi94 on August 04, 2018, 06:23:18 pmQ5 forms a constant current sink which discharges the output capacitor when the output voltage needs to be reduced and also drains away the small amount of leakage through the main MOSFET.The diode and Polyswitch protect this circuit against a battery being reverse connected to the output, As far as I can tell, the rest of the power supply might briefly tolerate this kind of accident depending on the Current setting.The extra voltage added by the battery to the voltage drop across the main MOSFET could easily exceed its Safe Area Operating Rating.Q2 is off while the power supply is in CV mode leaving C1 disconnected from the CC op-amp's feedback path. In this state, the op-amp has full open loop gain.If the power supply is suddenly overloaded by a short circuit across the output, the CC op-amp will very quickly lower its output and take control of the MOSFET's Gate from the CV op-amp.At this point, the current that the CC op-amp will be sinking from Q4 will turn on Q2, connecting C1 into the op-amp's feedback path resulting in stable current regulation. The CC op-amp goes form open loop to Miller Integrator.When the op-amps are configured for a positive reference, it becomes an advantage for the CC side.It causes loading of the CV Pot's wiper which I try to avoid as it affects linearity and possibly setting stability.Although this concern does not apply to the the micro-controller version, I have not altered much.

author: Audioguru
author_rank: Super Contributor
timestamp: August 04, 2018, 09:20:05 pm
content: The original Greek Kit and the project at Electronics Lab used a TL081 as the voltage regulating opamp. It has a problem called "opamp phase inversion" where the output suddenly goes as high as it can if an input voltage gets within a few volts from its negative supply voltage, which happens in this circuit when the power is turned off then the negative supply drops instantly. So T1 was added to short the output of the voltage regulating opamp to the input 0V when the negative supply disappears.But this version uses a TL071 as the current regulating opamp that has the same phase inversion problem as the TL081 so its output goes high when the power is turned off and the remaining charge in the positive filter capacitors allows unlimited current to the load, BOOM!When I fixed the original kit and project 12 or 13 years ago I selected MC34071 and TLE2141 opamps that have a high 44V rating, no phase inversion problem and have inputs that work down to their negative supply voltage so that the negative supply is not needed (except to allow the diode to reduce the output voltage when the current regulation is used).A couple of years ago a few Chinese companies copied the original circuit and all its problems. I don't think they built one and tested it to see all its problems.I don't think they sell their horrible copies anymore.

author: xavier60
author_rank: Super Contributor
timestamp: August 04, 2018, 10:52:54 pm
content: I found the discussion here,http://www.electronics-lab.com/linear-lab-power-supply-digital-meter/.Does someone have a link to the schematic?In the design that I have been working with, zener diode D8 is meant to cause Q4 to stop supplying pull up current when the control rails drop. It would also help if the negative regulator has a larger input capacitor so that the positive rail drops first at power down.Is this it?https://www.instructables.com/id/Superb-Lab-Power-Supply/

author: radoczi94
author_rank: Contributor
timestamp: August 05, 2018, 09:03:10 am
content: @xavier60:I found this:http://www.electronics-lab.com/project/voltmeter-ammeter-lcd-panel/Is the type of the MOSFET critical, or it's interchangeable with another types?How about a negative reference for the CV and a positive for CC?@AudioguruThe oldest version I found of this schematic, is in a polish magazine (ELEKTROinzert) from '96. I tought, it is a good one, but turned out to be a transistor graveyard.

author: xavier60
author_rank: Super Contributor
timestamp: August 05, 2018, 10:59:59 am
content: Quote from: radoczi94 on August 05, 2018, 09:03:10 amMOSFETs can be substituted in many situations except in very fast switching applications where the MOSFETs switching and conduction losses have been carefully match to the job.There would be many parts that would happily switch a fan. I tend to use whatever I already have in stock even if it's massively over-sized for the job but still switches reasonably quickly. I would use an IRLML0030 because I already have some. Its current rating is much higher and although it is rated at 30v instead of 50v, there should be only 12v present anyway. Both parts are dirt cheap on ebay.You could use a negative reference for the CV and a positive for CC. Because I don't expect that you are going for 20 amps output and if you use a 10 milliohm shunt also, R1 can be increased to over 50K which wont load the CC Pot much anyway.

author: xavier60
author_rank: Super Contributor
timestamp: August 05, 2018, 11:06:09 am
content: I have been assuming that the Pots are 10K because my blue ones are 10K.

author: Audioguru
author_rank: Super Contributor
timestamp: August 05, 2018, 12:23:54 pm
content: Quote from: xavier60 on August 04, 2018, 10:52:54 pmThe Instructables uses the Chinese kit that is a copy of the original old Greek kit that has many parts overloaded, is not reliable and has poor performance. I am surprised that the Instructable author never tested the maximum output voltage and max current which will be much less than its 30VDC/3A rating.Here is my latest schematic and parts list that work well:

author: radoczi94
author_rank: Contributor
timestamp: August 08, 2018, 08:37:47 pm
content: I saw those chinese kits all around ebay, with a red soldermask, and a single SPT. They were too good to be true.Alright, I just bought a few mosfets, transistors and opamps to experiment with. I think I will try 2 or 3 schematics and then decide, wich one gets into the design state. Then, the cad will be started, but for now, release, the magic smoke! Will bring some news and experiences.

author: xavier60
author_rank: Super Contributor
timestamp: August 08, 2018, 10:37:34 pm
content: Quote from: radoczi94 on August 08, 2018, 08:37:47 pmWhat MOSFET did you get? The one that I'm using, psmn020-150w, was originally intended for a Buck regulator project. By chance it also has a very good SOA rating making it suitable as a series pass element.Keeping the loop formed by the input capacitor, MOSFET, shunt and output capacitor as tight as possible will lessen the chance of oscillations. Also the Gate resistor needs to be placed close to the MOSFET.

author: radoczi94
author_rank: Contributor
timestamp: August 09, 2018, 04:54:51 am
content: Thanks for the tips! It's IRFP250NPBF 214W. It has the best bang for buck ratio at the local store.

author: xavier60
author_rank: Super Contributor
timestamp: August 14, 2018, 10:30:08 pm
content: A single op-amp could be used to buffer the CV Pot, not necessarily a 741, something that will be stable at unity gain.I read in another thread that placing dropper resistors in series with Pots accentuates drift due to the usually high temperature coefficient of Pot's resistive element.I haven't done much research on voltage references, other members are likely to know more. I used a TL431 controlling a MOSFET in mine. I believe that even an LM317 can be reasonably stable when used at low power because of the resulting low thermal gradients across the die.The attached schematic is for illustration, some parts have been omitted to reduce clutter.On second thoughts, the CC Pot should have a divider feeding it because of its small voltage range.

author: radoczi94
author_rank: Contributor
timestamp: October 28, 2018, 10:19:38 am
content: Hi!Made a schematic based on the ELV22532 supply. I hope it's easily readable, tried to make the smallest mess possible.Check out the attached pdf file.I didn't made the board or any testing yet. Right now, I just want to have a supply that just works, afraid of changing out the opamps, because the circuit is compensated to the LM324 (as far as I know, the LM358 is the dual version of LM324). There are some parts that dont have value, well, I didn't calculated them yet.(About the compensation, the more I read about it, the less I know)One thing I don't really understand is the diodes between the BE of the TIP142's. That are they wanted to do with those? To avoid the reverse biasing of the BE diode?Any suggestions are welcome.

author: xavier60
author_rank: Super Contributor
timestamp: October 29, 2018, 12:12:18 pm
content: D9 and D10 seem to be for limiting reverse B-E bias of the TIP142's.Involving IC3B would do nothing useful. Only the loading of the 1K resistors might do something.

author: radoczi94
author_rank: Contributor
timestamp: October 29, 2018, 12:29:00 pm
content: Do you see any kind of crippling factor other than the probably slow opamps? I plan to solder in sockets for the opamp compensation caps, this way, I can start experimenting with other opamps to make the control system faster. I'm a little bit afraid of the voltage drop across the series output diode, but don't know if it's a big problem, or should I ground the control supply after the diode.

author: xavier60
author_rank: Super Contributor
timestamp: October 29, 2018, 12:54:32 pm
content: Stability is difficult to predict. Tweaking the compensation capacitors should fix it if necessary.Shifting ground to the other side of the diode will put the diode inside of the CC loop and cause problems.What is D9(1-2) for?

author: radoczi94
author_rank: Contributor
timestamp: October 29, 2018, 01:03:19 pm
content: I put that dual diode in for protection purposes. In case of somebody leave a battery or cap on the output when switching off the power supply, or connects the battery backwards. The power supply will have better chances of surviving this way, I think. Just had some leftover TO220 diodes lying around in the drawer, tought, I could use them for protection.

author: xavier60
author_rank: Super Contributor
timestamp: October 29, 2018, 01:10:00 pm
content: D9(1-2) won't protect for reverse polarity. D9(2-3) and the fuse should protect for reverse polarity.You could parallel the 2 diodes for better current rating.I don't think that applying voltage to the output while powered down will hurt it. Ill see what it does to my bench supply tomorrow.

author: radoczi94
author_rank: Contributor
timestamp: October 29, 2018, 01:21:10 pm
content: Oh, I see now, what you asked. I tought, if I put the D9 1-2 diode in series on the output, it could protect the series pass transistors against if some idiot leaves a cap or battery on the output.

author: xavier60
author_rank: Super Contributor
timestamp: October 29, 2018, 11:29:32 pm
content: I did some testing with my bench supply.Connecting a 12v battery to the output with the correct polarity and the bench supply powered off did not cause any problem.Connecting a 12v battery to the output with reverse polarity and the bench supply powered on caused it to current limit.I had the CC set to 1 amp. Higher current settings would cause a much higher dissipation in the MOSFET because of the added voltage across it.When I made the last tidied up PCB, I made provision for a BC548 plus some resistors and diodes that sense reverse polarity and clamp Gate drive. I have now added theses parts. So now when more than 2v reverse voltage is applied to the output, the MOSFET is turned off.I guess eventually the output capacitor will explode. I could replace it with ceramics but the capacitance varies too much with changing voltage.

author: radoczi94
author_rank: Contributor
timestamp: October 30, 2018, 06:12:35 pm
content: Hm, interesting. I don't know what will happen to the series pass element if it is a transistor. What will hapen to the circuit if it is on, and set to a voltage lower than the battery voltage? I guess, the CV opamp pulls down the gate, the mosfet closes and nothing happens.

author: xavier60
author_rank: Super Contributor
timestamp: October 30, 2018, 09:02:14 pm
content: Quote from: radoczi94 on October 30, 2018, 06:12:35 pmThe CV op-amp will turn off the TIP142's. There should be no problem.Btw, I don't understand the terms "open" and "close" with respect to transistor conduction states.

author: radoczi94
author_rank: Contributor
timestamp: October 30, 2018, 10:20:33 pm
content: Quote from: xavier60 on October 30, 2018, 09:02:14 pmSorry, the cause is probably the lack of my english knowledge. In hungarian, we open (turning on, forward biasing) and close (turning off, reverse biasing) a transistor or a diode. Thanks for the correction, I did not knew about this, will not use again.