LAB PSU Schematic Explaination - Page 1
Forum section: electronics / beginners

author: xoom
author_rank: Regular Contributor
timestamp: March 19, 2016, 11:25:53 am
content: Hi,im trying to fully understand this schematic, but im stuck:)most confusing part is that PSU positive rail is as ground.. and all my understanding disappearsI colored some rails for better visual understanding and marked some control circuits..My questions is:1. Control circuit that switches relay for other transformer tap. How it senses voltage on output. How to calculate what threshold is?2. As i understand for current limiting control R22 is shunt resistor.. but again im confused how it measures voltage across it.and there is more questions but for now i just want to understand this concept.My point it to make Power supply with higher output voltage (30V@3A) and some more improvements.. like more stable voltage ref. (maybe some digital controls)

author: mmagin
author_rank: Frequent Contributor
timestamp: March 19, 2016, 01:04:11 pm
content: Wow. That's like the "how not to draw schematics" example in The Art of Electronics.And the way the voltage feedback circuit involving the 741 has a +12/-6 volt supply is extra confusing.I think I'm not awake enough this morning to do more than observe a couple things:1. that relay control circuit probably can be thought of as a comparator with hysteresis?2. The shunt for the current limit is R22+R15. It provides a voltage to the - input of U2B, and the front panel pot provides a user-set voltage to the + input (referenced to 'ground', which is the positive output!). So if the voltage across R22+R15 exceeds the voltage from VR2, the output goes low, and through the 'diode-OR' current is pulled until the pass transistor brings the output down to the limit.You probably cannot increase the output voltage without a different transformer (or rewinding the T-5/6/7 winding). I assume the T-8/9/10 winding is probably much lighter gauge. I don't think the stability should be horrible for a cheap supply, as long as that TL431 clone isn't that bad (could try replacing it with the real thing) and R3/VR3/VR4 and the 741 aren't too affected by temperature.

author: amspire
author_rank: Super Contributor
timestamp: March 19, 2016, 01:13:52 pm
content: To understand the circuit, notice that there are two different ground symbols used. The 0V "ground" that of the control circuit is just that - the control circuit 0V and nothing more.The chassis ground, or the mains earth ground is totally separate and is isolated from the power supply output.From outside the power supply, the control circuit ground means nothing to you at all.Now to explaining the circuit a little.First, the control system has its own power supply. That makes design much easier - you can change the voltage at will to to main supply without any affect on the control circuit. You can measure the voltages across a current sense resistor and have the voltage of the sense resistor right at the 0V potential of the control circuit (there is a +/- supply for the control circuit, so 0V is in the middle).Secondly, the thing that the control circuit is doing is controlling the base current of a NPN power transistor that is at a potential of about 0.7V above the output potive rail voltage. So if you choose the positive output rail as the 0V point of your control circuit, driving that pass transistor is simple.Thirdly, if the control circuit has its own independent supply, and is has its 0V at the positive output rail potential, it is easy to scale the supply to any output voltage you want. This way, they can take this design and tweak it a bit to be a 30V supply, a 60V supply, a 100V supply - whatever they want. You will probably find that this supply is part of a family, and if you get the schematic of the 30V version, you may be able to just change the components of your supply to make it 30V. That will include needing a new transformer, new mains filter capacitor, perhaps a new power transistor and the output current will be half that of the current supply. You probably have to make a new meter scale ..... Is it worth it? Probably better of selling your supply on eBay and buying the model you want.Now to your questions. The pre-relay circuit is a comparator with input connected to the positive output. The other input goes to a voltage divider that has one end connected to 12V above the positive output, and the other end to the negative rail. When the voltage across R35 is less then the supply output voltage, the relay turns on halving the voltage from the transformer. This will happen when the supply is below 15K * 12V/24K = 7.5 volts out.The current control has one end connected to the more positive end of the shunt. The other end of the shunt is of course control circuit ground. The other current control opamp input goes to voltage divider that has one end at control circuit 0V (the output) and the other end at a reference that is 2.4V above the output. So when the voltage across the shunt equals the voltage on this divider, current limiting starts.Why did they pick the positive rail for control circuit ground? They actually could have picked the negative rail and used a PNP power transistor or a P-Channel power MOSFET. They chose a NPN power transistor and the positive rail because NPN power devices and better, cheaper and more plentiful then PNP power transistors. That is the only reason.

author: Kleinstein
author_rank: Super Contributor
timestamp: March 19, 2016, 01:16:59 pm
content: This supply is a more or less classic circuit using the floating regulator. So the regulating circuit is powered by the upper transformer. Ground for this is tied to the positive output terminal, and the output stage is running as a common collector circuit.The voltage controlling amplifier looks a little odd: Normally one would have have a resistor from ground to the inverting input of the OP and a pair of antiparallel diodes instead of C1. They also seem to use one of the zero adjustment pins for compensation. In this respect the circuit is unusual and depends on the OP type used here.

author: homebrew
author_rank: Frequent Contributor
timestamp: March 19, 2016, 01:39:50 pm
content: Interesting, I get a different tripping point. But maybe I'm confused now also ...Let's see:In the end this thing works as any other voltage comparator. R5 and R35 form a voltage divider. Try to imagine that we would select ground to be the negative output rail. Then the voltage on top of the divider would be V_out + 12V. Hence the output of the divider would be v = (R5 / (R5 + R35)) * (V_out + 12V). When this falls below V_out, the relay will energize. Hence solving this equation leads to a tripping point of Vtr = (12V / (1 - R5/(R5+R35))) and hence to 19.5V.Hopefully that is not complete bullshit. If so, feel free to correct me ...

author: xoom
author_rank: Regular Contributor
timestamp: March 19, 2016, 01:44:02 pm
content: Thank you for replay.@mmaginMaybe that's why its so hard to understand for meI will use different transformer with .. 13.2V X 2 @3.2A Maybe its not reach 30V DC but its ok.. problem is that it don't have second secondary for powering control circuits.. and using same secondary for controls and power i think is bad.. as its have 'ground" on positive rail.. also i use MAX6043 for ref. voltage (2.5V) so TL431 is no needed any more..Other question what about voltage control.. its reference to negative rail of PSU. How it behaves?@amspireThank you for more explaination. Im not owning this Power supply. I want to make it my self and using this schematic as reference which i want to modify to my needsNow just need to read again your post and try to understand more betterP.S. Sorry for my bad English

author: homebrew
author_rank: Frequent Contributor
timestamp: March 19, 2016, 01:44:18 pm
content: Yeah, forget it ...I mixed up the resistors. When swapping R5 and R35, my equation also leads to 7.5V...

author: amspire
author_rank: Super Contributor
timestamp: March 19, 2016, 01:56:43 pm
content: Quote from: xoom on March 19, 2016, 01:44:02 pmAs I said, just find the schematic of the 30V version of the same supply. Then you can copy it directly. If it is an HP supply, there will be other models with the same basic circuit, but with different output voltages. You have just picked the circuit of the 15 volt model. when you building it, don't try too hard to find a transformer with all the windings. Get one big transformer for the power circuit, and a small transformer for the control circuit. The WL431 reference IC is more commonly known as the TL431 reference - very common. Might even find one in scrap PC or USB power supplies.

author: xoom
author_rank: Regular Contributor
timestamp: March 19, 2016, 02:11:42 pm
content: That's what im thinking to do with second transformer.. I sticked to this schematic becouse it looks simple (but still little confusing for me becouse in analog circuits im not so good) And yes i know TL431 but i have MAX6043 voltage ref. chip, so i want to use it. Also opamps i have ISL28207 so thinking to use themAnd want to challenge my self a little to make this PSU by modifying this schematic and calculating new values for resistorsIts more learning then need of this power supplyAlso as here is negative supplies regulated only by zeners i want to change them to voltage regulators and thinking about that -6V rail to get rid of it(i guess i could do it). and just supply only + - 12V

author: amspire
author_rank: Super Contributor
timestamp: March 19, 2016, 02:29:35 pm
content: The opamps in the circuit look like LM324 quad opamp, and the uA741 which comes from the 1970s. A bit too crusty, I think. The compensation circuits are designed to suit the chosen opamps, but your Intersil opamps may actually work. It is possible the supply control circuit may oscillate if you substitute parts.

author: xoom
author_rank: Regular Contributor
timestamp: March 19, 2016, 06:10:32 pm
content: Looks like im really stupid.. trying to redraw that comparator part in more understandable way, but allways come to a dead endthat negative rail comes like mystery ..R19 and R20 are for hysteresis (im right?) so left with R5 and R35 which is divider as reference for comparator.I'm looking at amspire formula but not understand why its calculated like thatwhen im looking at homebrew formula i see more familiar formula for divider Uout = (R2/(R1+R2))*Uin

author: saturation
author_rank: Super Contributor
timestamp: March 19, 2016, 07:07:44 pm
content: This is the schematic for a widely made clone "Mastech" 1803D, well discussed on this forum. ~$50 new or a $30 used. Overall, its fairly reliable 1980s era tech. You can save time and maybe money buying a full working unit and mod it from there.http://www.ebay.com/itm/MASTECH-HY1803D-VARIABLE-LINEAR-DC-POWER-SUPPLY-REGULATED-T2-C5-/291679603533?hash=item43e9758b4d:g:0WMAAOSwnLdWr8hs

author: Kleinstein
author_rank: Super Contributor
timestamp: March 19, 2016, 07:47:30 pm
content: The divider is between the reference and output. Its easier to see it like comparing currents thus the trip voltage is at 12 V * R5 / R35.This type of circuit needs the second transformer or extra isolated winding. Depending on the OPs used the auxiliary supply can have other, preferably lower voltages. It's similar to the system used in many HP supplies. Its well possibly to use this type as a basis for a own design with digital control - the control part also powered from the extra transformer. However the plan still has a few points that should/could be done better (e.g. current reading more accurate, better compensation for voltage control, add output enable).Power supplies can be rather cheap - so even buying the part can be more expensive than getting a cheap Chinese one and modify it to a lower current to make it more durable. So building your own supply is not about saving money but more something to learn unless you have parts to reuse.

author: Iwanushka
author_rank: Frequent Contributor
timestamp: March 19, 2016, 11:34:26 pm
content: Quote from: Kleinstein on March 19, 2016, 07:47:30 pmIn our country anything that has label "LAB" is very expensive, this clone is at around 50+euros 30v goes for 100e (single channel) or something like that, and we have plenty of "free" transformers, so that thing can be made for 1/5 -> 1/10 and so on... of the retail price, and shipping from other countries would be like buying second PSU... if not two of them ;/

author: Kleinstein
author_rank: Super Contributor
timestamp: March 20, 2016, 10:11:38 am
content: The self build solution can still be cheap, if you can start with some used / recycled part. For a lab supply something like an old (broken) Hifi amplifier would be a very good starting point. It just gets expensive when you need to buy things like the case, transformer and heat sink.Today the price for the cheap Chinese parts you get from ebay should not be so different in most parts of the world. Just a few less developed countries have difficulty in delivery or high customs duties and a few more developed ones insist on proper CE/UL/FCC markings so you may not get all crap send directly from china there.

author: xoom
author_rank: Regular Contributor
timestamp: March 20, 2016, 10:19:44 am
content: I have transformer. Have heaps of heatsinks.. its more about learning.. i want to fully understand this circuit.. I tryed to emulate in not very good simulator (crocodile physics) but i cant.. it not works like thatIf i put zero volt rail or ground rail i cant lift it like here.. only thing i can think of is to put negative voltage on negative rail and change it .. if anyone can redraw that part (comparator) in more understandable way (like for beginner) i would really appreciate it

author: xoom
author_rank: Regular Contributor
timestamp: March 20, 2016, 11:19:59 am
content: Ok done some progress i guessHere ground symbols are same as in schematic. AndRedprobe is on Negative rail becouse only this one i can vary with signal generator. So i configured it to go from 0V to -30V.And here is what i get:At about -9V (which would be +9V in normal operation) relay kicks in.. and when the voltage comes back to 0V at about -6V (+6V) it de energize relay.If im doing it right)

author: BigEarl619
author_rank: Newbie
timestamp: May 16, 2017, 10:51:52 pm
content: Hello everyone,I found this topic on the mastech power supply. What is the point of the DW3 zener diode that is connected to R10 then R10 to +12 volts and anode is connected to the OR circuit, seems like it is missing a connection from cathode to base of Q1 so it doesn't go above the zener voltage of 2V.If not, what is the purpose of this zener diode if not used a ref volt/voltage stabilizer,Thanks.

author: Kleinstein
author_rank: Super Contributor
timestamp: May 17, 2017, 04:24:41 pm
content: The DW3 zener in the HY1803D circuit is there to prevent the output to turn on when the regulated 12 V supply is not yet high (e.g. about 4,x V) enough to allow the OPs to operate. So it is more to prevent / reduce a spike on turn on.

author: FabioLucca
author_rank: Newbie
timestamp: August 14, 2017, 01:41:22 pm
content: This scheme is very similar (it seems almost a copy) of a source scheme of the December 1982 issue of Elektor (which I am assemble), read the articlehttp://www.retro.co.za/zs1ke/projects/PrecisionPowerSupply/PrecisionPSU-Elektor-Dec-1982.pdfthat will be clearer for you, I also took time to understand the operation and to make the source work, (I had to insert an extra resistor in the power circuit).Herehttps://dicasdozebio.com/2014/06/12/projeto-fonte-de-alimentacao-cc-profissional-0-40v-0-4a/has the process of assembling this source (Elektor) by a Brazilian.My assembly worked (schematic Elektor magazine), in the Proteus simulator the circuit works but I want to improve the circuit with digital control and measurement with ADC so I need to study a lot. Obviously it's cheaper to buy the ready-made source or a DC-DC converter but what fun does it have? I'm in for learning. I apologize for the bad english of google translate.

author: xoom
author_rank: Regular Contributor
timestamp: September 15, 2017, 10:17:00 am
content: Little update on old project

author: 001
author_rank: Super Contributor
timestamp: September 18, 2018, 10:28:17 am
content: Hi!Can I replace D11 D12 with LEDs to indicate voltage mode and current mode?

author: xoom
author_rank: Regular Contributor
timestamp: September 18, 2018, 10:59:12 am
content: Quote from: 001 on September 18, 2018, 10:28:17 amThats what i did

author: 001
author_rank: Super Contributor
timestamp: September 18, 2018, 11:03:49 am
content: Quote from: xoom on September 18, 2018, 10:59:12 amSimply replacement? Is color have matter? Why original circuit uses 1A diodes?

author: xavier60
author_rank: Super Contributor
timestamp: September 18, 2018, 11:05:42 am
content: Quote from: xoom on September 15, 2017, 10:17:00 amNice job. How did you get the enclosure?

author: xoom
author_rank: Regular Contributor
timestamp: September 18, 2018, 12:20:28 pm
content: Quote from: xavier60 on September 18, 2018, 11:05:42 amI found it in metal scrap yard.. actualy found 3 of themQuote from: 001 on September 18, 2018, 11:03:49 amYes, color doesn't matter .. maybe becouse they are popular and cheap

author: xavier60
author_rank: Super Contributor
timestamp: September 18, 2018, 02:47:45 pm
content: I hope those blue Pots aren't the same asmine. They started going intermittent open circuit.This will cause the output voltage to increase in some power supply designs.

author: xoom
author_rank: Regular Contributor
timestamp: September 18, 2018, 04:00:38 pm
content: i have another PSU with those pots.. so far so goodhope they dont start to die

author: ferdieCX
author_rank: Regular Contributor
timestamp: September 19, 2018, 07:39:07 am
content: Quote from: xavier60 on September 18, 2018, 02:47:45 pmThis is a quite common design error.

author: xavier60
author_rank: Super Contributor
timestamp: September 19, 2018, 07:55:46 am
content: Quote from: ferdieCX on September 19, 2018, 07:39:07 amI would rather that the Pot adjusted the reference voltage into the error amplifier.With the addition of bleed resistor, the power supply becomes fail safe to a Pot failure.

author: 001
author_rank: Super Contributor
timestamp: September 19, 2018, 02:42:13 pm
content: Quote from: xavier60 on September 19, 2018, 07:55:46 amCan You draw mod for this shematic?

author: Kleinstein
author_rank: Super Contributor
timestamp: September 19, 2018, 03:28:24 pm
content: When replacing the diodes (D11 / D12) in the regulator with LEDs, chances are the LEDs can see quite a bit of reverse voltage, possibly more than specified for most LEDs. Chances are it can still work, though outside the specs. The diodes only need to conduct a small current (about 1 mA), just enough for LEDs. With slightly smaller value for R10 the current could be increased a little if needed.Changing the circuit, so that the reference voltage is adjusted instead of the divider is a good idea. This is not just because of possibly bad contacts, but also to have an essentially constant regulation loop. So regulation can be faster at high voltage and less ringing at low voltageIt likely needs an extra buffer (e.g. 1 more OP) at the upper end of R3 and the pot as a divider before the OP.There are a few more odd points in the circuit (e.g. C1, no resistor at the inverting input of voltage regulator, DW3 not fully avoiding spikes, not fast current limit). So the general concept is OK, but the details could have been solved better.

author: 001
author_rank: Super Contributor
timestamp: September 19, 2018, 04:02:01 pm
content: Thanx!Is it clone of something wellknown circuit? Any impruwements?

author: Kleinstein
author_rank: Super Contributor
timestamp: September 19, 2018, 08:28:21 pm
content: The circuit is one classical type of regulator. Hard to tell if it is a clone of another circuit - there are lots of lab supplies (something like 75%) that use a similar type of circuit, with the floating auxiliary supply for the regulator. Especially in the old days it was quite common to adjust the divider as shown, though it is not really good.The main points to improve on this circuit would be:1) add a fast (usually fixed) current limit, as the present current limit via the OP is rather slow to react. The speed of the current limit is a 2 sided point: a fast reaction can also be confusing. Ideally one would choose slow / fast or have an independent (higher) fast limit.2) A better lockout for insufficient auxiliary supply, to prevent spikes on turn off / turn on.3) adjust from the reference side and thus constant loop gain. This comes naturally when using digital control via DAC is used.So modern, digital controlled ones usually make it right.4) LEDs to show CV/CC mode active - usually a comparator from the outputs of the 2 regulating OPs. Directly using LEDs for the diodes is tricky.5) Some windup limiting, especially for the voltage (e.g. diodes across the OP inputs would be a 1st step).6) possible 2 power transistors parallel - Chinese supplies tend to be rather optimistic with power handling capabilities7) some adjustments to the frequency compensation to alow faster reaction and reduced output capacitance (e.g. 100 µF instead of 470 µF).

author: Kleinstein
author_rank: Super Contributor
timestamp: September 21, 2018, 07:53:17 am
content: The transition from CV to CC mode is a tricky part. In the simple circuits (like shown) it take quite a bit of time until the CC regulator comes out of saturation. Some 100 µs are a little on the long side and 2 µs is already rather fast. There are a few tricks to speed up this time down to the µs range. However there is a downside to the CC mode to take over so fast: to counteract fast load transients the CV sometimes need to give short pulse of higher current. So quite often a short time higher current is actually good. Ideally one would have 2 current limits: an accurate one, that is a little slow and may take some 10-100 µs to engage and an additional fast one that is a little higher (e.g. 150%+1A). The common solution is to have a fast, but fixed limit at something like 1.5-3 time the maximum current and than a moderately slow enacting adjustable limit.There is also the charge from the output capacitor, that can deliver quite some current on a short, with no control. A 100 µF cap charged to 20 V can deliver 20 A for 100 µs in case of a "short". So there is limited use of having the current limiting much faster than this, though it gets more moderate at lower voltage.

author: YaakovK
author_rank: Newbie
timestamp: May 20, 2020, 01:29:47 pm
content: Hello Xoom,Just wondered if you can help me to find a schematics of the HY3003D model?Many thanks