Is 550uF too big for a power supply that has CC limit? - Page 1
Forum section: electronics / projects

author: bloguetronica
author_rank: Frequent Contributor
timestamp: January 29, 2019, 01:08:59 am
content: Hi,I'm designing a linear power supply that has both voltage and current control. Through both simulation and experimentation, I've concluded that 270uF is marginal in order to stabilize the power supply. So, I had to double the capacitance just to make sure.However, by seeing the most recent video on EEVBlog, I've concluded that I might have impaired the CC mode of my supply. Should I be worried?The attached schematic is just the output stage. There is also a pre-regulator stage, and you can see the discussion here:https://www.eevblog.com/forum/projects/is-there-any-variable-dc-dc-converter-suitable-for-pre-regulation/By the way, this project had a few corrections, in order to filter some instability caused by the interaction of the pre-regulator with this final stage. Also, the CC mode was unstable, so I had to add a capacitor (C12) near Q1. The new boards are not made yet, but they are now under way. Lets hope this works, because I've fried the whole control section of my "old" output stage board (a mystery).Kind regards, Samuel Lourenço

author: bson
author_rank: Unknown
timestamp: January 29, 2019, 05:43:47 am
content: The IC4B based sense amplifier lacks external compensation.

author: Kleinstein
author_rank: Super Contributor
timestamp: January 29, 2019, 07:14:44 am
content: How much capacitance is acceptable at the output depends on the current range and application. 500 µF is not good, but about normal for many 3 A supplies. The capacitor C12 is odd - it slows down the CC mode a lot and has good chances to make it oscillate with an inductive load. Both the current and voltage sensing are using differential amplifiers in a way that is not good for accuracy and depending on tolerances can cause stability problems. So the whole concept is not that good.Normally a supply with an emitter-follower output stage can work with a much smaller ouput capacitor. So with compensation at IC4B it should work with a much smaller (e.g. 10 µF) cap.

author: Zero999
author_rank: Super Contributor
timestamp: January 29, 2019, 12:23:04 pm
content: It's all but impossible to design a PSU with constant current and constant voltage and have good load regulation for both. A capacitor will improve the transient response of the CV mode, at the expense of CC. No capacitor an a large inductor will improve the CC mode, at the expense of CV mode.Normally we optimise for CV mode and accept that current surges beyond the current limit can occur. I've toyed with the idea of building a CV/CC power supply with a three position switch which adjusts the filtering to be optimum for CC, CV or the best of both, because no PSU I've seen offers this, although I've never got round to it.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 29, 2019, 12:39:44 pm
content: Delete C12, add a resistor in series with C11 and tweak their values; do the same with IC4B, an R+C from out to -in.Don't expect it to go terribly fast with that 2.7k (R5) supplying a darlington. I don't know offhand what C and ESR would be necessary to compensate this combination when tuned for best performance, but 100s of uF is probably not unreasonable.Why two different DACs? Seems a waste of a BOM line. Similarly, the different op-amps and comparator used could be simplified. But these don't affect the correctness of the circuit, in any case.Tim

author: bloguetronica
author_rank: Frequent Contributor
timestamp: January 29, 2019, 12:56:12 pm
content: Quote from: bson on January 29, 2019, 05:43:47 amThe compensation is done externally, via shunts in J5.Quote from: Kleinstein on January 29, 2019, 07:14:44 amI was recalling incorrectly. According to the simulations, 100uF was still marginal. A single capacitor of 150uF up to 270uF might be sufficient, after all. When you say 500uF is not good, is it too big or too small?Quote from: Zero999 on January 29, 2019, 12:23:04 pmThe CV mode takes priority here too. The CC mode is not vwery precise, and it is only there to protect the load. But I might be defeating the purpose if I use a too large filter cap.Quote from: T3sl4co1l on January 29, 2019, 12:39:44 pmHi Tim,One DAC is to control voltage, and the other is to control current. I think those resistors that you are suggesting could affect the gain, especially around IC4B (the loop is closed externally). What do you have in mind? Anyway, I've tried putting a 1K resistor in series with C11, and saw no improvement (the OPA705 is already unity gain stable). C12 seems to do the job.Kind regards, Samuel Lourenço

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 29, 2019, 01:05:29 pm
content: Well, trying a single point doesn't mean anything if you're very far from the local minima. Try 10k, 100k, 1M even, and different C values. A random search of the space is not necessary, you just have to get close enough to discover, and follow, the gradient.A resistor in series with a capacitor has no effect on DC gain.Tim

author: bloguetronica
author_rank: Frequent Contributor
timestamp: January 29, 2019, 01:17:48 pm
content: Quote from: T3sl4co1l on January 29, 2019, 01:05:29 pmBut wouldn't larger resistors (in series with C11) make IC7 unstable again (and defeat the purpose of C11)? The issue here is not local instability. This instability was caused by IC6, IC7 and Q1 combined. Each sub-system is stable by itself (tested that). C12 is there to slow down the response and make the system stable as a whole (the CC mode is now constant with any load). I don't need a fast response in CC mode, but I do understand your objection.Kind regards, Samuel Lourenço

author: Kleinstein
author_rank: Super Contributor
timestamp: January 29, 2019, 01:44:25 pm
content: C12 is not just making the current regulation slow, it also adds phase shift and thus makes stability difficult. So C12 is definitely a bad idea.If one really wants it slower it would be a larger cap for C11. Usually with this kind of circuit current regulation is already relatively slow - if one makes it too slow the simulated capacitance may be more than the real 500 µF. One problem it the output of IC7 swinging rather high, well above the active zone that is at some 1-3 V. In the circuit the OP has only 5 V supply and this already helps, but there is still some delay before the current limit sets in.A resistor in series with C11 adds a little more higher frequency gain and phase boost. If not too much it can help with stability. One may not need this if C12 is left out.

author: blackdog
author_rank: Frequent Contributor
timestamp: January 29, 2019, 02:12:44 pm
content: Hi,C12 is there to slow down the response and make the system stable as a whole (the CC mode is now constant with any load).Wrong!As others have indicated, remove this capacitor.The very common mistake when designing power supply is to make it very slow to get stability.You just need Speed...........All extra RC times, all extra active components like opamp and transistors within a control loop make this loop inherrent unstable (slower).These components delay this loop and cause the phase spmargin to get smaller and smaller, when you reach 180-degrees you have built a transmitter.And the next component that will cause you a headache is Q1 in your schematic.This transistor gives you extra loop gain that is hard to compensate.It is clear that you miss a piece of knowledge regarding loopgain, phase margin, phase margin, etc.Without this knowledge you can't design a good power supply.Good information about this is available on the TI website:https://training.ti.com/ti-precision-labs-op-amps-stability-1I hope this helps,Kind regarts,Bram

author: bloguetronica
author_rank: Frequent Contributor
timestamp: January 29, 2019, 02:13:14 pm
content: Quote from: Kleinstein on January 29, 2019, 01:44:25 pmIn the previous circuit, there was no C12, and in CC mode the circuit oscillated and the current read incorrectly, with spikes of 1.5A when the limit was set to 500mA! Already tried a much bigger C11 (1uF!) and a resistor in series. The capacitor improved, but the resistor didn't helped.Anyways, I see conflicting opinions regarding the need of C11 without C12. Taking C11 out solves the current spikes in the totally opposite direction (current is set to low, still oscillating).Kind regards, Samuel Lourenço

author: bloguetronica
author_rank: Frequent Contributor
timestamp: January 29, 2019, 02:22:26 pm
content: Quote from: blackdog on January 29, 2019, 02:12:44 pmThe fact that Q1 ruins the phase margin is a hint, but I can't get rid of it. Hence, that's why C12 is needed, to dampen the response (I don't see any oscillation at the amp's output with it). Makes sense, and that's why any combinations of C11 plus/minus a 1K resistor in series wouldn't solve this. Now, C11 doesn't need do be huge, although without it the circuit will oscillate, due to the fact that the op-amp is unstable (a different issue here!).I can only make the necessary changes if I have hard evidence that they will work. As far as I'm concerned, the CC mode was working fine, after the modifications (one of them was the inclusion of C12), until the 5V regulator blew up. I don't like C12 either, but theory doesn't make practice (also it is not as bad as if C12 was there right at the op-amp's output).Kind regards, Samuel Lourenço

author: bloguetronica
author_rank: Frequent Contributor
timestamp: January 29, 2019, 02:41:25 pm
content: Ok, I think there was an issue with my question, and I got misunderstood. It is not that my circuit was unstable. It worked fine with a 270uF output cap. I don't have stability issues. I just wanted to know of 550uF was too much.Kind regards, Samuel Lourenço

author: Kleinstein
author_rank: Super Contributor
timestamp: January 29, 2019, 02:52:27 pm
content: The IC7 with C11 still has unity gain at high frequency. So C11 can not provide the full compensation for the CC loop. So C12 may indeed be needed. So there is a chance to keep C12 (though maybe a little smaller) as the main compensation for the CC loop and make C11 small enough or provide enough series resistance so that at the higher frequencies it is C12 doing the job and C11 has still finished it's job.

author: bloguetronica
author_rank: Frequent Contributor
timestamp: January 29, 2019, 02:57:10 pm
content: Quote from: Kleinstein on January 29, 2019, 02:52:27 pmThe OPA705 is unity gain stable. I could try using a 10K resistor in series with C12, which would effectively double the gain at high frequencies, but I don't see this as a substitution for C12.The issue here is that, I was initially going to use two 47uF caps. But now I found a 270uF cap that has the same footprint, and at some point I've decided to use two of them, as a replacement (the more, the merrier, right?). However, I was alerted by Dave's video that a too big output capacitance might impair the CC limiting of the power supply.Nevertheless, I saw that a single 270uF cap would be more than enough. And perhaps, I should replace it with a 150uF cap (different footprint).Kind regards, Samuel Lourenço

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 29, 2019, 03:45:12 pm
content: The remark about unity gain is not about stability of the amp itself, but about its place in the loop. Namely, even if the amp is severely overcompensated (say by putting 1000uF from out to -in), it still has gain of 1 from +in to out, which means the loop gain has not dropped to an ideal dominant pole as you might've been expecting with such a large cap.The solution is to apply filtering to +in so it drops at the same rate.Compare this, which has G = 1 at high frequency,with this, which has G --> 0 at HF.For an error amp, R2 and R4 --> infty, so that DC error is as nearly zero as possible.For a pole-zero compensation of course, a resistor is placed in series with C2 and C4.The zero itself means HF gain levels off; the point of doing this, versus leaving it alone, is you have control of the gain and frequency where this happens, rather than being stuck with wherever G=1 rolls it off at.Tim

author: blackdog
author_rank: Frequent Contributor
timestamp: January 29, 2019, 05:09:05 pm
content: Hi bloguetronica,For example, assume that IC7 is set as a 1x amplifier and has 60 degrees of phase space.Of these 60 degrees little remains, this because of the following components in de current loop: IC6, R7+C12, Gain Q1 and Q4 your power stage, which will eat all phase margin.45 degrees is the absolute minimum you need for a good en stable power supply, this under all conditions.That at a certain capacitor value at the output connector, and the power supply does not generate, does not mean that it is stable...You will have to test this circuit at various output voltages and also at various currents.Test it dynamically with short pulses, pay special attention to the transition area when your power supply changes from CV to CC.Kind regards,Bram

author: xavier60
author_rank: Super Contributor
timestamp: January 29, 2019, 05:22:44 pm
content: From my experience, designs that have high impedance output stages, also called transconductance or voltage to current transfer function are easy to compensate for good CV and CC stability using what TI calls "Type II Compensator".This typically applies to so called floating type designs which tend to give very good performance.With a low impedance or voltage follower type output stage, in CC mode the loop gain would increase steeply with lowering load resistance.http://www.ti.com/lit/an/slva662/slva662.pdf

author: xavier60
author_rank: Super Contributor
timestamp: January 29, 2019, 05:46:27 pm
content: Have a close look at a typical LM723 implementation. Although the output stage is voltage follower in CV mode, it becomes transconductance in current limiting mode.

author: floobydust
author_rank: Super Contributor
timestamp: January 29, 2019, 07:34:52 pm
content: 550uF is too much.Prove it by setting any PSU to say 15V at 10mA and testing some LED's, as a real world example.Energy is stored in the output capacitor which will be charged at 15V which dumps into the LED load down to 3V. You thought the LED was 12V but it's a 3V part. The CC mode transition takes some time as well.So the LED might just look like a strobe light (bright) flash happened, or the LED might be dead now due to the impulse.Some lab power supplies I can't even test LED's, they surge so bad. This is my criterion, a decent PSU is at most ~100uF but if you work with discrete semi's and small parts, 550uF is too big.

author: bloguetronica
author_rank: Frequent Contributor
timestamp: January 29, 2019, 09:58:23 pm
content: Quote from: T3sl4co1l on January 29, 2019, 03:45:12 pmOp-amp stability is a science in itself. I confess that it is hard for me to understand.Quote from: blackdog on January 29, 2019, 05:09:05 pmTesting is planned for sure. Lets hope I don't fry another board. I'll have to test this using a 150uF output capacitor, see if it is stable, and do some improvements. I hope there is no instability due to the interaction and feedback with the DC-DC pre-regulator, and for that R15 and C16 will have to serve (those were added too).Quote from: xavier60 on January 29, 2019, 05:46:27 pmThanks! The LM723 implementation is no different that the typical implementation that employs a series sensing resistor that will pull the BE junction of a transistor, which will in turn pull down the output of the error amplifier. I've used a 2.7K here to avoid shorting the amplifier output. Their approach is simple and effective, but only works if you want a fixed current limiting (something that I may have to do if this doesn't work).Quote from: floobydust on January 29, 2019, 07:34:52 pmYup, even 270uF is enough to damage a LED in CC mode. Tested that yesterday. I'll have to consider the minimum value of 150uF just to keep the supply stable.Kind regards, Samuel Lourenço

author: bloguetronica
author_rank: Frequent Contributor
timestamp: January 29, 2019, 10:58:58 pm
content: Hi,After seeing a few videos, it seems that I'll have to test the transient response of IC7 to see if there is any ringing. I'll try to test for both transients on the inverting and non-inverting input. I'll also check the stability of IC4B.Kind regards, Samuel Lourenço

author: David Hess
author_rank: Super Contributor
timestamp: January 30, 2019, 02:24:01 am
content: 10 to 100 microfarads per amp is good for a simple but well designed constant voltage and constant current power supplies. Anything higher indicates problems with the frequency compensation. Usually these problems come about because of additional phase lag from multiple cascaded stages within the control loops or excessive or uncontrolled voltage gain in a level shifter.High performance constant voltage and constant current power supplies can do much better than this with only say 0.022 microfarads per amp but use faster output transistors and have provisions to prevent integrator windup in the error amplifiers. They also may use a class-ab output stage with some active pull-down capability.

author: bloguetronica
author_rank: Frequent Contributor
timestamp: January 30, 2019, 03:09:05 am
content: Quote from: David Hess on January 30, 2019, 02:24:01 amI think that this other power supply, that has the corresponding schematic attached, might have stability issues too. I ran into oscillations caused by a misplaced ground connection (www.eevblog.com/forum/projects/usb-controlled-precision-power-supply-(or-voltage-reference)/msg2007782/#msg2007782). Normally, this supply is stable, but the constant current load that that you see is being supplied via a SMPS and causes some oscillation, as long as it is not grounded to earth. I'll have to do some tests tomorrow.Never had such issue with the FAU200 model, that uses a simple buffer to control the pass transistor.Kind regards, Samuel Lourenço

author: David Hess
author_rank: Super Contributor
timestamp: January 30, 2019, 04:19:33 am
content: Quote from: bloguetronica on January 30, 2019, 03:09:05 amMaybe I am reading the schematic wrong but it looks to me like the remote sense to the voltage feedback loop is backwards, but it reminds me of a circuit I was studying a couple weeks ago from the LT1010 datasheet shown below. The OPA703 is slow enough that it should be able to control the output through a TIP31 emitter follower without problems.The first example shown below is a good example of what I was talking about. D2 clamps the current control operational amplifier to prevent integrator windup. C1 and to a lessor extend C2 maintain stability despite the current control loop being in series with the voltage control loop with two relatively fast operational amplifiers. This design requires no output capacitance but has what is effectively a fast full class-ab output stage in the LT1010. Note that the 2 ohm resistor in series with the output of the power stage aids stability when a capacitive load is present and this is an advantage of having the current shunt in series with the output.The second example is from National Semiconductor is the low output capacitance design I was thinking of. D2 takes advantage of the external compensation feature of the LM301A current control operational amplifier to clamp it preventing integrator windup preserving fast response. Q2 makes the output stage more like class-ab with the ability to pull the output down.The third example uses 100 microfarads of output capacitance for a 1/2 amp output current to control transient response because no clamping of the operational amplifiers is used. In practice I think that capacitance could be 22 microfarads without problems but they wanted extra stability.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: January 30, 2019, 04:44:06 am
content: Quote from: Zero999 on January 29, 2019, 12:23:04 pmCould it be a good idea to put a MOSfet in series with the ouput capacitor and disconnect it in CC mode?Another point of attention are the sense lines. If these are connected to the outside world to compensate voltage drop over external wiring they can get attached to external voltages and this may lead to excessive current through the 1n4148 diodes.If Vsense_Pos gets grounded somehow it looks like your power supply wil output the maximum voltage it's capable of. Owtch.

author: bloguetronica
author_rank: Frequent Contributor
timestamp: January 30, 2019, 01:17:39 pm
content: Quote from: David Hess on January 30, 2019, 04:19:33 amThe remote sense is not backwards, that much I can guarantee. The resistor networks that you see are a means to provide internal lead compensationfor both the positive and ground leads. The schematics you've shown seem to be completely different applications.Quote from: Doctorandus_P on January 30, 2019, 04:44:06 amNo, that would actually cause instability. You don't want a capacitor connecting and disconnesting when the supply is constantly jumping between CV and CC modes.Quote from: Doctorandus_P on January 30, 2019, 04:44:06 amAny supply with remote sensing is capable to do that. Some even will output the maximum when the sense wires are not connected. This one will output 10% more of the intended voltage in this case. However, I can't guarantee cases of bad usage (the cases you mentioned). The diodes are there just for EDS protection.Kind regards, Samuel Lourenço

author: David Hess
author_rank: Super Contributor
timestamp: January 30, 2019, 04:20:53 pm
content: Quote from: bloguetronica on January 30, 2019, 01:17:39 pmIf the sense inputs are relatively high impedance, then medium value resistors can be used between force and sense at the power supply to prevent failure if a sense line becomes disconnected.

author: bloguetronica
author_rank: Frequent Contributor
timestamp: January 30, 2019, 07:12:39 pm
content: Quote from: David Hess on January 30, 2019, 04:20:53 pmIndeed! And that is the purpose of R18 and R19. They were chosen as a sort of compromise. The output voltage will not get higher than 10% when the sense is disconnected on both sides. However, they don't have such a low value that may affect precision due to lead resistance (think of a shorted R18 and R19 that would defeat the purpose). They also wont dissipate too much current if the sense leads are switched.However, if the sense leads are switched, or if one sense wire gets shorted to the opposite polarity, the load will probably be damaged despite the supply being not.Kind regards, Samuel Lourenço

author: bloguetronica
author_rank: Frequent Contributor
timestamp: January 30, 2019, 07:46:31 pm
content: I think that R5 might be causing some delay and, therefore, instability. I wonder if adding a capacitor in parallel will improve things. Notice that I can't simply short out R5 because I don't want to short the IC4B amplifier's output.I employed a similar solution today to one of my supplies. See this post:https://www.eevblog.com/forum/projects/improving-the-fau200-power-supply/Of course, I'll probably have to add some capacitors elsewhere in the loop (near IC4B), as described in previous posts.Kind regards, Samuel Lourenço

author: bloguetronica
author_rank: Frequent Contributor
timestamp: January 30, 2019, 11:37:59 pm
content: So, I've been doing some mods to the FAU201 power supply too, which is somewhat similar to the one in the OP. Definitely, I'll have to put a capacitor between the output and the inverting input of the op-amp, if I want to use a capacitor in parallel with R5. The last capacitor (in parallel with R5) will increase the high frequency response, so the former is required in order to decrease the high frequency noise.In the case of FAU201, a 10nF from the op-amp output to the inverting input would greatly reduce the HF noise, although it was only a slight improvement comparing to the effect of a 3.3nF cap in the same place.Edit: Added some measurements before and after the mods. They are full of noise because of the cable that I've had to use. The scope probe doesn't fit well between the banana plugs. Even though, you can notice the improvement.P.S.: Unshielded cables make good antennasKind regards, Samuel Lourenço

author: xavier60
author_rank: Super Contributor
timestamp: January 31, 2019, 08:28:07 am
content: I have learned much from experimenting with this circuit,https://dangerfromdeer.com/2016/04/06/bench-power-supply-build-part-ii/#jp-carousel-898Notice that the author gets it totally wrong with the compensation.After adding proper compensation and other changes, it is now very stable. I did some experimenting with the output capacitor last night.CV mode is stable with a 1uF MLCC in series with 0.5Ω. CC mode is stable with no output capacitor because the output stage itself is a current source, controlled by the error amplifiers.I have mostly solved the CC op-amp's windup problem. The CC op-amp is allowed to slew at its full speed until it has taken control of the output stage.The schematic is on this page,https://www.eevblog.com/forum/projects/linear-lab-power-supply/300/I don't believe that it has turned into an over complicated monster yet.I recently changed the sharing resistors to 50mΩ, reducing dropout to 1.65V @ 4A.BTW: the Darlington at the output is not an Emitter Follower. It is driven by the voltage drop across the 2.2K caused by the current sourced by the PNP transistor, Q1.

author: bloguetronica
author_rank: Frequent Contributor
timestamp: January 31, 2019, 08:47:21 pm
content: Quote from: xavier60 on January 31, 2019, 08:28:07 amI must say that I've learned plenty in the last few days, while reading the comments, seeing some videos and making the mods. Definitely, increasing the speed and compensating (increasing phase margin) is the way to go.I've took new measurements regarding the AC noise at the output of the FAU201 power supply when loaded in the same conditions. This time, I've used the scope probe and a short piece of wire. One of the measurements was done after the mod, the other before. The noise reduction is incredible. Both capacitors are instrumental to increase phase margin.I'll apply the same principle to the project in the OP, in an effort to reduce the output capacitance and take out or reduce some capacitors that are masking the issue. Type I compensators (TI terminology) near IC4B and IC7 are definitely needed. Perhaps, I should use a type II compensator near the latter op-amp.Thanks for all the help so far!Kind regards, Samuel Lourenço

author: xavier60
author_rank: Super Contributor
timestamp: February 01, 2019, 12:16:22 am
content: My main point is, it is difficult to make Current Control work properly with a Voltage Follower output stage.

author: chickenHeadKnob
author_rank: Super Contributor
timestamp: February 01, 2019, 02:56:22 am
content: Quote from: Zero999 on January 29, 2019, 12:23:04 pmWell the HP 6632B and related family members have a fast/normal mode switch , located inconveniently on the back. Quoted from the user manual:Fast/Normal OperationA switch on the rear of the unit lets you switch between operating in either Fast mode or Normal mode.When set to Fast mode, this switch disconnects the output capacitor that is located inside the unit. Fastmode lets you improve or enhance certain operating characteristics; while at the same time degradingother operating characteristics.1. In Fast mode, the programming time for voltage programming is faster than for normal operation,however, output noise is greater.2. In Fast mode, the absence of the internal output capacitor results in increased output impedance andtherefore, greater stability when driving inductive loads. Conversely, the addition of externalcapacitive loads in Fast mode will reduce the stability of the unit during constant voltage operation.3. In Normal mode, the internal output capacitor helps control peak voltage excursions away from thethe nominal value for sudden changes in load current. In Fast mode, larger peak voltage excursionswill show up at the output of the unit during sudden load current changes.Capacitive LoadingIn Normal mode, the dc source will be stable for many load capacitances, however, large loadcapacitances may cause ringing in the dc source’s transient response. If this occurs, the problem may besolved by increasing or decreasing the total load capacitance.In Fast mode, the dc source can maintain stability only for small capacitive loads. These limits are:Agilent 6631B 2.2 μF.Agilent 6632B and 66322A 1.0 μF.Agilent 6633B .22 μF.Agilent 6634B 0.10 μF.Looking at the schematic I am lost but it appears the capacitance that is switched is 100 mu Farad for the 6632b, 50 or 22 for the other modelsI think having such a feature enabled by a front panel switch on a hobby purpose built supply is not a bad idea, however it doesn't eliminate the need to design for minimum output capacitance though.

author: bloguetronica
author_rank: Frequent Contributor
timestamp: February 01, 2019, 03:03:08 am
content: Quote from: xavier60 on February 01, 2019, 12:16:22 amWhat do you mean? I think there is some confusion. IC7, the op-amp responsible for the CC regulation, is not a voltage follower. The circuit in the OP works, but is not very well tuned.There are two DACs: IC3 sets the voltage and IC5 sets the current limit. The only voltage follower here is IC4A, and it is needed as the IC3 DAC buffer. IC4B is the error amplifier, IC6 amplifies the sensed voltage in order to "read" the current and IC7 acts as a comparator of sorts that sets the CC mode when the current is exceeded.Kind regards, Samuel Lourenço

author: xavier60
author_rank: Super Contributor
timestamp: February 01, 2019, 03:40:18 am
content: Quote from: bloguetronica on February 01, 2019, 03:03:08 amI mean the output stage transistor Q4. I don't fully understand how everything works , mainly what I have discovered from experimenting and somewhat from what others have said.Having the output transistor functioning as a voltage follower makes voltage regulation easier because the transistor itself is performing voltage regulation in having the Emitter voltage follow the Base voltage. The CV error amp only needs to make small corrections.Current regulation is easier when the output transistor/s are configured to function as a voltage controlled current source or transconductance amplifier. The CC error amp only needs to make small corrections.When one has the choice of only one of the output stage types, the transconductance type is the best compromise, from my experience.It can be made to work well for CV and CC modes.

author: bloguetronica
author_rank: Frequent Contributor
timestamp: February 01, 2019, 09:27:56 pm
content: Today, I've been doing some tests, and concluded that a capacitor in parallel with R5, R7, or in parallel with any resistor that is located in series after the output of a compensated op-amp, is a bad idea. A capacitor from the output to the inverting input is generally a good idea and works well. In my case, when testing modifications to the FAU200 and FAU201 boards, I found out that a type I compensator consisting of a 1K resistor at the inverting input and a 3,3nF capacitor from the output to the former input works well. The noise generated by the DC-DC converter is around 60KHz, and gets reduced.Quote from: xavier60 on February 01, 2019, 03:40:18 amI've far more concerned with the CV performance. The CC mode is accessory in this project, and was only added as a safety feature in order to protect the loads.Kind regards, Samuel Lourenço

author: David Hess
author_rank: Super Contributor
timestamp: February 02, 2019, 01:30:10 am
content: Quote from: bloguetronica on February 01, 2019, 09:27:56 pmThis points to a difference between common power supplies which have an adjustable current limit and more specialized power supplies like those found in source-measurement instruments. Good constant current performance requires low output capacitance and low output capacitance also limits surge current when a lower voltage device is connected to a higher voltage output.

author: bloguetronica
author_rank: Frequent Contributor
timestamp: February 02, 2019, 02:37:37 am
content: Well David, my objective is to make this supply as versatile as possible, while stable. I'm planning to add just the needed capacitance and a little extra. But CV performance is primordial. It must be low noise and high precision in terms of voltage. The current setting is not nearly as precise.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: February 02, 2019, 09:56:58 am
content: I once tried setting the ouput of my power supply to 30V and the current limit to 20mA, and then connect a led.It should just light up with 20mA right?(Note: In those days a led really needed 20mA to be somewhat visible).In my case the led exploded. Didn't expect that.But over time you learn.The power supply was ok

author: Kleinstein
author_rank: Super Contributor
timestamp: February 02, 2019, 10:43:47 am
content: Ideally a LED would just get the set 20 mA. However with real life power supplies there is usually a much higher initial current. This extra current comes from the capacitor at the output, that can be quite large in some cases (e.g. up to 1000 µF for some 3 A supplies) and also from the time it takes for the current regulation to set in.Here one kind of has the choice of 2 evil. There are mainly 2 types of regulator circuits: those with an output stage that sets a current and those where the output stage sets a voltage with a low impedance.For the 1st. type a reasonable size capacitor in mandatory for stability, but the cross over to CC mode can be fast. If the regulator is fast and well tuned the capacitor can be relatively small - so no real need for 500 µF, but some 10 µF may be required.For the 2 nd type one does not need a capacitor at the output or least can use a very small one (e.g. 1 µF). However the onset of CC mode can be slow, especially if the voltage needs to drop. So there is a kind of simulated capacitance, that can be about the size required in the other case. Often there is also some delay before the current limit sets in. This is sometimes good (e.g. when testing a SMPS or fan) and sometimes bad (e.g. the LED).So it is really difficult to get a fast acting current limit. Normally one can live with the limitations. There is no need to connect a LED to the supply set to 25 V and hope for the current to be fast enough - set the supply to 0 V and only turn up the voltage after the load is connected.

author: bloguetronica
author_rank: Frequent Contributor
timestamp: February 03, 2019, 07:28:35 am
content: I can only assess the capacitance that is needed when I have the board on my hands and when I manage to solve the issues. Everything else is a guess. Anyway, it wont be 10uF, as it is too low for this design.Kind regards, Samuel Lourenço

author: Kleinstein
author_rank: Super Contributor
timestamp: February 03, 2019, 10:24:08 am
content: The circuit shown here uses a low output impedance power stage and could in principle work essentially without capacitor at the output. So 10 µF is not such a bad target for this type of circuit. The difficulty in this circuit is more in getting the current control work well, especially if the differential amplifier is not perfect.

author: bloguetronica
author_rank: Frequent Contributor
timestamp: February 03, 2019, 11:15:43 pm
content: Quote from: Kleinstein on February 03, 2019, 10:24:08 amWell, today I had an epiphany about that differential amplifier. I think I should try a 10K resistor in series with that capacitor in the feedback from the output to the inverting input. That, in conjunction with the 10K resistor that goes in series to the inverting input, will cause the AC gain to be -1. I think that the AC gain of that stage was set too low, to almost 0, if that makes sense. Perhaps then, it is possible to eliminate that extra capacitor named C12.Given that all conditions are stable (all op-amps compensated), then it is a matter of testing at full load, in order to find the minimum capacitor.Kind regards, Samuel Lourenço

author: bson
author_rank: Unknown
timestamp: February 04, 2019, 06:59:50 pm
content: Quote from: bloguetronica on January 29, 2019, 12:56:12 pmI haven't had time to follow this thread, but an error amplifier can't have the same or greater bandwidth than the response of that which it controls, be it a voltage regulator or VCO. The way to control its bandwidth is through external compensation, and that needs to be local to the amplifier. It can't be shared. If it has excessive bandwidth it will swing wildly when the downstream stage can't respond.Edit: oh, and a shunt is a bad compensation mechanism; it works by presenting a load that makes the error amplifier operate current limited. Since it's shared by the input of the next stage, any change to the input impedance will affect the error amplifier's bandwidth! And then there's the potential for power dissipation in the error amplifier when it has to drive a large shunt cap.

author: Kleinstein
author_rank: Super Contributor
timestamp: February 04, 2019, 07:07:49 pm
content: Quote from: bson on February 04, 2019, 06:59:50 pmThis limitation is not very strict: if the system to control is well known and well behaved it is allowed to have the controlled system to be slower than the regulator. However this is not the normal case. For a simple (dominant pole) compensation one needs one time constant to be much longer than the others - in theory this could be the system to control.Still it's likely no a good idea to have an OP without a local compensation - even if only to counteract varying speed of individual units.

author: bloguetronica
author_rank: Frequent Contributor
timestamp: February 05, 2019, 01:06:57 am
content: Quote from: bson on February 04, 2019, 06:59:50 pmThat makes sense.Kind regards, Samuel Lourenço

author: Marco
author_rank: Super Contributor
timestamp: February 06, 2019, 10:13:04 am
content: A switching+linear power supply would have an advantage in getting fast CC with an emitter/source follower, it could use a smaller/faster output transistor.

author: bloguetronica
author_rank: Frequent Contributor
timestamp: February 08, 2019, 04:41:32 pm
content: I officially declare this project to be a failure. Assembled the board, and meanwhile the f*king regulator decided to deliver 24V to the whole 5V bus rail, so I had to replace many of the ICs.Second try, managed to compensate IC4 successfully using a 10nF capacitor in the feedback loop (a 3,3nF one would do, although marginally). However, when loaded, I was having interferences caused by the DC-DC pre-regulator, causing the output voltage to drift a few 10s of mV. Even soldering a 10uF capacitor in parallel with C16 and replacing R15 with a 22K resistor wouldn't solve the issue. Disconnecting the Vret line to the DC-DC made the issue go away. Meanwhile disconnected the 24V line feeding the regulator and pfff. Again!So, there are some frailties:- The DC-DC pre-regulator idea must go away;- The INA180A1 can surge the 5V line if the supply is disconnected, taking the regulator with it. Then, when the regulator is powered, it fries itself and everything on the 5V line;- The CC must go away, all of it, despite the fact that it works. A simple SC protection should work.The good news:- Apparently, a single 47uF capacitor at the output is enough to stabilize the supply, provided that IC4B is compensated using a 3,3nF to 10nF capacitor;- The temperature limiting works great.Learned the lesson. Project ditched. Thread closed!Kind regards, Samuel Lourenço