YALPS! Yet Another Linear Power Supply - Page 1
Forum section: electronics / projects

author: pietr
author_rank: Contributor
timestamp: May 06, 2013, 09:35:47 pm
content: Hi everyone!Like with many others, Dave's videos have tickled my engineering senses and have switched my mind from a passive EE lurker mode to a 'let's create something' mode. After watching Dave's LT3080 PSU videos, I decided a nice beginner project would be to create my own linear power supply. I did some digging around on the forum and on some other sites and based on some very interesting comments (e.g. those by free_electron), I decided not to go for a LT3080 solution, but to go for a classic opamp and transistor approach instead.I based my design on the Agilent E3610A (with a few changes). Here are the basic specs I'm aiming at:* Output voltage: 0 - 15V* Output current: 0 - 3A (over full voltage range, this is different from the E3610A)* Digital (MCU) controlled (this is also different from the E3610A)* Automatic switching between transformer tapsI understand about 95% of the Agilent E3610A schematic (see PDF in attachment), but I still have a couple of question I hope you guys can help me with:* What is resistor R1 good for? It is a 2K 3W resistor placed in parallel with the big input capacitor C2.* What is resistor R46 good for? It is a 390 ohm 3W resistor placed between the transistor output and the negative output terminal.* How critical are the values of the resistors and caps in the small feedback loops of the op-amps (R6, C6 and C12) and how should I determine them? I suppose these components are used to get a stable control loop?* Why does the high-voltage (120V output) E3612A have two 500V 0.1uF caps between the output terminals and earth? I suppose this provides some additional filtering?Edit: just to clarify: the part numbers in the text above are referring to the HP schematic, not my own schematics.I have included my schematics as attachments. I'm curious what you guys think of my design...Edit 3: updated schematics to latest version (3 Sep 2013)

author: ddavidebor
author_rank: Super Contributor
timestamp: May 06, 2013, 09:54:27 pm
content: I can't see the images (my problem) but r1 based on your description is for discarge the caps, so if you touch it after some minutes, you'll remain alice.

author: c4757p
author_rank: Super Contributor
timestamp: May 06, 2013, 10:03:22 pm
content: Quote from: ddavidebor on May 06, 2013, 09:54:27 pmIf you start as Bob, do you become Alice after touching it?

author: kizzap
author_rank: Unknown
timestamp: May 07, 2013, 12:21:09 am
content: Quote from: c4757p on May 06, 2013, 10:03:22 pmOnly on TuesdaysAs to the OPs questions:R1 to me sounds like what ddavidebor said, a drain resistor so that when the PS is switched off, the main caps drain and don't kill the service techsIm assuming you mean R17 (not r46 which is a 1k resistor for the io panel header....). As to it's purpose, I am not 100% sure, possibly some form of minimum load resistor?C6 is a decoupling Cap on the power input on the MCU, I'm therefore going to assume that you are talking about C9, C10, C12 and C13 on the DACs in the second PDF? If so, they are low pass filters, which are used to help change the PWM control signal from the MCU to a voltage level for the control circuit.And again I cannot find whatever components you are talking about here, but I would assume that they are a form of filtering. The more noise that goes into a system, teh more noise you will have on the output.Just a tip: For better help, make sure when referencing components, that you have the correct component names. It helps....

author: rdl
author_rank: Super Contributor
timestamp: May 07, 2013, 05:19:06 am
content: I'm pretty sure the part numbers in the questions are referring to the HP schematic which is in the attached pdf , not to the schematic in the png image which is attached separately.

author: pietr
author_rank: Contributor
timestamp: May 07, 2013, 06:44:14 am
content: Yes, I should have clarified the component numbers are referring to the HP schematic.So R1 is to save any service tech from a shock due to C2, that makes sense... Perhaps R46 has a similar purpose, i.e. to discharge the output capacitor C3?

author: kizzap
author_rank: Unknown
timestamp: May 07, 2013, 08:52:22 am
content: ah, how did I miss that schematic....R1 is definately there to drain C2, I still maintain R46 is something to do with minimum load regulation.readhttp://www.alphascientific.com/technotes/technote3.pdfespcially the part about Frequency Compensation. Might give you somewhere to start reading.And the 500v 0.1uf are there to filter too I suspect.-kizzap

author: pietr
author_rank: Contributor
timestamp: May 09, 2013, 04:00:52 pm
content: Thanks for the advice. I read the article but I think I will wait until I have something build up on a bread board before given an attempt at stabilizing the control loop.For now, I'm trying to replace transistor Q3 (in the power sheet), which controls the base current of the main power transistor Q2. In the current schematic and the HP schematic, a 2N4036 (http://www.farnell.com/datasheets/39769.pdf) is used for this transistor, but I can't get those at a decent price from my supplier. So far I've found the following alternatives, but I don't know if they're suitable replacements because I don't know which parameters are important...2N4037 -http://www.farnell.com/datasheets/1702564.pdf-> this one seems pretty close in specs, here is a comparison:http://www.centralsemi.com/leadedpdf/2n4036.pdf2N4235 -http://www.farnell.com/datasheets/1702568.pdf-> this one has a much lower transition frequency, is that OK?2N4032 -http://www.farnell.com/datasheets/1702562.pdfCan anyone give an indication on which parameters are important for this transistor?

author: ConKbot
author_rank: Super Contributor
timestamp: May 10, 2013, 09:21:34 am
content: Dont forget a diode in anti-parallel with your output transistor, so if your output voltage is higher then the input cap voltage it wont pop your output transistor. Ive already had to buy a replacement 2n3055 for my $60 cheapie bench power supply when I was using it to charge a (large) battery, and when I was done, I flipped the switch to turn it off, and then started to disconnect stuff. It didnt have an output disconnect, so generally the power switch worked fine for turning on and off. Well after that, it didnt work right, and a bit of trouble shooting and I figured out that it was the transistor mis-behaving. I still need to bodge in a diode, but it worked just fine after replacing the 3055.Also, and resistors directly across the output help when you turn down the power supply output voltage, especially since your outputs are controlled by a DAC, large steps with a minimal load will take quite a while to settle, even with only a 470uF on the output. In an older HP service manual I dug though, there was a whole "down programming" circuit, for dropping the output voltage quickly when you were remotely controlling the power supply.Perhaps use some of the unused op-amps to drive a BJT+current limiting resistor as a pull down the output if the voltage control amp is trying to drive the output transistors into cutoff instead of biasing it into a linear region.

author: pietr
author_rank: Contributor
timestamp: May 12, 2013, 07:47:42 pm
content: Great, thanks for the tips! I'll definitely add the diode to protect the output transistor and I'll look into adding this 'down programming' circuit. I think I'll first build up (parts of) the PSU as such and see what the down programming performance is without specialized circuitry...

author: pietr
author_rank: Contributor
timestamp: May 12, 2013, 08:10:30 pm
content: I just noticed in the 2N6056 datasheet that the device already includes a protection diode anti-parallel with the transistor. I suppose this internal diode will suffice, and I don't need to add another one...Edit: here is the datasheet:http://www.svntc.com/TPDF/329.pdf

author: ConKbot
author_rank: Super Contributor
timestamp: May 13, 2013, 06:18:13 am
content: Agilent has some vague literature about what a down programming circuit does but no schematics.If youre just hand operating the power supply and have no intentions of computer controlling it or trying to ouput a waveform or the like, then you'd probably be fine with a pull down resistor, 470uF of output capacitance isnt huge, so it shouldnt take long to pull down. Just throwing the idea out there because it never hurts to have little tidbits like that on occasion for weird stuff that can pop up.

author: pietr
author_rank: Contributor
timestamp: May 27, 2013, 07:34:16 pm
content: So I bought some components and started building up a few circuits on a breadboard and I've run into the first problem: I was using an SN74F125N bus buffer (http://www.farnell.com/datasheets/1455119.pdf) to drive my 5V reference voltage by the PWM output of the MCU, but apparently this bus buffer causes a significant voltage drop, since I'm only getting about 4V on the output! After carefully checking the datasheet, this turns out to be normal for this component.Unfortunately, I can't seem to find a buffer in a DIP package that will let my reference voltage pass through unharmed. Any recommendations for such a chip?An alternative would be to simply use a proper DAC, such as the MCP4922 (http://www.farnell.com/datasheets/701582.pdf). Would such a DAC (with an external reference voltage) typically outperform a PWM + low-pass filter solution in terms of output ripple?

author: SeanB
author_rank: Super Contributor
timestamp: May 27, 2013, 08:24:01 pm
content: Use the HC or C version, they do rail to rail at low current outputs.

author: Paul Price
author_rank: Super Contributor
timestamp: May 27, 2013, 09:19:24 pm
content: Use only 8 pins of a 555 instead of 20 if you go with a 74C/HC/HCT245. This will do the job with 8 pins and you regain a port pin for use on your MCU. Use schottky diodes to gain another .5V or so at the output.There have so many good suggestions, here, but what exactly is your final schematic? This would help to make sure everyone is on the same page to offer parts recommendations.

author: edavid
author_rank: Super Contributor
timestamp: May 27, 2013, 09:55:15 pm
content: Quote from: ddavidebor on May 06, 2013, 09:54:27 pmNo, this is a low voltage supply, there's no safety issue. It may be there to keep the output from spiking too much on turnoff, when the control circuit drops out.

author: Monkeh
author_rank: Super Contributor
timestamp: May 27, 2013, 10:28:47 pm
content: Quote from: edavid on May 27, 2013, 09:55:15 pm240V is low voltage, too. There are safety issues with large capacitors when charged, whether the voltage is 6V or 600V.

author: edavid
author_rank: Super Contributor
timestamp: May 28, 2013, 04:14:31 am
content: Quote from: Monkeh on May 27, 2013, 10:28:47 pmNo way. It's only 10000uF. That is not why the bleeder is there.

author: ConKbot
author_rank: Super Contributor
timestamp: May 28, 2013, 04:40:20 am
content: Quote from: SeanB on May 27, 2013, 08:24:01 pm+1 74Fxxx logic was the bane of my existence for a few weeks at work for a while when we had a logic control board that was temperamental at best in a system we were working on. High input currents causing fan out problems, 5v outputs on systems that was actually 3.7V (well it would peak upto 5v and then drop to 3.7...) Replacing with HC series stuff as needed made everything play nicer.Well to be fair a pair of linear regulators for the 5V rail that were paralleled with a 0.1ohm resistor and conditionally oscillated at 40MHz depending on the load, when we were troubleshooting in the field with a 20MHz scopemeter caused more trouble. But the F/HC issue was more of a pain to fix. TIP: just because you plug something in, noise shows up, unplug it and the noise goes away, doesnt mean that item is the problem.Also tell the green EE that just because the SOIC-8 regulator can handle 1.5A, it doesnt mean it can handle a 5V 1A supply when its fed from 12V

author: SeanB
author_rank: Super Contributor
timestamp: May 28, 2013, 04:03:01 pm
content: Quote from: ConKbot on May 28, 2013, 04:40:20 amIt can handle it, so long as you consider operating for under 2 seconds before going into thermal shutdown is operating correctly.

author: pietr
author_rank: Contributor
timestamp: May 28, 2013, 06:51:25 pm
content: Quote from: Paul Price on May 27, 2013, 09:19:24 pmThe final schematic hasn't changed very much: only a few additional components and modifications. Anyway, I updated the schematics in the first post to the latest version.I don't quite understand the 555-circuit you showed. How will this save me an MCU pin? I only require a single pin per PWM output channel with the 74F125-based design. Will the 555-circuit give me an accurate PWM output voltage from my voltage reference?About the 74HC125: its specs seem much better than the 74F125, but there is still a voltage drop, especially when pulling some current (e.g. a 0.2 volt drop when taking 6ma), I wonder if this will influence the analog output voltage after the low-pass filter. The output of the filter is connected to a high-impedance input, so it won't draw a lot of current at all, but the filter itself will cause some current to flow through the buffer. I guess it's worth to try this out...However, I'm still thinking about using a proper DAC such as the MCP4922 (http://www.farnell.com/datasheets/701582.pdf), but I'd like to know how its performance will compare to the PWM-based solution. Will do some reading on this now...

author: edavid
author_rank: Super Contributor
timestamp: May 28, 2013, 07:29:12 pm
content: Quote from: pietr on May 28, 2013, 06:51:25 pmThere shouldn't be any DC error, but you could easily make your filters higher impedance, and/or parallel multiple 74HC gates, or just switch to a 74AC part.

author: Paul Price
author_rank: Super Contributor
timestamp: May 29, 2013, 12:46:35 am
content: I think things are getting a little confused here. Why are you trying to PWM buffer a reference voltage? PWM itself serves to create a varying voltage that depends on duty cycle. 0 duty cycle =0 V and 100% duty cycle = whatever reference voltage you have connected to the VDD terminal of the digital buffer (74HCT245, for example) you would like to use.If you take the outputs of the HCT buffer and connect each to a single RC filter to integrate the PWM signal to whatever precision(PWM ripple voltage) you desire and each integrating capacitor connects to a + input of a voltage follower op-amp, then the filtered PWM voltage is very lightly loaded and accurate. You then have a chance to output the reference voltage that you want here set by PWM. This is a simple open-loop multi-channel approach to D to A conversion.In MCU monitored circuits I have built, I close the loop and monitor the desired effect of the PWM control (for instance, the output voltage) and adj. the PWM accordingly to achieve the desired results using (10 or 12-bits) A2D inputs of the particular MCU that's watching the show.Am i getting too technical?

author: pietr
author_rank: Contributor
timestamp: May 29, 2013, 07:16:14 am
content: Yes, this is exactly what I had in mind. It's just that I was worried about the effects the buffer's voltage drop would have on the output voltage. Suppose I have a reference voltage of 5V (so that's connected to the VDD of the buffer) and I want a 2.5V output, hence I need a 50% PWM duty cycle. Suppose the output has already converged to about 2.5V. When the PWM input signal is high, the buffer output will be high and the filter's capacitors will begin to charge. This creates a current through the buffer, which will make its output voltage be 4.8V (for instance) instead of exactly 5V. Then suppose the PWM input signal is low, so the buffer's output is also low and the filter's capacitors will begin to discharge. This again creates a current through the buffer, which makes the output voltage be 0.05V (for instance) instead of exactly 0V. At a 50% duty cycle this will result in an output voltage of (4.8 + 0.05)/2 = 2.425 volts instead of 2.5V.So the difference in pull-up and pull-down resistance (is that the correct term?) of the buffer might cause a non-linear duty-cycle to output voltage relation. However, as you've said, I could correct for this by 'closing the loop' and monitoring the output voltage with the MCU and adjust the PWM duty cycle to get as close as possible to the set output voltage.

author: Paul Price
author_rank: Super Contributor
timestamp: May 29, 2013, 01:22:02 pm
content: PWM always sets a voltage level in the circuit we described here by always slightly pumping up the integrating cap during the high interval of the PWM pulse and slightly draining this cup of electrons during the 0-V portion of the duty cycle. It doesn't matter if the high and low outputs are asymmetrical as PWM achieves the desired output by dithering and integration.

author: pietr
author_rank: Contributor
timestamp: September 03, 2013, 11:22:05 am
content: A few months have passed and I've finally found the time to prototype this design. The digital part seems to working as expected, but the analog power design is oscillating like crazy. I've built up the the current control, the voltage control and the transistor driver (see attached schematic) on a breadboard (see pics), using potentiometers to set the current and voltage (so the digital part is not at play). I measured the output voltage and voltage control op amp output signal with my scope (see attached screenshots: yellow is inverted output voltage, blue is voltage control op amp output voltage) and it clearly shows the oscillations. The circuit has a 68 ohm resistor as a load and is set to constant voltage mode (my multimeter indicates about 0.5V output voltage). If I change the constant voltage setting using the potentiometer, the oscillation frequency changes (which is expected, I suppose).Any idea why the circuit is oscillating this badly? I've used very similar components as in the original Agilent PSU, so I would expect this circuit to be more stable. Can this heavy oscillation just be caused by using a breadboard instead of a proper PCB, or is there something more fundamentally wrong with the circuit?

author: sync
author_rank: Frequent Contributor
timestamp: September 03, 2013, 12:51:34 pm
content: Which op amp do you use?The E3610A uses two pass transistors in parallel. You only one. Maybe this causes instability.Try shorting R11. The voltage error amp compensation of the original E3610A has some problems. Seehttps://www.eevblog.com/forum/testgear/hp-agilent-e3610a/Try about 1uF and 75 ohm in series across R20. Look at the E361x schematic. There is R10, C10 across the voltage pot.

author: pietr
author_rank: Contributor
timestamp: September 03, 2013, 09:54:09 pm
content: Quote from: sync on September 03, 2013, 12:51:34 pmAn LF412CN, which is the dual version of the LF411 used in the E3610A.Quote from: sync on September 03, 2013, 12:51:34 pmMaybe, but I don't have another 2N6058 to test this.Quote from: sync on September 03, 2013, 12:51:34 pmThis improves the response a lot. It makes the output stable for voltages less than about 0.60V, but now I get a different kind of oscillation for higher voltages (see attached screenshot, notice the different voltage scales as opposed to the screenshots in my previous post).Quote from: sync on September 03, 2013, 12:51:34 pmThis doesn't seem to work. Doing this causes a lot more oscillation and turning the voltage adjust pot doesn't influence the output voltage anymore. I'm using a electrolytic 1uF cap, maybe I should use a tantalum instead (but I don't have one laying around for the moment)?

author: sync
author_rank: Frequent Contributor
timestamp: September 03, 2013, 10:28:18 pm
content: Quote from: pietr on September 03, 2013, 09:54:09 pmTry lower values for C2. Maybe a couple of 100pf. Keep R11 shorted.QuoteThe voltage pot should still working. Maybe i described it poorly. Look a the attachment.Edit:Does your original circuit oscillates only at low voltages?At 0.5V only 5uA flow through R20. Maybe this is to low for breadboarding. In the E3610A there is a constant current through the resistor (voltage pot).

author: pietr
author_rank: Contributor
timestamp: September 09, 2013, 08:58:11 pm
content: No progress so far. Here's what I've tried:Quote from: sync on September 03, 2013, 10:28:18 pmLowering C2 doesn't seem to have any measurable effect.Quote from: sync on September 03, 2013, 10:28:18 pmThe attachment describes how I interpreted it. I double and triple checked the connections I've made, but adding this resistor and capacitor over R20 totally messes up the response: the output oscillates wildly. I tried using a 1uF tantalum instead of a 1uF electrolytic, but this doesn't seem to have any effect.Quote from: sync on September 03, 2013, 10:28:18 pmNo, the original circuit oscillates at basically any voltage except 0V. After shorting R11 the output is stable for low voltages only. I've tried using 3.3K and 10K for R14 and R20 respectively (10 times lower than their original values) but this seems to slightlylowerthe voltage at which the output becomes unstable (about 0.4Vinstead of 0.6V).I'll do some more measurements and experiments tomorrow. Thanks a lot for the help so far!

author: megajocke
author_rank: Regular Contributor
timestamp: September 11, 2013, 04:02:47 am
content: Shorting out the 1M5 resistor is probably a good start, but that 1 MHz oscillation needs to be fixed.There does not seem to be any decoupling caps close to the opamp on the breadboard. That might be a problem. Also, I'm a bit suspicious about those 1N400x diodes. They might have a little too much capacitance at the opamp inputs.Is the output capacitor really properly connected? The previous scope shot showed the output voltage (I assume) shooting up to 16V in less than a microsecond which doesn't make much sense if the capacitor is properly connected.

author: pietr
author_rank: Contributor
timestamp: September 11, 2013, 05:01:01 pm
content: I managed to fix the oscillations by adding a 100nF capacitor from the op amp's non-inverting input to ground. I noticed that when I probed this input with my scope, the oscillations disappeared up to about 1.5V on the output. The cap seems to fix the oscillations completely.Quote from: megajocke on September 11, 2013, 04:02:47 amI added a decoupling cap shortly after taking the picture of the breadboard.Quote from: megajocke on September 11, 2013, 04:02:47 amYou are right. If I remove the 100nF cap that fixes the oscillations and remove those diodes as well, then the oscillation is still gone. I guess I will leave the diodes and the cap to ground in (because the diodes are there to protect the op amp) and see if the circuit is still fast enough to get a decent transient response.Quote from: megajocke on September 11, 2013, 04:02:47 amWell the cap is connected straight across the output, without much resistance in its path, so I guess it can charge up very quickly.Thanks for the help! Now I'm on to test the constant current part...

author: sync
author_rank: Frequent Contributor
timestamp: September 12, 2013, 03:17:31 pm
content: Quote from: pietr on September 11, 2013, 05:01:01 pmThis will make the control loop really slow.QuoteReplace the diodes at the op amp inputs with small signal ones like 1N4148. And in the E3610A the diodes are connected differently. Like the attachment.D27, D28 should also signal diodes. 1N400x are too slow.

author: pietr
author_rank: Contributor
timestamp: September 12, 2013, 07:15:38 pm
content: Quote from: sync on September 12, 2013, 03:17:31 pmI will try replacing them with 1N4148's. Note that my voltage control circuit is slightly different from that of the E3610A, by design. The reason is simply that the E3610A uses an analog pot to set the voltage, while my circuit is designed to take a 0-5V signal (from an MCU) as input to control the output voltage. One thing I only noticed now is that R13 in the E3610A schematic is rated for 3W, even though there will be hardly any current flowing through it (if I'm not mistaken). I wonder why it needs to be a 3W resistor... I suppose this resistor plays a different role in the E3610A schematic than R15 does in my design (where it is only used as part of the op amp input protection).What would be the advantage of connecting the diodes like in your attachment?Quote from: sync on September 12, 2013, 03:17:31 pmThese are already signal diodes, for some reason the part numbers didn't show up on the schematic.

author: sync
author_rank: Frequent Contributor
timestamp: September 13, 2013, 02:18:45 pm
content: Quote from: pietr on September 12, 2013, 07:15:38 pmYes, i noted that already. It's maybe a problem. Your circuit is high impedance compared to the E3610A. Perhaps it's prone to oscillation.QuoteThe E3611A uses the same 3W resistor. When a voltage source (battery charging, paralleling PSUs) is connected to the output and the voltage is turned down then the current flows from S+ through CR6, R13, R37 (= 0 ohm) to the negative terminal. At 40V (E3611A) there will be about 1.5W dissipated in R13.In your design this doesn't happen of course.QuoteThis is how HP done it since ages. I'm not an EE and perhaps wrong or miss something. If the diodes are connected like in your circuit then the 40V from above will going from S+ through R8, CR6, R13, R37. There would be about -20V at the op amp inputs - ouch!CR7 is (only?) for discharging C10 if the output are shorted, etc. The PSU is already protected against reserve polarity by CR3.Your circuit may oscillates because of the diodes capacitance and a additional cap to ground help. In the HP design the diodes goes directly to ground (S+). This maybe prevent your oscillation problem.

author: pietr
author_rank: Contributor
timestamp: September 17, 2013, 07:32:51 pm
content: Quote from: sync on September 13, 2013, 02:18:45 pmI see. You are absolutely right!Quote from: sync on September 13, 2013, 02:18:45 pmI don't see why there would be -20V at the op amp inputs. CR6 would ensure there is max 0.7V (1 diode drop) across the inputs, right?Strangely I can't seem to reproduce the results I previously had when leaving the diodes out. The circuit still oscillates with the protection diodes removed. It also oscillates with 1N4148 diodes instead of the big 1N4004's. Changing the diode connections like in the E3610A circuit doesn't help either.I now use a 2.2nF capacitor from R15 (attached to the non-inverting op amp) to ground, instead of the 100nF one I used in my previous post. This seems to be enough to make the circuit stable. I will have to measure whether it messes up the transient response. I also changed R14 to 10K and R15 to 33K, which should provide the same amplification but is lower impedance (but still high enough not to load the PWM low pass filter too much). I'll update the schematics in my opening post shortly.

author: sync
author_rank: Frequent Contributor
timestamp: September 18, 2013, 06:55:46 pm
content: Quote from: pietr on September 17, 2013, 07:32:51 pmQuoteMaybe the breadboarding causes all that trouble.QuoteWith this capacitor you basically have a RC low-pass. R is about 26k (R14 and R20 parallel + R15). With 2.2nF the time constant is 57us. I don't think you can get the performance of the E3610A with it. But experiment with it's value in your final build.QuoteDo you mean R14 and R20?

author: pietr
author_rank: Contributor
timestamp: September 18, 2013, 08:04:55 pm
content: Quote from: sync on September 18, 2013, 06:55:46 pmRight! I was only thinking about the voltagebetweenthe op amp inputs, not their absolute value. I should indeed prevent the inputs from getting out of range.Quote from: sync on September 18, 2013, 06:55:46 pmYes, indeed. I must have been breathing too much solder smoke while writing my previous post...

author: pietr
author_rank: Contributor
timestamp: April 15, 2014, 07:25:53 pm
content: It's been quite some time, but I found some time for this project again. I've had some time to think about the problem described by sync above, and I've come to the conclusion that this problem does not occur in my circuit. If 40V is placed between the output terminals in my circuit, the current will go through R8, D7, R15 and R20 (I'm referring to the components in my power.png schematic, not the Agilent schematic). Since R8 is only 1K and R15+R20 is 34k, there will only be about -1.14V at the op amp inputs, which is OK. Even if I would lower R20 by an order of magnitude to 3.3K, there would still only be -9.30V at the op amp inputs.Idohave another problem, or there is at least something about the E3610 schematic I don't understand. In the E3610 schematic, you can see the current shunt resistor R2 on the output path. Between the current shunt and the positive output terminal, there is a ground connection (the +S label). Now, doesn't this mean that current can flow from this ground connection into the output terminal? Such current does not pass through the current shunt, and hence the total current out of the output terminal will deviate from the current set point.Why isn't this a problem in the E3610 power supply? Is it simply because the current flowing from the ground connection to the output terminal is too small to cause any significant offset, or is there a flaw in my reasoning?

author: sync
author_rank: Frequent Contributor
timestamp: April 15, 2014, 08:41:20 pm
content: Quote from: pietr on April 15, 2014, 07:25:53 pmRight.QuoteThere is a small current. I measured it over the current shunt while nothing is connected to the output. It's about 0.5mA.

author: pietr
author_rank: Contributor
timestamp: April 15, 2014, 08:56:45 pm
content: QuoteOK, so my reasoning was correct. But how can this 'current leakage' be calculated and controlled? I suppose my circuit can potentially leak much more current, because all the digital logic I use to drive the main power circuit draws a relatively large amount of current?

author: sync
author_rank: Frequent Contributor
timestamp: April 15, 2014, 09:31:47 pm
content: Your digital logic doesn't matter. It must have it's own separated power supply (E3610A's reference and bias supply). So the current of your logic doesn't flow through the shunt.

author: pietr
author_rank: Contributor
timestamp: April 16, 2014, 07:49:44 am
content: But what theoretically prevents the current from going through the load? Does it just come down to board layout? That is, should the board layout ensure that the digital logic ground path to the reference and bias supply has lower resistance than the path through the load?