YALPS! Yet Another Linear Power Supply

[pietr]

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 taps

I 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)

[ddavidebor]

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.

[c4757p]

so if you touch it after some minutes, you'll remain alice.

If you start as Bob, do you become Alice after touching it?

[kizzap]

so if you touch it after some minutes, you'll remain alice.

If you start as Bob, do you become Alice after touching it?

Only on Tuesdays

As 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 techs

Im 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....

[rdl]

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.

[pietr]

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?

[kizzap]

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.

read http://www.alphascientific.com/technotes/technote3.pdf espcially 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

[pietr]

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.pdf
2N4235 - 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.pdf

Can anyone give an indication on which parameters are important for this transistor?

[ConKbot]

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.

[pietr]

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...

[pietr]

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

[ConKbot]

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.

[pietr]

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?

[SeanB]

Use the HC or C version, they do rail to rail at low current outputs.

[Paul Price]

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.

[edavid]

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.

No, 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.

[Monkeh]

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.

No, 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.

240V is low voltage, too. There are safety issues with large capacitors when charged, whether the voltage is 6V or 600V.

[edavid]

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.

No, 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.

240V is low voltage, too. There are safety issues with large capacitors when charged, whether the voltage is 6V or 600V.

No way.  It's only 10000uF.  That is not why the bleeder is there.

[ConKbot]

Use the HC or C version, they do rail to rail at low current outputs.

+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 

[SeanB]

  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 

It can handle it, so long as you consider operating for under 2 seconds before going into thermal shutdown is operating correctly.

[pietr]

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.

The 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...

[edavid]

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...

There 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.

[Paul Price]

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?

[pietr]

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.

[Paul Price]

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.