Precision circuit that is CC and CV circuit or power supply

Question

I want to design a CC/CV (constant current/constant voltage) circuit in which I can set the voltage limit or current limit with DAC's 0 to 5V range. I know how to design variable CC circuits and CV:

schematic

^{simulate this circuit – Schematic created using CircuitLab}

Here is my problem, I need to design a fairly accurate programmable constant voltage and constant current circuit (the output needs to be 0.1% and within 100uV of the DAC input) the constant current portion also needs a similar accuracy and to be able to source 200mA at 0V to 7V.

I also have temperature req's and noise req so I'll be building this with low tempco opamps low noise op amps. I'm not so worried about that now. Right now I'm trying to find a good circuit topology, which in all the literature that I have this type of circuit is not covered. I don't want to use a DC to DC because of the ripple.

What is a circuit topology that I can use to build a precision CC/CV circuit?

(I could also use a precision LDO if needed) Bonus points if it can source and sink current, I can build the rails around the circuit.

Interesting question. Even assuming temp-stabilized, I'm still stuck on the difficulty of ***accurate*** and 0.1%. How exactly will this be calibrated to NIST traceable standards and then maintained against drift over time? Precision is one thing. Accuracy yet another. — jonk, Apr 12 '17 at 19:32
@jonk I don't know if this needs to be nist tracable, but I need to think about that, Thanks — Voltage Spike, Apr 12 '17 at 21:54
@laptop2d I've specialized in temperature *measurement*. People ***want*** *accuracy* as well as *precision* as well as *repeatability* between instruments. You get nothing but pain from customers if they buy 5 instruments and measure the same freeze point with all 5 and get different values. Whether or not they are willing to ***pay*** for that, is another matter. But it's what they want. Had to maintain NIST traceable standards, usually only one step away (we didn't hire NIST, but hired someone who did.) — jonk, Apr 12 '17 at 22:08
@laptop2d For temperature, you pay for *a few* exact calibration points (say 5 or 6) and in between you have to figure out the worst case deviations in order to provide specifications to the customer. If you support a wide-range of currents, you will have your own non-linearity issues to figure out, as well. This 0.1% thing is achievable. So is 0.01%. But I have to tell you that those 0.1% pieces of equipment were expensive and the 0.01% pieces were nearly sky-high (at the time, 20 years ago.) I used them, too, because we needed precision current to operate electron beams for IC wafer masks. — jonk, Apr 12 '17 at 22:15
Usually my requirements are in the uV or nV range so I don't think a set point accuracy in the 10uV range will be a problem. We'll probably calibrate the circuit. A 10 Ohm 0.01% resistor is expensive but it can save you a lot on calibration and give you accuracy. — Voltage Spike, Apr 12 '17 at 22:26

score 8 · Answer 1 · answered Apr 12 '17 at 16:15

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If you want precision, then your CC source doesn't cut it, what with transistor alpha an' all.

The classic way to do this is with two loops

schematic

^{simulate this circuit – Schematic created using CircuitLab}

Both the voltage and the current feedback are scaled and referred to ground, and compared against your DACs, and the comparisons OR'd into the output control, suggest a darlington for convenience. Whichever loop is 'over' pulls down the collector and regulates the output.

Note that stability needs to be maintained, ie designed for, so the comparison is done with lowish gain. If high gain is needed for precision, add an integrator in the loop. I would guess that such would have to come after the control OR, otherwise the inactive integrator would saturate and take a long time to recover when needed to take over.

With your low voltage and current requirement, a linear supply is all that's needed.

answered Apr 12 '17 at 16:15

Neil_UK

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I realize I could substitute an LDO or mosfet in for the BJT but that will change some aspects of the design. If your using a BJT for the current source won't the op amp compensate for whatever the BJT is doing? – Voltage Spike Apr 12 '17 at 16:22
are you referring to my circuit, or your circuit, I don't understand the question? – Neil_UK Apr 12 '17 at 16:30
I was referring to this: "If you want precision, then your CC source doesn't cut it, what with transistor alpha an' all." – Voltage Spike Apr 12 '17 at 16:53
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@laptop2d I'm pretty sure that he means that the base current of the BJT in the constant current supply you proposed skips the current sensing resistor. An extra current will flow from the opamp to the load, inversely proportional in magnitude to the transistor amplification factor (alpha), and the opamp has no way of compensating for it. – jms Apr 12 '17 at 17:22
I see it now, thanks for that. I suppose you could put the current sense after the BJT and do differential current sensing like this post suggests – Voltage Spike Apr 12 '17 at 17:33

score 3 · Answer 2 · 2017-04-12T17:09:59.530

3

OK, so what i suggest is actually two loops: current loop over voltage loop. That means, you have a current command (which will be a limit), then a voltage command. The output to the DAC is max(voltage command/current loop output). So as long as the current limit is not reached, current loop is saturated and doesn't interfere. The only thing you have to do is to measure the voltage and the current, which is pretty basic stuff.

As per circuit- well, you didn't say a word about voltage/current requirements. So maybe actually simplest way is an emitter follower for power amplification of the DAC and a very small current sense resistor for the current measurement.

Depending on your application and available digital components, i may suggest sigma-delta ADCs to measure the current. Some have very nice, very accurate built in PGA, so you will be able to tune the system very nicely.

So schematic is below. U3 is your microcontroller. In a sense the whole system is similar to one from the other answer, but the current loop should be easier tunable, but will have lower bandwidth.

Sorry, the INST- instrumental amp; also forgot a resistor on base, but you get it.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

Several more words about system behavior. If everything is done correctly, current loop will start from zero and will ramp up the voltage slowly up to voltage command. But if system is normally working in CC mode, there are some special cases. If load is suddenly disconnected, then reconnected, for some time it may be under current higher than the limit. So it may be important to detect disconnected load and reset the current PID loopm

edited Apr 12 '17 at 17:09

answered Apr 12 '17 at 16:09

I kind of did say a few things about voltage and current requirements. Do you think you could draw an example circuit with the tool? – Voltage Spike Apr 12 '17 at 16:21
Yes, i missed the numbers. OK, working on the circuit. – Apr 12 '17 at 16:22
Done! 123456789 – Apr 12 '17 at 16:47
Laptop, do you know how to arrange current loop in the microcontroller? – Apr 12 '17 at 17:20
Yeah, one problem is the software engineers that I work with are not good with real time stuff so I don't know if I would want to use 'firmware in the loop' – Voltage Spike Apr 12 '17 at 17:27
You can always replace it with opamps. But then tuning will be a disaster – Apr 12 '17 at 17:38
Actually, (funny story) Now that I look at this circuit, I built a similar circuit to detect a load and was able to calibrate a load to an 0.01% sense resistor to within 5mΩ (with cables on the load), but it was a CV circuit only and the sense resistor was in the feedback loop of OA1. Just have to have the right calibration technique – Voltage Spike Apr 12 '17 at 17:44
Don't forget to tell what you decide to do and how it works out. – Apr 12 '17 at 17:46
<=0.05% Ratios SIPs or SMD are essential – Tony Stewart EE75 Apr 12 '17 at 19:08

Tony Stewart EE75 · Answer 3 · 2017-04-12T17:29:51.153

2

- depending on system specs, otherwise I would NOT use DAC but rather 10 bit PWM (1024)

- I would choose <=0.1% Vref and choose linear high side CC and CV

schematic

CC loop inverted by hasty schematic (sorry)

choose k=0 to 1 for CC= x to max

edited Apr 12 '17 at 17:29

answered Apr 12 '17 at 17:15

Tony Stewart EE75

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Tony, could you please explain the circuit? – Apr 12 '17 at 17:21
It's a classic LDO design read any datasheets. OA1 is Isense amp from 100mV drop and OA3 is Vreg while OA2 is Imax reg. with down control only. The key is dont use an expensive DAC or less accurate one. – Tony Stewart EE75 Apr 12 '17 at 17:22
Damn you are smart! Did you draw all this by yourself and no schematics?? – Keno Apr 12 '17 at 17:24
after 40 yrs I can do this in my head.. just now here – Tony Stewart EE75 Apr 12 '17 at 17:25
So why would you draw it? – Apr 12 '17 at 17:25
just for newbies ...to learn .. with critical details like R arrays avail with 0.05 or 0.1% tolerance cheap – Tony Stewart EE75 Apr 12 '17 at 17:25
I once did a design for Lexmark with similar tolerances but 4 regulated outputs from -1.5kV to 5kV PWM is way better. Total BOM cost was $16 incl PCB and connectors for target volume 10k units/day circa Y2K for laser printers, with no EMI and no shield needed. – Tony Stewart EE75 Apr 12 '17 at 17:33
Cool, do you think I could ditch the PWM\integrators and throw in a 16 bit dac signal instead? Also, is OA2 in the right feedback? Seems like the terminals should be reversed. – Voltage Spike Apr 12 '17 at 17:37
Well you are very cool. – Apr 12 '17 at 17:37
@laptop2d you could if EMI won't affect signal... CC loop inverted by hasty schematic (sorry) 14 minutes ago – Tony Stewart EE75 Apr 12 '17 at 17:44
Surprised no one asked how did I design KV DC-DC regulated to tight tolerances with NO EMI ! using 24Vdc in – Tony Stewart EE75 Apr 12 '17 at 17:51
If you really want to tell us, you can. – Apr 12 '17 at 17:55
anyone who is interested will ask – Tony Stewart EE75 Apr 12 '17 at 18:45

score 1 · Answer 4 · answered Apr 12 '17 at 18:45

1

After doing much more searching I found one more circuit from eevblog that I thought I'd like to add to the list because it's interesting. Instead of using diodes in a 'max' like configuration, it uses a mosfet and a diode to switch from CV to CC.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

answered Apr 12 '17 at 18:45

Voltage Spike

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curiously almost identical to mine with diode crossover and IA config, but I have doubts with 0.1% accuracy – Tony Stewart EE75 Apr 12 '17 at 18:46
Not quite, yours doesn't use a mosfet and a diode to select between voltages it uses an op amp which would have an offset. This one would give you a more direct feedback path with no offsets for voltage feedback when your not in CC mode. Plus this one uses a BJT not a mosfet for the main current limiter. – Voltage Spike Apr 12 '17 at 18:50
yes it has an extra Q (FET ) instead of diode with high gain causing potential stability issues, – Tony Stewart EE75 Apr 12 '17 at 18:54
I thought I'd throw it on to the list – Voltage Spike Apr 12 '17 at 18:56
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There are at least two errors in the schematic: OA1 is inverted (voltage and current feedback should be negative), and M1 should be a P-channel enhancement mode part (it should conduct when the output of OA3 is low, and it should block when it is high). Currently M1 is a N-channel enhancement mode MOSFET. – jms Apr 14 '17 at 17:37

Precision circuit that is CC and CV circuit or power supply

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