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Can I use the same resistors for current shunt AND transistor balancing? Answered

In my pursuit for a LPSU design, to get more HFe gain, higher current capability, decrease Vce dropout, improve thermals, etc. I would use 2 (or more) pass transistors. This requires load balancing resistors, so I thought I'd get clever and use these balancing resistors to also measure current for my current limiting capability! See pictures and explanation below.

IMPORTANT: The second schematic implys the use of 100R resistors, that is a mistake, they are supposed to be 0.1 ohm, or around that figure.


Details:
The first picture below is a simplified diagram of how I currently have current feedback implemented. I generate a fixed (user adjustable) voltage across a 500R resistor via a current sink and an error amp (currently LM358) drives output pass transistor(s). Because the op amp drives the output to ensure the inputs are the same, the output current can be precisely regulated. It maintains a precise voltage differential across the shunt by means of the pass transistor.

I don't want to take the naive approach with multiple resistances in the path of the current flow from the unregulated supply to the output, for obvious reasons. I could instead just measure the Vdrop on one of these balancing resistors, but that assumes that the current through those resistors is identical, which is not the case. So instead, what if I expand a bit on the first schematic, leading to the last schematic? Would that work? I want to understand the mathematical analysis behind it to prove it does work. 

Discussions

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iceng

2 years ago

Still I would use two op-amps, otherwise there is an de-amplification of the unbalanced condition which may even thermally oscillate.

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-max-iceng

Answer 2 years ago

thermal oscillation? Not sure how that works. :/ Anyway I did calculate the "transfer function" of my idea to be ((V1 + V2) - I*R) / 2, which is unsurprisingly exactly what I expected and what simulation suggested. An average that is offset by a certain about determined by the current source and the resistances. It was much easier than I thought once I ignored everything but R1, R2, and I_set.

I have found some linear power supply designs that actually do almost exactly this, too! But I no longer need this solution BC I choose to move the shunt to the low side, and may just try to use a large enough transistor to avoid having to use multiple parelell transisors.

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steveastrouk

2 years ago

100 Ohm balancing resistors ???? They would usually also be in the mOhm range. Draw the safe operating area graph for your circuit, and then consider the values for the current sharing resistors.

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-max-steveastrouk

Answer 2 years ago

lol, that is a mistake. They are supposed to be 100 milliohm.

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steveastrouk-max-

Answer 2 years ago

The idea's OK, but my experience with emitter degenerating/sharing is they really work, so 50.0% of the current is in one leg

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-max-steveastrouk

Answer 2 years ago

If I want to make my supply set via an arduino, I think I need a DAC. I want at least 11 bits of resolution, or 0.01V precision from 0 to 15V. (1500 steps, that is.) Do there exist cheap current output DACs? I have a bunch of these huge fancy looking 16 bit DAC71's but they require dual rail supply, a logic rail, compensation components and circuitry, and have a hardcore parallel interface, requiring additional shift registers for use with arduino or whatever. They are anchent relics of the past like most of my junk!

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steveastrouk-max-

Answer 2 years ago

There are any number of them out there. A multiplying DAC is perfect for this sort of application. Get something 12 bits and there are lots of options. Get something with 8 bits, and cascade them into something with 16 bits of resolution.

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-max-steveastrouk

Answer 2 years ago

Gees, DACs are more expensive than I thought... :-/ 12 bit current DACs start at like $10 on mouser. There is not a huge selection. Can you elaberate a little on how multiplying DACs? Would using 2 lead to effective 16 bit resolution, or do I need four of them to give me 12 bits? If that's the case then 3 are needed to acheive 11 bits, since I only need 1500 steps.

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-max-steveastrouk

Answer 2 years ago

Thanks for the resource! I ended up changing my design entirely listening to suggestions on this forum thread I started at EEVblog.com.

http://www.eevblog.com/forum/projects/anything-wro...

As a result of moving the current shunt to the low side, I can use much cheaper and more relevant voltage output DACs to set the output voltage. I am not a fan of this design requiring a negative supply (although so did my original design) or the low side shunt, making it impossible to make a dual rail tracking supply from it, but oh well. I think it is much better performing than my original design.

Do you don't mind to take a look at it, what do you think? (if you want some explanations for a few of my design decisions, you can read my last couple posts on the forum.)

linear power supply design3.png
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steveastrouk-max-

Answer 2 years ago

Nah, I don't like it with the shunt there either. Use a ZXCT series high side current measurement chip instead.

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-max-steveastrouk

Answer 2 years ago

What's wrong with the current shunt? Adding a ZXCT would increase BOM costs and make the loop bigger (more components to consider in the transfer function) Also, if I add it before the pass transistors, it will not take into account the additional base current.

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steveastrouk-max-

Answer 2 years ago

BOM costs ? Its a couple of bucks. BOM costs worry you when you're making dozens, not onesies. And you wouldn't need the other supply.

I don't like ground not being ground.

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-max-steveastrouk

Answer 2 years ago

Well then my excuse is that I want to try to use parts I have. I don't see the need of using a dedicated chip to measure current. I have a tendency to kill chips so I would have to buy like 5 so I have some extras for prototyping.

I agree, I don't like "ground not being ground" but I have had several suggestions on the EEVblog forum to move the current sense to ground, and as it turned out, it greatly simplified how to drive the current set with a DAC. (requiring only a current limiting 10Kohm resistor.) I may use that chip to part to feed an ADC so the arduino can 'know' what the output current actually is.

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-max-

2 years ago

I verified this works in simulation, with my "real" schematic even when some transistors were disconnected and not contributing at all to the output current. However I am still trying to figure out how to solve this math problem with so many variables.

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iceng-max-

Answer 2 years ago

Real is as the New Year :-)