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I've been looking around for a good CCS schematic, and decided to try making a simpler one even if it means it'll behave worse. I need a 12V (would come from an ATX power supply, more than enough current from the 12V rail) circuit that can handle at least up to 1A on the load (2/3A would be ideal). Would be for general purpose, electronics testing/etc, nothing too serious or in need of high precision. Made this in a simulator, and it seems to work:

schematic 2

The load is represented as a pot to make things easier.

The NPN transistor is a TIP35A, power NPN. Will have a decent heatsink and a cooler pointing in its general direction.

The PNP would be a general purpose transistor, and the adjustment pot would have calculated resistors in each side to limit its throw, of course. (depending on the resulting hFE of the "darlington pair")

My question is: Am I missing something important here? I would add an LM317 as a current limit (or a high-watt resistor on the circuit at least). Would this supply vary too much with resistance changes? (It had no change on the sim, but again this app is no LTspice). I have all the components so I plan on breadboarding it to test it with low currents, but I want to be sure this can work...

One more thing: I have a (totally overkill) ST P40NF0 3L MOSFET which I guess I could use (by replacing the driver NPN with a PNP and the power PNP with this MOSFET, and connecting the PNP differently of course). This FET has a very low on resistance which I guess would make my supply more efficient. Would this be correct?

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    \$\begingroup\$ What you call "comparator + mosfet circuits" are probably op amp + mosfet circuits. They're similar things, but not the same thing. And if you've been trying to use a comparator where you should be using an op amp, perhaps that's your problem. \$\endgroup\$ – Hearth Mar 17 at 18:14
  • \$\begingroup\$ @Hearth - I was not, I used both an LM741CN and then an LM308N just in case it was the opamp. Little mistake, nothing else. I think I have it narrowed down to using a different MOSFET with a higher Vgs than the IRF530 mentioned in that circuit, that's all. But first I want to know if my circuit is even remotely viable; if not, i'll try again with one of these circuits. The one I used was a response to someone else asking here in stack exchange and there was no answer, so I have no idea if the circuit actually works... \$\endgroup\$ – Matias Gonzalez Mar 17 at 18:24
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    \$\begingroup\$ The TIP35A is an NPN, not a PNP... Can you add a link to the datasheet of the part you want to use? \$\endgroup\$ – Vladimir Cravero Mar 17 at 18:26
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    \$\begingroup\$ Please don't use slang in your questions, such as "idk", "whipped up", and "honkin". It makes it difficult for people whose first language is not English, or who don't tend to communicate via SMS, to understand what you are saying. \$\endgroup\$ – Elliot Alderson Mar 17 at 18:41
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    \$\begingroup\$ You say \$1\:\text{A}\$ is what you want, then you say that \$\frac23\:\text{A}\$ would be ideal. Which is it? Also, do you need to be able to support a full compliance of \$12\:\text{V}\$ across the load? What's your source power supply? Is it an unregulated one? Or a car battery system? What kind of linearity do you want with respect to the controlling potentiometer? (Is halfway on the pot supposed to mean half-max current setting?) And are you expecting meters to show you the set current and compliance voltage across the load? I don't think there's enough information. \$\endgroup\$ – jonk Mar 17 at 18:47
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In the circuit you have drawn, you are missing the single one thing electronics engineer usually crave for: negative feedback.

Silicon in the real world tends to do a number of unpleasant things that we would not like it to do, like heating up, or behave in unpleasant manners.

Your circuit is quite straightforward, assuming the BJTs are ideal the current through the power device will be proportional to the current into the base of the N BJT. Your control pot should rather be a series resistor into the base, because right now you are controlling the circuit with a voltage signal, which means that the dependency is exponential, which is less than ideal.

The worst thing that can happen is due to the early effect: the current through the power device is not independent of the Vce, but rather varies with it, so it varies with the load, wheter it is resistive, or a voltage load.

If that is acceptable for your application, it is something only you can decide, depending on the inaccuracy you can accept.

Let me suggest you a rather simple topology, that can work very well, and leverages the mighty negative feedback:

schematic

simulate this circuit – Schematic created using CircuitLab

R1 needs to be small, it is called a shunt resistor. For 3 A full scale, something in the 100 mOhm range is ok.

OA1 needs to be an opamp with very low offset, and low drift in temperature if you'd like your circuit to be nice versus temperature.

R2 is the load, and that's about it.

How does this circuit work? The voltage across R1 is proportional to the current through it. The opamp will work hard to keep the plus input and the minus input at the same voltage, so if you use a 100 mOhm resistor for R1, and you provide 100 mV to the input node, you will get 1 A through R1 (and R2).

A few remarks:

  • The (ideal) gain of the circuit is just 1/R1, you might want to pick a 1% part for that
  • The power transistor needs to be able to dissipate 12 V x 3 A = 36 W, design accordingly
  • This circuit is not fast, at all. If you need a fast, accurate current source then it becomes somewhat trickier to design it.
  • OA1 needs a low offset because even 1 mV of input offset means 10 mA of output current offset.
  • You might want to add a resistive divider on the input node, so to use higher voltages to control the output current
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  • \$\begingroup\$ I'll definitely try it. Thanks for the answer \$\endgroup\$ – Matias Gonzalez Mar 17 at 19:06
  • \$\begingroup\$ To be pedantic, the emitter degeneration resistor in the original schematic is a form of negative feedback. It's just not going into anywhere near as high-gain a system as an op amp. \$\endgroup\$ – Hearth Mar 17 at 19:27
  • \$\begingroup\$ @Hearth thanks for pointing that out. The resistor you mention is the load, in OP circuit... It does not provide any "helpful" feedback in the sense that emitter degen resistors stabilize the op point wrt transistor characteristics, while here we are interested in the DC gain of the whole thing, which is not "stabilized" (PVT) by the load itself. \$\endgroup\$ – Vladimir Cravero Mar 17 at 20:01
  • \$\begingroup\$ @VladimirCravero Oh, I thought the output was at the collector. Okay, then, yeah, that's not negative feedback. \$\endgroup\$ – Hearth Mar 17 at 20:03
  • \$\begingroup\$ Might want to look more closely at the (-) and (+) input assignments of your opamp. Also, given this is an ATX supply the OP has access to some negative rails (good thing) for the opamp to use. \$\endgroup\$ – jonk Mar 18 at 22:13

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