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I am designing a simple current limiter, which charges a large (4.7mF) capacitor with a charge current of (roughly) 500mA from a supply voltage from about 10-20V - see the below circuit.

Transistor current limit with extra transistor

My dilemma is that I already have a bunch of MMBT2222A and it would be nice to use this part without another line item. Whilst it can happily push 500mA through the collector, it only has a power rating of 350mW, which will be massively exceeded when the capacitor is first charged as the voltage across the collector will be about 10V and hence the power will be about 5W for the first 200ms or so.

In the circuit above, I have added an extra 2N2222A in parallel (with the idea of adding more as required) with the existing one, but I feel this strategy if fraught with danger - mismatched gains will cause unequal currents and defeat the whole purpose of the exercise.

So, is there a clever way to achieve this circuit? Or, do I just have to bite my tongue and put a transistor in place with a better power rating?

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    \$\begingroup\$ limit or constant current charge? \$\endgroup\$
    – JIm Dearden
    May 25, 2013 at 7:38
  • \$\begingroup\$ @Jim, good question. Current limit in this instance as the goal was more for the protection of the supply. A constant current charge in this instance would be equally as useful. \$\endgroup\$
    – Damien
    May 25, 2013 at 9:48
  • \$\begingroup\$ Reading from below you need to charge the capacitor as quickly as possible but limit the maximum charging current to 500mA. There was an old trick for charging the capacitor bank for flash guns using a bulb as a non linear resistor. Initially the resistance is low going higher as current is drawn through it and then low as the current reduces. A couple of 12V 6W filament bulbs in series should suffice. \$\endgroup\$
    – JIm Dearden
    May 25, 2013 at 15:04

2 Answers 2

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Use a small emitter resistor to counter difference in transistor gain like this:

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Seems quite reasonable and simulates OK with R2 = R1 = 1R5. What values did you have in mind? \$\endgroup\$
    – Damien
    May 25, 2013 at 9:46
  • \$\begingroup\$ I didn't have any in mind because I didn't bother to find the datasheets and do the calculations. My gut feeling says somewhere around 1 ohm, 1R5 should do fine. The trick is to add enough resistance to equal currents among the two transistors and at the same time improve output spec of the circuit. Did you simulate with different gains for the transistors? \$\endgroup\$
    – jippie
    May 25, 2013 at 9:57
  • \$\begingroup\$ ... Just simulated with a BC338 in place of the 2N2222A as a proxy for different gain. Without the resistors, it's split 159mA vs 341mA. With the 1R5 resistors, it narrowed considerably to 220mA vs 278mA. \$\endgroup\$
    – Damien
    May 26, 2013 at 0:12
  • \$\begingroup\$ I've selected this as the answer because it most directly answers my question. However, I'm going to have to bite the bullet and go a different transistor as the power handling isn't enough, even with quite a few in parallel. \$\endgroup\$
    – Damien
    May 26, 2013 at 0:16
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If all you want to do is simply limit the charging current then all that is required is a resistor.

Assuming Vc = 0 and Ic (the initial charging current) is 500mA then R = 12/0.5 = 24 ohms.

enter image description here

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  • \$\begingroup\$ Fair point, but this is only true at 12V - I want to be able to go to 20V. The 20R I have at the end of the circuit was a bit of an afterthought to try to limit the power dissipated by the transistor. It does, however, slow the charge of the cap which is undesirable for my purpose. \$\endgroup\$
    – Damien
    May 25, 2013 at 9:36
  • \$\begingroup\$ Have added the second circuit. It has the advantage that the resistance between the supply and anything connected across the capacitor is low but in the event of a short circuit output it will limit the current and not overheat. (Bulbs are meant to glow hot) \$\endgroup\$
    – JIm Dearden
    May 25, 2013 at 15:14
  • \$\begingroup\$ Nice trick and it does answer the question, but it's not quite the form I was looking for :) \$\endgroup\$
    – Damien
    May 26, 2013 at 0:13

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