Circuit http://www.play-hookey.com/analog/current_mirrors/images/current_mirror_re.gif

I'm trying to analyse a mirror current circuit with unequal emitter resistors, without neglecting base currents. So far, I had no luck with the math.

I found this on a website :

However, if R1 ≠ R2, then the currents through the two transistors will adjust themselves so that the emitter voltages of the two transistors are essentially equal.

It boggles my mind, since I know that Vbe varies with Ic(Ebers-Moll equation). Since Ic1 and Ic2 are different, shouldn't Vbe's be different as well? How would the transistors adjust currents.

Any help is appreciated.

Image and quote taken from:play-hookey

  • 1
    \$\begingroup\$ Note the quote says "essentially equal" not "equal". You are correct that the Vbes will be different : now compare the magnitude of that difference to the actual emitter voltages and decide whether it's safe to ignore for yourself. \$\endgroup\$ – Brian Drummond Sep 23 '16 at 11:00

That quote is approximating the transistors as having fixed B-E drop. Yes, there will be some B-E voltage variation with current, and between parts for that matter. With discrete parts, you generally make R1 and R2 big enough so that these variations don't matter. Current mirrors on a chip are made with well-matched transistors and usually no emitter resistors, and have a different set of tradeoffs.

Basically, the current thru each side is inversely proportional to each's emitter resistance. Note that this also ignores the base current of both transistors appearing only on the input side of the current mirror. If you really care about the last 2%, then you probably need a different topology in the first place.

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