1
\$\begingroup\$

Circuit
(source: play-hookey.com)

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

\$\endgroup\$
1
  • 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\$ – user_1818839 Sep 23 '16 at 11:00
2
\$\begingroup\$

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.

\$\endgroup\$
0
\$\begingroup\$

You are correct. If the collector current is different in each transistor then VBE is different in each of the transistors.

  1. But the difference is small enough that it doesn't substantially affect the answer so it was neglected.
  2. Additionally, making the assumption VBE1 = VBE2 keeps all the circuit equations linear rather than equations involving exponentials. That makes the math much easier.

DETAILS:

The base-emitter voltage difference in the two transistors will vary logarithmicaly with their ratio, so you will get very small changes in Vbe even with fairly substantial changes in emitter current.

From the basic equations for the BJT...
https://inst.eecs.berkeley.edu/~ee105/fa14/lectures/Lecture10-BJT%20Physics.pdf

We have
Ic = Is * (e^(Vbe/Vt) - 1) ≈ Is * e^(Vbe/Vt)

Therefore

Vbe ≈ Vt * ln(Ic / Is)

For two matched (Is1 = Is2 = Is) transistors Q1 and Q2...

Vbe1 - Vbe2 ≈ Vt * ln(Ic1 / Is) - Vt * ln(Ic2 / Is)

Vbe1 - Vbe2 ≈ Vt * ln(Ic1 / Ic2)

From the form of this equation we see that the difference in base emitter voltage varies with the logarithm of the ratio of the collector currents. Vt is usually small (like 26mV at room temperature).

So having IC2 = 2.718 x IC1 only results in a 26mV difference in base emitter voltage.

If your circuit was biased so that there was substantially more than 26mV across R2 (like say 1V), then making the assumption that VBE1 = VBE2 only leads to about 2~3% error in the calculation compared to a more detailed analysis.

But in the real world even matched transistor pairs are only matched to within a few percent anyways, so making a calculation more accurate than that usually serves little purpose.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.