Common Emitter's voltage gain.

[alphacat]

I was asked what is the voltage gain of Q17.
(all transistors operate in the forward active region).

My answer is: -gm17 * (ro17 || ro13B).

Am I right please?

Thank you.

 

[alphacat]

I'll explain how I reached it.

av = -Gm * Rout.

Gm = gm17 / (1 + gm17 * RE)

Rout = (ro17 || ro13B) * (1 + gm17 * RE)

 

[audioguru]

Your attachment doesn't work so we don't know what you are talking about.

 

[alphacat]

Thanks for letting me know about it, and sorry for that.

Here is the circuit:

 

[audioguru]

I don't know what is the output impedance of the current mirror.

 

[BrownOut]

You are leaving out the effects of the emitter resistor. The final gain is divided by that resistor.

Edit: I looked again, and you have RE hidden in your gm calculation. That's good. I think you can simplify your result to:

(ro17||ro13B)/RE, ingoring the very small effect of base current.

 

[alphacat]

You are leaving out the effects of the emitter resistor. The final gain is divided by that resistor.

Edit: I looked again, and you have RE hidden in your gm calculation. That's good. I think you can simplify your result to:

(ro17||ro13B)/RE, ingoring the very small effect of base current.

Thanks for the help.

Could you explain please how you reached (ro17||ro13B)/RE?

It is known that av = -Gm * Rout, and according to my calculations:

Gm = gm17 / (1 + gm17 * RE)

Rout = (ro17 || ro13B) * (1 + gm17 * RE)

 

[BrownOut]

Use the approximation: 1 + gmRE ~= gmRE;

since gmRE >> 1 typically.

For example, if IC = 100mA and RE = 100 ohms, then gm = 100/25 = 4ma/mv (at room temperature), and gmRE = 400V/V, which is much greater than 1. The approximation works for most reasonable values of IC, but may fail at very low currents.

However, your original calculation was correct. A+