Common Emitter's voltage gain.

Status
Not open for further replies.

alphacat

New Member
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.

 
Last edited:
I'll explain how I reached it.

av = -Gm * Rout.

Gm = gm17 / (1 + gm17 * RE)

Rout = (ro17 || ro13B) * (1 + gm17 * RE)
 
Your attachment doesn't work so we don't know what you are talking about.
 
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.
 
Last edited:

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)
 
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+
 
Last edited:
Status
Not open for further replies.
Cookies are required to use this site. You must accept them to continue using the site. Learn more…