Small-signal voltage gain
The common-emitter circuit has an emitter bypass capacitor.
(a) Derive the expression for the small-signal voltage gain Av(s)=Vo(s)/Vi(s). Write the expression in a form similar to that of equation (Av=(-Gm*Rpi*Rc)(Rs+rpi+(1+Beta)Re)*((1+s*taua)/(1+s*taub)).
(b) What are the expressions for taua and taub?
The circuit is attached.
I need a cookbook version of Bipolar AC analysis instead of the obtuse, bloated technical text book used to teach the class. Please explain following in your response.
- How did you find the small signal voltage gain exactly?
- Why is the equation cited so important?
- What are the time constants, and why are they important?
Attachments
Solution Preview
The method of solution is that we will find Wpi and Wzi for any capacitor in the circuit. Then
Av(s)= Av0*(s+Wz1)*(s+Wz2)...(s+Wzn)/((s+Wp1)(s+Wp2)...(s+Wpn))
where Av0 is the small signal gain and Wzi=1/tauzi and Wpi=1/taupi.
Well, first of all, according to the picture, because we just have one capacitor, namely CE, in the circuit, we expect to have only one pole and one zero. Now, we find the pole and the zero of this transfer function.
First let's find the zero. We have:
Wz= 1/(RE*CE)
where RE is the total resistance in the ...
Solution Summary
The method of solution is that we will find Wpi and Wzi for any capacitor in the circuit.