This circuit diagram is given in my book :
simulate this circuit – Schematic created using CircuitLab
and said that from the above figure , they calculate \$ \frac{V_{out}}{V_p} \$ while considering \$r_{o2}\$ and write down the answer is:
$$
\frac{V_{out}}{V_p} = \frac{1+g_{m2}r_{o2}}{1+\frac{r_{o2}}{r_{o4}}}
\approx\frac{g_{m2}r_{o2}}{1+\frac{r_{o2}}{r_{o4}}}
$$
No detail calculation was given and I tried my self but could not reach the solution. How can I do this kind of calculation?
can someone show me step by step solution for this so that I can do this kind of problems by my self in future. thanks in advance.
Best Answer
The output voltage Vout consists of two portions:
1.) Vout1 caused by the drain current Id of the transistor (voltage drop across r_o4),
2.) Vout2 caused (direct way) by the voltage divider between r_o4 and r_o2 .
Calculation:
1.) Transistor is in common base configuration with Vout1=gm*(r_o2||r_o4)
2.) Vout2=Vp*[r_o4/(r_o4+r_o2)
3.) The sum of both parts Vout=Vout1+Vout2 gives the total signal output voltage. Divided by the input Vp we get the given gain expression (after some mathematical manipulations).