# Determining gain, KCL/KWL (Basics)

Below is a pi model schematic of a cascaded amplifier. The goal is to find the gain with Rp as load.

So the gain would be: $$\ A = \frac{V_{out}}{V_{in}}\$$. So i set out to find $$\V_{out}\$$ in terms of $$\V_{in}\$$ as $$\V_{out}=i_{Rp}R_p=\frac{300}{500}\beta i_{b2}R_p \$$

I believe the answer is very simple, but I can't seem to express $$\i_{b2}\$$ in terms of $$\i_{b1}\$$ and thus $$\V_{in}\$$ because of the current $$\i_1\$$.

• Simply try $\Large i_{b1} = \frac {V_{in}}{r_{\pi 1}}$
– G36
Feb 13, 2020 at 15:46
• Yes, but how do I express $i_{b2}$ in terms of $V_{in}$. Current $i_1$ is bugging me (the node marked with an asterix). Feb 13, 2020 at 16:00
• Are you need a symbolic expression for a voltage gain or just a number (voltage gain)?
– G36
Feb 13, 2020 at 16:05
• I need an expression. See, $i_{b2}=i_1-\beta i_{b1}$ and thus $i_{b2}=i_1-\beta \frac{V_{in}}{r_{\pi 1}}$ but $i_1$ is unknown. Feb 13, 2020 at 16:10
• In that case, you need to use any circuit analysis techniques you know and solve the circuit as we do with the DC circuit, Or we can notice that the Q2 is an emitter follower and we know Zin for a voltage follower. So the IB2 current will be equal to $\Large I_{b2} = i_{b1}*\beta * \frac{R_1}{R_1 + r_{\pi 2} + (\beta +1)*R_E||R_P}$ See here: electronics.stackexchange.com/questions/476659/…
– G36
Feb 13, 2020 at 16:19

Well, we have the following circuit:

simulate this circuit – Schematic created using CircuitLab

I used Mathematica to write a code using KCL and KVL. The code I came up with is the following:

FullSimplify[
Solve[{Iin == I1 + I2, I3 == I2 + β1*I2,
0 == I4 + I5 + β1*I2, Ie == I5 + β2*I5, Ie == I6 + I7,
I1 == Ia + Iin, Ib == I3 + Ia, 0 == I4 + Ib + Ic,
Id == Ic + β2*I5, Id == I6 + I7, I1 == (Vin)/R1,
I2 == (Vin - V1)/R2, I3 == (V1)/R3, I4 == (V2)/R4,
I5 == (V2 - V3)/R5, I6 == (V3)/R6, I7 == (V3)/R7}, {Iin, I1, I2,
I3, I4, I5, I6, I7, Ia, Ib, Ic, Id, Ie, V1, V2, V3}]]


Using your values I got the following code:

FullSimplify[
Solve[{Iin == I1 + I2, I3 == I2 + β1*I2,
0 == I4 + I5 + β1*I2, Ie == I5 + β2*I5, Ie == I6 + I7,
I1 == Ia + Iin, Ib == I3 + Ia, 0 == I4 + Ib + Ic,
Id == Ic + β2*I5, Id == I6 + I7, I1 == (Vin)/100000,
I2 == (Vin - V1)/1530, I3 == (V1)/R3, I4 == (V2)/2200,
I5 == (V2 - V3)/359, I6 == (V3)/300, I7 == (V3)/200}, {Iin, I1, I2,
I3, I4, I5, I6, I7, Ia, Ib, Ic, Id, Ie, V1, V2, V3}]]


I left $$\\text{R}_3\$$ as it was because I need to set it equal to $$\0\$$ and than I will get an error message because then I will divide by $$\0\$$. Later I will take the limit.

Entering the code I got the following result:

{{Iin -> Vin (1/100000 + 1/(1530 + R3 + R3 β1)),
I1 -> Vin/100000, I2 -> Vin/(1530 + R3 + R3 β1),
I3 -> (Vin (1 + β1))/(1530 + R3 + R3 β1),
I4 -> -((Vin β1 (479 + 120 β2))/(
3 (1530 + R3 + R3 β1) (893 + 40 β2))),
I5 -> -((2200 Vin β1)/(
3 (1530 + R3 + R3 β1) (893 + 40 β2))),
I6 -> -((880 Vin β1 (1 + β2))/(
3 (1530 + R3 + R3 β1) (893 + 40 β2))),
I7 -> -((440 Vin β1 (1 + β2))/((1530 + R3 +
R3 β1) (893 + 40 β2))),
Ia -> -(Vin/(1530 + R3 + R3 β1)),
Ib -> (Vin β1)/(1530 + R3 + R3 β1),
Ic -> -((2200 Vin β1)/(
3 (1530 + R3 + R3 β1) (893 + 40 β2))),
Id -> -((2200 Vin β1 (1 + β2))/(
3 (1530 + R3 + R3 β1) (893 + 40 β2))),
Ie -> -((2200 Vin β1 (1 + β2))/(
3 (1530 + R3 + R3 β1) (893 + 40 β2))),
V1 -> (R3 Vin (1 + β1))/(1530 + R3 + R3 β1),
V2 -> -((2200 Vin β1 (479 + 120 β2))/(
3 (1530 + R3 + R3 β1) (893 + 40 β2))),
V3 -> -((88000 Vin β1 (1 + β2))/((1530 + R3 +
R3 β1) (893 + 40 β2)))}}


Now using your gain formula we will end up with:

$$\text{G}:=\frac{\text{V}_3}{\text{V}_\text{in}}=\lim_{\text{R}_3\to0}-\frac{88000\beta_1\left(1+\beta_2\right)}{\left(1530+\text{R}_3\left(1+\beta_1\right)\right)\left(893+40\beta_2\right)}=$$ $$-\frac{8800\beta_1\left(1+\beta_2\right)}{153\left(893+40\beta_2\right)}\tag1$$

• lol, I could write down the voltage gain expression by inspection. So there is no need to write down all these equations.
– G36
Feb 15, 2020 at 16:49
• @G36 Well, impressive that you're able to do that, not everyone is 'that' advanced. Feb 15, 2020 at 16:50
• Yes, but "your method" is inefficient, too many equations. Why you do not use more efficient method?
– G36
Feb 15, 2020 at 16:54
• and +1 for a effort and time.
– G36
Feb 15, 2020 at 16:55
• @G36 If you want to know all the voltages and currents, I think my method is the way to go. Yes indeed if you only want the voltage at the output you can use the 'easy' method. Feb 15, 2020 at 16:56