Having an issue with a small homework assignment. The question asks what the overall voltage gain is for the three stage amplifier in the attached picture below.

We were told to assume Hfe = 200

Here's the calculations and formulas I have thus far:

Zin Base (1) = Rswamp(R4) x Hfe = 1k ohms x 200 = 200k ohms
Zin Base (2) = Rswamp(R8) x Hfe = 100 ohms x 200 = 20k ohms
Zin Base (3) = Rload x Hfe = 100 ohms x 200 = 20k ohms


Av (1) = R1//R2//Zin Base (1) = 10k//5.6k//200k = 3.53k ohms
Av (2) = R5//R6//Zin Base (2) = 12k//1k//20k = 882 ohms
Av (3) = Have not figured that out, kind of assuming R7//R8//Zin Base (3)

Now the back of the book gives me a gain of 1679, but various other calculations I've messed around with do not give me that answer.

Can anyone help clear this up for me?

 

Voltage gain of a CE amplifier with CE capacitor is equal

AV =( RC||RL ) / re = 40 * Ic * Rc||RL

RL = R5||R6|| (β+1 * R8) ≈ 880Ω

Ic = 10V/1K = 10mA

Av1 ≈ 40*10mA * 1K||880 = 40mA * 468 = 187 V/V

The voltage gain of a Q2 stage is equal:

Av2 = (Rc||RL) / (Re + re) ≈ Rc/Re = R7/ R8 = 10 V/V (9.35V)

And the gain of last stage is equal:

Av3 = RL/( re + RL ) = R9/ ( re + R9 )

re = 26mV/Ic - but we don't know Ic of a last stage, so we can assume Av3 = 0.9V

The overall voltage gain is equal:

AV = 187 * 10 * 0.9 = 1683 V/V

 

Voltage gain of a CE amplifier with CE capacitor is equal

AV =( RC||RL ) / re = 40 * Ic * Rc||RL

RL = R5||R6|| (β+1 * R8) ≈ 880Ω

Ic = 10V/1K = 10mA

Av1 ≈ 40*10mA * 1K||880 = 40mA * 468 = 187 V/V

The voltage gain of a Q2 stage is equal:

Av2 = (Rc||RL) / (Re + re) ≈ Rc/Re = R7/ R8 = 10 V/V (9.35V)

And the gain of last stage is equal:

Av3 = RL/( re + RL ) = R9/ ( re + R9 )

re = 26mV/Ic - but we don't know Ic of a last stage, so we can assume Av3 = 0.9V

The overall voltage gain is equal:

AV = 187 * 10 * 0.9 = 1683 V/V

Nice, that makes a bit more sense. The only thing I am curious about (and I do understand there is no Ic of the last stage) is why automatically assume 0.9V for Av3? What's the reason for it?

 

is why automatically assume 0.9V for Av3? What's the reason for it?

Well becaues last stage is a emitter follower (CC amplifier)
http://en.wikipedia.org/wiki/Common_collector#Characteristics
So the voltage gain is close to 1. In practical circuit around 0.8...0.98

 

And of course there is a current flow trough Q3, Q4 but we don't have enough information from your diagram to determine the magnitude of a Ic current.

 

Definitely makes sense now thanks for the help mate!