Bipolar transistor amplification biased

This exercise is about a BJT transistor common emitter circuit, with a biasing resistor between the base and the collector. I am supposed to compute the signal amplification of the circuit $V_o/V_s$ ($V_o$ being at node1). I am given that the transistor has $\beta = 50$ and $r_{\pi} = 1.1$ for the $\pi$ equivalent model.

• What work have you done so far? Also, your value of $r_{\pi}$ seems low. Is that supposed to be in k$\Omega$? – Null Jan 24 '15 at 16:55
• Yes I'm sorry it is. I have drown the equivalent circuit with the pi model i just don't know where to put Rb, as part of the input or output? – user3401 Jan 24 '15 at 16:59
• Do you have some knowledge about negative feedback? – LvW Jan 24 '15 at 17:02
• not really, i am a second year student, just getting started with electronic – user3401 Jan 24 '15 at 17:03
• General: Pi model would need dragging out of memory and dusting off but this is general approach: Assume Z_AC source = 0 for DC. For model used, establish DC operating point. Establish Rbe internal from current.Solve. – Russell McMahon Jan 24 '15 at 20:44

There are, in principle, two ways for solving the task. Please understand that I do not intent to present the solution for you. However, I will try to give you some hints to find the solution by yourself.

1.) Application of the superposition theorem

Supplementing the transistor Pi-model with the external resistors results in a circuit which contains two sources: A voltage source Vin and the controlled current source of the Pi-model. Thus, after applying the superposition principle you get two equations for finding the two unknown quantities: Vo/Vin and Vbe. (Please note that the transconductance of the Pi-model can be expressed by the two known values for beta and Rpi).

2.) Application of the general gain formula for negative feedback

The closed-loop gain is Vo/Vin=Acl=AoHf/(1-AoHr). (Note that Ao will be negative).

Gain without feedback Ao=Vout/V(base) (simple gain formula for common emitter);

Forward (damping)factor Hf=V(base)/Vin for Vout=0;

Return (feedback) factor Hr=V(base)/Vout for Vin=0.

Both factors are calculated using simple voltage divider rules.

• This transistor circuit exploits the same idea (parallel negative feedback) as the op-amp inverting amplifier. – Circuit fantasist Aug 18 '17 at 6:04

You can take the Rb and connect it directly between the base and VC and denote it as Rb+Rc. Then the circuit simplifies a great deal and then you can use the two variable method specified by LvW.