I have the next circuit


simulate this circuit – Schematic created using CircuitLab

\$V_{BB}\$ and \$V_i\$ represent the output voltage of a previous stage of a circuit. Assuming that \$\beta = 200,r_x = 0, V_A = \infty\$, what would be the input resistance of the circuit (i.e. the resistance seen from the base of Q1), the output resistance of the circuit (i.e. the resistance seen from the node \$V_o\$) and voltage amplification \$A_v = \frac{v_i}{v_o}\$=?

I'm having trouble dealing with R2. I thought of expressing these parameters as follows:

\$R_i = r_{\pi 1}+\beta \cdot (1000\Omega //r_{\pi 2})+\beta \cdot 50\Omega\$

\$R_o = 50\Omega //(\frac{r_{\pi 1}}{\beta}+1000\Omega//\frac{r_{\pi 2}}{\beta})\$

Is this ok? And regarding the amplification, I have no idea of how to calculate it. I hope you can give me a hand.

  • \$\begingroup\$ For Ri you are not accounting for the current through R2 from Q1. \$\endgroup\$ – jippie Feb 20 '16 at 19:36
  • \$\begingroup\$ @jippie What is missing in the expresion of \$R_i\$ I wrote above? \$\endgroup\$ – Tendero Feb 20 '16 at 19:38

One way to figure the apparent resistance of a point is to set the point to a fixed voltage and calculate the current. By Ohm's Law, the resistance is the voltage divided by the current.

In this case, the dynamic resistance is probably more relevant. That's the change in voltage divided by the resulting change in current.

Set the base of Q1 to 5 V, for example, and work thru the circuit to ultimately find the current going into the base. You can make some simplifying approximations, like the B-E drop of both transistors is 700 mV, that won't change the answer much.

This may seem difficult, but it's really not if you do it one step at a time. For example, you know the emitter of Q1 and base of Q2 are therefor at 4.3 V, and the emitter of Q2 at 3.6 V. That tells you the currents thru R1 and R2 directly. Using the gain of the transistors, you can go from the current thru one pin to the others easily. Ultimately, you arrive at the current going into the base of Q1.

Now do the same thing again for 5.5 V, for example. By subtracting the two operating points, the dynamic resistance is the voltage rise (500 mV in this example) divided by the current rise.

After you do this a few times, you realize that β and β+1 are pretty much the same thing, and that the impedance on the emitter is reflected in the base times β. That may seem a little sloppy, but keep in mind that β can vary considerably between transistors, even from the same batch. Good circuits work from some minimum β, to infinite β. It therefore matters little whether the emitter current is 50x or 51x the base current.


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