I'd like to calculate the input impedance of the following circuit, assuming that Q1 is "active". I know that the impedance of the voltage divider is \$\frac{R_1R_2}{R_1+R_2}\$ and the impedance of the emitter follower is \$\beta R_3\$, where \$\beta\$ is the gain, but it's not clear to me how the impedance of the whole circuit can be calculated. I'm not just looking for a recipe. More importantly, I'd like to know why that recipe leads to correct answer, knowing the definition of impedance, i.e. \$\frac{\Delta V}{\Delta I}\$.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Do you know the small-signal equivalent circuit for Q1? \$\endgroup\$ – The Photon Apr 7 at 15:23
  • \$\begingroup\$ @ThePhoton, Unfortunately no. I'm a physicist trying to learn some electronics. \$\endgroup\$ – apadana Apr 7 at 15:24
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    \$\begingroup\$ I don't have time to write up an answer, but you can google "hybrid pi bjt model". If you're only interested in low-speed behavior, you can remove all the capacitors from the model and substitute it in for the BJT in your circuit. Then you can analyze the circuit to find the input resistance. \$\endgroup\$ – The Photon Apr 7 at 15:27
  • \$\begingroup\$ Due to the fact that the bipolar transistor is a highly nonlinear device to simplified the circuit analysis we are using the highly linearized BJT's small-signal model, thus try to read this ittc.ku.edu/~jstiles/412/handouts/… it should help you understand why the small-signal model work. \$\endgroup\$ – G36 Apr 7 at 15:30
  • \$\begingroup\$ @G36, I think my question at the level I'm interested in is quite basic. I'm assuming that the impedance of the emitter follower here is almost constant and known. \$\endgroup\$ – apadana Apr 7 at 15:40

I will not go through the calculation but I can show you how to build the equivalent small-signal or linear circuit from which you can calculate the input resistance.

As underlined by the authors of comments, you need to resort to a small-signal model of the bipolar transistor. The transistor is a highly nonlinear device and you can linearize its behavior around a given operating point. One simple low-frequency version is the hybrid-\$\pi\$ model shown below:

enter image description here

Now simply insert this invariant model into your circuit by respecting the pinout and you now have a linear circuit where you can apply all the classical theorems like Thévenin, superposition and so on if needed:

enter image description here

You see that the \$V_{cc}\$ is kind of folded to the ground - if my poor English is clear enough - implying a grounded collector and the upper connection of \$R_2\$ also at the 0-V line. This is so because whether we talk about a resistance or an impedance determination, we talk about small-signal values. Small-signal value means that, at some point, there is a low-amplitude stimulus injecting a signal in the circuit for analysis. This stimulus will propagate in the circuit to form a response, the output signal you want to analyze. Some wires in the circuit will see this stimulus propagating but some will not. Typically, the \$V_{cc}\$ line is well decoupled by a big capacitor meaning that despite a modulation applied to the circuit, you won't observe anything with an oscilloscope if you probe this line: its ac voltage is 0 V and a 0-V source is similar to a short circuit in the ac representation. This is why you short the \$V_{cc}\$ to ground in ac analysis.

To determine an input impedance (or a resistance in your case), simply install a test generator \$I_T\$ (our stimulus) and determine the voltage \$V_T\$ across the source (our response). The resistance you want is simply \$R_{in}=\frac{V_T}{I_T}\$. It would work in a similar way if you would have energy-storing elements in the circuit like inductors or caps. but it would then become an impedance defined as \$Z_{in}(s)=\frac{V_T(s)}{I_T(s)}\$.

In this example, you can immediately see that \$I_T=i_b\$ and you have two resistances in parallel directly connected across the current source. You could easily temporarily disconnect them and determine the intermediate input resistance without them, \$R_{int}\$. The final result would simply be \$R_{in}=R_1||R_2||R_{int}\$. This is already part of the fast analytical circuits techniques or FACTs that I keep preaching on this friendly site : ) Good luck with your exercise!

  • \$\begingroup\$ It's very generous of you to write such a long answer. Thanks. \$\endgroup\$ – apadana Apr 7 at 16:12
  • \$\begingroup\$ With pleasure! Hope this helps. \$\endgroup\$ – Verbal Kint Apr 7 at 17:37

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