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I've always had troubles with input and output impedances. It is not very well explained anywhere else I could look on the web so...

What is the most efficient and rigorous way to calculate input and output impedances?

A detailed step-by-step guide is what I'm looking for here, hoping the time it may take you will be beneficial to many others here. As a practical example, why not apply it to this simple common emitter circuit (AC and DC bias).

enter image description here

I'm assuming that would be starting from the definition of impedance $$Z_{in}=V_{in}/I_{in}$$ and vice-versa, but as soon as there are different paths for the current as seen above, I don't really know how to proceed.

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    \$\begingroup\$ It is unlikely that the most efficient way is the most rigorous way. Are you asking for the most efficient rigorous way or the most rigorous efficient way? The most rigorous way I've seen for the CE amplifier circuit is here: users.ece.gatech.edu/mleach/ece3050/notes/bjt/CEAmpSu10.pdf \$\endgroup\$ – Alfred Centauri Sep 19 '14 at 11:06
  • \$\begingroup\$ I think the discussion in "Art of electronics" is nice. (If you are serious about electronics you should own a copy.) The most efficient (time-wise) might be to stick it into LTspice and see how it responds with different source and load impedances... but this won't really teach you anything. \$\endgroup\$ – George Herold Sep 19 '14 at 13:59
  • \$\begingroup\$ I agree with Alfred, "rigorous" and "efficient" are conflicting goals. \$\endgroup\$ – The Photon Sep 19 '14 at 16:09
  • \$\begingroup\$ Also, the small signal impedance should be based on phasors (often indicated with lower case symbols v and i) rather than quasi-static values (often indicated with capital symbols V and I). \$\endgroup\$ – The Photon Sep 19 '14 at 16:12
  • \$\begingroup\$ I had the feeling they would be different, we can understand "and" as being two options instead of a unique one. I would like to know both sides since I believe we should be able to switch from the most efficient to the most rigorous when in doubt or when it doesn't work. \$\endgroup\$ – user42875 Sep 19 '14 at 21:44
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For output impedance start by understanding the right hand picture below: -

enter image description here

It's the right hand picture I'm interested in - it shows several curves for a BJT and these curves are based on a certain base current (omA to 4mA) and for each current (held constant), the voltage across collector-emitter is increased from 0V to 20V. As this voltage is increased, the collector current is measured.

Collector current rises sharply then settles down to a near horizontal line. Take the example when Ib = 0.2mA and, just concentrate on the flatter part of the curve (this is the normal working area for a common-emitter BJT amplifier).

Maybe at 2 volts across collector-emitter, the collector current is 10mA - note this down! At 20 volts, the collector current appears to be about 11mA.

Now, turn this into an effective resistance: -

\$R = \dfrac{20V - 2V}{11mA - 10mA}\$ = 18kohms.

This is sometimes called the collector's compliance and represents the effective series output resistance of the collector if the colelctor were regarded as a voltage source. For an AC voltage on the collector, this is in parallel with R3 in the OP's circuit and if R3 were (say) 4k7, the output impedance of the circuit would be 18k||4k7 = 3.73kohms.

An approximation generally used is to say that the output impedance equals R3 (not a million miles off but this is generally accepted as an approximate rule of thumb). Depending how rigourous you want to be you might take into account that for several different base currents, the "flat" slope of the collector current curve changes and here's a picture that demonstrates it: -

enter image description here

Point Va is called "the Early voltage" and is a useful way of determining how the slope changes with increased base current. Do you need to go this far? Sometimes although I never have.

Input impedance is usually approximated to R1||R2 - this doesn't take into account base currents that tend to lower the actual input impedance but, for a general type of circuit R1 and R2 are normally chosen so that about ten times the base current flows through those components. So this approximation is about 90% accurate. More accuracy means taking into account the BJT's Hfe (current gain) and the collector current. If collector current is 5mA and Hfe is 200 then base current is 25 micro amps. This will rise and fall with the undulations of the input signal so it can be accounted for but, like I said before, most folk, in most applications, will say input impedance is R1||R2.

C1 adds to the input impedance and at really low frequencies this will be significant but usually C1 is chosen so that it is generally regarded as a short circuit for AC signals meaning input impedance is still R1||R2.

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  • \$\begingroup\$ Nice answer, thanks. I believe it is more an efficient method than a step by step rigorous one but it is definitely a good one. As I have commented in the other answer I'm confused at why we can use the superposition theorem (DC on the one hand, then AC) since there is a non linear element in the circuit, though. \$\endgroup\$ – user42875 Sep 19 '14 at 21:54
  • \$\begingroup\$ Ah because I knew the answer nearly backwards I don't think of it in steps! It's all one cacophony of info. The transistor is biased to be largely linear so superposition works well. Personally I'd use something like LTSpice if I want a more exact answer but, there's no substitute for knowing the reasons. \$\endgroup\$ – Andy aka Sep 19 '14 at 23:13
  • \$\begingroup\$ @Andy aka, I agree to everything you wrote. Nevertheless, I have a question: I am really not sure how the Early voltage is defined: With IB as a parameter or with VBE (as in your case). We can find both definitions in the literature and I am curious if you have a finite answer or explanation? It is really funny, some authors use IB and some other use VBE - without any explanation or justification. And one thing is clear: In both cases the value of the Early voltage differs. \$\endgroup\$ – LvW Feb 28 '16 at 16:58
  • \$\begingroup\$ @LvW No I don't have an obvious answer other than it's like a resistor in parallel with the collector-emitter connections. \$\endgroup\$ – Andy aka Feb 28 '16 at 18:02
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Just a short answer: According to my experience the following may be helpful.

  1. At first, connect a voltage source Vin to the respective node and convince yourself how many different pathes are available which allow a current to exist. During this step you consider all dc sources as signal ground.

  2. Based on Ohm's law (and/or other laws, if applicable) you can write down the various currents. Then you apply Kirchhoff's current law to find Iin (add all the currents).

  3. Isolate Vin and solve for the conductance g,in=Iin/Vin. It is very convenient to find at first the conductance (and NOT the impedance) because the wanted input impedance consists of several parallel sections (and the various conductances are just added).

  4. Now it is simple to write: r,in=Ra||Rb||Rc....

  5. Example: The output conductance at the emitter node in common-collector configuration consists of three different ways: Current through (a) the ohmic emitter resistor, (b) the emitter-base region and (c) the emitter-collector region (in many cases to be neglected).

6.) Comment: In case there are capacitors which you have considered as a short, the result applies to frequencies only which are far above the corresponding corner frequencies.

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  • \$\begingroup\$ For 1) it sounds like it's a result of the superposition theorem, but can we still use it since the transistor is non linear? Could you apply this method to that circuit? \$\endgroup\$ – user42875 Sep 19 '14 at 21:57
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    \$\begingroup\$ No - it's not superposition. That would be necessary in case of more than one signal source. I forgot to mention under 1.) that, of course, the input signal source must be zero (input node grounded!) if you calculate the output impedance. And - don't forget that input/output impedances always are small-signal parameters. This strictly assumes LINEAR operation. \$\endgroup\$ – LvW Sep 20 '14 at 7:29

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