# How to differ output & input resistance of small-signal model BJT?

I understand how input and output resistance of small-signal model for bipolar transistor is defined and how draw one. What I don't understand quite yet is how actually input and output resistance of BJT are separated.

How do you know which resistances should be included for input and which for output resistance?

Because, if you ask me, models like this one could be identified as one whole resistor, whose value would be considered as input and output resistance simultaneously.

*For the record: ru - base-collector resistance, rpi - base-emitter resistance, ro - collector-emitter resistance, rb - base contact resistance, rc - collector contact resistance, rex - emitter contact resistance.

• these are models. They are drawn that way because it's handy to have them separated into multiple components. Of course you could just reduce the equivalent circuit's complexity, but then it would lose its purpose. – Marcus Müller May 18 '17 at 16:27
• I don't quite understand your question. Rin is a equivalent resistance see from the input source terminal. So if the base is the input we mainly see rpi. But if the emitter is the input the input source well now "see" rpi/beta. Rout is a resistance "seen" from load perspective into the BJT (towards the transistor) with input short. – G36 May 18 '17 at 16:33
• yes, Zin(b) is mainly Hfe (r_π +Ze(f)) for emitter load, Ze and Zout(e)=(r_π + Zb(f)/Hfe) for base load Zb. is same but includes source and load Z and depends if you are interested in DC load regulation for input bias or AC load regulation where Hfe is the small signal gain and hFE is the DC gain – Tony Stewart Sunnyskyguy EE75 May 18 '17 at 16:38

Because, if you ask me, models like this one could be identified as one whole resistor, whose value would be considered as input and output resistance simultaneously.

Your model has three terminals, B, C, and E. A resistor has only two terminals. You can't reduce a three terminal device to two terminals and consider it "equivalent".

If you are using the BJT in common emitter configuration, then the input resistance is, by definition,

$$R_{in}=\frac{{\rm d}V_{be}}{{\rm d}{I_b}}$$

And the output resistance is, by definiton

$$R_{out}=\frac{{\rm d}V_{ce}}{{\rm d}{I_c}}.$$

You can work out what these values are from the model you showed.

If these two derivatives don't have the same value (and they don't), then you can't say the input resistance is the same as the output resistance.

The concept of input and output impedance is not just replacing a circuit with a single resistor.

To find input impedance, you connect a power supply to the input, say Vin and find how much current(say $I_in$) passes through your input $V_in$. You get the ratio Vin/Iin and call it as input impedance a.k.a the impedance that the input voltage source Vin feels. You mathematically (using KVL and KCL) represent this ratio in terms of circuit parameters (Thevenin's equivalent). Whatever circuit parameters (resistors and capacitors) are in this equation, they become part of input impedance.

Similarly, you find output impedance. Few circuit parameters may be a part of both inputs as well as output impedance at the same time. This varies from circuit or circuit.

Input impedance is defined as impedance as looked into input terminals when output terminals are open circuited while output impedance is defined, as impedance looked into output terminals, when AC input is short circuited. Both will not be same in case of BJT. Consider image below..

Thus, even if you reduce equivalent circuit to somewhat similar i shown in pic, ip and op impedance will still largely differ owing to difference in conditions in which they're defined.

• You should clarify, the output resistance is defined with the AC input shorted. – The Photon May 18 '17 at 17:23