BJT amplifier circuit

I'm preparing for an exam and I have this problem where I need to find the input resistance as it is seen by the voltage generator ug.

After using H-parameters to find an equivalent circuit I came up with this equivalent schematic:

equivalent scheme using H parameters

Can anyone explain what it means to find the input resistance (denoted by Rul) in my schematic as it is seen by voltage generator ug? What is the difference between "normal input resistance" and this input resistance?

  • \$\begingroup\$ There are two definitions of gain. The first is versus the generator itself. The other includes the internal generator impedance with Rg versus Vb. The impedance are then Rul =(Ug/Iul) or (Rul-Rg)= Vb/Iul. \$\endgroup\$
    – Antonio51
    Jan 21 at 17:49

2 Answers 2


I had initially said that the resistor \$R_E\$ needed to be included. However, the capacitor in parallel most likely shorts it for signal frequencies of interest.

If you use an common-emitter amplifier without feedback, then its DC input resistance is \$R_b + R_\pi\$ (I'm thinking of the BJT small-signal pi-model). However, if you add some sort of feedback, like in your example (it's series-series feedback), then this will change the nominal input resistance of your BJT.

So, you could say there is an "open-loop" type of input impedance, which I already mentioned for the BJT. However, when you use any kind of feedback, this changes the input impedance. If you use series feedback at the input, that tends to increase the input impedance, while shunt feedback tends to lower it (e.g. a transimpedance amplifier uses shunt-feedback at the input as well as the output).

  • \$\begingroup\$ I didn't include the Re resistor because of short-circuiting Cp capacitor. I think I should short-circuit that capacitor because its impedance is 0. \$\endgroup\$ Jan 21 at 17:09
  • \$\begingroup\$ You're right, but the 2nd part of my answer still holds. \$\endgroup\$
    – Designalog
    Jan 21 at 17:10

For reference ...

Here is an example of "calculating" input impedance (with Rg or not).
The impedances are Rul =(Ug/Iul) or (Rul-Rg)= Vb/Iul.

enter image description here


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