I just want to know what exactly input impedance is with respect to the diagram below. And the second thing is obvious, how do we calculate it.enter image description here

  • \$\begingroup\$ What have you tried? is this homework? there are hundreds of books that explain this in detail \$\endgroup\$
    – S.s.
    Dec 11, 2020 at 13:55
  • \$\begingroup\$ what i understood is it maybe the parallel combination of the resistor R_1, R_2, but they are using some other additional resistance, which i do not understand. \$\endgroup\$
    – Sayan
    Dec 11, 2020 at 13:58

1 Answer 1


It's fairly simple (now that you've posted a schematic): -

enter image description here

  • Both 100 kΩ bias resistors are considered to be in parallel for input impedance because the 10 volt DC voltage is regarded as a short to AC signals
  • The emitter load is in parallel with the 5 kΩ emitter resistor and capacitors are regarded as shorts to AC when the circuit is operating mid-band AC.
  • The effect of the combined emitter resistance (5 kΩ||3 kΩ) is seen at the base as an impedance of 5 kΩ||3 kΩ multiplied by β. You probably know this but \$h_{FE}\$ = β.

So you have 50 kΩ in parallel with 187.5 kΩ = 39.5 kΩ. That's a fairly decent calculation for the AC input impedance mid-band. Trying to calculate it to any greater depth is missing a certain point about single transistor circuits - calculations are at best an approximation due to such large variations in β and its temperature dependency.

If you want to know what it is at low frequencies you need to take into account the capacitors shown in the circuit.

  • \$\begingroup\$ @ andy ...If you want to know what it is at low frequencies you need to take into account the capacitors shown in the circuit- if We are considering a circuit with very high input frequency , shorting the capacitor would be enough right? \$\endgroup\$
    – Sayan
    Dec 11, 2020 at 14:10
  • 1
    \$\begingroup\$ @Sayan yes it would but always check. At 1 kHz the 0.04 uF input capacitor has a reactive impedance of about 4 kohm so it will add to the 39.5 k and make it about 39.7 kohm. \$\endgroup\$
    – Andy aka
    Dec 11, 2020 at 14:13
  • 1
    \$\begingroup\$ @Sayan, for a better understanding: The multiplication with hfe - as explained by Andy aka - is a consequence of negative feedback caused by the emitter resistor RE. From system theory we know that the input resistance of a circuit with a negative feedback voltage (caused by RE) will be larger in comparison to such a circuit without feedback. \$\endgroup\$
    – LvW
    Dec 11, 2020 at 15:47

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