# Input and Output Resistance of an Amplifier

I attempted to find the input resistance by using the thevenin equivalent circuit. I shorted all the capacitors and changed the circuit to its small signal equivalent.

I found Rb to be 13.33kOhms because it is the equivalent resistance of 120k and 15k.

I used the equation rπ = β × re and found re=(25mV/Ic)

I found re to be 4.537 ohms by using the relation

Ve=1/10 *Vcc. Ve= 1.2V
1.2=Ie *Re
Ie=1.2/Re === Ie=1.2/220 ===Ie=5.45mA

but Ie is approx = Ic,
so re=25mv/ 5.45mA = 4.537Ohms,
thus rπ = 4.537β.

Then I found Rin= [(13.33*10^3)*4.537β]/[13.33*10^3 + 4.537β]

But I have no idea where to start finding Rout

• Rout is simply R1
– S.s.
Commented May 13, 2020 at 8:44
• You found Ve=Vcc/10. Can you explain HOW you arrived at this equation?
– LvW
Commented May 13, 2020 at 9:01
• i made if out from a data sheet i found. Is it right? Commented May 13, 2020 at 9:33

But I have no idea where to start finding Rout

Look at the standard graph of a BJT's characteristic: -

Choose a base current - say 60 uA

Look at the change in collector current ($$\I_C\$$) versus the change in $$\V_{CE}\$$: -

• change in $$\I_C\$$ is about 0.25 mA
• change in $$\V_{CE}\$$ is about 5 volts

The slope has a dynamic resistance of 5 volts / 0.25 mA = 20 kohm.

This is the effective "raw" output impedance of a BJT operated in common emitter. That raw impedance is in parallel with R1 as far as AC signals are concerned hence, if R1 is significantly smaller than the innate collector impedance, then a decent approximation to the circuit's output impedance is R1.

At higher frequencies than a few kHz, we also take account of the miller capacitance (internal capacitance between collector and base).

• @user248817 have you done with this question and answer now or, do you need something clarifying? Commented Jun 18, 2020 at 11:59