How do you actually derive the Output Impedance of Common Emitter ?

Thread Starter

silv3r.m00n

Joined Apr 15, 2010
51
My understanding of output resistance/impedance is that it LOWERS the voltage available to a Load R(L) when the Load draws current I(L).
So if the output resistance is Ro and open voltage is Vo

The voltage available to Load V(L) = Vo - I(L).Ro

Now with this common emitter configuration the output at collector has an output impedance
Ro = R(C) || R(L)



I have been trying for several days, but can't mathematically derive it ...
Can somebody show me the mathematical proof of it ?
 

Thread Starter

silv3r.m00n

Joined Apr 15, 2010
51
the eeeguide.com website is quite good.
things are quite well explained.
 

sparky 1

Joined Nov 3, 2018
331
One approach is to first look at a non amplifier impedance then look at an amplifier impedance

messing with leds on house current, an AC source.
1)

How to set up a Common Emitter amplifier, An example
now using both AC signal and DC source where the gain is 10. Will the amplifier circuit maintain the phase and amplitude ?
2)
 
Last edited:

Hugh Riddle

Joined Jun 12, 2020
51
My understanding of output resistance/impedance is that it LOWERS the voltage available to a Load R(L) when the Load draws current I(L).
So if the output resistance is Ro and open voltage is Vo

The voltage available to Load V(L) = Vo - I(L).Ro

Now with this common emitter configuration the output at collector has an output impedance
Ro = R(C) || R(L)



I have been trying for several days, but can't mathematically derive it ...
Can somebody show me the mathematical proof of it ?
For small signals at low frequencies, you can treat the transistor is a pure current source which varies with base-emitter voltage by 25mA/V at 1mA (increasing proportionally with) emitter current. That also means the maximum voltage amplification is about 40 times the DC voltage across the load. These quantities are inversely proportional to absolute temperature. The whole stage's output impedance is thus set by the load components (use Norton's theorem).

At higher frequencies, the transistor's collector-base capacitance reduces the voltage amplification. Hoping these simple rules of thumb are helpful.
 
Top