Question in relation to CE amplifier voltage gain.

thatoneguy

Joined Feb 19, 2009
6,359
Do you have a slightly bigger version of the image?

I'm reading it as the voltage is applied at the output (after the capacitor), is that correct?
 

Audioguru

Joined Dec 20, 2007
11,248
The gm and re of the transistor changes when the transistor's current changes.
When the collector voltage increases because the transistor's current is reducing then the gm is also reducing which reduces the voltage gain.

If the transistor is amplifying a sine-wave then the top portion is compressed and the botton portion is expanded causing severe distortion like this simulation:
 

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Thread Starter

ECE101

Joined Nov 25, 2011
29
Do you have a slightly bigger version of the image?

I'm reading it as the voltage is applied at the output (after the capacitor), is that correct?
Vin is to the left, before the input coupling cap C1, V out is to the right after C2.

The gm and re of the transistor changes when the transistor's current changes.
When the collector voltage increases because the transistor's current is reducing then the gm is also reducing which reduces the voltage gain.

If the transistor is amplifying a sine-wave then the top portion is compressed and the botton portion is expanded causing severe distortion like this simulation:
Ah I understand, thanks for the explanation!
 

crutschow

Joined Mar 14, 2008
34,412
The relative effect of the change in re (and gm which is determined by re) with collector current is determined by the value of re as compared to the emitter impedance at the frequency being measured. If the emitter impedance is much larger than re, then the change in re will have little effect on the circuit gain.
 

Thread Starter

ECE101

Joined Nov 25, 2011
29
From a mathematical point of view:

Gain = -gm*Rc

gm = Ic1/Vt
Rc = 8k

When Vo = +15

AV = -((15/8*10^3)/25*10-3) * 8*10^3
= -600


When Vo = -10

AV = ((10/8*10^3)/25*10-3) * 8*10^3
= 400



Is this correct?
 
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