Common - Emitter

Thread Starter

oracle_dfx

Joined Sep 19, 2004
4
Hi, Would anyone know why the output signal of the common emitter amplifier is out of phase by 180º? If available, pls include the computations that mathematically prove the inversing of the signal.. Thanks!

mozikluv

Joined Jan 22, 2004
1,435
check the books authoured by Hayes & Horowitz, Keith Brindle, Don Canon to say a few.

hgmjr

Joined Jan 28, 2005
9,027
Hi oracle_dfx,

If you haven't read it already, there is a pretty good introduction to common emitter transistor operation in the Volume III Semiconductors on this website. Well worth reading.

Good Luck

dragan733

Joined Dec 12, 2004
152
Originally posted by oracle_dfx@Feb 23 2005, 08:14 PM
Hi, Would anyone know why the output signal of the common emitter amplifier is out of phase by 180º? If available, pls include the computations that mathematically prove the inversing of the signal.. Thanks!
[post=5558]Quoted post[/post]​
Let's see how were the calculations, here

Thread Starter

oracle_dfx

Joined Sep 19, 2004
4
dragan thanks for the calculations,, helped a lot..^^

hgmjr

Joined Jan 28, 2005
9,027
Here is a 5 page analysis of the CE transistor amplifier that establishes the fact that this configuration does produce an output voltage signal that is an inverted version of the input voltage signal.

I have zipped up the 5 files that were previously posted individually to both reduce the size of the posted file as well as make it less cumbersome to download.

[attachmentid=466]

duqamoq

Joined Mar 23, 2005
3
It's all very well to go through the rigorous analysis, but it's also important to have a good conceptual understanding of this phase inversion. The CE transistor gets its collector bias from Vcc through a load resistor. The collector side of the resistor is the output. When the transistor is turned on harder (through an increase in base potential) more current flows into the collector through the load resistor, increasing the voltage drop across the resistor. Since the top of the resistor is at a fixed voltage (Vcc), the collector side has nowhere to go but down. Hence a rise in base potential causes a fall in collector potential (all with respect to ground, of course).

hgmjr

Joined Jan 28, 2005
9,027
Good point, duqamoq.

An explanation of the basic factors that account for the inversion in the CE stage makes the mathematical expressions easier to understand and appreciate.

Xeeshan Qureshi

Joined Jan 13, 2009
14
It's all very well to go through the rigorous analysis, but it's also important to have a good conceptual understanding of this phase inversion. The CE transistor gets its collector bias from Vcc through a load resistor. The collector side of the resistor is the output. When the transistor is turned on harder (through an increase in base potential) more current flows into the collector through the load resistor, increasing the voltage drop across the resistor. Since the top of the resistor is at a fixed voltage (Vcc), the collector side has nowhere to go but down. Hence a rise in base potential causes a fall in collector potential (all with respect to ground, of course).
thanx a lot for the answer, this precisely was my question.

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