common emitter amplifier

Discussion in 'General Electronics Chat' started by mentaaal, Jun 4, 2007.

  1. mentaaal

    Thread Starter Senior Member

    Oct 17, 2005
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    Hey there, i was reading up on common emitter amplifiers in allaboutcircuits and i was wandering if one of you could explain this to me: How is it that the output current is out of phase with the input? I understand how the voltage across the transistor is inversely proportional to the input voltage. But say for example if the input voltage increases, the voltage across the transistor will decrease allowing more voltage to be dropped across the load. And more output current in general. I notice that this happens except that it is 180 degrees out of phase according to the text.

    But how is it out of phase? does the current not flow in the same direction across the transistor? Are the respective currents not in phase into the emitter? And is the voltage waveform across the load in phase with the input voltage? And is the voltage waveform across the load in phase with the input voltage?

    Thanks alot guys!
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
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  3. rampart

    Member

    May 30, 2007
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    While it is true that the voltage across the transistor i.e. Vce is 180 degrees out of phase with the input voltage signal, I don't think the same is true for the base and the collector currents. An increase in the input voltage causes the Ib(base current) to increase which in turn enhances the collector current (Ic) correspondingly (upto a certain point,notwithstanding!) in accordance with the formula Ic=B*Ib, where B stands for beta(current gain). Thus more voltage drop occurs across the load resistor (or collector resistor) which means a low drop takes place across the transistor i.e. low Vce. Similarly, a decrease in the inpute voltage results in an increase in Vce. Thus the input voltage and the drop across the transistor (Vce) are 180 degrees out of phase with each other. As far as the base current and the collector current are concerned, they are in phase with each other as indicated by the formula Ic=B*Ib. This formula says that as the base current increases, the collector current also increases. So I dont think it is right to say that the two currents are out of phase.
    Thanks,
    Fahad
     
  4. mentaaal

    Thread Starter Senior Member

    Oct 17, 2005
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    Thanks for the reply and for clearing that up! From what you said.. could i tentatively ask that the voltage waveform is also in phase with the source? If this is so why bother calling it an inverting amplifier at all? I mean fair enough the voltage across the transistor is inversely proportional but who cares when the votage across the load is directly proportional? Unless i have this completely wrong which wouldnt be the first time....

    Cheers!!
     
  5. rampart

    Member

    May 30, 2007
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    In transistor circuits, the output voltage is considered to be the voltage across the transistor i.e. Vce, not the voltage drop across the collector resistance ( which I have incorrectly referred to as the load resistance...sorry for that). The load is coupled with the transistor through a coupling capacitor so that Vce appears across the load. Thus the load signal is the same as that appearing across the transistor.
    Thanks,
    Fahad
     
  6. mentaaal

    Thread Starter Senior Member

    Oct 17, 2005
    451
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    .....ah ok then fair enough. guess i am not far enough to understand the whole coupling capacitor thing yet but ok. So if you were wiring the circuit as i said like the simple illustrative circuit in the common emitter amplifier section of allaboutcircuits, then the voltage and current would be in phase across the load?

    but if you coupled the load with a coupling capacitor then the votage would be inversed... ok i think i have got it! and the capacitor is used to stop the dc and just pass the ac signal right?

    cool thanks for the reply helped me alot!
     
  7. rampart

    Member

    May 30, 2007
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    The load is coupled through the capacitor directly with the collector of the transistor and thus the load and transistor are in a sense in parallel. Thus the same voltage appears across both of them. So an "inverted" voltage appears across the load (parallel elements have the same voltage). In the next post I will try to "graphically" (or visually?) explain it to you.
    Thanks,
    Fahad
     
  8. rampart

    Member

    May 30, 2007
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    In the attached file, you can see that the transistor and the load are in parallel. (Yes, the capacitor only allows AC to be passed to the load and blocks DC.)
    Forgive me for the "rough" diagram. I just drew it hastily only for your understanding. It might not work properly.
    Is the thing clear now?
    (sorry...the file i have in my computer does not load. Maybe my connection speed is very slow.Please try this :-
    http://www.answers.com/topic/common-emitter
    In the circuit shown here, the load is attached with the end labelled Vout. So you can see that the load and transistor are in parallel. Also note that Rc(collector resistor) and Rl(load resistor) are two different elements.)
    Thanks,
    Fahad
     
  9. mentaaal

    Thread Starter Senior Member

    Oct 17, 2005
    451
    0
    Thank your very much for going to all the effort to answer my question and yes i understand it perfectly now! I made sure to bookmark that page. I have ordered some things from maplin and am going to have to try and make one of things! i have a small circuit mounted mic and and little speaker from past failed kits. Not the kits fault just my crap soldering skills!

    Cheers!
     
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