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Another transistor question

Discussion in 'Electronic Basics' started by zalzon, Aug 3, 2003.

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  1. zalzon

    zalzon Guest

    Please take a look here. Tell me if you can why putting an emitter
    bypass capacitor on a common emitter amplifier makes the signal the
    way it is. (i.e. amplified and clipped). Exactly why does sending
    the AC to ground at the emitter increase the gain?

    http://www.anycities.com/user/msc/normal.html

    thank u in advance to anyone who helps me understand this.
     
  2. The emitter resistor develops an AC voltage component (in addition to
    its DC drop) that subtracts from the base to emitter input signal.
    This subtraction is a form of negative feedback that reduces the
    effective stage gain of the transistor.

    Bypassing the AC current that would have passed through the emitter
    resistor by providing a very low impedance capacitive path, reduces
    the voltage this AC current produces at the emitter, so much more of
    the input signal voltage appears across the base emitter junction.
    This larger input signal produces a larger output signal, to the
    point, in this case, of saturating the transistor on or off (or both)
    at the peaks of the output signal.

    Put an additional resistor in series with the bypass capacitors, and
    you can adjust the effect between no bypass and full bypass, and get
    any intermediate effect you want.
     
  3. zalzon

    zalzon Guest


    Is this because of the phases of the signal waves. i.e. the phase of
    the signal wave in the base is out of step with the phase of the
    signal wave at the emitter-base (without the bypass capacitor). Do
    the waves cancel/muddle each other out thereby reducing the signal
    voltage at base?

    Does my description below accurately portray what's happening :

    A bypass capacitor is putting the AC signal at the emitter at ground
    level. So the difference in potential between base and emitter
    (ground) makes the signal voltage appear larger.

    This greater signal voltage than makes more or less DC current to flow
    through from emitter to collector thereby amplifying the signal
    voltage even further.

    Is that all correct? I think I would be devastated if i saw the
    word... NO :0
     
  4. John Jardine

    John Jardine Guest

    No emitter cap' and the emitter voltage follows sympathetically whatever
    voltage is fed to the base.
    This emitter voltage is across the emitter resistor,
    Therefore the emitter current also follows in sympathy.
    As the emitter current is pretty much the collector current, then the
    collector voltage must also follow suit.
    As the collector resistor is a bigger resistance than the emitter resistor
    then the voltage across the collector resistor must be a bigger version of
    the emitter voltage. You've already fixed the current (a.c.) that can flow
    down the transistor by using the emitter resistor, so a value of collector
    resistor can be chosen to give whatever voltage gain you want (or even
    voltage loss).

    Add an emitter cap' and you lose the ability to develop a voltage across the
    external emitter resistor *but* you still have a small resistance inside the
    transistor at work. This resistor value is = 25 / emitter ma's.
    (In your case, the quiescent transistor emitter current is set about 0.45ma,
    so this internal resistor is 25 / 0.45 =56ohm.)
    Thus the input signal voltage across this internal emitter resistance sets
    the emitter current and hence the collector current and in turn the colector
    voltage and overall voltage gain.

    So ...

    Without an emitter cap' your circuit has a fluctuating emitter current of
    100mVac / 1500ohms = 67uA. This turns up across the 10k collector resistor
    as a fluctuating voltage of 67uA X 10kohms = 0.67Vac. Thus the stage has a
    voltage gain of 0.67Vac / 100mVac =6.7 times.

    With an emitter cap' the fluctuating emitter current is 100mVac / 55ohms =
    1.8ma. This tries to force a fluctuating voltage across the 10k collector
    resistor of 1.8ma X 10k = 18Vac!. Thus the voltage gain is massively higher
    (consequently clipped at the rails), as can be seen on the scope display.

    There's other knock on effects caused by this high gain and in practice it's
    better to leave some undecoupled external emitter resistance.
    Essentially, go for quality rather than quantity.

    regards
    john
     
  5. zalzon

    zalzon Guest

    Thanks john your reply cleared up some issues. Its starting to make
    sense.

    Who would have thought learning electronics could be so complex :(
     
  6. zalzon

    zalzon Guest

    Thank you all for your reply, I'm going through each message twice
    trying to follow the examples & calculations laid out.

    You people must all be pros cos you seem to understand it like the
    back of your hands.

    Thanks once again.
     
  7. zalzon

    zalzon Guest


    No its not the math at all. Its the lack of clear, proper
    explainations of what's going on in a circuit.

    Sometimes when i read explainations, i get the impression the author
    of a textbook/message is talking to himself. He has lost conciousness
    of the fact that he's explaining it to a beginner.

    If a beginner could understand all that techno bable on electronics,
    he would not be asking for an explaination in the first place.

    Too many instructors/authors just fly over the essential or difficult
    points assuming its all obvious when its obviously not. It may be
    obvious after spending 3 yrs in the field but its not to a beginner.
     
  8. A E

    A E Guest

    You want clear explanations? Head for the used book store, and find books from
    the golden era of electronics hobbyists: the 1960s. Understanding electronics,
    for example, the Schaum outline series. I think that most books these days exist
    mainly to make money through university sales, but back then, it was for the
    passion of the thing.
     
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