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A circuit question on emitter amplifiers

Discussion in 'Electronic Basics' started by zalzon, Jul 26, 2003.

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  1. zalzon

    zalzon Guest

    Shown below is a 2 stage emitter amplifier.

    I am unable to figure out why that capacitor is placed in the middle
    between the two stages. Also how did they get 2250 for the load
    resistance? Please help me someone.

    http://www.anycities.com/user/msc/stages.html
     
  2. It's called "ac coupled." You can DC couple amplifier stages,
    too. But it can eat up your supply voltage headroom pretty
    fast. You want to recenter the average voltage at the collector
    of the stage's transistor to about midway, so what they are
    doing with the AC coupling is to allow the signal oscillations
    to pass through the capacitor while blocking the steady DC level
    of the 1st stage output. This makes it easier to set the input
    bias point for the next stage and to better isolate each stage's
    design and allow the recentering of this signal average at the
    output back to about midway, again.

    The load resistance seen by the 1st stage is approximately (not
    exactly) looking at the 1st stage collector resistor in parallel
    with each of the two input biasing resistors of the 2nd stage.
    That works out to 2330 ohms. It's a little lower than that
    because the emitter resistor of the 2nd stage, multipled by the
    beta, is also in parallel. So assuming a beta of 100, this
    means 2330 in parallel with 68k (100*680), which works out to
    2250 ohms.

    Also, it's typical to use a larger emitter resistor to ground,
    in parallel with another series resistor+capacitor, to set the
    AC gain while lowering the required DC quiescent current. But
    that's not in the diagram you mentioned.

    I'm a hobbyist, not a designer. This is *not* something I'm
    practiced at, so take all this with a grain of salt.

    Jon
     
  3. zalzon

    zalzon Guest

    What does 'eat up your supply voltage headroom' mean? Please I'm a
    novice.
    I assume this is the Q point (quiet point). I read about it but could
    not understand what its significance was. What really does it
    represent?
    Now these are the two most important questions I have :

    1) Why do they want to block DC?

    2) Would not the first capacitor on the extreme left already be
    blocking the DC from the power supply? Why the need to put another
    one in the center and at the load end?
    What happens if the capacitor is not there? I cannot seem to get my
    head around why it makes it easier to set the bias point. All I
    understand is that :

    - a capacitor charges up and discharges
    - it passes AC but blocks DC

    In the context of the above, why/how does a capacitor help to set the
    bias point for the next stage?

    Sorry I'm all messed up.
     
  4. zalzon

    zalzon Guest

    Ah. That's an excellent explaination Mr Black. I just need to
    clarify one thing :

    If AC voltage goes into a transistor's emitter, is the output both DC
    and AC at the collector?

    I was under the impression that one capacitor at the start is all
    that's needed to get rid of the DC for good. That is why I could not
    understand why 3 capacitors were needed !

    So in conclusion, the capacitors are there to strip off DC everytime
    the current goes through a transistor! Is this correct?
     
  5. zalzon

    zalzon Guest

    Thanks for your message. Most of it made sense to me except the first
    part about the q point. I suspect I need to read about it a little
    more for it all to sink in especially the DC stuff at the collector.

    Anyway, I loaded up electronic workbench (a software to simulate
    electronic circuits) and rebuilt your schematic of a circuit. I used
    the oscilloscope in the software to check out the amplification. It
    works. Compared to the source, the signal coming out the collector
    was amplified.

    I then built on a second stage exactly like the first (with the
    capacitor inbetween) and it was amplified further. I checked the
    current at various places using the multimeter in the software..etc

    I also was able to followed your calculations quite allright.

    Thanks once again!
     
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