CE Amplifier bypass capacitor

Discussion in 'General Electronics Chat' started by HarrisonG, Aug 30, 2016.

  1. HarrisonG

    Thread Starter Member

    Aug 1, 2016
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    I can't understand the current flow in a CE amplifier with bypass capacitor. If there is no capacitor, then the AC signal will pass through the emitter resistance and cause additional voltage drop there, making the emitter more positive. This will decrease the base and the collector currents and the voltage at the collector will increase (when it is supposed to decrease). This is the reason to bypass the Ac part of the signal with an emitter capacitor.
    But, since the capacitor is supposed to have very little reactanse at the working frequency, in order to be like short circuit for the ac signal, wont that mean, that for that part of the amplification process, when the AC source is positive at the base, the emitter will be grounded at 0volts and the whole bias current through the emitter resistance will decrease since now it is approximately 0,1v/RE, non like the previous 1v/RE? I dont understand this, I ask for your help.
     
  2. ericgibbs

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    Jan 29, 2010
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  3. Jony130

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    Feb 17, 2009
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    No, CE capacitor will not change the DC bias condition. DC current will not flow through the capacitor, only AC current part of a signal will be shorted to ground.
     
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  4. HarrisonG

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    Aug 1, 2016
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    Oh, so it's like there are two separated, or parallel circuits. The AC one, which current can pass freely through the capacitor and from its prospective, it will look like the emitter is almost shorted to ground because of the significantly small reactance of the bypass capacitor. And the DC circuit, which current cannot pass through the capacitor and the only path it can take is the emitter resistor.

    Thanks, guys!
     
  5. HarrisonG

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    Aug 1, 2016
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    Upon further inspection, I see that it doesn't even follow Kirchoff's second law if the DC component were to be affected by the AC signal in this case. Shortening the emitter to ground via AC signal means that the current through RE will decrease, because now there is a shorter path to the emitter of the transistor. And what would happen is that the emitter will become more negative and this will increase the base current (Vcc-(Vbe+VE)/Rb). But that is if both the emitter and the capacitor currents were to flow through the base to Vcc, which isn't true, because the current through the capacitor will flow to the AC source and not through the base to Vcc. And now if we remove that current from our equation for the base current, we see, that there is some inconsistency because the current is far less then what it was calculated to be. There were obviously two separated circuits with separated voltage drops and such.
     
  6. Jony130

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    Did you ever herd about superposition ?
     
  7. HarrisonG

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    Aug 1, 2016
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    Isn't that from physics? The ability of a particle to be at more than one place at the same time? Just joking. I haven't heard of super position in electronics, but I hope I was right about the separated circuits. Was I?
     
  8. Jony130

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  9. HarrisonG

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    Aug 1, 2016
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    I remember now. At the time I was a bit confused with Superposition because it was hard to me to understand the point and the calculations behind it. But now, years later as you shared with me this link I finaly learned it - There might be a circuit with more than one Voltage source, which drive their currents in an opposite direction to one another through a common resistor. And we can youse the Superposition theorem to solve for the current through that resistor.
     
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