Blocking capacitor value for CE amplifier

Discussion in 'General Electronics Chat' started by Gordon Freeman, Sep 6, 2016.

  1. Gordon Freeman

    Thread Starter New Member

    Sep 6, 2016
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    Hello. I wanted to ask about choosing a blocking capacitor value for common emitter amplifier. I know that at working frequency, the reactance of the capacitor should be less than 2ohms. But as far as I know, different sounds are composed of different frequencies and what should I do if my microphone has a frequency range for example 300 to 3kHz. What is the working frequency then? Or I should take 3kHz as the working frequency and hope that at 300Hz it wont cause too much of a rise in the capacitor reactance?
     
  2. DickCappels

    Moderator

    Aug 21, 2008
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    Often DC blocking capacitors are chosen such that the low frequency roll-off (1/(2Pi x R x C), for audio, is 1/10 or less of the lowest frequency of interest.

    The exact cutoff frequency is a function of your application and its specifications.
     
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  3. Gordon Freeman

    Thread Starter New Member

    Sep 6, 2016
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    Ah, so this way we can assure that at the lowest frequency of interest we will have no more than 2ohms. And as the frequency rises the ohms will fall.
    Thanks!
     
  4. #12

    Expert

    Nov 30, 2010
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    I don't know where you're getting that 2 ohm number. It seems out of range for an audio amplifier.
    If you can post any kind of schematic we might be able to point to a better way.
     
  5. Gordon Freeman

    Thread Starter New Member

    Sep 6, 2016
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    From here: Screenshot_2016-09-06-21-24-36.png This is where at the time I was searching for a way to calculate the blocking cap value. Although the thread maker in there isn't me :D
    But I get it. Because it must behave like a shortage and since capacitor's reactance decreases with rising frequency, then I should calculate it for my lowest working frequency.
     
  6. #12

    Expert

    Nov 30, 2010
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    Yebbut...you aren't making a gigahertz amplifier with a 50 ohm impedance and nobody said, "common emitter".
    Back up about 2 steps.
    Design the rest of the "common emitter" stage, then figure the capacitor size.
     
  7. Gordon Freeman

    Thread Starter New Member

    Sep 6, 2016
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    But how if that's not the way?
    Im not doing the stuff these guys do. Im planning on maybe building CE amplifier to amplify a microphone signal just for practice. And I know that the dc blocking capacitor needs to.behave like a short between the source of the ac signal and the base of the transistor which means it must have resistance close to 1 at working frequency.
     
    Last edited: Sep 6, 2016
  8. #12

    Expert

    Nov 30, 2010
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    OK. At 300 Hz a capacitor which has an impedance of 1 ohm is 530.5 uf. Probably the next higher value that you can buy is 680 uf.
     
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  9. Gordon Freeman

    Thread Starter New Member

    Sep 6, 2016
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    Okay. My textbook says that together with the base resistors, the capacitor acts like a high pass filter. And so it's value must be calculated to pass the frequency we are interested in using this: C>=1/2pf*(R1||R2)
    Is this the way?
     
  10. Gordon Freeman

    Thread Starter New Member

    Sep 6, 2016
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    Oh, #12, I thought your comment was ment to prove me wrong, but it seems the night time is making jokes of me and I have read it the wrong way which I am sorry about :D, but back on the point: So it is not wrong to actualy calculate the value of the cap to have 1 or less ohms at the lowest working frequency? And I can go that way with CE amplifier?
     
  11. Veracohr

    Well-Known Member

    Jan 3, 2011
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    It's not wrong, it just might be overkill. If the input impedance of the amp is 10k ohms, you could get away with a capacitor with an impedance of up to 1k ohms in your range of interest, if the source has very low impedance.

    If you're intent on getting 1 ohm max, you may end up with a much larger (physically) cap than is necessary.
     
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  12. HarrisonG

    Member

    Aug 1, 2016
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    Gordon Freeman? Dr Gordon Freeman? What are you doing here, I thought you have a world to save..
    Nvm. What are you looking for is best demonstrated in the AC analysis of a common emitter amplifier. The coupling capacitor together with the input impedance of the amplifier form a high pass filter. Here, take a look:
    images.png
    RB is simply the voltage divider resistors in parallel(they are in series, but for the ac current they are in parallel)-R1||R2. Next to in on the right side is the input impedance of the transistor ZinQ. ZinQ is calculated via multiplying Hfe by the internal resistance of.the transistor re and you find it by dividing the emitter current by the thermal voltage which is always 26mV:26mV/Ie. Now you calculate ZinQ=Vin/Iin= re*Ie/Ib = re*((Hfe+1)*Ib)/Ib = re*Hfe+1.
    Together, Rb and ZinQ in parallel form the input impedance of the amplifier. When you find it, you just use the cutoff frequency equation to solve for C. f=1/2пRC= C=1/2пfR.
     
    Last edited: Sep 7, 2016
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  13. Gordon Freeman

    Thread Starter New Member

    Sep 6, 2016
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    Thanks!
     
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