Transistor capacitance multiplier

Discussion in 'General Electronics Chat' started by Veracohr, Oct 9, 2014.

  1. Veracohr

    Thread Starter Well-Known Member

    Jan 3, 2011
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    I just learned of this capacitance multiplier circuit, but it doesn't seem to simulate as described. Comparing it to a simple RC filter using the same values, there's no difference in filtering. The only thing it does is act as a buffer when heavily loaded, but it still doesn't reduce the ripple by any larger percentage.

    What am I missing here? This shows it without a load, but with a load it functions as I said.

    schematics.png waves.png
     
  2. alfacliff

    Well-Known Member

    Dec 13, 2013
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    you also need a main filter cap before the transistor. then the circuit will help reduce ripple some. no main filter cap, none at all. why do the waveforms show sine waves? no rectification?
     
  3. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Your circuit don't work because no current is flowing through transistor (transistor is cut-off). Also this circuit is not "real" capacitor multiplet (the circuit is not really compare to a real capacitor). Because how can voltage follower multiple the capacitance? This circuit will only repeat the base voltage at the output nothing more. For 10k ; 1uF and F = 120Hz, Vin ripple = 0.2Vpp, the ac voltage at capacitor equal to Vc = 20uA * 1.3k = 26mVpeak.
    So the output voltage will also have 26mVpeak of ripples.
     
  4. alfacliff

    Well-Known Member

    Dec 13, 2013
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    it can work, because the base being held with less ripple than the emitter causes the emitter to base voltage to stabilise the output. there is also a voltage regulator circuit that is similar, just has a zener across the cap. if the emitter voltage drops, the transistor turns on harder.
     
  5. ronv

    AAC Fanatic!

    Nov 12, 2008
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    You just need to add a load to Vo.
     
  6. Veracohr

    Thread Starter Well-Known Member

    Jan 3, 2011
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    I just made the sine wave to have some kind of ripple to test it out. I didn't feel like going to the trouble of more, I was just simulating out of curiosity to see it work (which it didn't).

    I saw the circuit as part of a larger circuit my teacher has been describing. Unfortunately he hasn't made the schematic available to us, just showed it in class, and I don't remember the exact values or how exactly it was connected. In the circuit where it appears, it's used as a sort of enhanced power supply decoupling right at the main circuit. The circuit I simulated is exactly what appears in quite a few links I found on Google.

    Like I said, adding a load doesn't really make it work. The ripple was maybe reduced a tiny bit, but my understanding is that it should act like the capacitor value is multiplied by the transistor beta, which in the simulation model is 300. It didn't come even close to that.
     
  7. ronv

    AAC Fanatic!

    Nov 12, 2008
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    Yes, I see now. You need to balance the R & C to get the desired results. Larger R & C lower ripple but longer time constant to a final value.
     
  8. Veracohr

    Thread Starter Well-Known Member

    Jan 3, 2011
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    Well I'm an idiot. Turns out I just wasn't using a heavy enough load. It seems the degree of ripple rejection increases as the load increases. Which, now that I think about it, makes sense; the more current the load draws, the more current the transistor base steals from the capacitor.
     
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