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Diode capacitor VCF

Discussion in 'General Electronics Discussion' started by pyrohaz, Dec 24, 2012.

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  1. pyrohaz

    pyrohaz

    33
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    Oct 28, 2012
  2. bob91343

    bob91343

    10
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    Mar 24, 2010
    The diodes are switches. When they are forward biased, they act like short circuits, connecting the capacitors to ground. Otherwise the capacitor is out of the circuit.

    You can do the same thing with a switch.
     
  3. pyrohaz

    pyrohaz

    33
    0
    Oct 28, 2012
    I thought of that, but the filter frequency is infinitely variable between two points depending on whether the voltage is 0 or 5v. I was reading up on how the AC impedance of a diode can change depending on its DC current, would someone kindly explain this to me?

    Thanks, Harris
     
  4. bob91343

    bob91343

    10
    1
    Mar 24, 2010
    Well then it's a lossy variable filter. The diodes, if gently biased, act as resistances and thus will sort of add the associated capacitor into the filter. Not a great system but hey if it works..
     
  5. pyrohaz

    pyrohaz

    33
    0
    Oct 28, 2012
    Hey, thanks for the reply. Is there anyway that I could work out the equivalent resistance of the diode?
     
  6. bob91343

    bob91343

    10
    1
    Mar 24, 2010
    Sure. Look at the V-I curve of the diodes (or plot one yourself with a voltmeter, ammeter, and power supply) and look at the slope of the curve. The incremental value is the equivalent resistance (neglecting high frequency effects). V/I=R. Or more precisely, delta V/delta I = delta R and you are seeking delta R.

    Of course you need to take measurements at the operating current of the circuit, which is pretty small.
     
  7. pyrohaz

    pyrohaz

    33
    0
    Oct 28, 2012
    If I was to measure this circuit in real life, could I apply a voltage to the 330k resistor and measure the voltage drop across the resistor, subtract that from the applied voltage and calculate the current.

    Using kirchoffs laws, which state that the current is the same through out a loop (there are no branches), then divide the voltage drop across the diode by this calculated current to find the equivalent resistance? You mentioned earlier about the resistance possibly being different at higher frequencies, since this is an audio low pass filter where the maximum frequency is around 22kHz (And a lo-fi one at that!) is this effect negligible?

    Cheers,
     
  8. bob91343

    bob91343

    10
    1
    Mar 24, 2010
    You don't have to measure the current. Seeing the 330k resistor, that swamps any resistance the diode may have so you just subtract the approximate diode drop (say 600 mV) from the applied voltage and divide by 330k. So if you apply 5V, you subtract and get 4.4 V and then divide by 330k to get about 13 microamperes. And yes, at 22 kHz the capacitive effects are insignificant.
     
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