Diode Clamping Capacitor Question

Discussion in 'General Electronics Chat' started by cabbage_breath, Oct 2, 2013.

  1. cabbage_breath

    Thread Starter New Member

    Oct 2, 2013
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    Hi!

    I am doing a lab for diode clamping I have a circuit that looks like this:

    [AC Signal 6vpp 1khz] -> 100uF electrolytic -> 100k resistor in parallel with a reverse biased 1n914 -> ground

    The cathode of the diode is connected to the positive side of the capacitor.

    On the oscilloscope which is measuring the voltage across the diode/resistor there is only the .7v clamp when the capacitor has a low voltage rating.

    WHY?

    Why does the voltage rating of the capacitor affect this circuit?
     
  2. #12

    Expert

    Nov 30, 2010
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    post a schematic
     
  3. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    It sounds like this circuit:
    http://upload.wikimedia.org/wikipedia/en/thumb/d/d2/Positive_Voltage_Clamping_Circuit.svg/220px-Positive_Voltage_Clamping_Circuit.svg.png

    The resistor in parallel with diode. The diode is connected in reverse so that when voltage is decreasing and goes to negative peak, the diode will provide the path.

    Wiki calls it Positive Unbiased Clamp
    In the negative cycle of the input AC signal, the diode is forward biased and conducts, charging the capacitor to the peak positive value of VIN. During the positive cycle, the diode is reverse biased and thus does not conduct. The output voltage is therefore equal to the voltage stored in the capacitor plus the input voltage gain, so VOUT = 2VIN
    http://en.wikipedia.org/wiki/Clamper_(electronics)#Positive_unbiased
     
  4. cabbage_breath

    Thread Starter New Member

    Oct 2, 2013
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    That's the circuit.

    The real question is why does changing the voltage rating of the capacitor affect the voltage waveform across the resistor?
     
  5. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    From Wiki:
    In the negative cycle of the input AC signal, the diode is forward biased and conducts, charging the capacitor to the peak positive value of VIN. During the positive cycle, the diode is reverse biased and thus does not conduct. The output voltage is therefore equal to the voltage stored in the capacitor plus the input voltage gain, so Vout = Vcapacitor+Vin

    As you change the cap, the Vout changes accordingly, assuming you keep Vin the same for all cases of different capacitors.
     
  6. cabbage_breath

    Thread Starter New Member

    Oct 2, 2013
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    I might be missing something. Isn't the voltage rating of the capacitor in no way related to the voltage across it in any given circuit?
     
  7. cabbage_breath

    Thread Starter New Member

    Oct 2, 2013
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    How are capacitors of the same capacitance but various "voltage ratings" different?
     
  8. #12

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    The voltage rating of the capacitor has nothing to do with the output voltage (as long as the voltage rating is sufficient so that the capacitor does not get damaged). The capacitor does not have "gain".
     
  9. cabbage_breath

    Thread Starter New Member

    Oct 2, 2013
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    exactly. So why did this circuit behave differently when i put in a 50v cap vs a 15v cap?
     
  10. #12

    Expert

    Nov 30, 2010
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    Maybe one of them is leaky. Maybe the impedance of the source took longer to charge the capacitor. Then, you never said how much voltage was being applied, so maybe you were applying .7 volts. I can't tell from here. I also can't tell how it behaved differently because you did not say.
     
  11. shteii01

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    He said 6 volts peak to peak at 1kHz.
     
  12. cabbage_breath

    Thread Starter New Member

    Oct 2, 2013
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    I posted the voltage in the first post. 6Vpp.

    My current theory is that the ESR of the caps is different; I assume that larger caps with larger voltage ratings have larger equivalent series resistance. This would mean higher potential drop across the capacitor. If the voltage across the diode/resistor was less (b/c of a higher ESR on the cap) then the waveform display on the oscilloscope (forgot to mention that this is all taking place on a scope) would show a more pronounced clamp since the 0.7v drop from the diode would...

    nevermind that is starting to sound wrong.

    i don't really know what to think
     
  13. cabbage_breath

    Thread Starter New Member

    Oct 2, 2013
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    Good point, how were they different.

    the caps with larger voltage ratings did not show a flat clip at the negative peak.

    the caps with smaller voltage ratings did.

    I forget what the time of the clip was but it was probably something around .5ms?
     
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