Waveclipper with capacitors as Voltage sources.

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Homework Statement




A triangular wave generator produces a waveform oscillating at 1kHz. Draw a circuit that will clip the waveform, with peak values now at 5V. You may use a DC power supply witou outputs of +10 V & -10V as well as resistors, capacitors and standard signal diodes. Indicate the components used and their respective values.


Homework Equations



v=iR
i=C*dv/dt
V_c=V_o(1-e^-t/RC)


The Attempt at a Solution



Okay guys, I designed the circuit successfully enough, here's a picture of it:

http://imageshack.us/a/img209/4081/circuit1.png [Broken]


The problem is that I have almost no idea where to get started with the actual calculations. We had a very poor electronics teacher last year so I'm essentially playing catch up this year. Any help as where to get started would be greatly appreciated!
 
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Answers and Replies

  • #2
rude man
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Your circuit has serious problems. The capacitors can only charge, not discharge, so they and the diodes will look like open circuits and there will be no clipping. Vout = Vin.

Why are you thinking of using capacitors?
 
  • #3
CWatters
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and without the capacitors it will clip at about Vf = 0.7V not 5V.

How about setting up a pair of voltage dividers with a voltage of + and - 4.3V. Then if the signal voltage exceeds 4.3+Vf or -4.3-Vf it will be clipped.
 
  • #4
rude man
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and without the capacitors it will clip at about Vf = 0.7V not 5V.

How about setting up a pair of voltage dividers with a voltage of + and - 4.3V. Then if the signal voltage exceeds 4.3+Vf or -4.3-Vf it will be clipped.

He'd need a very low-resistance voltage divider, requiring oodles of power from the 15V supplies.

Is there a way around that?
 
  • #5
gneill
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Consider that the problem doesn't appear to limit you to using just two diodes...
 
  • #6
rude man
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Consider that the problem doesn't appear to limit you to using just two diodes...

You're thinking lotsa diodes from + and - supplies? :uhh:
 
  • #7
gneill
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You're thinking lotsa diodes from + and - supplies? :uhh:

Not quite. Lotsa diodes across the signal. No supplies. Treat each stack like a zener, so use a current limiting/voltage dropping resistor.
 
  • #8
rude man
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Not quite. Lotsa diodes across the signal. No supplies. Treat each stack like a zener, so use a current limiting/voltage dropping resistor.

Good shot. The resistor already exists. Hope it's not too small ... but if it is, the only alternative is an active circuit which the OP is not allowed ...

Of course, there is also the dVD/dT problem of -2n mV/deg C where n = no. of diodes ...
 
  • #9
rude man
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I'd use 2 diodes, 4 resistors and 2 capacitors ...
 

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