Positive Diode Clamp on PWM

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drkidd22

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I have a PWM signal going into a HPF. I had to add a diode to clamp the signal at the output of the HPF. Why does the PWM signal starts to 'drift' down when the clamp diode is not present post the HPF? It's still 24V out, but it centers at 0 with +12 and -12.
How does one pick the correct clamping diode? In the simulation with the diode connected the voltage swings between -1.2V and 24V.
 
In steady state (the far right side of your graphs), the HPF blocks DC which is 0Hz. That leaves only an high frequency components (AC if you want to call it that) which makes everything center around 0V.

In the simulation with no diode, at the very beginning the cap is not charged to the DC bias yet so it doesn't block DC yet. As it charges up to the bias voltage, it blocks more and more DC which causes the graph to eventually drift down and center around zero (reaching steady state).

The same thing is happening in your circuit with the clamp diode. The diode basically keeps resetting the circuit back to the power-on state of the diodeless circuit. Every time the cap bias voltage charges up enough to block enough DC to make the voltage go just a little negative, the diode forward biases which shorts out R1. This drains the DC bias voltage across the cap back to almost 0V and resets you to be in the same situation as the left side of the graph for the clampless circuit.

You pick a diode with a reverse breakdown voltage that you want to clamp to.
You pick a diode that can handle the current.
You pick a diode that is fast enough for the expected signal spikes so it can turn on and clamp the voltage before it gets too high (some systems have very fast and powerful voltage spikes).
 
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In your case the diode is doing the work in forward bias, not reverse bias so thats why reverse breakdown voltage doesnt matter as long as its higher than 24V

Reread my first post. Was editing.
 
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Thanks, odes forward voltage have an effect in this configuration?
Yes. Forward voltage determines how low below 0V the circuit will go. -1.2V in your case. You circuit is actually working as negative clamp in forward bias, not a positive one in reverse bias/breakdown.

Please reread everything. Was editing and am slow because one hand is eating pizza.
 
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haha ok, thanks. I think I can understand better now.
 
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