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I need help solving this circuit. It's supposed to be a waveform clipper and I already simulated it, however I want to know how the circuit works and so far I haven't been able to analyze it properly:

Diode Clipper

D4 is a 9.1 [V] Zener while D5 is a 5.1 [V] Zener, I put on the specific models according to that. The rest of the diodes are rectifier Si Diodes.

Thank you

EDIT: I corrected the graphic as I had forgot to close a port of the circuit, it looks like this, I apologize for the mistake.

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This circuit isn't going to do very much other than clip the positive-going peak to anywhere from 0V to about 3.8V (depending upon the input signal amplitude) and remove the negative-going signal.

D3 & D4 aren't going to do anything - they might as well not be there.

Here's why:

1) D5 clips the positive-going signal to about 5.1V and clamps the negative-going signal to about -0.65V.

2) D1 & D2 pass the positive-going signal through to R2. If the input signal is significantly higher than 5.1V, D5 conducts and clips the signal to 5.1V. If the input signal is only somewhat higher than 5.1V, the voltage divider formed by R1 & R2 reduces the signal.

In other words, the output voltage is about (Vin / 2) - 0.65V for Vin ranging from 0V to about 8.8V.

You can calculate the voltage at which this happens and by how much, taking into account the forward voltage drop introduced by D1 & D2.

3) Because D5 has clamped the negative-going input signal to -0.65V, the series combination of D3 & D4 never conducts. The negative half-going input signal is therefore completely attenuated.

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  • \$\begingroup\$ I have to run out the door. I'll leave it to you to answer - you seem pretty competent and I like your answer style better than mine. But I'll point out what I was adding to your answer as a comment - use it as you craft your answer: For your point #4, you are still ignoring the effect of R1 & R2. My (really quick, in the head) calculations show that D5 doesn't start to conduct until Vin reaches about 8.8V. That's based upon calculating the current through R1 as (5.1 - (2* 0.65V) / 10k, then using that current to calculate the voltage drop across R1. That was based on the original Question \$\endgroup\$ – Dwayne Reid Apr 4 '15 at 20:00
  • \$\begingroup\$ Yes, pretty muh all said above makes sense when looking at the simulation, plus I just started solving the problem again with pencil and paper, thanks a lot again, and sorry for my clumsiness! \$\endgroup\$ – HCalderon Apr 4 '15 at 20:10
  • \$\begingroup\$ @DwayneReid thank you. But I am little confused here. Please answer if you can. I don't mind you following my style if you feel that would be better. :) \$\endgroup\$ – nidhin Apr 4 '15 at 20:18

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