I am stuck at understanding how voltages add up in a voltage clamper, reading through Learning About Electronics - Diode Clamper and All About Circuits - Voltage Clamper, and also YouTube a few tutorials. If I have a simple clamper circuit with an AC input signal series with a capacitor and parallel to a forward or reverse diode and a load, after the first positive part of the input cycle charging the capacitor, once the diode is forward biased, the polarity of the capacitor's voltage would always be opposite to the polarity of the AC input signal (hence, DC-AC instead of DC+AC). The clamper circuit should shift the output signal horizontally through time, instead of vertically on the voltage.

Take the image as an example of a positive clamp circuit:

  1. After initial positive input signal, input polarity reverse and capacitor is charged in the negative direction, thus voltage negative to positive from left to right
  2. Input signal goes positive again but capacitor is charged in the opposite polarity and would cancel out instead of adding up the voltage

sample voltage clamper

Clearly I am mistaking somewhere in the voltage clamper and hope someone can clarify.


1 Answer 1


The Diode as a current rectifier will have resistance < 10 Ohms or 0 compared to 10k with a 0.6V drop (@ 1mA).

So the cap charges up for negative swings with the diode conducting thus R1 ends up with a + Voltage as it slowly decays the cap and ends up with a POSITIVE steady-state half-wave ripple voltage. You know this is true as the Diode blocks negative peak waves.

Since RC = 1ms is the 63% response time and f = 1kHz you end up with R1 sees a half-wave rectified sine wave but the 1st pulse height is only a little over half of the final peak voltage as the cap charges up.

To add more clarity using Falstad.com sim enter image description here The Cap. is shown in red as diode conducts with the
Polarity chosen so that Vcap + Vdiode = Vsource The diode voltage is ~ 500mV due to I=C dV/dt with a power diode.

  • \$\begingroup\$ What I still can't understand is how the voltage get added or "shift" since either the diode and input AC increases when the other decreases and their value with opposite polarity will always add up no bigger than the input 5 or -5 volts assuming 0 voltage drop over the perfect diode, and can't pull -5 to 0 and 5 to 10V. \$\endgroup\$
    – KMC
    Mar 5, 2020 at 23:12

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