What is the difference between half wave voltage doubler and full wave voltage doubler? (Using Practical Diode with forward bias voltage Vd)

To be specific:

i)Do the two differ in their principle of working in anyway?

ii)How do the two differ in input/output waveforms?

iii)What is the effect of change of Capacitance values in each of the two cases?

And finally, Which is more efficient and why?(Also if any advantages/disadvantage of either over the other)


simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ So, you certainly have a schematic for both. Add it to your question, and ask based on a comparison of the two! Also, you can research all of your questions using Google to a fairly large extent, so do that and reduce your question to the things you didn't understand \$\endgroup\$ – Marcus Müller Sep 10 '17 at 7:47
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    \$\begingroup\$ "How do the two differ in input/output waveforms?" - this question could be answered by a simulator, no need to get humans involved. \$\endgroup\$ – Dmitry Grigoryev Sep 11 '17 at 9:49
  • \$\begingroup\$ An important difference is that the source can be ground referenced in the half wave doubler like a microwave oven transformer for instance. \$\endgroup\$ – KalleMP Sep 15 '17 at 21:52

The working principle of the first (I assume no load and ideal diodes):

The first circuit is a clamper followed by a peak detector.

When V1 is negative, C1 is charged through D2. At the end of the first half period, the voltage on C1 is positive from right to left, and the value is |Vpeak(V1)|. When V1 is positive (second half period), the voltage at the node C1-D1-D2 is brought to V1 + |Vpeak(V1)|. D1 turns on, and there will be some charge-sharing between C1 and C2, therefore the voltage of C2 will increase each cycle. After "many" cycles, the voltage of C2 will reach the peak voltage of C1, i.e. 2*|Vpeak(V1)|

The second circuit is made of two-half wave peak detectors (one positive and one negative), in series (C4-D3 and C3-D4).

When V2 is positive, it charges C4 to |Vpeak(V2)|, through D3. When V2 is negative, it charges C3 to -|Vpeak(V2)|, through D4.

Some differences:

  • The first circuit shares the ground with the input. The second doesn't.
  • In the first circuit, the maximum voltage might be reached after many cycles, because C1 is charged and then subsequently discharged on C2. In the second circuit, the full double voltage is reached in one full cycle. In other words, the second might provide a better voltage stability and lower output impedance, given the same capacitor values.


simulate this circuit – Schematic created using CircuitLab

| improve this answer | |
  • \$\begingroup\$ Your explanation helps! Can you please provide waveforms of output in both cases. Also why are the named as "Half-wave" and "Full Wave" respectively? Does the double the latter voltages in both the half-cycles and the former only gives positive half cycle with doubled peak value? \$\endgroup\$ – G2G Sep 10 '17 at 13:23
  • \$\begingroup\$ Without load, both circuits will provide a voltage (Vdouble with respect to the ground), which is double with respect to the sine peak voltage. However, the initial transients will be different. In the circuit on the left, the voltage will progressively grow up to 2*Vpeak. In the second one, Vdouble is reached in one cycle (provided that adequate diodes/capacitors are used). If you click to "simulate this circuit" above (be sure to put the capacitor values, as I have deleted them because they were unrelevant for the discussion), you will be able to see the waveform! \$\endgroup\$ – next-hack Sep 10 '17 at 13:50

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