Talk:Voltage doubler

Clamping voltage
This article says "The negative peaks of the AC waveform are "clamped" to 0 V (actually −VF, the small forward bias voltage of the diode) by the diode."

Why state it that way instead of just saying clamping is at -VF?

ICE77 (talk) 00:20, 20 August 2015 (UTC)


 * Because nominally the circuit is clamping to 0V. With an ideal switching element it would actually clamp to 0V.  The forward volt drop of the diode can be looked on as an error that needs to be taken into account. SpinningSpark 10:58, 22 August 2015 (UTC)

Ideal components are nothing but fiction. Let's stick to the real world. I think it's a more practical approach.

ICE77 (talk) 06:03, 27 August 2015 (UTC)


 * Just about every competent textbook ever written about electronics starts by analyzing ideal circuits. They do this because that is the more general case, and gives the greatest insight.  Deviations from the ideal can be discussed, but particular component types must then be specified. SpinningSpark 14:55, 27 August 2015 (UTC)

Diodes and switching
"The switching elements are simple diodes and they are driven to switch state merely by the alternating voltage of the input." - Is it just me? I don't really like that description. I would sooner describe the diodes in a simple passive circuit as "steering" devices rather than "switching" devices. Anyone else? 31.125.76.2 (talk) 10:59, 29 July 2020 (UTC)
 * I've moved the above comment from the GA review which is a closed discussion. SpinningSpark 11:43, 29 July 2020 (UTC)

Greinacher circuit
"As with a bridge circuit, it is impossible to simultaneously ground the input and output of this circuit.[6]"

The reference [6] is to p107 of Ryder's 1970 text Electronic Fundamentals and Applications. It's an excellent text, but it says nothing at all about grounding either bridge or multiplier circuits. It doesn't even mention Greinacher, and its only Greinacher-like circuit isn't a doubler but a quadrupler which (rightly) is usually credited to Cockroft and Walton. The reference should be removed.

Delon circuit
"The Delon circuit uses a bridge topology for voltage doubling"

It's a full-wave circuit but not in a bridge topology. It can only resemble a bridge if the diodes are viewed as one leg and the capacitors as the other. That's a tortuous misinterpretation, which requires that the input appears between the junctions in the two legs and the output between the top and bottom - the opposite of accepted bridge topology.

It seems likely that The Delon circuit uses a bridge topology is a wrong assumption based on where it's widely seen: in computer power supplies, many of which include an encapsulated bridge rectifier. To accommodate low-voltage mains, for reasons of economy they use half the component in a Delon doubler - but not the whole bridge, because the Delon circuit doesn't use a bridge topology.
 * It looks like a bridge topology to me. Constant314 (talk) 08:54, 29 May 2024 (UTC)