# Voltage Doubler Circuit with a Duty Cycle Less than 50%

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#### ngmedaboard

##### Member level 3
If one were to build a voltage multiplier circuit constructed of a chain of "doubler" such as that shown below and inputing a wave with a 50% duty cycle, one could expect to double their voltage at each stage. That we know. But what if the duty cycle were say 10%. Would that reduce the voltage multiplier effect of each stage from X2 to X1.1? Does that make sense?

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#### checkmate

No. The diodes are used for charge transfer, and the on/off durations just have to be sufficient for complete charge transfer, after which it CEASES. Furthermore, charge transfer is highly non-linear. Trying to control it via duty cycle will be unreliable.

#### checkmate

This is only a 7 times multiplier. If you say that the inductor output is only 70V, then it can barely reach the displayed output voltage. Besides, the feedback network is regulating the first stage output at about 85V. So we are probably looking at around 100V from the inductor output.

#### ngmedaboard

##### Member level 3
I calculate the feedback network is regulating the inductor output to 68.5V. That comparator is going to trip at 2.5V and the divider ratio is 2.5M/68.5M. So when there's 68.5V there the comparator will see 2.5V.

But according to this wikipedia article, each cap and diode stage should be doubling the voltage of the previous stage.

https://en.wikipedia.org/wiki/Voltage_multiplier

So starting with 68.5 through 7 stages shouldn't we see a lot more than 475V to 600V? Your insight in explaining this would be greatly appreciated.

#### FvM

##### Super Moderator
Staff member
each cap and diode stage should be doubling the voltage of the previous stage
So you think the multiplication factor would be 2 raised 7 (= 128)? It isn't, it's just 7, as said. I don't read this erroneous believe
from the Wikipedia article, by the way.

### ngmedaboard

Points: 2

#### ngmedaboard

##### Member level 3
FvM, sorry if this is redundant but you're telling me this type of circuit takes the input voltage x the number of stages to make the output voltage? So if it were a single stage it would be a doubler but beyond that it's no longer doubling the input voltage, it's adding it at each stage.

If such is the case the circuit would make a lot more sense to me. you basically have a line of rectified DC voltages and in parallel a bunch of AC couplings in which you add your original AC wave to each rectified DC voltage.

FvM, I'd appreciate you confirming you confidence in this and also if anyone else agrees I'd also appreciate your confirmation.

If so, perhaps I should ask for the Wiki article to be updated.

#### FvM

##### Super Moderator
Staff member
I didn't read the article word by word, but it's not celar to me where you may have found the idea of "doubling the voltage of the previous stage".

Either if the article is right or wrong, there's possibly a misunderstanding about "doubling" operation. What the two
diode building block actually does is generating a DC voltage equal to the peak-to-peak value of the AC waveform (ideally, in practice
it's reduced by two diode voltage drops). n stages are simply producing n times the peak-to-peak voltage - with no load.

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