# Voltage Multiplier

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1. Apr 8, 2016

### Hassan Raafat

in Voltage Tripler and Quadrupler , why don't current divide between two diodes ( D1 , D3 ) or ( D2 , D4 ) ?
they are forward in the same cycle , right ?
I understand that charging happens in a very little time , so two cycles already is not shown on oscilloscope but in BoyleStad is explained that in every half cycles (in the first two) one diode only works ...

2. Apr 8, 2016

### cnh1995

I am no expert but here's what I think..
Assume the diodes are ideal. Current flows through C1 and D1 because this path offers less impedance than C1-C3-D2-C2 path. There are three caps in the second path, hence reactance offered is more. Also, when D1 turns on, it shorts out C2, D2 and C3.

3. Apr 8, 2016

### Svein

This is a negative voltage quadrupler.
• C2 and C4 are smoothing capacitors, storing energy between pulses
• C1 and C3 are coupling capacitors, connecting the pulses from Vi to the diode junctions.
Now look at the capacitor/diode combinations: C1/D1 rectifies the negative peak pulse voltage and stores it on C1. The pulses are still present, but they are now referred to the voltage across C1. The same argument goes for C3/D3, except that the voltage across C2 is added to the rectified voltage across C3.

In the same way, the negative pulse is rectified across D2 and stored in C2. The resulting voltage across C2 is the sum of the voltage across C1 and the negative pulse amplitude. Finally, The negative pulse is also rectified across D4 and stored in C4. The resulting voltage across C4 is the sum of the voltage across C3 and the negative pulse amplitude. All in all - the voltage on C4 is 4 times the pulse amplitude minus 4 diode drops.

4. Apr 9, 2016

### Hassan Raafat

I think that your last sentence is confusing me , C4 has 2 times and also C2 has 2 times so when I took the voltage on the bottom from right to left I have 4Vs - 4 Diode drops

5. Apr 9, 2016

### Hassan Raafat

If I understood u well , u mean that current will choose the shortest path to pass through (assuming all diodes are ideal) and so to have less capacitors in the path to have less impedance , right ?

6. Apr 9, 2016

### cnh1995

If D1 is turned on, it will short out D3, C3 and C2. So, D3 won't conduct. If D3 were on, then surely there would be some voltage drop across C3 which means D1 would also be on. But, if D1 is on, D3 can't be on. So, D3 is off in the first half cycle.

7. Apr 10, 2016

### Svein

Sorry, several incorrect statements:
1. It is a negative voltage doubler.
2. The statements regarding the capacitors are correct.
This is a simulation of the circuit. The blue trace is the output of the pulse generator. The red trace is the voltage across D1. The green trace is the voltage at the output (junction D4/C4).

8. Apr 10, 2016

### Averagesupernova

It is certainly more than a doubler. The actual measured output may not be quadruple of the peak voltage but in any text book I have seen this would be more than just a doubler.

9. Apr 10, 2016

### Svein

Oops - we are talking about two different things here. I am used to thinking about generating a voltage from a pulse train - and then the circuit doubles the pulse amplitude. But if you are talking about a sine wave AC input, the circuit quadruples the sine wave amplitude (the sine wave amplitude is half the peak-to-peak voltage).

10. Apr 10, 2016

### Svein

Here is a simulation with a 5V amplitude sine wave at the input. The output is then 4 times the amplitude (and 2 times the p-p voltage).

11. Apr 11, 2016

### Averagesupernova

There really is no difference. It's just how the input is spec'd. An AC voltage that is typically driving a voltage multiplier in a power supply is always spec'd as RMS. The attachment in post #7 is no different. Typically we don't spec 0 to 5 volt 50% duty cycle waveform as 2.5 volt RMS but it still multiplies it in the same manner.

12. May 7, 2016

### Hassan Raafat

Thanks a lot , this images really helped me to understand it ... I'm grateful for that my professor