half wave voltage doubler using diode

[i_engineer]

hi all,

i want to make a half wave voltage doubler, at input side of rectifier is 9v ac

can anybody please tell me what are equation/formula for find out proper value of capacitor

thanks,
i_engineer

 

[FoxyRick]

The formula you probably need is this one, to calculate ripple voltage from a load current:

V_ripple = I / fC*( 2n³/3 + n²/2 - n/6 )

Where:
I = load current, amperes
f = frequency (50Hz, 60Hz?)
C = capacitance, farads (identical for every stage to preserve our sanity)
n = number of multiplier stages (that is 1 for a doubler)

Rearranging to solve for minimum capacitance for a maximum ripple voltage gives:

C = I / fV_ripple*( 2n³/3 + n²/2 - n/6 )

 

[badddys]

it consist of a clamper and a peak detector..there is no actual formula to calculate the capacitor value but both should be 100microfarad as per your setup..i am guessing you will use 1N4007 diodes.

 

[FoxyRick]

it consist of a clamper and a peak detector

It does indeed.

there is no actual formula to calculate the capacitor value

There is so! How else do we know what value to use? If it makes a difference, it can be calculated.

but both should be 100microfarad as per your setup..i am guessing you will use 1N4007 diodes.

Do you know something the rest of us don't? 100uF will give 2V ripple at 10mA load, with a 50Hz supply. What if we need to draw 100mA???

 

[badddys]

dont get angry buddy..i am just a learner..trying to learn things..but how come a 9v ac will give 2v ripple?? ripple factor for half wave is around 1.21.. i am just asking..

 

[FoxyRick]

Lol - I was at most mildly indignant, not angry :lol:

One should be careful of giving advice in such a definite way though, as a beginner - other beginners might think the information is correct and base designs on it.

We are not talking about a half-wave rectifier, but about a voltage doubler. The half wave ripple factor of 1.21 you are referring to only applies to a simple, single diode half wave rectifier with no smoothing capacitor. Add anything else and you have to recalculate the whole thing.

Often, as here, you will see ripple voltages given as a peak to peak value, since that tends to be more useful. It means that the voltage drops by two volts at worst, from the full DC output, as the capacitors discharge into the load.