Multiplier voltage query

Thread Starter

Doppler2902

Joined Sep 15, 2020
12
Hi guys, I need to do a multiplier voltage with a load that use 0.6W of power (From 12Vrms increase to 64 V DC with ripler voltage of 0.5Vpp). The question is, how I calculate the value in farads of the capacitors that I need? Thanks for the help!
 

Papabravo

Joined Feb 24, 2006
14,701
Hi guys, I need to do a multiplier voltage with a load that use 0.6W of power (From 12Vrms increase to 64 V DC with ripler voltage of 0.5Vpp). The question is, how I calculate the value in farads of the capacitors that I need? Thanks for the help!
I don't think there is a way to do what you want to do. Capacitors are not used to multiply voltage. How about a schematic diagram of what you have in mind.
 

Thread Starter

Doppler2902

Joined Sep 15, 2020
12
I don't think there is a way to do what you want to do. Capacitors are not used to multiply voltage. How about a schematic diagram of what you have in mind.
This is my schematic:
1600208505606.png
I assume that the capacitance value is 620uF but I amn't sure how to calculate it
 

Papabravo

Joined Feb 24, 2006
14,701
Capacitors and diodes together can do the job. Do you get the same voltage or a different voltage if you use a different value?
You do get different values, but the effect is slight. If you use values from 100u to 1000u you go from about 60V to 64V so I don't think the actual value is that critical. For that same range of capacitors the discharge current is in the range of 10 mA to 10.6 mA. C3 & C4 are in series so the total capacitance across the load is half their value. There is also a point of diminishing returns where you can make the capacitor too big. I'm guessing the calculation will involve the time it takes to charge C3 and C4 versus the time it takes to discharge them through the load.

Does that help?
 
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Papabravo

Joined Feb 24, 2006
14,701
If I change the value, the voltage changes
Slightly. For 100u the output is about 60 Volts. For 620u to 1000u it goes from just under to just over 64 Volts. The value is not that critical. Even if it was you can't get that size capacitor from the E48 or E96 series, so you get close and call it a day.
 

Thread Starter

Doppler2902

Joined Sep 15, 2020
12
I've tried do this: I know that C = Q/V where Q = i * t but I don't know how to obtain the time, I think it's (2)(60Hz) (depends of frecuency signal). Anyway, thanks :D
 

AnalogKid

Joined Aug 1, 2013
8,706
Fact: During the time of 1/2 cycle of the power line, the capacitor voltage can sag no more than 0.5 V (post #1).

Calc: Half-cycle to half-cycle, the load energy, in watt-seconds, is 0.6 W x the input voltage period / 2.

Calc: The total energy in a capacitor, in watt-seconds, is 1/2 C V^2 >> 0.5 x (capacitor value in farads) x (capacitor voltage squared).

With those two equations, you can calculate the theoretical minimum value of the two output capacitors to meet the ripple requirement.

ak
 
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