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power losses

cheniour

Member
how I Can calculated the power losses in each stage of the voltage multiplier circuit
Diodes are 1PS79SB30, and the capacitors are 47-μF ceramic components 0.5 volt input and 44 HZ
VOLTAGE MULTIPLIER.png
 

cheniour

Member
sorry but in my project I'm in the measurement phase and I face several difficulties also in the internet I don t find any information about this voltage multiplier sorry again
 

cheniour

Member
I'm master degree my specialty is automatic and my project energy harvesting topic I don't have a good background in electronic
 

KeepItSimpleStupid

Well-Known Member
Most Helpful Member
What I can suggest is to download the free simulation package from Analog devices/linear technology called LTSpice.
I haven't used it, but see what it's capable of doing. There are some very capable people on this forum that should be able to help.

At low voltages and power levels, there is a diode parameter called Rds(on). Diodes have a voltage drop depending on temperature.
The most common material is Silicon. The usual number used is between 0.6-0.7V at room temperature A schottkey diode is a metal semiconductor junction which has a lower voltage drop. The https://www.nexperia.com/products/d...iodes-and-rectifiers-if-lt-1-a/1PS79SB30.html is Schottkey diode..
 

ronsimpson

Well-Known Member
Most Helpful Member
The voltage multiplier circuit is not function so the power loss there is near zero.
You picked some good diodes that only loose 0.36 volts. But that is a great deal of the 0.5V 44hz power source.
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Can't find it now. There is a paper on what you are trying to do. They use a 1:10 transformer to change the source from 0.5V to 5V. Now there is enough voltage to work with. Some one makes a IC that includes the diodes and boost PWM to get power to charge a super cap.
 

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