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Capacitive Power Supply

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Adwaittronics

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Hi
I want to design a capacitive power supply , with i/p voltage as 90V to 270V and o/p voltage as 24V , how can I calculate maximum current of the circuit?
I calculated the capacitance of the X-rated capacitor by the formula X = 1/(2* pi*f*C) where f=50-60Hz.
Then calculated the current by the formula I = V/X , where V is the input voltage.
Is this the correct way to calculate maximum current of the circuit ?
In all the above calculation output voltage is not in picture , so am I right or wrong kindly explain.

NOTE : for all the i/p voltages from 90V to 270V ,I want the o/p voltage to be 24V.
 

hi,
Read thru this PDF, note the lack of safety isolation for this type of circuit.
E
 

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  • TxrLessPsu2.pdf
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well yes, your load and the capacitor effectively form an potential divider......what capacitor size you need depends on the load current, as well as the mains level.
 

Adwaittronics said:
I calculated the capacitance of the X-rated capacitor by the formula X = 1/(2* pi*f*C) where f=50-60Hz.
Then calculated the current by the formula I = V/X , where V is the input voltage.
Is this the correct way to calculate maximum current of the circuit ?
In all the above calculation output voltage is not in picture , so am I right or wrong kindly explain.

NOTE : for all the i/p voltages from 90V to 270V ,I want the o/p voltage to be 24V.
Click to expand...

The capacitive voltage divider is completely unregulated.
For a CONSTANT load, varying the input voltage by a 3:1 ratio, the output voltage will also vary by a 3:1 ratio.

The way to regulate the output voltage is to have a shunt regulator (zener) in parallel with the output. It will consume various levels of current to maintain the output voltage.

But beyond a few milliamps of load, it becomes impractical and ineffective.
 

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