Help with capacitive dropper capacitor value

[Athosworld]

I want to make a capacitive dropper and i have a 203 film capacitor that i will use for it. What will be the voltage drop on it. I tried to use an online calculator but it doesn’t work.

 

[augustinetez]

Hopefully this will help you work it out:-
https://electronics-fun.com/ac-voltage-drop-using-capacitor/

 

[rjenkinsgb]

i have a 203 film capacitor that i will use for it. What will be the voltage drop on it.

That's like saying "I have a 1K resistor. What will be the voltage drop across it?"

It depends on the rest of the circuit and the other resistances, currents and operating voltages!
(And frequency, with the cap).


You calculate the reactance of the cap, at the frequency involved - generally 50Hz or 60Hz, for a transformerless power supply.

for a 20nF cap at eg. 50 Hz:

The formula is: 1 / (2 x pi x F x C)
F is frequency in Hz, C is capacitance in Farads.

You have 20nF; 20 x (10-9) Farads.

That works out in round figures to 160 KOhms reactance.

You can consider that cap as a 160K (virtually) lossless resistor, in that application at 50Hz.

On a 220V AC supply, for 12V output, that would give you a maximum current of around 1.3 milliamps.

Note that the capacitor in any such supply should rated for AC mains connection, not a generic capacitor. "Y" class caps are an appropriate type. The AC input must also be fused as a suitable low value.

Any such capacitive dropper supply MUST have a shunt regulator, such as a zener diode after the bridge rectifier, to limit the voltage. I'd always use a zener that can handle far higher current & power dissipation than you expect to need - eg. 10x power minimum is not unreasonable, to handle voltage "spikes" that can cause high current through the capacitor.


And, DO NOT USE CAPACITIVE DROPPERS IN ANYTHING UNLESS IT HAS NO USER CONTROLS, EXTERNAL CONNECTONS (other than the AC wiring) OR ANY REASON FOR ANYONE TO EVER TOUCH IT - AND AT LEAST TWO SEPARATE FORMS OF APPROPRIATE HIGH VOLTAGE INSULATION AROUND ALL COMPONENTS.

Capacitive dropper power supplies are LETHAL in any other circumstances. They are not what any experimenter should ever be using in prototypes or gadgets.


You can get isolated PSU modules very cheaply - but I'd still always advise using a separate "Wall wart" or Power brick type supply, with just the low voltage DC connection from that in to your project.

Keep mains AC totally away from things that only need low voltages.

You don't have to worry about what parts you can touch, then, and it avoids problems with connections from other grounded devices or test gear.

 

[Lightium]

Wouldn't you know it? I wrote a program for just this thing for a dc-dc convertor I made.

#include <stdio.h>
#include <math.h>

double cap1, cap2, z1, z2, frequency, Vin, Vout;

int main(void)
{
printf("Enter capacitance of capacitor 1: ");
scanf("%lf", &cap1);

printf("Enter capacitance of capacitor 2: ");
scanf("%lf", &cap2);

printf("Enter frequency: ");
scanf("%lf", &frequency);

printf("The reactances are:\n");
z1 = 1 / (2 * 3.141592654 * cap1 * frequency);
z2 = 1 / (2 * 3.141592654 * cap2 * frequency);
printf("Z1 = %.6e\n", z1);
printf("Z2 = %.6e\n", z2);

printf("Enter input voltage: ");
scanf("%lf", &Vin);

Vout = (z2 * Vin) / (z1 + z2);
printf("Output voltage = %.6e\n", Vout);

return 0;
}

 

[Nigel Goodwin]

I want to make a capacitive dropper and i have a 203 film capacitor that i will use for it. What will be the voltage drop on it. I tried to use an online calculator but it doesn’t work.

If you have to ask, then you shouldn't be doing it - as you don't understand it enough to do it safely.

In fact, best if you don't do it at all, even if you do understand it.

It has it's place, but should only be used in certain specific cases, where doing so makes sense.

 

[Tony Stewart]

Do you understand that C dropper works like an R dropper by impedance ratios?
If the step-down ratio is large like 100:1 then you can approximate the power by applying full power across the input C.

So if X(f)=1/(2pi*f*C) and neglecting ESR, P(f)= V²/R but substituting X(f) forR, we get;

P(f) = V² * 2πfC so 203 film for 50Hz 230V, 1/X= 72e3 * 2e-8 = 144 e-5 = 1.44 mW ( puny )

You will need a bulk thru-hole film part that can handle the ripple current.