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enter image description here

I've been trying to build the cap drop power supply as the major requirement in the project is to reduce cost.In the reference circuit 5.6 zener diode was present but out of curiosity I removed it and I found following results.

voltage across c2 = 8 volts. I'm not getting how to analytically solve this problem and get the voltage.

Any help is appreciated.

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A problem with this circuit is that it is not transformer isolated and therefore dangerous. The 230 VAC line voltage just adds to this.
Another problem is that capacitors used like this are prone to failures due to line spikes unless the voltage rating is greatly derated. A line-rated capacitor like an X2 would be best.

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  • \$\begingroup\$ I just looked up the two app notes from Microchip. They recommend 250 VAC on a 120 VAC line. No a good idea in my humble opinion. \$\endgroup\$ – Robert Endl Mar 21 '16 at 15:16
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Simplified tremendously a capacitor looks like a resistance to AC. (actually its an impedance but i'm simplifying). The conversion is 1/(2*pifc). In your case the capacitor looks like a 16k resistor. Then it's a simple voltage divider

230*(800/(470+800+16000)) = 10.5v

There are going to be some losses unaccounted for and the diode will drop something. 8v sounds about right.

You can not replace the 800 ohm resistor. This would be dangerous.

schematic

simulate this circuit – Schematic created using CircuitLab

The voltage regulator would look like a variable resistor depending on the load. If there is every a condition where the is no load the regulator would look like a near open circuit. This would expose the regulator to extremely high voltage.

schematic

simulate this circuit

The two circuits pictured above are safer. They both limit the voltage to the regulator even if the regulator is unloaded. They also place two 100k resistors across the big capacitor that drain any residual voltage. My favorite is the circuit with the zener diode.

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  • \$\begingroup\$ why you have not considered the impedance of c2 in parallel with 800 ohm ? \$\endgroup\$ – BRT Mar 21 '16 at 18:50
  • \$\begingroup\$ @BRT Because of D1 the capacitor C2 ends up on the DC side of the circuit. On the DC side the capacitor behaves differently. \$\endgroup\$ – vini_i Mar 21 '16 at 22:28
  • \$\begingroup\$ @ vini_i thanks for clearing my concept but what if I have a ldo with output voltage 3.3 volts .how can I model it as a resistor and find the capacitor voltage. \$\endgroup\$ – BRT Mar 22 '16 at 18:51
  • \$\begingroup\$ @BRT i'm not sure what your referring to, could you clarify? \$\endgroup\$ – vini_i Mar 24 '16 at 2:48
  • \$\begingroup\$ @ vini_i let's say that in the place of 800 ohm in above circuit I connect a voltage regulator of output voltage of 3 volts and continous current of 3 ma at output.what voltage will capacitor will be charged to.as capacitor will be connected to the input of regulator how can I model it as the imedance connected with capacitor. \$\endgroup\$ – BRT Mar 24 '16 at 20:01

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