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Capacitive Power Supply

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me_guitarist

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capacitive power supply

Hi guys,

I used this capacitive power supply to run some LED from 240VAC.

R1=2W10R Metal Oxide Resistor
C1=1.5uf 450VAC(x2 Type)
C2=220uf 35V
D1=24V
D2=1N4007

The output around 35VDC. Everything running smoothly. But when I switched off the main 240VAC, there is still around 250VDC inside C1 & I get electric short. How do I discharge C1?

Pls help. Thanks in advance.[/img]
 

FvM

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capacitive supply

Apart from the problem of of providing a safe protection against accidental contact for your circuit, I think you shouldn't operate it open circuit (at 35 V output), cause this overloads D1. As another remark you could double the output current efficiency by using a full wave rectifier.

Finally C1 can be discharged by a high ohmic resistor, e. g. 1 MOhm.
 
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me_guitarist

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working of capacitive power supply

Hi FvM,

Thanks for yr input.

U mean put a 1/4W 1M resistor parallel to C1?
 

yanzixuan

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capacitive power supply design

me_guitarist said:
Hi FvM,

Thanks for yr input.

U mean put a 1/4W 1M resistor parallel to C1?

1/4w 400k~800k resistor parallel to C1;
or
1/2w 200k~400k resistor parallel to C1;
 

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capacitive power

yanzixuan said:
me_guitarist said:
Hi FvM,

Thanks for yr input.

U mean put a 1/4W 1M resistor parallel to C1?

1/4w 400k~800k resistor parallel to C1;
or
1/2w 200k~400k resistor parallel to C1;

I tried 1/4W 1M is ok.
Thanks yanzixuan & FvM.
 

me_guitarist

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400k resistor

FvM said:
Apart from the problem of of providing a safe protection against accidental contact for your circuit, I think you shouldn't operate it open circuit (at 35 V output), cause this overloads D1. As another remark you could double the output current efficiency by using a full wave rectifier.

Finally C1 can be discharged by a high ohmic resistor, e. g. 1 MOhm.

Why is that D1 overload? As it should output 24VDC right?

This circuit is proper isolated.
 

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capacitive power supplies

With no load connected, D1 is dissipating about 1.2W, if not using a power Z-diode, this means thermal overload.
 

yanzixuan

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full wave capacitive power supply

me_guitarist said:
FvM said:
Apart from the problem of of providing a safe protection against accidental contact for your circuit, I think you shouldn't operate it open circuit (at 35 V output), cause this overloads D1. As another remark you could double the output current efficiency by using a full wave rectifier.

Finally C1 can be discharged by a high ohmic resistor, e. g. 1 MOhm.

Why is that D1 overload? As it should output 24VDC right?

This circuit is proper isolated.

Rc=1/(2*pi*f*c)=1/(2*3.14*50*3*10^-6)~~1.06k;

R1=10.

Iin~~240/(Rc+R1)=220ma;

With no load connected,

Pd1=((Iin*VD*T/2)+(Iin*Vth*T/2))/T=Iin*(Vd+Vth)/2=0.22*(24+0.7)/2=2.717w
 

max0412

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capacative power supply

me_guitarist said:
This circuit is proper isolated.

Definitely not,there is no isolation!

These regulators are meant to be in an enclosure with no possible way a user can come into contact with it. As long as this is the case they have there uses .
 

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capactive power supply

max0412 said:
me_guitarist said:
This circuit is proper isolated.

Definitely not,there is no isolation!

These regulators are meant to be in an enclosure with no possible way a user can come into contact with it. As long as this is the case they have there uses .[/quote

The circuit was inside an enclosure. Its away from user touching.

I ran this circuit for a week & its ok until now.
 

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calculation of capacitive power supply

Rc=1/(2*pi*f*c)=1/(2*3.14*50*3*10^-6)~~1.06k;
Isn't correct. You apparently understood x2-type as two caps in parallel. But X2 is a type specification of AC filter caps. So C is 1.5uF and the current is only half. Also the Z-diode voltage drop has to be considered in current calculation, so I think, an estimation of 1.2 W(@230 V) is rather exact.
 

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rc=1/2*pi*f

FvM said:
Rc=1/(2*pi*f*c)=1/(2*3.14*50*3*10^-6)~~1.06k;
Isn't correct. You apparently understood x2-type as two caps in parallel. But X2 is a type specification of AC filter caps. So C is 1.5uF and the current is only half. Also the Z-diode voltage drop has to be considered in current calculation, so I think, an estimation of 1.2 W(@230 V) is rather exact.

If I put 2W for ZD1 will it sufficient?
 

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full-wave rectifier current calculation

Yes, also the high Z-diode power dissipation is only occurring in no load case. It has to be considered only, of the load can be disconnected somehow.
 

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capasitive psu

FvM said:
Yes, also the high Z-diode power dissipation is only occurring in no load case. It has to be considered only, of the load can be disconnected somehow.

Can I put C1 before R1? To suit my application.
 

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