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?
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.
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.
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.
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 .
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.
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.
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.