Driving the relay causes large ripples in PSU.

[Sceadwian]

A half wave rectifier completely ignores half of the AC cycle, so during half of your AC cycle ALL of the power supplied to the regulator is coming from the capacitor. This increases the ripple current on the cap which shortens it's life and makes them more susceptible to voltage dips on the mains line because it can only charge during half of the available AC cycle.

Your diagram also doesn't show any form of isolation, which if your switch mode IC ever fails will expose your circuit (and anything electrically connected to it) to the mains voltage (usually a bad idea) A full wave rectifier will help with voltage dips, not sure if it would completly eliminate your problem or not. But a full wave rectifier would require issolation of mains voltage.

One other thing, there's no bleeder resistor depicted with C4 which means you're trusting the switch mode power supply to drain the capacitor. If you have the switch mode off for some reason the cap C4 will contain 170 volts.

 

[rainman1]

Thank you very much.

"This increases the ripple current on the cap which shortens it's life and makes them more susceptible to voltage dips on the mains line because it can only charge during half of the available AC cycle."

But if I would have a full-wave input, then the capcitor will charge only in one half of wave and will discharge in the other half.
So whats the difference?

"there's no bleeder resistor depicted with C4"
you mean that I should connect a resistor in parallel with C4 in order for C4 to discharge through the resistor?
Should this resistor be very large so no current will flow through it in active mode?

Thank you.

 

[Sceadwian]

Apparently you don't know how a full wave rectifier works =) It inverts the negative half of the incoming AC signal, the capacitor only ever see's a positive voltage. It does however require issolation of the mains voltage usually with a transformer because as it stand right now you're using the neutral wire of the mains as a ground for your DC parts, you can't do that with a full wave rectifier. If you use an isolation transformer that has a center tap on the secondary you only need two diodes for a full wave rectifier.

 

[Mike_2545]

If you use an isolation transformer that has a center tap on the secondary you only need two diodes for a full wave rectifier.

By isolation transformer do you mean 120 primary and something like 12 secondary with CT?

How's that?

Two diodes, not four?

 

[Sceadwian]

No, 120 to 120 rated for whatever current his control board needs, it would be relatively small, just something to isolate the mains from the circuits ground otherwise you can't use a full wave rectifier, a full wave rectifier's DC output will float over ground if it's not isolated. The isolation will also eliminate user or fire risk if the regulator IC fails. If the secondary of the isolation transformer is center tapped you only need two diodes, but the voltage will be half the mains voltage more than high enough for his needs. The only reason he doesn't need an isolation transformer right now is because he's using a half wave rectifier. Alternately he could just throw more capacitance on the input capacitor side as well, but large high voltage capacitors aren't cheap.

Rectifiers

I dug this website out of my old bookmarks, it explains pretty much everything the poster needs to know. GREAT electronics reference site for everything from the basics to the advanced

And yes, the bleeder resistor can be high for C4, a 1meg resistor will only waste 6mw's of power and will discharge the 10u cap to less than 10 volts in 30 seconds.

 

[Mike_2545]

That makes sence.

Thanks