Rocker Switch Query

[gEcky]

Hi

I'm constructing a PSU, stepping down 230Vac to 24Vac via a transformer. The power to the transformer is connected via a rocker switch rated at 230V-16Amp. It's a DPDT i suppose since it has 4 lugs and forms 2 isolated pairs. So I connected the AC "live" and "neutral" on each side of the switch with the transformer inputs at the opposite side. The circuit will thus be complete when the switch is on.

The circuit worked fine and i was able to on/off the PSU without leaving it on for more than a while to check the regulayed ouyput voltage. However when i left the PSU on for about a minute, the neon lamp of the switch blew. So did the fuse in included in the circuit, fortunately.

A check with a ohm meter shows that the 2 pairs of lugs on the swith is now shorted! 2 inputs shorted / 2 outputs shorted.

Could someone tell me what when wrong?
(Transformer is 24Vac-1A and the mains supply is 230Vac-13Amps-50Hz.)


Thank you in advance!

[audioguru]

gEcky,
Did your switch melt? Perhaps it was poor quality.
The transformer may have a short from primary to secondary which would cause your problem, and is VERY DANGEROUS.
Check that transformer and use a quality switch.

[gEcky]

Thanks!

there is no short between the 2 coils..... i think i'm not suppose to use the 2 pairs of lugs the way i did... Means i can only connect the "live" or "neutral" and not use both to open/close the circuit.

Another thing, at which point will it be good to place a LED to indicate power on?

[audioguru]

gEcky,
That's good that your transformer isn't shorted. Your switch just "failed".
A double-pole switch has 2 switching contacts. One for Live, and the other for Neutral.
An LED needs a current-limiting resistor in series. The value and power-rating of this resistor is determined by the voltage across it and the current through it. A good place to connect the LED/resistor is across the filter capacitor that follows the rectifier, but we don't know how much voltage is there since you didn't say whether the 24VAC winding is center-tapped or not. Another place for the LED/resistor is at the regulated output, of which we don't know its voltage or current.
Also, how much current do you want to pass through the LED?

[gEcky]

Hi! Thanks for the prompt reply!

That's good that your transformer isn't shorted. Your switch just "failed".
A double-pole switch has 2 switching contacts. One for Live, and the other for Neutral.

Does that mean that the blown switch should be a one time thing?

A good place to connect the LED/resistor is across the filter capacitor that follows the rectifier, but we don't know how much voltage is there since you didn't say whether the 24VAC winding is center-tapped or not. Another place for the LED/resistor is at the regulated output, of which we don't know its voltage or current.

The transformer i used is a center-tapped +/- 24V at 1A. The dilemma is that if I put the LED jus after the rectifier, a large resistor with high power rating may be required to protect theLED (I still have little clue to calculating R values for LEDs. I always use 220R for 5v suppplies ;D). If i put the LED after the regulators, the LED prob will not light when the regulated voltage is below, say, 3v.

(I mainly work with PICs and recently got a 3v RF transmitter so i need voltages at ~3+,5,9,12++ :) )

I'm jus afraid that i might leave the PSU on if there is no indication, besides the switch position, that the thing is still on. :) The bad experience I had with the blown rocker switch that had a neon lamp didn't help either.

[audioguru]

gEcky,
1) A quality switch should not blow-out.
2) I am still confused by your "+/- 24V" transformer. Is it 24V center-tapped, which is also 12-0-12? Then its rectified and filtered output is about 16.2V.
3) Your LED has about 1.8V (for red) across it. If you use a 220 ohm resistor in series on 5V, then Ohm's Law calculates the current to be 5-1.8 divided by 220 = 0.0146A. A resistor to provide the same current from 16.2V is calculated: 16.2-1.8 divided by 0.0146 = 986 ohms. Use 1K ohms. The power in the resistor is calculated: voltage x current, so 16.2-1.8 times 0.0146 = 0.21W. A 1/4W resistor will get quite hot, so use a 1/2W resistor.