Need help with Warning LED indicator for redundant power supply.

[ramussons]

Thanks Harald,
This is what I thought. So to do the LED indication I would need to put a diode on each of the regulators positive outputs, (2 regulators, 2 diodes in series on positive rail of output.)
ramussons states "Simply put LED with a Protection diode across the Isolation diode. This will glow when the PSU is down (being fed from the other PSU. Will need to adjust the series resistor for current limit)."
Could you guys explain this in laymans terms, not sure I am picturing this the correct way in my head. I am quite new to this exciting world of electronics, please don't be too hard on me.
Jason

Thought of something like this.

 

[ramussons]

Harald,
I tried the above circuit and it did not function as intended. Both leds stayed lit when either power supply was shut down. im not sure why, possibly the diodes I used? I used 1N5404 diodes for D1 and D2. I breadboarded it exactly as shown and checked it 3 times. If I separate each ground on the regulator output (not tied together) The respective led will dim when I remove the input positive or negative.
Thoughts?

Most likely because the Output Capacitors take some time to discharge thru' the LED's.

Shut off the PSU and wait for some time. The LED will (or should) finally turn off.

 

[Harald Kapp]

Most likely because the Output Capacitors take some time to discharge thru' the LED's.

Good point.

If that's the case, a zener diode in series with the LED can help to turn off the LED more quickly.

 

[Audisturbo1]

That's a good point about the caps. If I tried a zener should I use a 12v since the PSUs and LEDs are 12v?

 

[Harald Kapp]

No, 12V zener + 1.6V LED is more than the output voltage of the power supply and the LEDs will not turn on. Use an 8-10V zener. The resistor is calculated as (12V-(Vzener+Vled))/Iled.