Need circuit:(bi-color) red LED=off, green=on

[redrabbit]

I built four 317(LZ) based PSU's in one box, to power some DIY stompboxes.
I want to use a bicolored LED on each PSU, where red =OFF, and green=ON. (red also tells me the mains are active.)

Each PSU will have a on/off switch.

I have some three-lead bicolors already (anode common), but.....?

One option I thought of is to wire the LED to the switch, but that means the switch is post-transformer, and after the red LED's power.....the green's power is after the switch (at the output possibly ?).....
(or green is wired on a seperate pole of the switch....... but that does not tell me the psu is truly functioning, it just tells me the switch is functioning and the PSU might be working .)

I would rather have each switch pre-transformer, in which case I don't know how to wire such a LED circuit.

Any ideas??

=RR=

 

[PM650]

The only way to do what you want is to build a power supply just for the leds. The supply would have to have a small step-down transformer or be an off-line type capabable of dropping 165 volts or so. If you have the space, the standard step-down transformer supply will be less complicated and probably easier to build than an off-line one. (here's an off-line example) For small transformers, I suggest browsing mouser's online catalog, pages 1508-1523. Obviously, a single supply can be used to power all the led's, so it would be relatively simple to set up.

 

[N-Channel]

Switching

RR_

How about a dedicated supply for the LEDs, like PM650 suggests. I would use a 12V supply, less than an amp. Then you could drive each LED with the outputs of almost any dual op-amp. Let's call the op-amp driving the red LED the red op-amp, and the one driving the green LED the green op-amp.

Now, then, tie red's (+) and green's (-) inputs together, and then tie these to the junction of a resistor-zener reference (resistor going to +Vcc, and Zener to ground). Let's call this Junction 1. Make the resistor something like 5KW and the Zener's voltage anywhere from 3-5V. (Almost forgot- this assumes that all your power supplies' grounds are tied together, and to the LED's power supply ground.)

Next, tie red's (-) and green's (+) inputs together, and to a resistor voltage divider (say 10KW each resistor. Let's call this Junction 2. Now, take the upper end of this divider, and run it to the (+) output of your amplifier's power supply.

When a given device is off, the voltage at Junction 1 will be lower than Junction 2, and the red LED will be on, and the green one off. When the given device is turned on, then Junction 1's voltage will rise above Junction 2, extinguishing the red LED, and lighting the green LED.

Repeat this entire circuit for every device you wish to power.

Sounds like alot, but if you draw it out on paper, it is really quite simple. I have used a similar circuit for indicatin blown fuses on different +Vcc buses. Works well.

Hope this helps,

Steve

 

[poobah]

For openers, use a double pole - double throw switch.

Wire the green LED directly to output of the PSU (use a limiting resistor of course) ... done.

Use one pole of the switch to make/break the LINE connection to the primary.

Use the remaining pole, and the OFF state throw the apply mains to the RED led when the switch is in the off state.

You will need a resistor, 12 kOhm, and a diode, 1N4004, in series with the LED... this will give a 5 mA current. You should also place another 1N4004 in "anti-parallel" with the red LED (anode to cathode / cathode to anode); this will protect the LED from reverse bias caused by the junction capacitance of the rectifying diode.

Be sure to wire your switch in such a way as to prevent ant direct connection between the primary and the secondary.

If you can use 2 separate LEDs... there is an easier way to this with a SPST switch...