How to add led indicator to my 0-30v power supply

 

Put a 2k-10kΩ ½W resistor in series with the LED. Connect these across the output of the power supply, paying attention to the polarity. Select the value of the resistor to suit the desired brightness.

If you want to build a constant current source to drive the LED that will be slightly more complex.

You can also use an LM3914 dot/bar display driver to create a crude voltage level indicator.

 

Welcome to AAC!
What do you want your indicator to indicate? Output voltage? Output active? Mains input active? Current limit exceeded? Error? .....?

 

hi djr,
Look at these options.
E
https://www.ebay.co.uk/sch/i.html?_...TitleDesc=0&_odkw=digital+voltmeter&_osacat=0

 

Since your supply is capable of outputting zero to 30 volts (assuming DC) it would stand to reason you want an indicator to show that the supply is powered. Putting an LED on the output would mean you'd have to prepare for the full voltage so as to not burn out the LED. I'm guessing you desire just a simple indicator that it's on. For that we'll need more information about the supply itself.

I have a couple supplies scrapped from computers. They have multiple outputs ranging from 3.3V to 12V. I've used the 5.1V leg as my indicator. When it's present the 5V goes through a resistor to limit current to 15mA. But you have a variable output supply. There should be a few spots you can scavenge a steady voltage to use as an indicator. OR if you use the main power switch then you're talking about either 110 volts AC or 220 volts AC. If that's the case then you want a line from the switch to a resistor of appropriate size going to; and there are different options; two LED's or a single LED and a diode. IF you run on 110 VAC and the LED you want to use as an indicator is a standard 5mm LED then I'd guess you will want about 15mA going through the indicator. 110V ÷ 0.015A = 7,333Ω. In that case I'd opt for the closest standard resistor value (that I have in my inventory) of 6.8KΩ which would give me (110V ÷ 6800Ω =) 16mA. I'd also use another diode such as a 1N4004 which is rated for 400 volts at 1 amp. The 1N4003 is rated for 200 volts at 1 amp, which, if you're using 220VAC would not be big enough.

Either the diode will be in series with both the LED and diode pointed in the same direction OR the diode in parallel to the LED each pointed in opposite directions. Either way - you get an indicator that will operate on 110V. for 220 just double the resistance to maintain approximately the same current. LED's are current driven devices, hence, the need for a resistor. Keep in mind you'll also need a sufficiently sized resistor wattage. In the example I gave you'd need a 2 watt resistor MINIMUM; and you'd do well to use a 3 watt resistor instead. 15mA at 110V (0.015A x 110V = 1.65W). Too often people forget about calculating the wattage as well. Dropping 110 volts means producing a lot of heat.

 

Under "Low Voltage" applications you need to consider the forward voltage drop of the diodes but with 110 volts the difference is insignificant, and in this case can be ignored.

 

two LED's or a single LED and a diode.

I think one of your images is wrong - the parallel setup. In the parallel setup it should have an LED pointed in each direction. That would be "Two LED's". However, that same image (parallel but reversed diode) will also work. It will just not produce any light when the current runs opposite to the LED.

 

Be careful with high voltage disconnect the mains and remember to short capacitors with a resistor.
If you get tired of leds you can try neon bulbs.

 

Neon bulb! Why didn't I think of that?! Just a bulb and a resistor. When power is on - the neon bulb is on. Problem solved.

 

I think one of your images is wrong - the parallel setup. In the parallel setup it should have an LED pointed in each direction. That would be "Two LED's". However, that same image (parallel but reversed diode) will also work. It will just not produce any light when the current runs opposite to the LED.

Actually they are both 100% correct. The purpose of the non-light emitting diode is to protect the light emitting diode from very large reverse bias.

LEDs have a very low reverse voltage and tend to release magic smoke when over driven. Either way puts the reverse voltage across the diode and not the LED.

 

ErnieM That's exactly what I intended to indicate. Having the LED and Diode both oriented the same way protects against exactly that - the reverse voltage that comes with AC. You can have any version of all three drawings. With the diode parallel but reverse to the LED you have a current path so that the LED doesn't try to block the current. And having the double LED's, one is always lighting. Except for when the voltage is crossing zero, but that's just nit-picky.