Hi everyone,

I’m looking for advice on the best way to run a general-purpose red LED indicator from a 120VAC supply and another LED indicator from a 24VAC supply.

I want a safe and reliable solution and was wondering whether it’s better to use pre-made AC LED indicators or build a circuit using components such as a rectifier and current-limiting resistor.

For the 120VAC application, I understand that a standard LED cannot be connected directly to mains without proper protection. For the 24VAC application, would a simple rectifier and resistor setup be suitable?

Any recommendations for suitable components or circuit designs would be appreciated.

Thanks!

 

Can't do schematics right now, so an all-text response.

The classic minimal circuit for both voltages is one series resistor to limit the peak current, plus one diode in anti-parallel with the LED to clamp the reverse voltage. Anti-parallel (or reverse-parallel) means the LED cathode is connected to the diode anode and the diode cathode is connected to the LED anode.

LED peak current is determined using Ohm's Law. Don't forget to subtract the LED forward voltage (Vf) from the mains peak voltage. Also, calculate the peak power dissipated in the resistor (using either Watt's Law or Joule's Law) and use a resistor rated for twice that amount.

Note that this is a half-wave rectifier, so the LED will be flashing at 60 Hz. This might be noticeable, particularly in you peripheral vision. To double the flash rate to 120 Hz, replace the single diode with a diode bridge.

ak

 

I want a safe and reliable solution and was wondering whether it’s better to use pre-made AC LED indicators or build a circuit using components such as a rectifier and current-limiting resistor.

There's nothing safe about a DIYer connecting an LED to line voltage.

Any recommendations for suitable components or circuit designs would be appreciated.

This circuit will work for both mains and 24VAC. Just size the resistor appropriately to give a few mA.

For line power, I'd use a 1W resistor.

You could also replace D1 with another LED.

 

Thank you everyone for the responses and helpful suggestions. I appreciate you taking the time to share your knowledge and advice. It has given me some useful ideas to consider.

 

It has given me some useful ideas to consider.

You might find more ideas in this thread.

 

Use a commercial AC-rated LED indicator for 120 VAC. It provides safety, reliability, and simplicity.

 

red LED indicator from a 120VAC supply

For line power, I'd use a 1W resistor.

Many small resistors are not voltage rated for 170V, the peak of the power line voltage of 120v. You might want two resistors in series.

 

Many small resistors are not voltage rated for 170V, the peak of the power line voltage of 120v. You might want two resistors in series.

Good point, I appreciate you mentioning that. The voltage rating of the individual resistors is definitely something to consider, not just the resistance and power rating. Using two resistors in series would help share the voltage stress and give a better safety margin, especially with the 170V peak from a 120V AC line. Thanks for the reminder, it's a small detail that can make a big difference in reliability.

 

Many small resistors are not voltage rated for 170V, the peak of the power line voltage of 120v. You might want two resistors in series.

Most resistors are rated for 200V.

From Yaego for metal film resistors:


The resistors above that end with 'S' are the smaller variants that are rated for higher wattages:

Additionally, for AC voltages, it's the RMS voltage that matters for power calculations. If he rectified and filtered, that would be different, but OP was talking about AC voltages.

EDIT: SEI data:

 

Most resistors are rated for 200V.

Sorry you are right. I have not used through hole parts in a long time.

 

For line power, I'd use a 1W resistor.

Nope.

If the mains voltage is 120 Vrms and the resistor current is 20 mArms, R1 power dissipation is 2.4 W. -ish

I recommend a minimum of 5 W rating. Still uncomfortable to touch, but won't melt your fingerprints.

ak

 

If the mains voltage is 120 Vrms and the resistor current is 20 mArms, R1 power dissipation is 2.4 W. -ish

I said to size the resistor for a few mA.

Just size the resistor appropriately to give a few mA.

 

Just size the resistor appropriately to give a few mA.

4.1 mA (RMS) or less.

ak

 

For 120VAC operation choose R1 220 Ohm 0.5W, C1 220nF 400V, C2 220uF 6.3V.
For 24VAC operation components R1 R2 C1 can be replaced to 2.4kOhm 0.5W resistor.
Capacitor C2 eliminates Led flickering
Both diodes 1N4001

 

Capacitor C2 eliminates Led flickering

 

View attachment 369493

You're right.
I wrote the value of 220uF approximately; it can be increased 2 or more times. My circuit has an advantage – it has fewer components.
Now you could check in the simulator the above-mentioned circuit, where a diode is connected in parallel to the LED.