# Power indicator LED for 230VAC light circuit

I would like to add a (power) indicator light (LED) in box where a 230VAC light is switched on.

The issue I have is that there is no "neutral" line in the switch box (to put the LED over). It only opens or closes the phase line.

To me, the only way would be to put a light (LED) in series with main light.

What light/LED would that need to need be? It would need to be an LED that only has a small voltage drop but can handle the "high" current drawn by the main light (three lights of 4 watts).

See attached scheme as an illustration.

What would you recommend?

• So put a resistor in parallel with the LED. Calculate the resistor value so the voltage drop drives the LED, also consider AC… Jan 12, 2022 at 20:45
• @SolarMike Not sure how it will help? Jan 12, 2022 at 20:46
• Then you have a problem. The only option you have is to connect something in series to the circuit. But you can't connect anything in series to it, because it will load the line and interfere with it's main function. Jan 12, 2022 at 20:53
• No neutral line means little you can do without volt drop to the load and potential problems. Not a good solution. Anyway, if the load is also a LED lamp, what use is the added indicator (a bit like a solar powered flash-lamp - i.e. of no use whatsoever) because the main lamp will be the indicator. Jan 12, 2022 at 21:18
• The indicator would be on a switch which is outside the closed (no window) toilet space that has the main load, the (LED) lights. The reason to have it is exactly to know if somebody is in there and/or left on the lights. Jan 12, 2022 at 21:24

If you were to put a red LED in series, as in the circuit below, the minimum voltage drop would be ~1.8 V, reducing the efficiency of the circuit, as well as causing a slight imbalance in wave shape.

A better option would be to wind a few turns of AC mains wire through a small ferrite toroid core, and wind a few turns of finer wire through it, connected to the LED and a series resistor. A small toroid will saturate with the high, 50 or 60 Hz current in a household device at mains frequency, so the output to the LED will be far less than would be calculated from turns ratio. You'd need to experimentally find the value for the series resistor that lights the LED, but that doesn't burn it out, perhaps 100 Ω.

• Tip: when you use the CircuitLab button on the editor toolbar and Save and Insert on the editor an editable schematic is saved in your post. That makes it easy for us to copy and edit in our answers (or correct yours!). You don't need a CircuitLab account, no screengrabs, no image uploads, no background grid. Jan 12, 2022 at 22:13
• Don't forget that the power dissipated by the LEDs will be $\frac 1 2 3V_f I + \frac 1 2 V_f I = 2 V_f I$ (where the halves are due to only conducting for half the cycle. The diode forward voltage drop, $V_f$, will probably be 0.7 V. You'll need to allow for heat dissipation. Jan 12, 2022 at 22:43
• @koen.bulcke 15W at 230V means a current of about 65mA, I doubt you'll have any trouble finding diodes capable of handling that. Jan 13, 2022 at 16:03
• but why then not simply putting two LED's in parallel (but flipped to each other) in serie with the load? Jan 13, 2022 at 18:29
• LED's can dissipate only limited current. A common red LED could handle no more than 20 mA. The 1N4007 diode can handle 1,000 mA (i.e., 1A). Each diode drops about 0.6 V at room temperature, so 1.8 V is dropped across 3, which would be <2 W at 1 amp, for half the cycle; the other diode would dissipate ~.6 W (all wasted as heat). This presumes the load is <200 W. Jan 13, 2022 at 21:23

For the record, I have created the circuit (using 8x 1N4007 diodes + 1N4007 1 diode in "reversal" as proposed by @DrMoishe and it works for the moment flawlessy.

Total forward voltage drop is about 5.3V over the 1N4007 diodes (8x0.65V), which is enough to feed the LED (operating range: 3V-12V).

Components used:

https://www.aliexpress.com/item/32479437247.html

https://www.aliexpress.com/item/1005001382197271.html

https://www.aliexpress.com/item/1005003704434945.html

and a small box to put the PCB in.

Thanks y'all for the help!

Why not buy a small 50 Hz current transformer which will bring the currents down by the turns specified on the secondary. This has low losses and is safe. Then use a simple fullwave diode rectifier to drive your LED.

Use 50 W halogen miniature bulb or 100 W incandescent (filament) bulb in series (where you are placing led in diagram). When in series with 15 W load, it will consume 1-2 W only. Con is it will glow little only. I calculated it. For 50 W halogen bulb: 15 W load wattage will be reduced to 8.8 W and halogen bulb will glow with 2.65 W. I do not know if 100 W miniature halogen bulb is commercially available or not. But if available then 15 W load wattage will be reduced to 11.26 W and halogen bulb will glow with 1.69 W.

Did experiment myself with 15 W load and 50 W miniature halogen bulb as indicator. Halogen bulb was glowing right enough to indicate load is on as shown in pics attached. Moreover this indicator shows three states. If indicator dim on = load is on. If indicator full glow on= shortcircuit at load end. If indicator off=load is off. This 50 W miniature halogen bulb cost 25-34₹ in India. If 100 W miniature halogen bulb is available anywhere, I guess that will also glow enough and will be better option.

When load is 9watt, halogen bulb of 50 watt do not glow. Then halobulb of 25watt max will work. I found these halobulb require 1.69 to 2.65 watt to glow.

• And another con is that the volume will be at least 1000 times that of an LED without considering the bulk of the lampholder. If you have a better idea you can always edit your answer. Feb 17 at 16:05
• Sorry not neon. It is miniature halogen bulb, which is small. Called mirchi bulb in india. It is 14mm x 43mm. I ordered it at Amazon. I will do this experiment and revert back. Feb 19 at 15:09
• Experiment done and was successful. See main answer above Feb 21 at 9:26
• There is no rule against using eg Rupees for prices BUT most readers will be unfamiliar with them. You are liable to get a better reception if you convert prices to say \$US or Euro or ... . This is NOT anti-Indian, just trying to suit your main target audience. Feb 21 at 12:28

Instead of putting 8 or 4 IN4007 diode, you could use a 5v zener diode with 500 mA current carrying capacity.

This has the advantage of replacing all the diodes in the multi-diode solution with a single zener diode. In the reverse-biased-zener direction the zener provides enough voltage to light the LED. When the zener conducts it places about 0.8V reverse bias across the LED. Most LEDs will tolerate higher that this as a reverse voltage.

• @Velvet Each of the concepts he offered are different and multiple answers are acceptable, although unusual. This has the advantage that each can be voted on on its merit. The neutral addition is not a marvellous idea as the OP is asking because he has not got a neutral. The halogen bulb solution is interesting (but not as good as an LED plus diodes IMHO). The zener suggestion is a good one. Replace s ALL diodes with a single device. Feb 21 at 12:09
• @RussellMcMahon Thanks for clarification. I once got told to post two conceptionally different answers into the same answer box. The downvote was removed as soon as I complied. It didn't come from a moderator though, just a high-rep active user. Feb 21 at 13:14

If you can extend neutral from load to board through pipe in which half is going to load,then it can be done as shown in screenshot.

• This is not a marvellous answer BECAUSE the OP said"he issue I have is that there is no "neutral" line in the switch box (to put the LED over). It only opens or closes the phase line." ie he KNOWS there is no neutral line AND if he could easily have added one he would have. This answer will be downvoted and do you no good. The others are good. I recommend that you delete this answer and keep the other two. Feb 21 at 12:17