replace main  power indicator by LED  - Page 1
Forum section: electronics / beginners

author: JacquesBBB
author_rank: Frequent Contributor
timestamp: October 16, 2016, 07:16:40 pm
content: When repairing old devices, it is very often that the power indicator isdead and the bulb is not easy to find.I just want to replace them by leds which will thus be on the main (240 v).But searching on the web for an easy way to implement that, I saw a very large variety of solutions,and I am thus wondering what would be a simple and reasonably safe solution ?My first guess would be to put two leds in opposite way in parallel with 100 kR resistors in series on both sides.--- Res 100kR ---- + LED- --- Res 100 kR---\ /- LED+The current should the be only 1.2 mA ( 1.7 mA peak), and the dissipative power in the resistor about 150 mW .So 1/4 w resistors should be OK. Say 1/2 W resistors to be safe, but I see no reason to put 1W resistors.Is this OK ?Thanks

author: Ian.M
author_rank: Super Contributor
timestamp: October 16, 2016, 08:14:20 pm
content: The 50Hz flicker between the two LEDs will be annoying unless the original bulbholder has a *VERY* good diffuser. Try a bridge rectifier round a single LED with the 100K resistors feeding it. The bridge rectifier never sees any significant voltage so even 30V diodes would be fine.You may need a Zener in series with the LED to provide some 'standoff' in some applications so it doesn't glow due to capacitive leakage currents when the supply is supposedly switched off and there is no real load on the circuit.

author: TimFox
author_rank: Super Contributor
timestamp: October 16, 2016, 10:00:32 pm
content: To run low-current LEDs from an AC line, I have done the following (forgive the snoopy graphics):AC --- [capacitor C1] --- [resistor R1] --- [LED, 5 mA] ------ AC[reverse diode]For 120 V, 60 Hz, I used a capacitor C1 with reactance approximately 25 k (100 nF at 60 Hz). The resistor was less than that, maybe 2 k, and the reverse diode was a normal 1N4005 rectifier diode (in parallel with the LED) to keep the current through the chain approximately sinusoidal.Pay attention to insulation requirements since the LED is connected to the line. Use a film capacitor (with appropriate voltage rating) for C1.

author: JacquesBBB
author_rank: Frequent Contributor
timestamp: October 16, 2016, 10:56:23 pm
content: Thanks Tim,This circuit is probably better than the one I made with resistors, as it will dissipate much less power.But why do you still put a resistorWhy not just doAC --- [capacitor C1] -- [LED, 5 mA] ------ AC[reverse diode]with a smaller capacitor ?

author: JacquesBBB
author_rank: Frequent Contributor
timestamp: October 17, 2016, 05:22:41 am
content: If I am not wrong, this would be enoughAC (240 V ) --- [capacitor C1 47nf] -- [LED, 5 mA max] ------ AC[reverse 1N4007]and the power consumption will be just limitedto the diodes ( < 10 mW ) and the loss in the ESR of the capacitor which will be negligible with a 5 mA current.Is this OK ?

author: Ian.M
author_rank: Super Contributor
timestamp: October 17, 2016, 06:11:45 am
content: You are wrong. If switched on at the instant of peak mains or if operating during a high voltage line transient, the cap will pass a current spike in the tens of Amps range which may well be enough to vaporise the LED's bond wires. You *ALWAYS* want a resistor in there, typically with very approximately 10% of the line voltage across it.

author: JacquesBBB
author_rank: Frequent Contributor
timestamp: October 17, 2016, 07:41:24 am
content: Thanks Ian for t his info, which I cannotDeduce from my spice simulation.But I an not sure to understand completely what you mean. Could you detail more or provide a reference ?From what you say, a 1k resistor in series will be OK, but I would like to understand more.

author: Zero999
author_rank: Super Contributor
timestamp: October 17, 2016, 07:49:34 am
content: Yes, 1k should be fine. There should be no need for SPICE simulation. Use Ohm's law to calculate the maximum current through the resistor when the mains voltage is at its peak. It just needs to be below the LED's maximum surge current rating. If you're unsure then a few 100mA should be fine, as it won't last for long. Another good idea is a bleeder resistor across the LED to discharge the capacitor when it's unplugged: 1M should do.Personally I would find the flicker annoying so I'd use a full wave rectifier but if half wave is good enough for what you want, then stick with it.

author: HackedFridgeMagnet
author_rank: Super Contributor
timestamp: October 17, 2016, 07:58:00 am
content: If you have a sim then try putting in a 240*(sqrt 2) square wave in the simulator, instead of a sinusoid, to simulate a top of cycle turn on.

author: JacquesBBB
author_rank: Frequent Contributor
timestamp: October 17, 2016, 08:25:10 am
content: OK,thanks for all, now I understand my mistake. For an ideal cap, the current isI = C dV/dtSo when one start from 0 to V0, dV/dt is infinite, and so is the current. It will thus be only limited by the resistor of the circuit.

author: stj
author_rank: Super Contributor
timestamp: October 17, 2016, 10:50:19 am
content: or use a neon lamp & resistor.neons come in lots of colours now.(just different phosphors inside the glass)

author: Ian.M
author_rank: Super Contributor
timestamp: October 17, 2016, 12:54:14 pm
content: Under LTSPICE, its possible to use a behavioural resistor as a time-switch. I like to use the European resistor 'block' symbol (misc\EuropeanResistor) for 'special' resistors like this switch. Simulation of peak turnon attached. It looks like your LED will see a peak of about 300mA.

author: setq
author_rank: Frequent Contributor
timestamp: October 17, 2016, 01:05:29 pm
content: Just stick a canned neon mains indicator in it. They cost very little.

author: StuUK
author_rank: Frequent Contributor
timestamp: October 17, 2016, 01:11:01 pm
content: Quote from: setq on October 17, 2016, 01:05:29 pmor just use something like thishttp://sinolec.co.uk/en/10-panel-mount-indicator-lightsfor less than a £1 why mess about?

author: stj
author_rank: Super Contributor
timestamp: October 17, 2016, 02:03:47 pm
content: Quote from: StuUK on October 17, 2016, 01:11:01 pmnice link - bookmarked.

author: Kjelt
author_rank: Super Contributor
timestamp: October 17, 2016, 02:38:27 pm
content: It sounds simple and trivial but even pro firms mess this up.Two years ago I replaced a couple of 230V bulbs in an Eaton M22 button with M22-LEDC230-W modules.They have a rectifier and couple of resistors and finally a white led. Within a half year both modules failed.Just blow up the led. So then I replaced them again with bulbs. Now I have re-ordered the M22-LEDC230-W modules and they appear different, hopefully they will alst longer as the previous device.

author: Seekonk
author_rank: Super Contributor
timestamp: October 17, 2016, 03:21:57 pm
content: Just picked up an outlet strip at a yard sale with a LED. I use AC outlets for everything at my camp 12V, 36V and 120V. Nice thing about this LED circuit is I can use it on all three powers and the LED still lights even with 12V.

author: JacquesBBB
author_rank: Frequent Contributor
timestamp: October 17, 2016, 05:18:53 pm
content: Quote from: stj on October 17, 2016, 10:50:19 amYes, that seems finally the easiest way. I did not realised before that one had to put all these components, with this bulky capacitor. I will look for a stock of cheap neon bulbs which I could then fit in these various power indicators.Indeed, on most of hte old devices I see, these are dead. Most probably the users do not bother to change them whenthey broke.In fact, for the one I just repaired, as it was a power supply, I did it the easy way by powering a led (with resistor) from the fixed 15V DC output instead of the input AC, as it was in the original design.Of course, it would not be desirable if the voltage regulation was very critical but it was not the case.

author: setq
author_rank: Frequent Contributor
timestamp: October 17, 2016, 05:43:24 pm
content: You should have the "pilot light" on the mains side of the transformer as if the transformer burns out, secondary or rectifiers burn out, a connection fails etc, it will still show as powered up. This is why you see integrated power light in switches in more modern equipment as it's straight after the mains filter/inlet.

author: JacquesBBB
author_rank: Frequent Contributor
timestamp: October 17, 2016, 09:42:31 pm
content: Thanks for this precision.So I will definitely invest in a set of neon bulbs for the next repairs. I understand now the philosophy of this power indicator.I had the opposite idea, which was to put the indicator if the output was there.But I will follow the general rule if there is one.

author: Brumby
author_rank: Unknown
timestamp: October 18, 2016, 12:58:23 am
content: Quote from: JacquesBBB on October 17, 2016, 09:42:31 pmI don't believe there is a general rule - but the logic expressed here is the same as my preference ... the indicator is on if the outputshouldbe there. This indicates the mains supply is present and if the device is not working, then the fault lies within it. These two points are helpful for safety and troubleshooting.