Author Topic: Powering an LED from international line voltage  (Read 1642 times)

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Offline kenyonshuttTopic starter

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Powering an LED from international line voltage
« on: May 03, 2024, 01:00:58 am »
Yo! I can’t believe I’m struggling with this, but I’m finding it really difficult to design a cheap circuit that can drive a 10-15mA LED, powered by a wide range of AC voltages (roughly 80-250VAC). It’s frustratingly challenging, given the fact that all kinds of products have “power present” indicators. That’s all I’m trying to do.

Some attempts:
  • AC feeds LED directly thru a resistor (resistor needs to be super high wattage, LED brightness changes with voltage)
  • Voltage divider with zener regulator (zener ends up conducting way too much current at high voltage
  • wide voltage range LDO like onsemi NCP785AH50T1G (lots of components, this family is only rated to 10ma which isn’t quite bright enough

Sure I could add a friggin transformer or switching power supply but I literally just want the LED to turn on, at a predictable brightness, when the device is plugged into the wall.

Thoughts anyone? Thanks!!

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Offline HackedFridgeMagnet

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Re: Powering an LED from international line voltage
« Reply #1 on: May 03, 2024, 01:22:03 am »
Capacitive Dropper may suit.
 
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Online NiHaoMike

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Re: Powering an LED from international line voltage
« Reply #2 on: May 03, 2024, 01:43:53 am »
How about use multiple smaller LEDs in series? Then you'll need less current for the same brightness, possibly to the point where a series resistor would become practical.
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Offline mariush

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Re: Powering an LED from international line voltage
« Reply #3 on: May 03, 2024, 05:46:47 am »
Single diode for half wave rectification, optionally a small capacitor to reduce flicker, a zener diode to cut off the voltage if it goes above around 200v DC or less (up to you where you want heat to be produced) , a HV linear current driver  like CL220 (5v - 220v DC, 20mA out) : https://www.digikey.com/en/products/detail/microchip-technology/CL220K4-G/4902728

« Last Edit: May 03, 2024, 05:54:23 am by mariush »
 
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Offline jonpaul

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Re: Powering an LED from international line voltage
« Reply #4 on: May 03, 2024, 06:12:32 am »
Use any VDE/TUV /CE approved led driver, cost ~ EU 5..15 depending on the LED power, V, I.

We use the Italian TCI   

Avoid the  Chinese drivers

j
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Offline kenyonshuttTopic starter

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Re: Powering an LED from international line voltage
« Reply #5 on: May 03, 2024, 06:49:05 am »
Single diode for half wave rectification, optionally a small capacitor to reduce flicker, a zener diode to cut off the voltage if it goes above around 200v DC or less (up to you where you want heat to be produced) , a HV linear current driver  like CL220 (5v - 220v DC, 20mA out) : https://www.digikey.com/en/products/detail/microchip-technology/CL220K4-G/4902728

Hey that thing is really cool- very much along the lines of what I'm looking for. thanks for the idea!
« Last Edit: May 03, 2024, 06:52:49 am by kenyonshutt »
 

Offline kenyonshuttTopic starter

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Re: Powering an LED from international line voltage
« Reply #6 on: May 03, 2024, 06:55:36 am »
How about use multiple smaller LEDs in series? Then you'll need less current for the same brightness, possibly to the point where a series resistor would become practical.

That's possible, but to keep costs low, I'd like to stick to only one LED. Thanks for the idea though
 

Offline kenyonshuttTopic starter

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Re: Powering an LED from international line voltage
« Reply #7 on: May 03, 2024, 07:06:33 am »
Capacitive Dropper may suit.

Looking into this more. So one drawback with this is that caps act linearly (obviously) so just a cap, series resistor, and LED would cause the LED brightness to vary depending on AC voltage.

I wonder if there's a way to set up a circuit like this...
line voltage - C1 in series - R1 in series - D1 in series - ZD1 to ground - C2 to ground - R2 in series - LED
with the intention that the C1 and R1 drop the voltage down to some 5 to 20 volts depending on applied AC voltage. then, through D1, that voltage charges up C2, limited by the zener voltage of ZD1, and then current from that is then trickled through R2 into the LED.

I do not know how to solve this kind of circuit on paper, so I might just grab some values and try it (keeping safety in mind).

Is this more complicated or costly than finding some dedicated HV linear current driver? *shrug*
 

Offline kenyonshuttTopic starter

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Re: Powering an LED from international line voltage
« Reply #8 on: May 03, 2024, 07:08:46 am »
not that i ever actually would, but i'm trying to design this indicator light into a product that could someday see mass production. therefore, cost and simplicity is a huge consideration.
 

Offline mariush

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Re: Powering an LED from international line voltage
« Reply #9 on: May 03, 2024, 07:10:00 am »
How about use multiple smaller LEDs in series? Then you'll need less current for the same brightness, possibly to the point where a series resistor would become practical.

That's possible, but to keep costs low, I'd like to stick to only one LED. Thanks for the idea though

It's not a bad idea ... you can get leds with high forward voltage for a few cents.

For example, LiteOn LTW-5630 has a 24v forward voltage and costs 5 euro cents if you get 10 : https://www.tme.eu/de/en/details/ltw-5630szn40/smd-white-leds/liteon/

Put three of these in series and now you've got a minimum of 72v DC needed to run, so now you need to drop less on the resistors or diodes or current drivers etc and maybe less current is also an option.
 

Offline soldar

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Re: Powering an LED from international line voltage
« Reply #10 on: May 03, 2024, 07:53:02 am »
How about the classic NE-2 indicator lamp?

https://en.wikipedia.org/wiki/Neon_lamp
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Online Ian.M

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Re: Powering an LED from international line voltage
« Reply #11 on: May 03, 2024, 08:12:00 am »
Its certainly not impossible to stabilise the LED current from a capacitive dropper, but the problem is: the capacitor, line voltage and frequency pretty much set the current, and if you want less you either have to burn it off as heat in a shunt regulator, or use a series regulator that has to stand-off a significant fraction of the peak mains voltage.   You probably wont like the resulting cost and complexity (especially vs a NE2 neon + dropper resistor)!

See attached LTspice sim for a slightly different take on the shunt regulator option using jellybean parts apart from the dropper capacitor.  It stabilises the LED current at about 15mA with under 2mA variation between 100V in and 260V in.  As its full-wave and tries to reduce flicker, there are considerable options to Muntz it if you don't care about flicker.
« Last Edit: May 03, 2024, 08:16:22 am by Ian.M »
 
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Offline mikeselectricstuff

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Re: Powering an LED from international line voltage
« Reply #12 on: May 03, 2024, 09:10:09 am »
Capacitive dropper plus some resistance to make the current more consistent- this will not be constant current but may be good enough - does brightness change over supply  voltage really matter  ?

Also, do you really need 10-15mA ? A more efficient LED/Color (e.g. white or green) may make the problem easier by reducing current, also consider multiple LEDs in series to reduce the current.

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Online Gyro

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Re: Powering an LED from international line voltage
« Reply #13 on: May 03, 2024, 09:17:46 am »
Use any VDE/TUV /CE approved led driver, cost ~ EU 5..15 depending on the LED power, V, I.

We use the Italian TCI   

Avoid the  Chinese drivers

j

EUR 5..15 to drive a 10-15mA indicator LED?  Seriously? :o


@OP: I would suggest something like a mains dropper, feeding into a small package, low current bridge rectifier [Edit: or discrete diodes] (flicker removal), then a Zener of, say, 4V7 followed by a small resistor to limit the LED current down from 4V7.

Assuming an efficient LED @ 10mA, a resistive mains dropper would dissipate 750mW at 80V and 2.5W at 250V - much too high in my opinion. A capacitive dropper is the way to go. Assuming a 470nF safety rated capacitor, this would give ~12mA at 80V (Zener essentially dissipating nothing) and 37mA to 250V. At maximum voltage, the dissipation of the Zener, after taking off the 10mA for the LED would be only 127mW, so a 250mW rating would be fine. Don't forget to put a low value series resistor (preferably fusible) of around 100R as the first component between mains live and the capacitor to cope with mains spikes.


EDIT: My capacitor reactance calculation was for 50Hz. At 60Hz, you would get away with a 390nF cap, maybe even 330nF.
« Last Edit: May 03, 2024, 09:28:28 am by Gyro »
Best Regards, Chris
 
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Re: Powering an LED from international line voltage
« Reply #14 on: May 03, 2024, 09:22:22 am »
Capacitive dropper plus some resistance to make the current more consistent- this will not be constant current but may be good enough - does brightness change over supply  voltage really matter  ?

Also, do you really need 10-15mA ? A more efficient LED/Color (e.g. white or green) may make the problem easier by reducing current, also consider multiple LEDs in series to reduce the current.
I agree. A capacitive dropper is the easiest way. An anti-parallel diode across the LED (or a dual LED in one package) is necessary to make an AC current flow. Not much current is needed for a modern LED so designing something which works from 80VAC to 260VAC should be possible.
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Online NiHaoMike

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Re: Powering an LED from international line voltage
« Reply #15 on: May 03, 2024, 01:15:30 pm »
That's possible, but to keep costs low, I'd like to stick to only one LED. Thanks for the idea though
Tiny SMD LEDs are dirt cheap. Design a small board with a whole lot of them on one side and some support components like a diode bridge on the other. (The main dropping resistor and/or capacitor remains external.) The tiny bit extra you spend on the LEDs is offset by the savings on the supporting circuit.
How about the classic NE-2 indicator lamp?

https://en.wikipedia.org/wiki/Neon_lamp
My experience with a NE2 and a modern LED in series (for a charge indicator on a high voltage capacitor module) is that the LED appears much brighter for the same current. The efficiency of those small neon bulbs is really low.
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Offline soldar

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Re: Powering an LED from international line voltage
« Reply #16 on: May 03, 2024, 01:41:29 pm »
If all you want is an indication that power is ON then NE-2 neon + resistor is difficult to beat in terms of simplicity and reliability. It is what is still used in switches and power strips. I have quite a few that are half a century old and working like the first day.
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Offline mikeselectricstuff

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Re: Powering an LED from international line voltage
« Reply #17 on: May 03, 2024, 01:52:29 pm »
If all you want is an indication that power is ON then NE-2 neon + resistor is difficult to beat in terms of simplicity and reliability.
Where do you get the surface-mount NE-2's from ?   :D
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Offline mikeselectricstuff

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Re: Powering an LED from international line voltage
« Reply #18 on: May 03, 2024, 01:53:32 pm »
That's possible, but to keep costs low, I'd like to stick to only one LED. Thanks for the idea though
Tiny SMD LEDs are dirt cheap. Design a small board with a whole lot of them on one side and some support components like a diode bridge on the other. (The main dropping resistor and/or capacitor remains external.) The tiny bit extra you spend on the LEDs is offset by the savings on the supporting circuit.
There are also LEDs with multiple die in series in one package, mostly white for lighting applications.
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Offline soldar

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Re: Powering an LED from international line voltage
« Reply #19 on: May 03, 2024, 02:25:43 pm »
Where do you get the surface-mount NE-2's from ?   :D
What a great business idea! :)
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Online TimFox

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Re: Powering an LED from international line voltage
« Reply #20 on: May 03, 2024, 03:36:43 pm »
I have used NE-51s and NE-2s since 1970.
The problem is that over the years electrode metal sputters onto the inside of the glass and darkens the display.
 

Offline xvr

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Re: Powering an LED from international line voltage
« Reply #21 on: May 03, 2024, 03:42:43 pm »
I'd peek some circuit from one of PE article. It can solve here.
 

Offline switchabl

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Re: Powering an LED from international line voltage
« Reply #22 on: May 03, 2024, 04:04:15 pm »
If all you want is an indication that power is ON then NE-2 neon + resistor is difficult to beat in terms of simplicity and reliability. It is what is still used in switches and power strips. I have quite a few that are half a century old and working like the first day.

There is nothing wrong with that but there are bidirectional/bipolar LEDs (two antiparallel diodes in a single package) that are basically a more efficient drop-in replacement.
 

Online coppice

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Re: Powering an LED from international line voltage
« Reply #23 on: May 03, 2024, 04:23:55 pm »
Capacitive dropper plus some resistance to make the current more consistent- this will not be constant current but may be good enough - does brightness change over supply  voltage really matter  ?
More specifically you need enough resistance to limit the capacitor charging current if you turn on near the peak of the mains. Otherwise you can pop the LED. You probably want a reverse diode across the LED, too. Watch out if the product might be supplied from a highly distorted mains, like some UPS solutions give you. The low impedance of the capacitor can put a high current through both LED and resistor with a lot of high frequency harmonics. Quite a few products with cap drop supplies can catch fire with a distorted waveform, as the resistor cooks. Use a large one.
 

Offline BrokenYugo

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Re: Powering an LED from international line voltage
« Reply #24 on: May 03, 2024, 06:05:19 pm »

@OP: I would suggest something like a mains dropper, feeding into a small package, low current bridge rectifier [Edit: or discrete diodes] (flicker removal), then a Zener of, say, 4V7 followed by a small resistor to limit the LED current down from 4V7.

Assuming an efficient LED @ 10mA, a resistive mains dropper would dissipate 750mW at 80V and 2.5W at 250V - much too high in my opinion. A capacitive dropper is the way to go. Assuming a 470nF safety rated capacitor, this would give ~12mA at 80V (Zener essentially dissipating nothing) and 37mA to 250V. At maximum voltage, the dissipation of the Zener, after taking off the 10mA for the LED would be only 127mW, so a 250mW rating would be fine. Don't forget to put a low value series resistor (preferably fusible) of around 100R as the first component between mains live and the capacitor to cope with mains spikes.


EDIT: My capacitor reactance calculation was for 50Hz. At 60Hz, you would get away with a 390nF cap, maybe even 330nF.

This is exactly what you see inside a cheap LED night light, with the addition of a transistor+LDR to short the current source during the day. They usually use a half wave rectifier and a small low voltage electrolytic for smoothing, but for an indicator I don't find 120hz very flickery. Very proven design.
« Last Edit: May 03, 2024, 06:07:26 pm by BrokenYugo »
 


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