LED indicator on AC

[Watth]

Hello EEVfolks!
In a project, I have a transfo to get 48 ~ 60 V AC from 230V 50Hz mains. (To power PDLC films).

As the load current (a bit less than 80mA) is way lower than the transfo ratings (>= 20A), secondary voltage is actually about 55V.
As the transfo has fuses on primary and thermistor on secondary, I want to add an LED to show if voltage has been cut.
I copied was inspired by this guide: https://www.eleccircuit.com/mains-voltage-indicator-with-a-led/.

I will use red LEDs.

What is your opinion on this circuit?
What values should I use for the secondary LED circuit?

Thanks for reading!

[mariush]

I'd just do 

AC ---- [ diode for half wave rectification  ===>| ] ----- [ capacitor to ground) ---- [ + led -] ---- [ resistor to limit current ] --- ground

See this simulator link : link to falstad.com circuit emulator

You can play with the capacitor and resistor values - for example if you change the capacitor to 1uF you will see the led brightness fluctuate because the capacitor can't supply the led enough energy.


Hold mouse over components, click on the switch to turn on/off power to the led etc

I suppose you could "improve" by adding a "zener" diode to cap the voltage to something like 5.6v or 12v or whatever zener diode is cheap and easy to source, then you won't drop as much energy in the resistor but you'll drop some on the zener diode.

[Watth]

Thanks a lot for your reply, mariush.
To minimize power on the LED, I will put the first LED on the secondary. What matters is that it is placed between the fuse and the thermistor.
Regarding the capacitor, except that it must handle a minimum of ~75V, do you have any advice? Which type should it be?

[ledtester]

You might consider just using a neon lamp - a lot simpler and safer:

https://www.digikey.com/en/blog/dont-rule-out-the-neon-bulb-as-an-ac-power-on-indicator

[Watth]

Thanks for your feedback, ledtester

The project will be in an enclosure that should be as small as possible, and the light should not be too intense, so I am restricted in the options that work for me.

On the primary (230V AC) side, It is indeed a solution I thought about, and as I will make purchase on tme.eu, there are available products: https://www.tme.eu/en/details/9sltb09sltsvful/neon-indicators/comelux/ : as I can adapt the 3D printed enclosure I design myself.
As for the light intensity, would a series resistor allow to lower it?

However, I need an indicator that works on the secondary 60V AC. Could a series resistor make it work on 230V AC?

Thanks for reading!

[PCB.Wiz]

Regarding the capacitor, except that it must handle a minimum of ~75V, do you have any advice? Which type should it be?

At 60vac, with a decent red led, you may not need a cap at all.
1-5mA should be enough and that’s 300mW max.

[ledtester]

Thanks for your feedback, ledtester

The project will be in an enclosure that should be as small as possible, and the light should not be too intense, so I am restricted in the options that work for me.

On the primary (230V AC) side, It is indeed a solution I thought about, and as I will make purchase on tme.eu, there are available products: https://www.tme.eu/en/details/9sltb09sltsvful/neon-indicators/comelux/ : as I can adapt the 3D printed enclosure I design myself.
As for the light intensity, would a series resistor allow to lower it?

Yes, a series resistor will lower the intensity of the neon bulb.

However, I need an indicator that works on the secondary 60V AC. Could a series resistor make it work on 230V AC?

I'm not sure what you're asking here. A neon bulb designed for 230V likely will not work on 60V.

However, a neon designed for a lower voltage -- like this one designed for 12V:

https://www.tme.com/ux/en-us/details/ni-2fyl-12v/neon-indicators/ninigi/

could be made to work at 60V with an appropriate resistor.

Update: Based on this discussion:

https://www.eevblog.com/forum/projects/lowest-voltage-neon-tube/

the 12V neon indicator may not really be a neon bulb but an LED. I would verify that it can work with an AC source. The product shot of this green "12VAC neon indicator" from the same company clearly shows a "DC" label on it:

https://ninigi.com/gb/en/product/ninigi/neon-indicators/ni-2fgr-12v/112494

[Gyro]

The inverse parallel diode wastes power (light output). putting a small diode bridge rectifier before the LED doubles the light output for a given current and avoids any motion flicker. Alternatively you can get 'AC' LEDs which contain a pair of inverse parallel (same colour) dies. They avoid the need for an external diode(s) altogether.                     

[liteyear]

The circuit is fine. Heed what PCB.Wiz said.

A neon bulb is the right tool for the job and saves on some calcs, some power and some components, but the LED is easier to tune and is more familiar to many these days.

The challenge with mariush's circuit is that that cap gets subjected to mains voltage. There will be an enormous in-rush, and if it fails it shorts your supply. Nothing that can't be managed - every rectifier is a variant of this - but it wouldn't be the first option I'd go playing with if you only want mW of power.

For the LED circuit, consider the role of the R and C. Together they limit current. At PCB.Wiz says, 5mA will probably be plenty. So you need R = 55/0.005 = 11kΩ. At 5mA, power in 11kΩ is 0.005*0.005*11000 = 275mW. So you could use a 1W resistor and call it a day.

But it's a bit wasteful. So suppose you added C = 0.33µF. Zc = 1 / (2*pi*50*0.33µ) = 9600Ω. Drop your R to 4.7k and you get total Z = sqrt(9600*9600+4700*4700) = 10.7kΩ. Similar sort of current but now the power in the resistor is just 0.005*0.005*4700 = 120mW, and a half watt resistor will be fine.

Also note you can replace the 1N4007 with another LED if you want to simplify you BOM. Now your 5mA is making light 100% of the time instead of 50% and just making stuff hot the other 50%.

[Gyro]

The circuit is fine. Heed what PCB.Wiz said.
...

Also note you can replace the 1N4007 with another LED if you want to simplify you BOM. Now your 5mA is making light 100% of the time instead of 50% and just making stuff hot the other 50%.

Huh? isn't that exactly what I said? - Light the LED on both half cycles. Assuming the OP is wanting a panel mount LED rather than SMD(s) shining through a plastic case (a fair bet considering the 20VA transformer 1N4007 etc). A dual die LED makes perfect sense, as does substituting the 1N4007 with 4 x 1N4148s - they won't want two panel LEDs.

[themadhippy]

. There will be an enormous in-rush,

So the current limiting resistor is magic as it only limits current to the led not the capacitor?

[liteyear]

Huh? isn't that exactly what I said?

1. No.
2. Your post arrived while writing mine.

[liteyear]

. There will be an enormous in-rush,

So the current limiting resistor is magic as it only limits current to the led not the capacitor?

Hmm, did you not follow mariush's link perhaps? The current limiting resistor is not in series with the capacitor.

[UnijunctionTransistor]

You might consider just using a neon lamp - a lot simpler and safer:

https://www.digikey.com/en/blog/dont-rule-out-the-neon-bulb-as-an-ac-power-on-indicator

I also believe this is the best suggestion.
And neon lamps also have this beautiful diffuse glow.

[gbaddeley]

Hi Watth,
In your first post, the secondary LED circuit is a simple approach. The capacitor is not required. Try R=100k. Adjust value if led is too bright or not bright enough.