Looking for an indicator light for 60-330V

I have a large 330V cap bank and I would like to add some kind of indicator light that shows when it is charged.

Ideally the indicator would:

• Not rely on external power.
• Be visibly lit from 60-330V
• Not dissipate excessive power at 330V

My best idea is a neon lamp with an appropriate resistor. However with a current of 0.6mA at 330V it would dissipate 200mW, and at 75V I'm not sure it would be visible.

Another option would be some kind of comparator powered by a coin cell, but it will eventually die, and create a safety hazard when it does.

Is there a better option?

• "Visible" is a highly subjective term. You'll need to define it much more precisely. Commented May 23, 2023 at 18:19
• @Finbarr I'm just looking for general approaches here. Ideally the indicator would not change in brightness between 60 and 330V. Less change = better. For most solutions I imagine the brightness can be linearly adjusted for reasonable values of "visibility".
– Drew
Commented May 23, 2023 at 18:40
• @user253751 5400uF, but it will be continuously topped off by a boost converter. Higher draw will waste batteries, but otherwise not cause any problems.
– Drew
Commented May 24, 2023 at 18:11
• Another option would be some kind of comparator powered by a coin cell, but it will eventually die, and create a safety hazard when it does. I solved a similar problem by a normally-on depletion mode FET as a crowbar, so that it immediately discharges the capacitor bank if control voltage is lost. Commented May 25, 2023 at 9:40
• What level of power is "too much" for you? From your question it seems that you consider 200mW too much for your application. What's the max wasted power you would like to achieve? Commented May 25, 2023 at 14:18

This absurd circuit uses a DIAC to create a pulsed current through the LED. This will be visible starting from around 50 V and consumes very little current.

C1 will be charged up to the DIAC breakdown voltage and discharged fast through the LED. A bigger capacitor produces lower flash frequencies but brighter flashes and my need a current limit resistor in series with the LED. The frequency rises with the voltage.

The shown values worked for me.

simulate this circuit – Schematic created using CircuitLab

• Interesting, thanks
– Drew
Commented May 24, 2023 at 2:52
• A pretty nice functional equivalent for the neon bulbs. Commented May 24, 2023 at 16:21
• I tried this with a neon bulb and instead of the LED and DIAC and it works fairly well. Using a 4.7Meg resistor and 200nF cap it strikes at ~75V and blinks at maybe 2hz. At 330V the blink rate is just at the edge of visible. Overall the indicator is fairly visible at all voltages, and it only dissipates 23mW. This is a good solution.
– Drew
Commented May 25, 2023 at 20:33
• The cap increases the visibility massively vs just the neon bulb alone.
– Drew
Commented May 25, 2023 at 20:34
• @Drew Clever! Nice to hear that :)
– Jens
Commented May 25, 2023 at 20:40

For years we used a Neon lamp (bulb) and resistor for that job. They turn on at 60 to 100V depending on which type. Turn off at a lower voltage. (limit the current with resistor) See https://en.wikipedia.org/wiki/Neon_lamp

• I like this. A neon lamp feels like high voltage and the brightness as "scale" for the voltage is maybe a nice touch too. Commented May 24, 2023 at 14:16
• Also, if it is supposed to stay lit as the capacitor discharges, the 60V lower limit for turn-off will be fine. Commented May 25, 2023 at 1:31

I suggest a high voltage MOSFET, made into a constant current sink like such:

It keeps the LED at a low voltage. Current is set by R1. Essentially all the power is dissipated in M1 (~1 W). It will work with whatever LED or NPN transister you bring to the table.

The current will be essentially stable from a very low supply voltage, all the way up to the breakdown of the MOSFET:

• With your switching current sink, doesn't it need a connection from ground, via a diode to the node between M1 and L1, so when the transistor turns of, the voltage in that node does not become negative infinity Commented May 24, 2023 at 14:37
• "The current will be essentially stable from a very low supply voltage" seems to be a detriment in this scenario. OP wants it to turn off at 60V. Commented May 24, 2023 at 14:43
• @BenVoigt I guess they are a bit ambiguous. They say it must be visible at 60..300 V. They don't disclose if it should be on or off below 60 V. Commented May 24, 2023 at 16:05
• @In any case this will draw at least ~70mW@60V and up to ~500mW@330V, definitely too much according on what the OP requires in his question. Commented May 25, 2023 at 14:56

You can try using a low-power AC(220AC)-DC(12/5VDC) board. Some of them can operate from 70V-390V DC. Consumption without load <0.05W. So you can connect one or two LEDs to the output in series. It is quite possible to achieve a consumption of 100-500mW by desoldering the no-load resistor and adjusting the converter feedback circuit. Basically, the board already has an LED indicator, so if their datasheet says that the no-load consumption is less than 0.05 watts, this is the best solution, even if the actual consumption is several times higher.

High-voltage buck regulator

The most efficient solution will likely be a dedicated buck regulator, especially since you don't need isolation. There are regulators that can handle high voltage inputs directly, such as the AL17050 or the MP9488:

VIN can be from 7.5 to 450 VDC which covers your full range, VOUT can be low enough for a single LED, and the efficiency should be at least 60%.

• Wow, I didn't realize there were reasonably priced switching regulators with such a large input voltage range. This is a great solution. Cheap, simple, and efficient.
– Drew
Commented May 25, 2023 at 19:58
• I bought the parts. I'll test this and the other solutions and report back.
– Drew
Commented May 25, 2023 at 21:07

The simplest and most versatile solution is probably a constant current circuit like the following. From 80 V to 300 V the LED current changes from about 3 to 5 mA. Power dissipation in Q1 will be about 1.5 watts maximum, and you would need to select a transistor that can handle the 300 V maximum.

simulate this circuit – Schematic created using CircuitLab

Here is another idea, which provides a substantial LED current over the range of voltages.

• OP asked for "not excessive power" and seems to think 200 mW is already a lot. Any linear solution will of course have excessive power on 300 V.
– pipe
Commented May 25, 2023 at 7:52
• Yeah, this dissipates far too much power. A neon bulb is superior.
– Drew
Commented May 25, 2023 at 21:00

You don't say what's the maximum power you may tolerate the indicator to draw. You just seem to imply that 200mW are too much.

So I want to discuss some general considerations in this respect that may help you choose between all the other solutions.

First consider that among common technologies the most low-power indicators are LEDs. You can turn on an high efficiency LED with 1mA and it will be quite bright. So it's difficult to go lower than that level of current.

For example, here is a high efficiency 5mm green LED datasheet.

It has a typical luminosity of 3000mcd@2mA:

which is 1500mA@1mA given the linear relationship between current and luminosity for LEDs, which is quite bright:

This means that any solution that draws 1mA continuosly needs to drop at least ~60V (let's neglect that ~2V of the LED). This implies at least 60mW of wasted power to turn the LED on (and up to 330mW when your caps are fully charged).

If you want less power draw you must either use some kind of DC/DC converter or power the LED with pulses, so that the average current will be lower.

Note that in this latter case you should use a visually blinking LED, so a ~1Hz frequency or less, otherwise you won't perceive the LED as blinking but you'd only see a dimmer LED.