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I have a DC high voltage (>270V) high current (>40A) circuit which needs to power a standard LED along the way to a heater. The LED is to indicate whether or not the heater is activated.

I'm struggling with dropping down the power to light the LED, whilst losing as little power as possible and maintaining an acceptable level of luminosity. Unfortunately I don't believe a potential divider will work, since the voltage source is variable (due to heater requirements) from around 30V to 300V.

What I need to be able is to have a stable low voltage signal powering a standard LED which shows whether there is voltage, with constant luminosity no matter the amount of voltage. Hope this makes sense.

Ideas I have looked at so far, without much success, are potentionmeters (power will be too high) and non-contact solutions (seem to only work with AC). A logic gate might work, but I wouldn't know how to incorporate a high power gate into this circuit.

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  • \$\begingroup\$ AC or DC? If AC then what frequency? "I'm struggling with dropping down the power to light the LED whilst losing as little voltage as possible." No, you want to drop as much voltage as required to limit the current to the LED. You mean 'I want to waste as little power as possible while maintaining a (reasonably) constant luminosity across the supply voltage range'. If this is just an indicator then you can tolerate a wide change in current and it can still be bright. Hit the edit link under your question ... \$\endgroup\$ – Transistor Sep 9 '19 at 15:18
  • \$\begingroup\$ So sorry should of mentioned DC \$\endgroup\$ – Matthew Clisby Sep 9 '19 at 15:19
  • \$\begingroup\$ And yes your correct apologies \$\endgroup\$ – Matthew Clisby Sep 9 '19 at 15:19
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    \$\begingroup\$ Sounds like the LED would just light up out of fear if you just hold it near that circuit. \$\endgroup\$ – pipe Sep 9 '19 at 16:02
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    \$\begingroup\$ See the answer with the DIAC in electronics.stackexchange.com/questions/430950/…. It flashes with a rate in proportion to the voltage but it is very low power. \$\endgroup\$ – Transistor Sep 9 '19 at 16:09
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Since you mention that you have a separate 28V rail available, you could use an opto-isolator with the HV rail as input, as shown in the schematic below.

schematic

simulate this circuit – Schematic created using CircuitLab

The opto-isolator must be chosen to be capable of the 28V output level. Please make sure that you size the resistors to achieve the necessary current on the input LED of the optocoupler, this current will be noted in the datasheet.

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  • \$\begingroup\$ Ill try this out and get back to you if it woks (or not) \$\endgroup\$ – Matthew Clisby Sep 13 '19 at 7:35
  • \$\begingroup\$ I'm a little confused about the opto-coupler. I can't seem to find one where the input can survive the high current of the HV rail. If you could point me in the correct direction that would be grand. the highest on RS is a 3A \$\endgroup\$ – Matthew Clisby Sep 13 '19 at 8:24
  • \$\begingroup\$ The optocoupler is wired in parallel, so the high current wil only go through the heater. The current into the coupler will be limited by the 100kΩ resistor, and should be enough to turn on the LED in the optocoupler, which is usually a few milliamps. \$\endgroup\$ – Elmardus Sep 13 '19 at 8:52
  • \$\begingroup\$ Perfect thanks. \$\endgroup\$ – Matthew Clisby Sep 13 '19 at 8:58
  • \$\begingroup\$ This page may provide some more information about the use of optocouplers: electronics-tutorials.ws/blog/optocoupler.html For your use, a transistor-output optocoupler is suitable, you can use a common type like the CNY17-3. One further note, from your questions it seems you have little experience with circuits like this. A 270V DC rail is seriously dangerous and can kill you easily. Let somebody experienced check your setup, these voltages and currents are nothing to mess around with. \$\endgroup\$ – Elmardus Sep 13 '19 at 9:05
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You can use opto-isolators to control low-voltage DC from high-voltage DC. Wikipedia says you can use these with massive voltage differences:

Commercially available opto-isolators withstand input-to-output voltages up to 10 kV.

This would require you to have another power rail at a lower voltage, though. Perhaps using a voltage regulator attached to the high voltage rail. This would depend on the application though.

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  • \$\begingroup\$ There is a 28V rail that can be used, it is currently being used to control other devices in the circuit. Opto-isolators seem like they work quite well except I cant seem to find any that can survive the high voltage and high current \$\endgroup\$ – Matthew Clisby Sep 10 '19 at 8:27
  • \$\begingroup\$ The opto-islotor should not conduct the main heater current. Instead, connect the input side of the isolator in parallel to the 270V rail, with a series resistor to limit the current in the isolator input. Then, use the output of the isolator to switch the 28V to the LED. If the isolator output is not capable of conducting the indicator LED current directly, you can use an extra transistor on the output to increase its current handling capability. \$\endgroup\$ – Elmardus Sep 12 '19 at 10:11
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An indicator LED does not draw a lot of current -- maybe 30 mA or so.

If you use a simple active regulator, like an emitter follower with a voltage reference, you will have to draw that 30 mA from the 300V source, so the circuit will have to dissipate about 9W. That's a lot for an LED, but not a lot compared to your heater, so maybe that will do for you. You can certainly get power transistors that can handle it in combination with appropriate filtering components on the collector.

If you want to draw a lot less power, you will need some kind of switching step-down circuit like a buck converter -- that's a lot of of expense for an indicator LED.

The common solution is to avoid this problem altogether by powering the LED from the heater's control circuit, which will have its own lower voltage supply.

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If you can have an independent power supply for the the indicator LED then you can use a Hall effect sensor to sense the high current flow and switch on the LED.

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I'm going with dumb low-tech solutions:

  • A "superbright" 5mm LED at 1mA will be quite visible in interior lighting.

Add a 270k resistor in series, or better two 150k resistors to spread the voltage. The resistors will dissipate about 1/4 Watt. That's tolerable.

  • Stick an analog voltmeter on your 270VDC line...

Use your 28VDC supply to light it with LEDs. It has the extra feature of indicating the voltage!

  • "Universal input 100-240V AC" switching power supplies can usually run on DC.

In the low power ones, mains goes through a filter and then a rectifier, so it makes no difference whether the input is AC or DC. 240V AC has a peak voltage of 336V. Thus a switching supply rated for 240VAC will tolerate 270V DC just fine. The input cap is probably rated for 400V.

Thus, a simple 5V switching supply followed by a LED and a resistor would probably work. In fact, a cellphone charger would also probably work. There will be a delay at turn off though, as the input capacitors discharge, the LED will stay ON for a few seconds after power is cut.

DO NOT use a non-switching power supply, of course. Transformers don't handle DC.

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