High Power Variable Potential Divider

Question

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.

Answer 1

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.

^{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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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.