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I'm wonder if I can power an LED from a 330V DC main power supply without loss of power in Joule effect in the resistors. (This is a challenge).

Suppose we want 30mA and 5V in the LED and the reverse Zener voltage is 12V.

schematic

simulate this circuit – Schematic created using CircuitLab

This method is horrible because we have at least 9W of power lost in R2.

I don't want to use converter (flyback or other converter.)

Can we use an NPN transistor to flow only 30mA using h gain?
I don't have any other method that comes to mind.

Any ideas?

JRE
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kodi
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  • Switching DC/DC converters come to a mind. But surely this circuit does not exist in a vacuum, is it? It is a part of some other application? Please provide some context. – Eugene Sh. Feb 11 '22 at 15:26
  • Yes I kwow but I don't want use it, (it's a challenge) – kodi Feb 11 '22 at 15:29
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    Then it is not possible, as any linear method that takes 30mA from 330V will take 9.9W. – Justme Feb 11 '22 at 15:33
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    Regardless of using a resistor or a transistor you will still dissipate the same amount of power. No getting around Ohm's Law here using that method. I agree that a switching converter is your best bet if you must use this 330V source. – jwh20 Feb 11 '22 at 15:36

2 Answers2

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I'll propose a low-tech solution:

Upgrade the LED to a model that produces a decent amount of light with about 500µA. That will probably mean a green high-visibility LED like this one. user1850479 recommends this 48V LED which should be 48x brighter with the same current!

Use a 660 KOhm resistor of suitable voltage rating (ie, not a 0805 SMD!)

This will use about 0.15W. Still not down to the level of a good switcher, but most likely cheaper.

bobflux
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  • indeed. 30mA is a lot in a LED. There are plenty that will light on 1mA and probably less. If you want brighter, use a few of them in series. simple. – danmcb Feb 11 '22 at 16:18
  • thank you for your answer, in my assumption I wanted to take a 30mA led, but you are right, it's better to use your solution. – kodi Feb 11 '22 at 16:23
  • 2020's LEDs are not the same as 1980's LEDs, with the one I linked, at 30mA you'll need sunglasses. If you use superbright 5mm LEDs with a tight beam lens, it's even more ridiculous, I've run them at 100µA for indoor use. But the tight lens means they're only bright when you're in front of them of course. – bobflux Feb 11 '22 at 16:26
  • @kodi Even better would be to use higher voltage LED modules. If you used this 48V LED, which is made of ~16 diode junctions in series, you need 16 times less current for the same brightness while dropping less voltage across the resistor: https://www.digikey.com/en/products/detail/lumileds/L130-27800CHV00001/5877557 – user1850479 Feb 11 '22 at 16:46
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A transistor used as a variable resistor isn’t special just for being a transistor: it’ll just get hot. The gain just tells you the ratio of the collector and base currents.

You will absolutely need a converter of some sort, but it doesn’t imply a switching mosfet and an inductor. You can be switching capacitance. You can be shining many series-connected LEDs without a dropper resistor onto a solar panel to power that one LED - it’ll surely waste way less than 9W! And so on. It doesn’t mean that the approach would be practical for mass production or even for a marketable product of course.

If you can get by with lighting many LEDs instead of just one, then you can connect a couple hundred LEDs in series and a much smaller dropper resistor wasting much less power. By setting the current correctly, the LEDs will together emit as much light as one LED would, and no more. In fact, for this sort of an application, you’d really want a small current source driving the LED string, like LM334 - since the current will be minuscule with that many LEDs, as they’ll operate at say 0.2-0.5% of the current the single LED would operate at. I guess 50-100uA is all you’d need.

If this is for a product then a switcher will be a solid solution, but not necessarily a switcher you can just copy from an application section in a data sheet. A self-resonant transformer converter, set up as a tank, with a transistor oscillator at one end of the winding, would probably do the job just fine. It can be very cheap in quantity in spite of the custom transformer - it’ll be a very low power high frequency transformer, after all. You can wind one yourself for a pot core, using a split bobbin for isolation. It can be reasonably efficient too.

But I think that perhaps you’re over complicating this: there are self-contained standard industrial indicators that can take such inputs directly, and include the necessary power conversion inside.

Kuba hasn't forgotten Monica
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  • Ok so if I want power Led or op amp from 330v, self-contained standard exists and that can take high voltage ? Do you have reference in mind? – kodi Feb 11 '22 at 15:51