Self powered High voltage indicator

Mea culpa then. I thought the spec was 5 - 25 kV, low current draw, no aux power supply and no recollection of AC or DC. So how did I do against that spec? :-)

The "spec" was to "indicate" the presence of voltage, 50V to 3kV.

Anything Larkin conjures up is purely his own private wet dream. ...Jim Thompson

No, that's what Habib said he needed. In the post that started the thread. And that's what his first circuit does, DC plus indication.

Your oscillator thing does neither.

That's simple. But it won't make enough light to be useful. You can't average more than 33 uA into the LED without a proper switcher, and a perfect switcher can't deliver more than 400 uW. That's where you *should* be doing math.

Actually, it would be better to dump the 1.5M straight into the LED and eliminate the hairball oscillator, but that doesn't provide the aux DC.

An indicator should be visible in room light. Yours won't be.

And Habib needs the DC to run some low power circuits. He said so at the top of the thread.

Drip. Drip. Drip. Your mental capacity >:-} ...Jim Thompson

Umm, where does the cap get its energy from? Have you invented the self-charging capacitor?

What are you some kinda 'tard? It slow charges thru the 1.5M until it breaks the UJT threshold, the UJT turns on the 2ma CS to discharge the 1uF through the LED + CS, from its + to its - terminals. Sheesh.

Your ignorance is abysmal. ...Jim Thompson

Actually, limp, limp, limp is a more apt description >:-} ...Jim Thompson

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You'd be more credible if you occasionally posted stuff that works.

You have blundered a lot lately.

That doesn't increase the average current through the LED, which is still under

And, at 3KV, it burns 6 watts to make about that same useless number of photons.

It's just wrong.

Flail away, but it's wrong. It can't make enough light to be a dependable indicator.

You're like a leftist... you'd like the world to think that your lies are true. Too bad that the world has some brains and can think for themselves. ...Jim Thompson

How are you going to get a decent LED indication from 33 microamps?

You're not.

You couldn't even manage to read the circuit description and observe the LED current in the simulation?? And you have the gall to call _me_ senile? You prove that senility does not require old age >:-} ...Jim Thompson

You are talking past each other as per usual, too busy exchanging insults.

The led is flashing, a C charges at 33uA and discharges periodically into the led at much higher current.

The issue is, whether the led flashes are likely to be sufficiently noticeable for reliable use as an indicator (a hazard indicator, it sounds like). If they are too short they will be too dim due to the response time of the eye. If they are too infrequent they can be missed.

My guess is that it will sort of work but indeed be a bit dim. I have a led on a USB:serial converter like that, you have to shield it from ambient light to see the dim short flashes from the data.

Right. 2 mA for 3 ms every 1.5 seconds. Nobody is going to notice that in room light.

Jim's circuit does nothing to increase the average current available through the resistor. In fact it decreases it.

A single 1.5M resistor connected directly to the LED would be more visible than his circuit, but still marginal in room light. With a really good LED, 1 mA average current might be OK, and arguably making that, say, 5 mA at 0.2 duty cycle would be more visible.

Some sort of blocking oscillator or other switcher could better use the available voltage, but the numbers are still bad. 1.5M dissipates 6 watts at

Do the math.

Both are the case with Jim's circuit.

New, improved version on the way... including aux supply >:-} ...Jim Thompson

Cool. Should be fun.