LED driver

I might design a box that has a 5000 volt power supply inside. I was thinking that I could have a warning LED (or two independent ones, for redundancy) that indicate that HV is present. It could be inside, shining through an overlay on the top cover, so it's visible with the cover on or off.

So, how do I drive an LED pretty bright, from 5KV down to the officially safe level of 48 volts?

Clearly a resistor won't work very well, with a 100:1 voltage range. Even 1 mA at 5 KV is 5 watts dissipation.

One could charge a cap through a biggish resistor and, when it hits maybe 30 volts, discharge it into the LED and get a flash. Rate would be about proportional to voltage. But it's still just as efficient as a simple resistor:

2.5 volts of LED drop from a 5KV supply is 0.05% efficient no matter how you time slice it.

So I need a more efficient 5KV-to-LED circuit (or 48V-to-LED), some sort of switcher. It needs to be low power dissipation, small, and preferably made from standard surface-mount parts. Maybe charge that cap to 30-ish volts through a resistor, but dump it into the LED with some efficiency, to make a bright flash, Still messy.

I've used depletion fets into LEDs as cap-charge indicators. They are cool, constant-current, constant rate discharge into the LED all the way down to about

5 volts, where the LED winks out. But a depletion fet string good for 5KV would get ugly, and efficiency is still 0.05%.

I could use a neon relaxation oscillator, but I don't trust neons, and they wouldn't work at 48 volts.

Maybe something with a lithium battery or supercap in it. Ditto messy.

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John Larkin                  Highland Technology Inc 
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Wasn't something like this addressed before in a similar thread... yep...

Subject: Re: Self powered High voltage indicator Date: Sat, 08 Mar 2014 14:13:59 -0700 Message-ID: References: ...Jim Thompson

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| James E.Thompson                                 |    mens     | 
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Jim Thompson

Clean slipped my mind; I must think about a thousand circuits per year, 100 or so in the last week. And the LED HV indicator will be an *easy* part of the upcoming project.

That referenced thread still didn't produce a neat solution.

I'm thinking that energy storage, supercap or battery, is the only reasonable way to go. If I start with a resistor from HV to the blinker, it needs to be about 20M to keep the power dissipation down, and then things get messy at 48V. A separate power supply for the LED lets me use huge resistors in the divider.

A constant-current semiconductor stack is just too much.

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John Larkin                  Highland Technology Inc 
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John Larkin

LOL- but it was good enough for that delusional dumb-ass camel-dung smoker from north Africa... good one.

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bloggs.fredbloggs.fred

Here you go...

No expensive parts, or risky voltage divisions.

However... I see no way to have ON:OFF at 0.5s:0.5s, except with extremely low LED currents, given your power budget, or with more expensive parts... coming next ;-)

In my linked example LED current is ~2mA for ~3ms, which you should easily see.

But denigrated by John Larkin. ...Jim Thompson

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| James E.Thompson                                 |    mens     | 
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Jim Thompson

R5 dissipates 6 watts at 3KV, 17 watts at 5KV.

Make R5 more reasonable, 25M maybe, and the blink period goes way down at 48 volts, ballpark 20 seconds, which wouldn't do as a safety warning. It may not even work at 48 volts. Bottom line, a resistive driver, 5KV to 2.5 volt LED, is very inefficient here.

The circuit could be simplified, too, if it worked. That's a heap of discrete parts.

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John Larkin                  Highland Technology Inc 
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And it's not enough to just put a light bulb on the low voltage side because...?!?

If safety is what you're after, lights burn out regardless. For that you need crowbars and interlocks. Hinges, levers, bits of metal. That sort of thing.

Tim

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Tim Williams

Stored energy in HV capacitors.

Hence using two independent indicators. And LEDs last a long time.

For that you

Eventually someone will take the cover off, and no microswitch is going to discharge 5KV on the caps. I guess some radical mechanical design could dischage the caps. PITA.

But when we work on it, with the cover off, we'd have to defeat the mechanical interlock, so we wouldn't know if things were hot. An LED would be nice.

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John Larkin                  Highland Technology Inc 
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John Larkin

Apparently this is 5 kVdc, since for 5 kVac, the obvious thing would be to use a voltage transformer.

Why do you insist on LEDs ? For highish voltages and low current, the obvious solution would be to use some form of gas discharge tubes, such as neon lamps. If you do not think one is not reliable enough, use two independent systems as you suggested.

With a gas discharge the current will be low and hence series resistor resistance very high and resistor dissipation low. Check the series resistor voltage handling, since you may need to put a few resistors in series to get a 5 kV total rating.

If you are going to shine through some isolation material, why bother with ELV limits (and besides the IEC ELV limit is at 60 Vdc).

And of course with any high voltage power supplies with some smoothing capacitors, make sure that you have some bleeder resistors to get away of the high DC voltage within a few seconds after the input power has been disconnected.

Reply to
upsidedown

You should use bleeder resistors across the capacitor anyway. Since the voltage is so high, you may need to use several resistor in series for sufficient voltage handling capacity. Connect the neon lamp across the lowest resistor (possibly with lamp specific series resistor). For extra safety, use two independent bleeder resistor string in parallel, each with their own lamp.

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upsidedown

The problem with neons is to make them light up from 5KV to 48 volts.

Customer considers 48 DC to be safe. That's way conservative.

Fast bleeders might dissipate a lot of power. And I want a visual indication of HV, cover on or off, primary power present or not. I sort of like my employees.

The problem may be fundamentally unsolvable without a battery or supercap somewhere. If I use a 25M resistor to limit the dissipation to 1 watt at 5KV, then when the voltage is 48, the maximum power transfer theorem gives me 24 uW, about 1 uA average, to work with. Even at 100% conversion efficiency into the LED, that's not much light. A 1% blink duty cycle could make a 10 ms blink, 2.4 mW, once a second, which might be OK, but the circuit to do that is a challenge.

There are multi-chip LEDs around, which could help some. A 24 volt LED would be about right.

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John Larkin                  Highland Technology Inc 
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John Larkin

Can you just hook the LED to the HV PS (current limited of course) and use fiber optics to bring the indicator to the front panel.

Think of it as the isolation used in the "Kill-A-Watt" meter, i.e. eyeball isolation.

Reply to
miso

It will take a long time for it to bleed from 5K to 48 volts. And I can't dump many watts into the bleeder.

Connect the neon lamp across

The neon would be off at 48 volts. And I may have to program the HV supplies from 5K down to to maybe 500 volts.

So, I want an LED that's on nicely from 5KV to around 48 volts, with primary power on or off. Not too much to ask.

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John Larkin                  Highland Technology Inc 
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John Larkin

Current limited how?

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John Larkin                  Highland Technology Inc 
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John Larkin

Modified to R5 = 22Meg...

No expensive parts, or risky voltage divisions.

A little watt-seconds analysis will show that a 1 Watt power limit will restrain flash rate... so something has to give.

Perhaps an auxiliary supply? Or something that only burns power with the enclosure open? ...Jim Thompson

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| James E.Thompson                                 |    mens     | 
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"John Larkin" wrote in message news: snipped-for-privacy@4ax.com...

I've never worked on equipment that dared to indicate this. And that's just the smaller models, in the single-megawatt range.

Checking and verifying a zero-energy state is the second step, after PPE, of any company's OSHA required safety procedure. If you're not doing that, you're stupid. If your customers aren't doing that, they're stupid. Glancing at a light will only encourage unsafe practice.

There is no safe way to work on 5kV. The clearance distance is something like 4 feet or more -- that's clearance of personnel away from the equipment. I don't think they allow heavy rubber gloves under those conditions, you need a Jesus Stick and that's it.

I don't know how the rules cover power supplies with maybe-less-than-lethal DC current capacity, like TVs. But stored energy is stored energy, whether it's glass-aquadag or huge electrolytics.

Tim

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Tim Williams

Did you see these two sketches from a few months back?

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Should do what you want at 500uA from 48V to 5kV (or 1mA if two independent redundant led drivers). The weird looking cc flasher I actually breadboarded and have used elsewhere but not the current limit chain.

piglet

Reply to
piglet

It would seem to me that a good safe design would have an automatic crowbar that would snuff the HV anytime mains are off ??

So, for test purposes, the mains must be on, so a self-powered indicator isn't necessary, just provide an auxiliary supply. ...Jim Thompson

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| James E.Thompson                                 |    mens     | 
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Jim Thompson

In actuality, then, crowbar plus blinking LED inside the box anytime mains are on >:-} ...Jim Thompson

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| James E.Thompson                                 |    mens     | 
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Let the LED flash? If you have 50msec on time and 500msec off time the current consumption drops to 1/10th. That would cut the bleeder dissipation to around 10%. I found that even way less than 50msec is very visible, sometimes nmore so than a constantly lit LED. But that depends on whether the safety standards and you customer allow it.

What you'd need in that case is a not so high powered bleeder, a zener, a cap and some blinky-blink circuit. One that deliberately quits when the zener current goes away and this could be set at 48V or wherever you want it to stop flashing.

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Joerg

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