LED driver

what about this one?

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I guess if it's too heavy to lift...

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umop apisdn
Reply to
Jasen Betts
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John Larkin Wrote in message:

Why not use some photodiode optocouplers in series with high value resistor from the HV supply to trickle charge a capacitor on the secondary side. Then on the secondary side use an SCR made out of a couple of bipolars to dump the cap charge into the LED when a threshold is reached. Something like this:

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on the secondary.

Reply to
bitrex

Do you know how much HV capacitance will be present? How long will there be a hazardous charge after power down? Are we talking seconds, minutes, or hours?

piglet

Reply to
piglet

Or use a flasher chip on the secondary like this:

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It should sit there until the supply (cap) hits 1.35 volts, then dump the cap through the LED.

Reply to
bitrex

Apparently it is an 80 W 24 V/5000 V DC/DC converter. With any decent switching frequency, a new charge is delivered every 10-100 us, depending on frequency and waveform, so not much storage capacitance is needed.

With 16 mA load current and assuming worst case 100 us charge update rate and assuming 1 % peak-ro-peak ripple requirement, a 32 nF would be required. If 1 % (0,16 mA) can be allocated to bleeder and indicators that would represent a 31.2 Mohm resistance, with in combination of the 32 nF capacitance would represent a 1 s RC time constant. For he 5 kV voltage to drop to 48 V, about 5 RC time constants are required i.e. 5 seconds.

With higher switcher frequency and larger ripple limits larger than 1 %, it takes much less than a second to reach 48 V.

Reply to
upsidedown

I'll go you two better.

1) why not just wire those 48 volts worth of LEDs in series and combine their outputs optically--shazzzzam!

2) Or, more seriously, let's look at this quantitatively.

John wants

Reply to
dagmargoodboat

Ahh! So why not just do what Xerox copier and laser printer manufacturers do and have a big HV warning label and mandate waiting 60 seconds (or similar cover-your-backside time).

piglet

Reply to
piglet

Hi,

I guess basically a modified Fet would work for this, with the 48V to

5kV signal on the gate to enable the Fet to power an LED through a resistor. The Fet gate should fully turn on at 48V and then withstand breakdown of the gate until 5kV! :)

That won't work of course for normal Fet gates only operating to 20Volts before breakdown, but there is no reason why a Fet couldn't exist with a gate that breaks down at any voltage that a diode breaks down, Some Fet's may be packaged so that their substrate is heatsinked to the PCB, and this backside of the Fet could maybe be used as the HV gate such as in this picture:

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the HV would go on the bottom of the p-type substrate and act as a less sensitive channel trigger due to the larger depletion zone.

Also another idea would be to take a HV diode, and turn it into a JFET, with one more ohmic contact, and this JFET could have a HV gate, and act as a variable resistor up to the diode breakdown voltage, but that isn't as applicable I think. I wonder if there are any JFET's that can handle high gate voltages (or mosfets too).

cheers, Jamie

Reply to
Jamie M

snipped-for-privacy@yahoo.com wrote:

Doesn't even need a switcher. Have the power switch turn off the 5KV, but leave the control circuits powered.

The control circuit can monitor the decay of the high voltage and turn off the warning led when the voltage falls below whatever voltage is considered safe. Then turn power off to itself.

If for some reason the 5KV doesn't decay as expected, the control circuit can issue a warning and set a flag that prevents turning on the high voltage when the power switch is on. One possible failure is a blown bleed resistor. For this reason, two should be used for redundancy. This will double the decay time but that's ok as long as it is monitored.

We can estimate the decay time. JL mentioned the powr drain will be approximately 80 watts. This is a curent drain of

i = p / e = 80 / 5e3 = 16 mA

Assume a ripple of 0.1%. This is 5e3*0.001 = 5V.

Assume a switching frequency of 25 KHz. This is a period of 40us.

With a simple flyback converter, the high voltage cap will be

c = I * dt / dv = 16e-3 * 40e-6 / 5 = 0.128 uf

As you calculated, the bleed resistor is 25M. The time constant is

t = r c = 25e6 * 0.128e-6 = 3.2 seconds

Five time constants take 16 seconds. The high voltage will be

e^-5 * 5e3 = 33.69V

With a 120nf cap, this is non-lethal. A higher switching frequency would result in a smaller cap and faster decay.

BTW, notice the different meanings of 'e' in the above equation. Google and my calculator still get it correct.

To delay the operator from opening the lid immediately after power is turned off, simply use longer screws to secure the lid. Use the ones with metric thread to make them more difficult to find shorter replacements, and use a secure head to make it more difficult to cheat.

Before I came up with the above idea, I spent some time on a power multivibrator. Unlike most multivibrators that are content if they simply oscillate, this one is intended to deliver power to a load. It uses the classic NPN/PNP pseudo-scr as the latching element, with an additional npn to discharge the cap through the led. Any small signal npn and pnp can be used.

The oscillator produces around 1Hz at 31V, and 100Hz at 5KV. The switching threshold is set by the voltage divider R3/R4. The high frequency limit is set by R5. Current into the led is set by R7.

The analysis sweeps the supply voltage from 30V to 10KV. R5 is set to

3.9K to show that oscillations cease just below 10KV. I verified the operation on a virgin installation of LTspice.

My only question is if the led will be bright enough at 1Hz. Larkin ignored my question about brightness so I have little to go on.

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Here is the PLT file

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Reply to
Steve Wilson

Kilovolts to 2 volts (a regular LED) is inherently inefficient. "Lighting" multi-chip LEDs come in voltages from 6 to 24. A 24 volt LED is just what you suggest, a bunch of chips electrically in series and optically in parallel. Maybe 10x the light for a given current.

That's a string of HV mosfets. Given that, it's easy. 200 uA into a 5 or 10-chip LED will be fine, job done.

It may not be possible with a resistor as the current limiter. The tau of the resistor and storage cap may get unreasonable.

The blinker is non-trivial, too. I suppose I'll have to change the rules.

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John Larkin                  Highland Technology Inc 
www.highlandtechnology.com   jlarkin at highlandtechnology dot com    
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Reply to
John Larkin

Given a power supply to light up the LED, sensing the voltage is trivial. A big series resistor and a gate-protected mosfet would work. Or same series resistor and a high-beta NPN transistor. Or a voltage divider and a cmos comparator.

I wanted to get the LED power from the HV supply only, but that looks difficult over the range of 5KV to 48 volts.

--
John Larkin                  Highland Technology Inc 
www.highlandtechnology.com   jlarkin at highlandtechnology dot com    
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Reply to
John Larkin

I was thinking you could bleed through a string of opto-isolators (with photo-voltaic output.) And then add the currents electrically to power the blinker. (with some series/ parallel combo to get the right V and I.)

George H.

Reply to
George Herold

The LED drive is about 7 mA for 70 us, which will probably be pretty dim. That's why I like the multi-chip LEDs... a better power transfer match to the available voltage.

Your circuit does have a steady-state hang state, which I'd rather avoid. The pseudo-SCR blinker thing is nontrivial to get to work reliably.

There's just not much energy to work with here, given a resistive limiter.

--
John Larkin                  Highland Technology Inc 
www.highlandtechnology.com   jlarkin at highlandtechnology dot com    
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Reply to
John Larkin

PV couplers are crazy inefficient, and we don't need isolation.

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John Larkin                  Highland Technology Inc 
www.highlandtechnology.com   jlarkin at highlandtechnology dot com    
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Reply to
John Larkin

Where's the hang state?

I had no problems getting it to work. There is a broad tolerance for component variation. I did mention the effect of R5 that limits the high frequency. Since it fails well above 5KV with the components shown, there should be no problem.

The .trans analysis sweeps from 30 to 10KV and you can see it functions cleanly over that range. The applied voltage starts low where the unit fails, then sweeps to the high value where it quits again. If there was a hang state, it should occur somewhere within the operating range. It certainly functions over the required range of 48V to 5KV.

I measured about 130us where the current is above 1 mA. With one of your high intensity leds, it should be visible at 1 Hz. You can control the current into the led with R7.

You said it is nontrivial to get to work. Where have you seen this circuit before? I'd be interested in a reference. Or maybe you are confusing it with a different circuit.

Anyway, I much prefer the concept of turning the high voltage off and leaving the control circuit on to monitor the voltage decay before it shuts itself off. This gives an added safety factor to ensure the bleeder resistors are functioning, and you can shut down the high voltage if you detect a fault.

Then you can have any blink circuit you wish without being squeezed for power or needing a separate temporary power supply.

Reply to
Steve Wilson

That's not safe though. If the control power fuse blows, the HV cap is still loaded, and the warning LED is 'off'.

Of course all these situations are for the tech working on an open unit. A customer couldn't open the unit quickly enough to face any shock hazard.

All very sensible. A cap-multiplier could be used to lower the ripple tremendously, without increasing the bulk storage.

I prefer that solution. Might be able to cut the bulk storage by a factor of three or so, which bleeds us to 48V in 5 sec. as I calculate it.

I was suggesting pretty much the same. John likely doesn't know how much brightness he needs, hence no answer. But, I get 45uA average LED current for your gadget at 5kV/100Hz. That's visually the same as 45uA continuous, which is decidedly not going to be bright enough.

Here's a complementary oscillator I prefer. The regeneration is easier to control, and it's less prone to latch-up:

5Kv --- .---------------. | | R3 | .---.i1 | 2.2k | | |200uA | | | V | | | '---' | LED V ~~> | | --- +---------' | | +------------+-----. | | | | | R1 R2 |/ Q1 | | +-----47K---+---3.3k----| R6 | --- C3 | | |>. 2M --- 470nF --- C1 | | | | --- 2uF --- C2 '>| R5 | | | --- 470nF |----3.3k--+-----' | | /|Q2 | === | | | '--------------+ .---' Z1 | ^ 48v .-. | R4 | | | 220 '-' === | === 48V-5kV Blinker Blink rate = 2.5Hz iLED(avg) = 120uA

The 120uA average LED current is a bit better. Packing that into visually distinct pulses helps a great deal too.

(It's strictly hacked together to illustrate the point--caveat emptor.)

Substitute a transformer primary as the collector load, and the circuit can perform as a flyback converter. That would increase the LED current by ~15, which is starting to get darn reasonable.

Of course my circuit requires a magic component that yours doesn't--a

200uA 5kV current source.

Cheers, James Arthur

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Reply to
dagmargoodboat

Blinker is trivial; buy a ("self") blinking LED - one part.

Reply to
Robert Baer

200uA into 10 LEDs is enough light, if concentrated. That would be slick. But even then blinking is really a LOT better than not--far more visible under all conditions, and especially appropriate for a warning light.

What's the Iq?

Cheers, James Arthur

Reply to
dagmargoodboat

Just for reference, blinking a T1-3/4 Cree single-junction high-efficiency white LED,

10 mA, 1 ms on, 1 Hz looks pretty good; 10 uA average current 1 mA, 10 ms on, 1 Hz looks about the same

10 mA, 10 ms on, 1 Hz looks really bright

--
John Larkin         Highland Technology, Inc 

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Reply to
John Larkin

So the blinking LED is your solution then? The FQD2N100 is rated at 1kV and cost less than 50c. Maybe some of them in a beanstalk to get the

5000:48 ratio down a bit, plus a resistor?

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LEDs can be amazing. I just bought a light set for my mountain bike. The front light is so bright that when I walked away from my bike on a trail and turned around I thought a motorcycle was coming. When I set it to flash I had zero in car drivers who failed to see me when pulling into the road. Some even rolled back, maybe thinking this was a police bike.

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Regards, Joerg 

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Reply to
Joerg

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