LED driver

Sadly, blackberries and poison oak are soul mates.

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John Larkin                  Highland Technology Inc 
www.highlandtechnology.com   jlarkin at highlandtechnology dot com    
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John Larkin
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Here's my take. There is nothing wrong with neon bulbs. They can have lifetimes of a million hours. They will outlast the equipment they are installed in.

Use them in a relaxation oscillator as an indicator. Two for redundancy. The brief flash is far more effective in getting attention than a continuous light. With a watt to work with, you can get a really bright flash. The real safety is a mechanical short across the high voltage. This could be a plunger that is pushed down by the lid and activates a lever. The lever drops a shorting bar between the 5KV and ground. The shorting bar is kept away from the contacts when the lid is closed and screwed down.

The lever can be assisted by a weak spring in case someone tries to defeat it by turning the box upside down. When the lid is opened, the shorting bar drops and removes the high voltage. Presumably a series resistor limits the current to avoid damaging the HV cap.

If the box is not opened, the two neon indicators will discharge the high voltage down to a safe voltage. The remainder can be a high value resistance to bring the voltage to zero. Or just forget about it. The shorting bar will discharge the cap if the lid is opened.

The procedure is as follows:

  1. Verify the neon bulbs are blinking.
  2. Turn power off.
  3. Unscrew the lid. Lift it slightly, perhaps 1/4".
  4. Verify the shorting lever has moved. You should hear a thud.
  5. Verify the neon bulbs have stopped blinking.
  6. If these conditions are met, open the lid.
  7. Place a grounding clip on the high voltage to keep it shorted to ground.

The neon bulbs are just used as indicators. They verify the high voltage is present before you turn off the power, and they give a positive indication that the shorting bar has discharged the high voltage cap.

Failure Mechanisms

  1. One neon bulb oscillator can fail.

- The other will still give the positive indication that high voltage was present and was removed when the lid was opened.

  1. Both neon oscillators can fail.

- The shorting bar will still remove the high voltage if the lid is opened.

- The power on procedure should include a check for blinking lights when power is applied.

  1. The linkage can jam.

- The neon bulbs will indicate high voltage is still present.

- Activate the plunger to try to clear the jam.

- If not successful, verify power is removed and let the unit sit until the neon bulbs stop blinking.

  1. High voltage circuit can fail.

- Neither bulb will blink when power is applied.

- With no high voltage, the system is safe.

Other Indications

  1. When high voltage is present, the circuit that uses it should operate normally.
  2. When high voltage is removed, the circuit that uses it should quit operating.

In this manner, there are three indications that high voltage was present, and that it is removed when the lid is opened. The key point is they give a positive indication that the status has changed.

I think it is a fairly safe assumption that if the shorting bar can reduce the high voltage from 5KV down to the level that the neon bulbs quit oscillating, the shorting bar will most likely reduce the voltage to zero. Thus there is no need to check for the presence of 48V on the high voltage cap.

Reply to
Steve Wilson

Depends on current supply from the 5KV source.

We have 60VDC HI-POT equipment, the operator sits in front of it during the test at arm reach to the control panel. But those are no more than 10 mA output. Still enough to tickle some one of course.

That equipment uses mechanical crow bar shorting switchings. They do have a 10 Watt 500M HV resistor for the meter and current sense.

Jamie

Reply to
Maynard A. Philbrook Jr.

We use 50K ohm 10 Watt wire would through the shorting switch. This works fine if you have a over current sense that is designed to kick the

5KV off at that time. I don't know how many Joules you are dumping but our larger machines uses 200 WATT resistors that plug into FRS 30 amp fuse holders with air crow bar switches.

Jamie

Reply to
Maynard A. Philbrook Jr.

I always try to emulate you, oh great master >:-}

On a more practical note, as others have noted, your approach is just plain dumb... AND unsafe. ...Jim Thompson

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| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
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Reply to
Jim Thompson

You are putting yourself in tight spot.

A storage cap with a blocking diode won't keep the LED on long enough as the voltage is decaying down at the critical point?

Jamie

Reply to
Maynard A. Philbrook Jr.

You don't necessarily have to power the LED from the 5 kV. How about a

1W-class LED shining on a solar cell?

Cheers

Phil Hobbs

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Dr Philip C D Hobbs 
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Reply to
Phil Hobbs

we use these for IR/DCR shorting switchings. Good for 5K and 30 amps, just R;

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Jamie

Reply to
Maynard A. Philbrook Jr.

What's dumb, or unsafe, about lighting up a couple of LEDs when voltage is present? Come on, don't give up, try to get a circuit to work.

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John Larkin                  Highland Technology Inc 
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Reply to
John Larkin

Is the 5kV the only power source?

You want a buck (or something) that works form 50V to 5kV?

George H.

Reply to
George Herold

No, I'm just playing with an interesting circuit challenge.

I did suggest a lithium battery or a supercap to power the LED when HV is present but primary power isn't.

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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

The primary power (IF we do the job) is 24 VDC. We'll need a HV supply, in the

80 watt ballpark. I may be able to use 2500 volts, instead of 5K, at which point powering an LED off the HV supply gets a lot easier.

I would like the LEDs to be on whenever HV is present, whether the 24 is up or not. It might take minutes for a bleeder to pull the supply down near 48 volts.

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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

Lemme see, getting a bit crude..a 5 meg resistor from supply to LED (to ground) means only a few volts for the LED, way under that stupid

48V bit. The (eg: white) LED will be bright enough to see in nominal illumination and that light can illuminate a CdSe or Si detector for an isolator type setup to generate more current in a separate circuit.
Reply to
Robert Baer

Your primary objective should be to get away with any dangerous voltages and only as a secondary issue how to indicate it.

To discharge a capacitor to 1 % of original voltage (e.g. from 5 kV to

50 V) you need about 5 RC time constants. Thus, in order to get below 48 V in 5 seconds the RC should be 1 s or to drop to that voltage in one minute a 12 s RC time constant is required.

With 1 uF capacitor, the bleeder resistor should be 1 Mohm resp. 12 Mohm and the steady state dissipation at 5 kV would be 25 W resp. 2 W.

If that is to much for steady state power dissipation, at least connect the bleeders when the AC power is lost, when the DC plug is disconnected from the power supply socket or when the power supply door is opened. In these cases, use RC time constants below 1 s.

If 1 uF storage capacity is too little, you should look at the mains ripple frequency and ripple amplitude. A half wave rectified mains will have 50/60 Hz ripple, an ordinary 6 pulse three phase rectifier will have only a few percent ripple at 300/360 Hz without _any_ storage capacitor at all and of course 40 kHz switcher, the storage capacitor requirement is minimal.

I have used neon lamps in tube transmitter and audio amplifier anode voltages and after switching off the power supply it takes several seconds of dimming, which is a sober reminder that the charge is actually discharging and one should wait.

Even after the lamp goes out or in your ideal case goes out at exactly

48.0 V, there are still some energy in the capacitor. You should still calculate, if there is still sufficient charge/energy to harm a people. While working with tube gear, I used a screwdriver to short the capacitor poles after the indicator was out.
Reply to
upsidedown

So you are building some CRT horizontal output class power supply, if you cant use more than 1 W to bleed of the capacitor.

Reply to
upsidedown

Those are three phase feed systems with 6 pulse rectifiers (or these days even 12 pulse for better PF), which seldom even needs any storage capacitors on the DC side, due to the low ripple voltage.

Apparently the OP is talking about a small (below 100 W) since he could only allocate below 1 W for any bleeder resistors, thus most likely a single phase power supply with some capacitor, unless some SMPS is used. Thus the steady state current at 5 kV would be only 20 mA, which just might be lethal. A partially discharged capacitor should no longer be capable of delivering truly dangerous currents for any significant times, but of course, these shocks are quite unpleasant, but at least you will live and perhaps the next time you are a bit more careful :-).

Reply to
upsidedown

So what is the problem ?

Assuming you have a 50:50 duty cycle H-bridge on the 24 V side, why do you need a big storage capacitor on the HV side ?

The storage capacitor x bleeder resistor RC time constant can be very short and hence the bleed current can be very low and you can drop the voltage below 48 V in hundreds of milliseconds.

I would suggest a fixed bleeder resistor (chain) across the capacitor and a separate neon lamp with resistor(s) to 5 kV to indicate presence of steady state 5 kV. My guess is that decay would be very fast, so the decay would be hard to detect anyway, so why create circuits with tens of failure prone components that are needed for a second or so after switch-off ?

In a proper switching mode power supply, if the storage capacitor is needed for only tens of microseconds, why would that capacitor be energized for minutes ?

The problem is of course different with 50/60 Hz iron core transformers, but it does not apply in this case.

Reply to
upsidedown

Your approach is unsafe at least for mains frequency power supplies, in which large storage capacitors are needed. The first thing is that you have a fail safe bleeder system and only after that it is time to think about indicator systems.

Now that it appears to be a high frequency DC/DC converter, for what do you need a big storage capacitor in a first place, so there is not much problem of properly bleeding it off ?

Reply to
upsidedown

This thread is just one sad example of trying to find some solution to a non-existing or at least easily avoidable problem :-).

The OP should supply sufficient background information so that anyone else would be able to give some educated guesses, what to do.

I understand that people are under various NDAs and must be very careful what can be said in the Usenet post.

I do not understand, why the OP could not have stated the power level (80 W) instead of everyone guessing something between 1 mW and 1 MW.

Specifying the input power source, e.g. 50/60 Hz single phase, three phase 50/60 Hz or DC-SMPS would have given a lot of information about the required circuit (if any).

In this case, the OP could have revealed these initial conditions in the first post, since the requirements are not that extraordinary, so that anyone could _not_ guess who the end customer is or what they are doing and hence break the NDA.

Since the OP apparently work as an consultant and hence must know how hard it is to get relevant information from the end customer, in practice by pulling out one finger nail at a time (figuratively speaking :-), I would expect that the OP would provide all the information from start that he has (without breaking the NDAs).

Reply to
upsidedown

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