High Voltage power supply regulation

I'm working on a high voltage power supply of about 9KV. I'm using a transformer that gets me up to about 1KV and then stages of capacitor diode voltage doublers.

I'd like to be able to do some kind of regulation of this supply, however just the cost and space for a high voltage (1 billion ohm) divider makes it a no go. This supply drives a capacitive load with no current flow (normally).

Any one of you super smart engineers have any ideas? Like a magic part that can sense field strength with no current flow? Wonder if I could get a mosfet to work with just an air gap from gate to the HV?

Thanks!

Rocky Lavine Rocky Test

Reply to
Rock
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Caddock will sell you a single-chunk high-voltage divider, and lots of people make high-voltage resistors. I've pushed a regular 0603 to destruction at about 1.6 kv, but I don't know what's realistically safe.

I can't see how any electrostatic thing could be DC accurate.

John

Reply to
John Larkin

Hello Rocky,

Why not add a simple rectifier (not cascading) to regulate the 1kV? That could be done with a bunch of 18M resistors which are standard issue (but not at 1%).

If that's still too much or you need non-adjust precision to less than

5% tap the transformer and create a smaller voltage just to regulate to.

Spooky...

Regards, Joerg

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Reply to
Joerg

What was the tolerance?

You could put a divider across the first or second stage. Probably get you with in 10%-20%.

Cheers

Reply to
Martin Riddle

I bought a couple of electrostatic voltmeters - they are just like in the textbooks, constructed much like an air spaced variable capacitor, but with very low friction bearings, and balanced. They aren't much good for feedback though, except manual feedback if you adjust the input power yourself. They still seem to have some more of those:

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If the power supply were only required to be stable for seconds or perhaps minutes then you might be able to make a capacitive divider, but any leakage resistance would cause it to drift.

If you make a grounded motor-driven rotating metal disc with a hole in it, and spin this disc between your high-voltage circuit and a small plate of metal connected to a virtual earth in an op-amp current-to-voltage circuit, then this ought to produce an AC output proportional to the electrostatic field. It's a pity about the rotating parts though. A hard drive motor should last a long time, but it uses a fair bit of power and makes a noise.

Inside the flyback transformer from a TV, there is usually a very high resistance potential divider for the focus electrode, I am trying to figure out how to use that to measure the EHT output on my high voltage supply, since I am using the flyback anyway so the divider is free, and it can handle the voltage. Maybe this would suit your needs too. The tap for the focus electrode itself is at too high a voltage to be much use to me unless I connect the wiper to a "low value" (1M?) resistor to ground, however I think this might increase the stress on the upper portion of the resistor string so I'd prefer to avoid that. If I can get at the bottom end of the resistive divider (separately from the return path of the windings in the flyback) and put a lowish resistor in series with that then I might have a solution.

Chris

Reply to
Chris Jones

On Mon, 01 Aug 2005 23:19:53 +0100, Chris Jones wroth:

That is called a "field mill". It's used often to measure the voltage of the atmosphere.

Jim

Reply to
jmeyer

** $4.20 in onesies from Mouser, rated at 15KV 1% is rather decent. Size: 0.340H x 1.580W x 0.1D. The 5KV rod-shaped resistors available go up to 5KV, cost $5.66 and are 0.880L x 0.165Dia.

However, if you regulate the 1KV signal, and the load is constant, then the 9KV would be regulated.

Sense with no current? It is called a capacitive voltage divider by some and a field strength meter by others. Fairly common. You can buy used HV meters that are (variable) capacitors. The voltage across their plates cause them to be attracted to each other, thereby moving the meter needle. That is how the DC ones work. There are also AC/DC meters that take no current. BUT. They all take *work* or power to move from zero to the indicated value. And need discharging when done.

Reply to
Robert Baer

I have the 4th edition. How about describing the section / chapter subject that includes that little gem?

Reply to
Robert Baer

If you have no objection to using thermionic devices and your HV supply can be arranged with the positive terminal grounded (two big IFs, I know), there is a description in "Radio Engineering" by Terman (Mc.Graw-Hill) of a method of using a valve with negative HV on the anode as a currentless voltage measuring device. [Page 149 in the Second Edition]

--
~ Adrian Tuddenham ~
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www.poppyrecords.co.uk
Reply to
Adrian Tuddenham

It's Sect 36 (in my 2nd edition) "Special Connections for Conventional Tubes", about five pages from the end of the chapter "Fundamental Properties of Vacuum Tubes".

Briefly the idea is to operate the valve with negative HV on its anode. That sets up a voltage gradient from anode to cathode which drives electrons towards the cathode. The level of positive voltage needed on the control grid to make grid current flow will be proportional to this gradient.

It seems to me that a triode with a top-cap anode would be the ideal valve for this sort of thing (line-output pentodes might work as long as the extra grids can be prevented from taking up voltage levels which distort the feld). Preferably the grid current should be sufficient to prevent cathode poisoning.

--
~ Adrian Tuddenham ~
(Remove the ".invalid"s and add ".co.uk" to reply)
www.poppyrecords.co.uk
Reply to
Adrian Tuddenham

I believe the resistance value you are trying to describe, assuming your vocabulary is United Statesean, is a gigaohm (G = 10^9). These are mfrd and sold.

How much cost and real estate are you willing to commit?

RL

Reply to
legg

Thanks; found it. Page 209; section 6-14, Fig 6-27 for the Fourth Edition.

Reply to
Robert Baer

I believe i gave costs and voltage ratings...

Reply to
Robert Baer

Right on the gigaohm, I'm using a FLUKE 1000 to 1, HV probe that is of that value.

This supply is jammed full now, and I'm trying to meet UL for it too so there needs to be a lot of space between parts on the business end of this thing. It's mostly surface mount too.

Cost wise, on the budget we can't afford any $2+ dollar resistors. If I regulate at the ~ 1KV point, we could prob. fit 6, size 1206 resistors to make a ~ 100 million ohm divider. The physical construction of the supply favors this as the higher voltage sections are farther from the switching circuits.

Rocky

Reply to
Rock

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