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Advice Required Re Kingshill HV Supply/Multiplier

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Ash Small

Thu Feb 24 2011, 12:02AM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

I've aquired a Kingshill Electronics board containing a CW multiplier, a transformer, and some other components. I've done some Googling, but can't find anything other than that they no longer appear to be in business (I think they were based in Gravesend, Kent, UK).

There are 24 diodes in the multiplier, and 62 X 8kV 4.7 nF capacitors, three on the lower stages, and two on the higher stages.

There are three leads, I assume power input, near the transformer, which I assume is around 500 Watt.

There is also a 'black box' at the output end, with a white HV lead coming from it.

Can anyone give me any advice, ie, what the black box is?, what input should I try?, what output can I expect?, should it be under oil? (there are no signs of oil on it), etc.

Thanks.

Proud Mary

Thu Feb 24 2011, 01:08AM

Registered Member #543 Joined: Tue Feb 20 2007, 04:26PM
Location: UK
Posts: 4992

Your black box might well be a filter to clean up switching noise in the output.

jpsmith123

Thu Feb 24 2011, 04:59AM

Registered Member #1321 Joined: Sat Feb 16 2008, 03:22AM
Location:
Posts: 843

Maybe the black box is a voltage divider network (or the very high resistance part of one)?

quicksilver

Thu Feb 24 2011, 08:27PM

Registered Member #1408 Joined: Fri Mar 21 2008, 03:49PM
Location: Oracle, AZ
Posts: 679

Do you know where did this originated from? Where or how was it used? It's application may give you further clues as to it's specifics, if it IS appropriate to use "as is" or whether it should be further insulated, etc.
Apparently the company itself is still in business:

Patrick

Thu Feb 24 2011, 08:36PM

Registered Member #2431 Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639

I dont know what that Black box is but its not the high voltage divider, that is clearly the blue planar HV R's paralled by the silver caps on that one side. the HV divider needs nothing else that would need to be where that box is.

it could possibly be a wave shape filter, or possibly a magic pixie fairies' home. Doing reverse EE by photo is like trying to decide a nuclear facility or cookie factory via spy satallite.

Ash Small

Fri Feb 25 2011, 08:32AM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

Thanks for your replies so far.

Quicksilver, the business directory you linked to is out of date. I've checked the Companies House online records, which show that Kingshill Electronics Products Ltd was dissolved on 19th December 2003. There is a company which advertises repairs to Kingshill equipment but I've not contacted them yet.

I know that the person I obtained it from obtained some surplus equipment from a university physics lab, so it may have come from there, that is my best guess as to it's history.

Patrick and Proud Mary, the output from the multiplier goes through the resistors (four paralleled pairs in series), then splits, feeding into the top of the resistive divider and the 'black box'. If this is a filter, how would it work?

The capacitors are marked 'MK 1'. Any idea what this signifies?

I've deduced that it was under oil, also that the two orange wires are the input to the transformer (it looks like they go to the voltage divider in the photo, but they don't) the red wire is a feed to a meter (from the resistive divider), the white wire comes from the bottom end of the multiplier/output of transformer (any ideas what purpose this serves?). The black wires also connect to an output from the transformer (Three output wires, does this mean the secondary is centre tapped?)

It appears that the maximum theoretical output is 96kV (24 stages with 8kV caps) and that the output is positive.

I will attempt to produce a full circuit diagram, however, I'm still puzzling over the secondary.

Any further advice, comments, corrections to the above assumptions will be welcomed.

Thanks.

Steve Conner

Fri Feb 25 2011, 09:35AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

I think the black box is some kind of surge limiting resistor, to stop it blowing diodes if the output arcs to ground.

All CWs have this, but the Glassman ones have it distributed through the stack, in the form of a resistor in series with each diode. Yours may have it in one big lump on the output. I think it's potted to increase the flashover distance: if the resistor ever flashed over that would defeat the purpose of it completely.

You can test this by measuring its resistance.

Alternatively it could be a series inductor as some others suggested, to clean switching noise. But the resistor chain there doesn't look hefty enough to be the surge limiter. Again you can test this by measuring its inductance.

It can't be a capacitor: where's its connection to ground?

Ash Small

Fri Feb 25 2011, 09:57AM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

Steve McConner wrote ...

I think the black box is some kind of surge limiting resistor, to stop it blowing diodes if the output arcs to ground.

All CWs have this, but the Glassman ones have it distributed through the stack, in the form of a resistor in series with each diode. Yours may have it in one big lump on the output. I think it's potted to increase the flashover distance: if the resistor ever flashed over that would defeat the purpose of it completely.

You can test this by measuring its resistance.

Alternatively it could be a series inductor as some others suggested, to clean switching noise. But the resistor chain there doesn't look hefty enough to be the surge limiter. Again you can test this by measuring its inductance.

It can't be a capacitor: where's its connection to ground?

It has a resistance of 500 kOhm (0.5 MOhm). The resistors between the multiplier and the black box have a total resistance of 250 kOhm.

Is it reasonable to deduce from this that output could be 125 milliamps or thereabouts? (96kV/750kOhm) or does this not follow?

Steve Conner

Fri Feb 25 2011, 10:02AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

No, that absolutely doesn't follow. That's the short circuit current, at which point the power supply's whole output would be dissipated in the surge limiting resistors, frying them in an instant. The actual available output current would be one-tenth or less.

A little known fact is that you can't draw arcs from these things. When pulling arcs, 50% or more of the output is dissipated in the limiting resistors, again frying them, because they're not rated for it. That's why the control electronics have a trip that turns the drive off for a second or two when it detects a ground arc. Glassman explain this on their tech site.

Besides, 96kV * 0.125A is 12kW, and you said yourself that the thing looks like it's rated for 500W. I'd guess the rating is 50kV, 10mA.

Ash Small

Fri Feb 25 2011, 10:21AM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

I'd realised the 12kW problem and was about to edit the above post, but thanks for your opinion, Steve.

Working on your assumption of 50 kV at 10 mA (500 W), that means the transformer output would be around 2kV at 250 mA (not taking into account losses in the multiplier).

Where can I find the equation required to work out the frequency that the transformer would need to operate at in order for the capacitors (2X4.7nF in upper stages and 3X4.7nF in lower stages) to take 250mA and output 10mA? (I hope this makes sense, the current passed by the caps will depend on frequency)

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