Re: High Voltage Circuit Question

On Wed, 24 Sep 2003 02:18:33 GMT, "Peter" Gave us:

First off, you should NEVER post a binary into a text ONLY usenet group.

Ok.... on to the supply section.

I see a multiplier after the transformer. That means the raw output of the transformer is 3400 volts divided by two, PLUS the diode drops.

The bottom resistor under the 100 Meg Ohm resistor is your metering take off. The meter is in the wrong place from what I can see. It should be across the bottom resistor, and the lower leg of that resistor should ALWAAY be tied to ground.

What you should get with a ten megohm DVM is likely one volt per kV. The HV 100M resistor, and the "bottom" resistor under it act as a voltage probe for safe voltage readings. The line under that resistor should always be grounded or it "float high". This is a very bad thing as it can damage your meter. The other little RC circuit. The pot appears to be the calibration resistor for the metering tap (probe point). So, tie those two "meter" nodes together, and move the meter up such that it is across the bottom, low value resistor in the voltage divider there. One node is the junction to the 100M, and the other ground, or right there at the bottom of that lower resistor. Do NOT leave that line as an open line (between the meter nodes). That is the only thing that appears to be wrong.

You need a real HV probe so that you can not only determine what it does put out, you can calibrate the metering stage of the circuit as well.

The low volt cap, and two diodes either got diagramed in wrong, or appear to me to have no function. The 10k resistor in series with the output is very likely a 1/2 watt carbon composition resistor, which is used in HV circuits as "arc suppression" resistors. They can also limit current.

Future diagram posts should be made to a binary group and referred to here, such as alt.binaries.schematics.electronic OR NOT here.

I hope this helps.

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It helped me out a bit.

I do use a high voltage meter and when I measure the top I get about 1800 volts instead of 3400 volts.

So that's where I'm confused because if you take the anode of D61 I should get a p-p of 3400 volts, then the cathode of D61 should be a DC 3400 volts (minus the diode drop)?

This is where my confusion comes in, I measured one that works and got something like 800 volts on the anode of C61 putting the scope probe on the high voltage probe. So my guess is, those two 3kv caps act as a doubler circuit but not sure how the bottom one C64 works. I"m use to seeing doubler, tripler and so on circuits in a diagonal pattern.

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On Wed, 24 Sep 2003 03:16:26 GMT, "Peter" Gave us:

Sounds like a diode is out. Mouser has HV diodes. 1/4 A or a 1/2 A

4kV version should work. Your diode drop may be higher as the number of element in the stack for a given vendor's diode differs. They are merely a stack of PN junctions in one package.

Well, the output lead would be where the DC should be read. There is no storage post multiplier so it will probably be quite high in ripple, not appearing as DC on scope (HV Probe of course), unless one notes that the scope is on AC or DC coupling.

You should measure the output of the xformer, and if it is half the expected voltage, yet not double that at the output, one of the diodes is likely bad, unless the current is high. That could indicate that one of the caps is bad. Ther are on the hairy edge of their design limits.

How it is drawn matters not, if the nodes are connected correctly in the diagram. Do a google on Cocroft-Walton multipliers, and you will see many different diagrams, but the hookups are the same, as is the action.

I would first suspect a diode. You should verify the secondary voltage on that unit or compare it with a working unit. The voltages on various nodes should match, or closely so, if the Xformer output is equal.

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First the disclaimers:

1) Do not be deceived by my posting to this engineering newsgroup: I'm not an electrical engineer (although I my Technikon registration is pending). I am, however, a former electronics technician. 2) I shall freely round numbers as convenience dictates.

Now the safety warning: Ought you be probing around this power supply?

Regarding the power supply: The power supply employs a "full-wave voltage doubler." I have personally verified that the BOOK (yes, book: not web page) "Handbook for Electronics Engineering Technicians", by Daufman/Seidman, covers such power supplies (as well as many other topics), but I haven't verified the depth of coverage. Still, it might profit you to consult the book (or your favorite text) should I leave any of your questions unanswered.

The secondary of the transformer is approximately 1,200 VAC. D61 & C63 form an extremely ordinary half wave rectifier/filter, producing approximately 1,700 VDC. D62 & C64 form an identical sort of pedestrian half-wave rectifier/filter, producing an additional 1,700 VDC. C63 & C64 are in series producing 3,400 VDC (1,700 + 1,700).

N.B. Either this power supply was designed to produce trifling (although still possibly unhealthy) amounts of current, or you're overlooking some capacitors somewhere.

Keeping in mind that I'm not actually an engineer (yet), and that its late (here), I shall venture the following:

- If D61 or D62 is open, the output will be almost naught (assuming C63 & C64 are low leakage capacitors), or at least certainly reduced.

- If C63 or C64 is open, the output will be a bit more than half the normal value, with quite a bit of ripple.

The two 1N4001 diodes & the 3 volt capacitor serve absolutely no purpose whatever: either they are detritus from some previous use the power supply was designed for, or (more likely) your drawing contains an error.

BTW, you never said what the power supply is used for (I'm guessing a physics/biology/chemistry lab), and if their was anything wrong with the power supply?

Good luck, and keep one hand in your pocket!

Cordially, Richard Kanarek

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Richard Kanarek

On Wed, 24 Sep 2003 08:02:55 GMT, Richard Kanarek Gave us:

There is only ONE multiplier, not two. Only ONE multiplication occurs. Two multiplier diodes are required for a single multiplication.

They are not "rectifier/filters" either. It is a Cocroft-Walton single stage multiplier. It performs rectification as a function of its operation. Without output storage caps, the ripple is fully conducted, and quite high. Likely over a hundred volts.

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