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Simple 2000V Microwave Power Supply

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Aussienitro

Mon Sept 28 2009, 11:47AM

Registered Member #2399 Joined: Mon Sept 28 2009, 09:29AM
Location:
Posts: 27

Hi fellas, Simple question, searched but the specific schematic here evaded me.

See attached picture. I need a 2kv power supply to charge a 2.5kv 1uF capacitor.

Few questions,

1. Is the picture fundamentally correct?
2. Where can I get a bridge rectifier capable of handling 2000 to 5000 volts? If no BR's are available what value diodes should I source?
3.Have I ommited anything that should be implimented?
4. A simple 240v lamp between the power outet and the transformer will show me when capacitor is charged, correct?
5. How can I measure the voltage at the capacitor? (multimeters I have max at 1000v, what should I look for?)

Thankyou.

Dr. Dark Current

Mon Sept 28 2009, 12:03PM

Registered Member #152 Joined: Sun Feb 12 2006, 03:36PM
Location: Czech Rep.
Posts: 3384

With 2000V AC input, you'll get ca. 2800V DC on the cap.
You can make the bridge rectifier out ouf 1N4007 diodes, 5 per string, but for such a small cap, halfwave rectifier is enough, use 10x 1N4007's in series.
Otherwise its OK.

Bjørn

Mon Sept 28 2009, 12:45PM

Registered Member #27 Joined: Fri Feb 03 2006, 02:20AM
Location: Hyperborea
Posts: 2058

3.Have I ommited anything that should be implimented?

You need a resistor in parallel with the capacitor to make sure the capacitor does not stay charged for a significant amount of time after the charger is disconnected.

4. A simple 240v lamp between the power outet and the transformer will show me when capacitor is charged, correct?

It will show when it is charging, not if it is charged, a vital difference. Your "charged" detector must go in parallel with your capacitor.

5. How can I measure the voltage at the capacitor? (multimeters I have max at 1000v, what should I look for?)

Use a voltage divider

, remember that the multimeter also has a resistance...

Also read this:

Dr. Slack

Mon Sept 28 2009, 01:18PM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

The lamp between the outlet and the MOT will current limit the charging process, a good thing. It will also hint at the value of the charging currnet, so will dim at the end of charge. As a MOT takes such a huge magnetising current, a primary lamp will not go out when charging is complete.

Even though a primary lamp will indicate that charging has finished, it doesn't hurt to monitor the capacitor voltage as well. When making a voltage divider, remember that the maximum voltage rating for normal resistors may only be 200v, so you should string several together in series to share the voltage between them, as well as to increase the resistance.

Aussienitro

Mon Sept 28 2009, 02:25PM

Registered Member #2399 Joined: Mon Sept 28 2009, 09:29AM
Location:
Posts: 27

Thanks a lot fellas, understanding most so far, just trying to work out the voltage divider.

I measured the resistance of the multimeter which gave a value of 1.002MÎ©. So I need to make up a 9.0MÎ© resister placed in series on the positive lead running into meter and negative lead onto negative terminal of capacitor, thus dividing the voltage by 10 at meter. Is that correct?

Bjørn

Mon Sept 28 2009, 02:47PM

Registered Member #27 Joined: Fri Feb 03 2006, 02:20AM
Location: Hyperborea
Posts: 2058

Yes, that is correct.

klugesmith

Mon Sept 28 2009, 04:36PM

Registered Member #2099 Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1716

Aussienitro wrote ...

I measured the resistance of the multimeter which gave a value of 1.002MÎ©. So I need to make up a 9.0MÎ© resister placed in series on the positive lead running into meter and negative lead onto negative terminal of capacitor, thus dividing the voltage by 10 at meter. Is that correct?

I think 1 MÎ© is uncommonly low for the resistance of digital or analog multimeters on 1000 VDC range.
So recommend that you test your meter-with-divider by measuring a known voltage, without changing the range selection on meter.
Also, pay attention to the power dissipation and voltage limit of your scaling resistor.
With the numbers you gave, if the meter were pegged the external R would have 9 kV and 9 watts.

Note to other commenters: the MOT will "fully" charge the 1 uF cap in a fraction of a second, so there's not much difference between a "power is on" light and a "capacitor is charged" light. I sure hope OP intends to disconnect the MOT and rectifier before a pulsed capacitor discharge.

Aussienitro

Mon Sept 28 2009, 04:51PM

Registered Member #2399 Joined: Mon Sept 28 2009, 09:29AM
Location:
Posts: 27

Klugesmith wrote ...

Yep I'll be sure to give it a test beforehand, I'll attach the test I did. The cheaper one being measured was intended to be butchered and become integral part of the supply.

Klugesmith wrote ...
Also, pay attention to the power dissipation and voltage limit of your scaling resistor.With the numbers you gave, if the meter were pegged the external R would have 9 kV and 9 watts.

You've lost me here. What resistors would you recommend?

Aussienitro

Mon Sept 28 2009, 04:52PM

Registered Member #2399 Joined: Mon Sept 28 2009, 09:29AM
Location:
Posts: 27

Klugesmith wrote ...
I sure hope OP intends to disconnect the MOT and rectifier before a pulsed capacitor discharge.

Wow, nope wasn't intending too. Whats the reason for this? Though it would be easy enough to do at the relay.

GeordieBoy

Mon Sept 28 2009, 06:55PM

Registered Member #1232 Joined: Wed Jan 16 2008, 10:53PM
Location: Doon tha Toon!
Posts: 881

> Whats the reason for this?

Reverse voltage and/or RF from the discharge is likely to fry the rectifier.

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