2
\$\begingroup\$

I need a DC 4.2Kv 300ma power supply, and am looking for advice.

I'm currently using a the guts of a microwave oven (transformer/diode/capacitor) to power a magnetron for a physics experiment. The transformer outputs 2.1kv into a diode/capacitor voltage doubler to produces a 50% duty cycle 4.2KV potential at 60hz. The resulting signal from the magnetron is noisier than an LA freeway at rush hour.

According to this paper changing the unit to a regulated HV power supply, controlling the filament heater separately, and reducing the filament current after it reaches operating conditions should allow it to phase lock onto an external signal and operate as a low noise amplifier.

For the HV side, my first instinct was to buy a commercial supply. I priced them, and after regaining consciousness thought I should ask here.

Question: Is there a way to smooth out the power from my existing supply or another inexpensive alternative I should consider?

Thanks in advance.

(Amateur: yes. Kid/teenager: no. Experience in electronics: Kit-building. Follows safety procedures/common sense: yes...Safety advice is always appreciated.)

Added additional details on safety and end use to comments.

\$\endgroup\$
13
  • 2
    \$\begingroup\$ Safety question: Are you experimenting under some sort of professional supervision? Otherwise it sounds risky. At least let professional to examine your setup. \$\endgroup\$
    – Eugene Sh.
    Sep 4, 2015 at 18:23
  • 2
    \$\begingroup\$ To reiterate @Eugene: Microwave magnetrons are incredibly, mind-bogglingly dangerous, insofar as I've heard. In fact, there's a saying I've heard: "Don't attempt any project that involves a magnetron". If anything goes wrong, you could easily end up with clouded and boiled corneas and all manner of disturbing injuries. Oh, and the magnetron itself contains easily exposed highly toxic beyrillium. Please do ensure that everything is checked by a professional before you attempt this, as Eugene says. \$\endgroup\$
    – 0xDBFB7
    Sep 4, 2015 at 19:09
  • 1
    \$\begingroup\$ I am always fearful when I see a question like this. People can cause so much damage with a magnetron if they don't know what they're doing. I was relieved to see the answer that you posted here. Your knowledge and attitude are reassuring :) One example I remember was a guy who had cut a 4" hole in the front and back of his microwave oven so he and his dad could dry lumber by passing it through... After doing this for a while, he realized it might be dangerous and asked about it online!! \$\endgroup\$
    – bitsmack
    Sep 4, 2015 at 20:23
  • \$\begingroup\$ Hmm, I had a look at the paper you posted on DropBox and after reading the introduction my bull-shit-o-meter starts to go crazy. Perhaps it is all legit but then again to transmit power the noise figure is unlikely to matter that much, the filament being 'back heated' if even possible is unlikely to have much effect. A whole genre of these 'wonderful' inventions relies on them being difficult to realise while people gaze at the designs with starry eyes. Even if you could bean ALL the power to just one 'collector' You would get a max of 750W of RF energy that will cook anything in the path... \$\endgroup\$
    – KalleMP
    Sep 4, 2015 at 20:51
  • 1
    \$\begingroup\$ ... also note the list of references used, just one by some other author. I would be interested if the next issue of this periodical included a retraction or at least some interesting comments from other readers. The Beryllium danger is probably over hyped by those that don't understand it. The ceramic insulating parts may be made of beryllium oxide and fine dust liberated by grinding these can be dangerous, the HV stored in one of those caps may kill you a lot faster. Take care. \$\endgroup\$
    – KalleMP
    Sep 4, 2015 at 20:56

1 Answer 1

1
\$\begingroup\$

The simplest solution to this problem is to replace the existing "Villard Circuit" pulse-type voltage doubler with a voltage doubler circuit that outputs solid DC. This modification is significantly less expensive than a new power supply. It requires only two appropriately rated diodes and two capacitors.

Two safety related modifications are also suggested. First, a small bleeder resistor should be connected across each capacitor to drain away charge when the device is not in use. This will reduce (but not eliminate) the possibility of an accidental discharge.

Second, a fuse or circuit breaker should be installed if not already in place.

Finally, to round out the updated power supply, consider adding a cooling fan and voltage/current meters.

Special thanks to @kalleMp for inspiring this answer. I am in their debt.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ Could you diagram the difference between a "Villard" doubler and whatever you found that you consider a "solid-DC-output" doubler? \$\endgroup\$
    – feetwet
    Jun 11, 2016 at 21:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.