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I have a power supply that can supply around -60 kV DC and 10 mA. This power supply provides energy to a device known as a fusor. A fusor utilizes a low current and a high voltage to ionize a gas and accelerate the plasma ions toward the center of the vacuum chamber.

However, my issue is that the fusor will pull a very high current (100 mA) for around 30 seconds to ionize the gas. Afterwards, it settles back to 10 mA. Unfortunately, my power supply can only supply up to 10 mA and cannot supply that 100 mA draw during the first few seconds.

To combat this issue, I was considering using an inductor. Would this work? Also, what would the specs of such an inductor be?

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  • \$\begingroup\$ You need to ask yourself if you need the fusor to have that 100ma of current. If you do, you're thinking wrong-- you need a capacitor to store the charge you need (probably a very big one!) \$\endgroup\$
    – Scott Seidman
    Apr 24, 2019 at 15:50
  • \$\begingroup\$ What is the design of the 60 kV supply? The HV supplies I've worked with if you try to pull too much current, the output voltage just drops without any damage being done (at least for short overloads). \$\endgroup\$
    – The Photon
    Apr 24, 2019 at 15:51
  • \$\begingroup\$ Greetings, it automatically draws 100 mA, I don't want it to take 100 mA, I want it to take 10 mA. Also, the power supply is a 60 kV 600 W electrostatic precipitator from EBay (CX-600A). \$\endgroup\$
    – t1r3d
    Apr 24, 2019 at 15:53
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    \$\begingroup\$ So, again, what is the design of your supply? Are you sure it isn't self-limiting? Any thing you do that limits the current to the load will have to reduce the voltage delivered to the load to do that. If the supply has that behavior already, you don't need any additional circuit. \$\endgroup\$
    – The Photon
    Apr 24, 2019 at 16:01
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    \$\begingroup\$ Your comment seems contradictory to me - "No, I am expecting a significant drop, the 60 kV is all that I need." \$\endgroup\$
    – HandyHowie
    Apr 25, 2019 at 7:01

1 Answer 1

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The physics lab manager wanted to build a CO2 laser. In short, he succeeded; we tested it while wearing special protect-the-eyes goggles. The laser was powered by 2,000 volt 0.1 amp CONSTANT CURRENT supply. Cooling water flowed, removing about 95% of the energy.

He asked my advice for how to control the power(current), and I suggested TRIACS, with pulse transformers to turn on the TRIACS at appropriate timing (phases) of the incoming 60Hz power. Thus his control system, to "regulate" the current, had 120 samples per second "bandwidth".

The laser worked; producing about 10 watts in 1cm*2 beam. Would set the standard lab screen-wire-and-asbestos pads ON FIRE in about 5 seconds.

What size output capacitor was used? I do not recall. But using the maths

dV/dT = I/C, and rearranging to get C = I / (dV/dT) or C = I * dT /dV

we learn, for 10 volts ripple, 1/120 seconds, at 0.1 amp

C = 0.1 amp * 0.00833 seconds / 10 volts

or

C = 80 microFarads, with 5,000 volt rating??

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  • \$\begingroup\$ Greetings, in my case, I am dealing with DC power. \$\endgroup\$
    – t1r3d
    Apr 24, 2019 at 16:16
  • \$\begingroup\$ Then you need a 100,000 volt-rated triode vacuum tube, as your series current controller. Varian-EIMAC may produce those. \$\endgroup\$
    – analogsystemsrf
    Apr 24, 2019 at 16:18

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