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I am assembling a power supply for xenon arc lamps. I use a simple welding inverter as a stable current source. I use the presented scheme.

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The ignition circuit generates pulses up to 25-30 kV for a short period of time. Large currents flow in this circuit, so the use of various inductive filters increases the mass of the device. The use of diodes will be accompanied by thermal losses, so I am interested in a simple and reliable EMI filter.

To protect a powerful source, I used a suppressor 1.5KE100A, but during the first power-up, the source burned out. The operational amplifier in the current feedback circuit (in welding machine) was damaged.

Consequently, the suppressor did not perform a protective function. What could this be related to? What other ways are there to protect the output of a powerful power supply?

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  • \$\begingroup\$ As an aside: would it be feasible to capacitively couple the HV arc ignition pulse, as is done in xenon strobes, e.g., by wrapping a wire around the tube, or with conductive paint? I might simplify circuitry and protection, but I'm not sure if that might stress the (expensive) tube envelope, though. Does the manufacturer's spec suggest that? \$\endgroup\$
    – DrMoishe Pippik
    Jul 23 at 6:24
  • \$\begingroup\$ The gas in the bulb is under very high pressure and the glass of the lamp is very thick. So external ignition will not work in my case. \$\endgroup\$
    – Typeof undefined
    Jul 23 at 7:22
  • \$\begingroup\$ The issue might be that the 1.5KE100A suppressor is not made to handle repeated high-power pulses. Yes, high-current and high-voltage components are bulky, but may be needed to keep the ignition pulse out of the supply. I'd add a series inductor between lamp terminal 2 and supply +, and a 0.01µF, 15 kV capacitor from lamp terminal 2 to ground. However, that capacitor is likely to be expensive (~US$60) and bulk, so simulate before you actually test. \$\endgroup\$
    – DrMoishe Pippik
    Jul 23 at 18:26
  • \$\begingroup\$ The inductor, I believe, should be rated for a current of more than 110 amperes, which will lead to an increase in the dimensions and weight of the product. I have a capacitor of about 6 kilovolts 1 microfarad. I'll try to add it to my scheme. \$\endgroup\$
    – Typeof undefined
    Jul 25 at 7:37

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