I was asked to design a circuit capable of converting 50 VDC to a short 70 kV 1 to 2 ms pulse.

I did some digging with different kinds of circuits like Cockcroft-Walton multiplier. I know there are some more experienced designers here, so I am kindly asking for advice of different kinds of approaches to this problem.

 

Into what load?

 

Into what load?

Into specific gas mix to create a spark with energy of 1.1 Wh. Some kind of dark magic I am not familiar with (chemistry I mean)

 

Automotive spark plugs actually fit inside those parameters very well. An ignition system might be a shortcut. It would only need 13.8V, though. Here’s a plot of a conventional ignition system voltage over time with ms divisions.

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Into specific gas mix to create a spark with energy of 1.1 Wh. Some kind of dark magic I am not familiar with (chemistry I mean)

As a ballpark calculation, to get 1.1 Wh from a 70 kV rectangular pulse lasting 2 ms, the current would have to be in excess of 50 A. Since the pulse probably isn't going to be anywhere close to rectangular, you're probably talking peak currents well above 100 A. Are those the kinds of currents you are expecting? I believe typical automative spark currents are on the order of 100 mA, give or take.

 

To express the energy of such a short spark in Wh would seem a strange choice of units – Joules would be much more logical.

 

Into specific gas mix to create a spark with energy of 1.1 Wh

Only by this way you can sent to mix of gases energy 1.1 Wh (4000 J):



ADDED:

High Voltage Power Supplies - High Voltage Power Supplies -- SL150kV
Description: Spellman’s SL150kV rack mount high voltage power supply
is designed for scientific and industrial OEM applications requiring 150kV at 1200 watts
in a compact cable connected standard sized rack.
Models are available in positive, negative or reversible polarity.
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Using capacitor below, set gap to 62 kV breakdown for 4000 J energy.

(CFO) SX210E20
Aerovox high energy quick discharge capacitor. 2.1 µF, 100kv.
Energy - 10,605 joules. Equivalent Series Inductance 50 nH max.
11" x 14" x 27" high. New, never used. Some rust due to moisture.

 

E = 1/2 CV^2

C = 2E / V^2

C = 8000 / 70000 / 70000 = 1.6e-6

So yes, charging a 1.6uF cap to 70000V has the desired energy.