Electronic – P-Channel Depletion Mode MOSFET for Negative Pulses

mosfetpulse

Would a p-channel depletion mode MOSFET make the most sense if the goal is to produce a high-voltage negative square pulse?

My reasoning:

  1. MOSFET, because we're talking about a kV square pulse with fast transition times,
  2. p-channel, because that puts the source at/near ground, so the gate can be ground referenced (and not floating on -1000 V, as would be the case in an n-channel circuit),
  3. depletion mode, because the load resistor must always be on the drain side, which would be -1000 V, so we'd like the MOSFET to be "on" most of the time to keep the output at 0 V, and then turn "off" briefly to take us to -1000 V and back.

The problem is that "p-channel depletion mode" MOSFETs seem to be scarce, which leads me to assume either I'm doing something wrong, or my application is unusual. Is my analysis of these types of MOSFETs incorrect, or are there really just not many manufacturers? Or is there a much better way of doing this that I'm missing?

(More details…) I need to produce a negative pulse with the following specifications:

  • amplitude: -1000 V (ref to gnd)
  • width: 100 microseconds
  • period: 20 Hz
  • risetime (leading edge, 10% to 90%): 1-2 microseconds or less

I'm just switching the potential difference (ref to gnd) on a chunk of stainless steel metal between 0 V and -1000 V quickly. I'm not driving a motor, or some huge inductive/capacitive load.

The circuit I'm thinking of is this:

enter image description here

Best Answer

The first problem with this approach is that having the transistor ON most of the time is VERY inefficient; you are dropping 1000V across Rl most of the time. Much better to turn the transistor ON for very short pulses. It'll shrink your 1000V supply needs for a start. (This means interchanging R and FET)

You could use an N type to pull down, with some isolation (opto, or pulse transformer) to its gate drive.

If R-C is too long to meet your rise time spec, you can use a totem pole (another FET in place ofthe resistor)

EDIT :

I just plugged a wild guess of 100pf for your lump of stainless, plus a relatively small interconnection into your stated requirements of swinging 1000V in 1 us.

Check these numbers but here's what I got.

100pf 1000V in 1 us
Q=CV = 100nC
I= Q/T = 100ma
P wasted in RL = IV = 100W.

This only considers the relatively fast slew when you turn on the FET. Turnoff will be slower... R = V/I = 10 kilohms RC = 1 us so the fall time to 10% will be about 2us, maybe acceptable.