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

  • \$\begingroup\$ making a tesla coil? \$\endgroup\$
    – vicatcu
    Feb 12, 2013 at 14:59
  • \$\begingroup\$ @vicatcu - switching the electric field near a group of atoms for a spectroscopy experiment. \$\endgroup\$
    – higgy
    Feb 12, 2013 at 15:22
  • \$\begingroup\$ Is the fall time (when probe returns to -1000V because the MOSFET is off) important? Same \$2 \mu s\$ requirement in both directions or something else? \$\endgroup\$
    – Phil Frost
    Feb 12, 2013 at 16:25
  • \$\begingroup\$ To clarify, the important transition time is when the probe goes from 0V to -1000V. The faster the better, but 1 microsecond is acceptable (while 10 is not). \$\endgroup\$
    – higgy
    Feb 12, 2013 at 16:36
  • 1
    \$\begingroup\$ You might want to consider an IGBT like an IRG4PH20K from international rectifier. It will hit the voltage and switching speed requirements. \$\endgroup\$ Feb 12, 2013 at 21:29

2 Answers 2


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)


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.

  • \$\begingroup\$ I agree it would be inefficient. The problem I see with any N-channel MOSFET is that I'd have to float the trigger pulse (to the gate) on the source (which for N type would be at -1kV, in this scenario). I assume that'd involve either a gate driver that is capable of floating on a -1kV, or a transformer or capacitor, which would ruin the flat pulse top for long pulse durations. \$\endgroup\$
    – higgy
    Feb 12, 2013 at 15:34
  • \$\begingroup\$ Agreed - as I said you would need some isolation. Optoisolation may be feasible without the drawbacks of a transformer. There are isolated gate drive solutions; I just think finding the right one is probably better than wasting 1000V power. If it turns out you need totem pole drive, you need to solve that problem anyway. \$\endgroup\$
    – user16324
    Feb 12, 2013 at 15:41
  • \$\begingroup\$ @higgy driving a high-side N-channel MOSFET isn't that hard. I have an H-bridge with a fairly detailed description which does exactly that. Or, as Brian says, an opto could work. What you are looking at building is essentially a higher voltage version of what I have there. \$\endgroup\$
    – Phil Frost
    Feb 12, 2013 at 16:22
  • \$\begingroup\$ How are you getting \$P_{RL} = 100W\$ without knowing the resistance of \$R_L\$? The highest power this resistor can dissipate is when the full \$1kV\$ is across it, but if we don't know the resistance, we can't calculate what that power will be. For an ohmic device like a resistor, \$P = E^2 / R\$. \$\endgroup\$
    – Phil Frost
    Feb 12, 2013 at 16:30
  • \$\begingroup\$ Sorry; 100W was a minimum value. Rl is determined at 10k by the 100ma required to meet 1us slew rate. Rl could indeed be lower if you wished! (and yes the edge driven by the resistor will be the slower edge) \$\endgroup\$
    – user16324
    Feb 12, 2013 at 16:43

I'm going to stretch my expertise a bit here and say, no, that is not the right approach. For a number of reasons:

  • I can't find any distributor selling such a device. N-channel is more common than P-channel, enhancement mode is more common than depletion mode, and 1kV is a higher voltage than most MOSFETs can handle. I see no reason such a device couldn't be fabricated, I'm just not sure anyone has bothered to do it.
  • If you need a fast transition from the high to the low voltage, the low-impedance path (the transistor) needs to be on the low side, not the high side.

This will be easier to implement if instead of thinking of it in terms of ground and -1kV, you think of it as ground and +1kV. If you want to stick the +1kV into a grounding rod so you get -1kV with respect to Earth, then by all means go ahead. As long as your power supply is galvanically isolated and the insulation in the transformer doesn't fail at those voltages, it won't make a bit of difference to the operation of the circuit. After all, the electricity can't see the labels on the schematic, so why should you care?

\$R_L\$ will have to be fairly large to avoid being vaporized by the high power resulting from the high voltage. \$ P = E^2/R \$, so if we want to do this with a \$\frac{1}{2}W\$ resistor:

\$ 0.5W = \dfrac{(1kV)^2}{R_L} \$

\$ R_L = \dfrac{1MV}{0.5W} = 2M\Omega\$ (absolute minimum)

This will put a bound on how fast the voltage can rise, since all the parasitic capacitances of the plate, transistor, etc, will resist a change in voltage. \$100 pF\$ is a reasonable guess, giving a time constant of

\$ \tau = 2M\Omega \cdot 100pF = 200\mu s\$

The 10% to 90% time will be somewhat longer, but our 100pF was just a guess anyway, so more complicated math isn't really justified. However, this does demonstrate why you need the transistor on the low side if you need that transition to be the fast one.

From there, it's just a matter of finding the right transistor. Mouser.com has some MOSFETs that could fit the bill. Some are a bit pricey, but would work. You will need a good gate driver to meet your transition time requirements. Power MOSFET Basics - International Rectifier is a good read.

You might also investigate IGBTs. I don't have much practical experience with them, but they are favored in higher voltage applications such as this. I think you can find a MOSFET that will work, but the IGBT might get you better performance at a lower cost.

Here's another thought that doesn't involve transistors at all: your metal plate is essentially a small capacitor. You could wind up an inductor and use its stored energy to drive your plate voltage quite high. Consider what happens when instead of a flyback diode is omitted and the inductor's energy has nowhere to go. Put your plate there instead, and it will go into the plate, driving it to a very high (or low) voltage. It will be harder to regulate the voltage, but you can dispense with the high voltage power supply, and your \$dV/dt\$ will be super fast. If you use a transformer instead of a simple inductor, you can even avoid the requirement of finding a switch or transistor that can block 1kV.

  • \$\begingroup\$ It is essential the the pulse have a "flat" top for at least 100 microseconds, which makes me think I should rule out the inductor method (or the spark gap method mentioned earlier.) If I settle on an N-type or IGBT, I need a driver that can float on -1kV. Any thoughts on that? These things exist? \$\endgroup\$
    – higgy
    Feb 13, 2013 at 17:37
  • \$\begingroup\$ @higgy why exactly can't you consider the plate at normally at +1kV, then quickly pulling it to ground? That's really the same thing, unless you are concerned about the voltage relative to something else. What's the something else? \$\endgroup\$
    – Phil Frost
    Feb 13, 2013 at 17:48
  • \$\begingroup\$ Good question- but there's a good reason. The -1kV pulse's job is to ionize atoms and launch the freed electrons towards a detector, which is positioned behind a grounded cylinder. (All in a vacuum chamber.) We only have a high-V feedthrough on the non-detector side of the atoms, so the pulse has to be on that side and negative to "push" the electrons towards the detector. (i.e. consider the direction of the electric field lines) \$\endgroup\$
    – higgy
    Feb 14, 2013 at 0:03
  • \$\begingroup\$ @higgy I guess by grounded you mean Earth grounded, but that doesn't really matter. All the bench supplies I have are galvanically isolated, so if I connect + to a ground rod, they generate a negative voltage relative to Earth just fine. I'd check that the transformers can withstand 1kV without breaking down, but if that all checks out, then I don't see any other problem. \$\endgroup\$
    – Phil Frost
    Feb 14, 2013 at 4:33
  • \$\begingroup\$ @higgy in case I wasn't clear, I was thinking the driver would be powered by a separate, also galvanically isolated, 12V supply, with it's negative output connected to the 1kV negative output. \$\endgroup\$
    – Phil Frost
    Feb 14, 2013 at 4:39

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