I’d like to design a high-voltage switch to toggle on/off a 600 V connection powering a small load (in particular, an electroadhesive clutch) based on an Arduino/Teensy control input.

With my basic EE knowledge, I was originally considering a common-source amplifier, similar to the one in the figure below, with HV = 600 V, R1 = 5 MΩ, R2 = 1 kΩ, Vin coming from an Arduino at 5 V, and the estimated load being two resistors R = 1.2 Ω and a capacitance C = 220 nF in series. I expect the load should consume <100 μA.

However, I have two problems with this common source circuit design:

  1. When the transistor is on (Vin > Vth) and thus Vout is low, this circuit essentially sinks an extra 600 V / 5 MΩ = 120 μA from the power supply. I eventually plan to lay out about 400 of these in a PCB, so that would add up to an extra 48 mA, which is quite significant for my high voltage power supply.
  2. Vout is noticeably less than HV because of R1. Assuming the load draws 50 μA, Vout would be (5 MΩ)*(50 μA) = 250 V less than HV.

My question is: is there a better high-voltage switch circuit diagram that consumes less current and doesn’t substantially reduce Vout relative to HV?

Because I plan to lay out an array of 400 of these switches, relatively common/low-cost components and a relatively small number of components would be a nice-to-have, albeit not high priority. I was also considering just buying 400 high-voltage relays, but since my expected load is < 100 μA I thought there might be a simpler circuit diagram that achieves the same functionality for me. Thank you!

Source follower schematic


2 Answers 2


You can make a high side switch using two NPN and one PNP transistor, although it can also be done with a PMOS and NMOS device. Here is a simulation for 350 volts, as the choice of devices limited it to 400V.

High side HV switch LTSpice simulation

For a capacitive load, a positive discharge may be needed at turn-off. Here's how that could be done:

High side switch with discharge

  • 1
    \$\begingroup\$ Q1 isn't necessary, as the load current is transient (capacitive load) in this case*. (When more load is required, it's a good idea.) Somewhat better performance is had with R2 moved "below" Q3 (series with emitter), and a smaller value; R3 can then be removed, and operation is also reasonably independent of V(V1). (*When using Q1, performance can also be improved by adding a B-E resistor in parallel with Q1, to shunt Q2's leakage current: lower off-state leakage this way, and somewhat faster turn-off.) \$\endgroup\$ Jul 9, 2022 at 11:01

There are several problems with high voltage drivers for this application:

  1. There are HV capacitive array drivers out there -- but only to modest voltages (100-200V?), such as for electroluminescent displays. Not enough, drat.
  2. As you note, the simplest solution sucks -- using a load resistor. I guess you don't need a very small resistor since it's just charging a capacitor, and I presume it doesn't need to switch very fast. (That said... is it really 220nF? That's a—to use a technical term—metric shitload of surface area?!)
  3. There are half-bridge drivers for higher voltages, but they are made for higher power levels (600V sure, but at some amperes of load -- maybe not by themselves as in gate drivers, but they're not really worthwhile unless that's what you're doing), and most require AC output -- the high side being powered by a bootstrap supply (e.g. IR2110). I assume you need, even if not DC operation, then long enough on-time that this is not a practical approach. (I.e.: close enough that it might come to mind, but not workable.)
  4. Making 400 of anything is going to be a right pain. This... doesn't help any, unfortunately, but more to say, sometimes there just isn't a pre-cooked array chip, and you have to do everything one at a time. I feel your pain. Any chance you have to use dual or triple transistors, resistor packs, etc., you know, see what's doable to help reduce parts count. (Mind the voltage ratings: SMT arrays aren't usually rated for much!)

As for solutions, there's the resistive loaded switch (OP); there's various ways to wire a push-pull/totem-pole/half-bridge, of which you'll want the least possible parts count (which @PStechPaul's answer has an example of; if not the least, then in that vicinity).

You don't have to use plain transistors; isolation can be helpful. There are optos with fairly high voltage ratings, which could switch the load directly (high side switch). They will be slow (the Darlington types especially); but again, it seems unlikely speed is a priority, so maybe this isn't a problem. Examples: LTV-352T, H11D1, TLP187, etc. Use a regular BJT or MOSFET for the low side. Because of the low speed, do leave ample time between high-side turn-off and low-side turn-on (a few ms, say). Biggest downside (besides speed if relevant): most of these are only rated 300V, so you need two in series to handle the full 600V. Blah.

You could also use a more regular opto (photodiode/BJT output), supplying drive voltage to a PNP/PMOS, from an auxiliary rail referenced to B+. That is, you have +600V, +595V, and whatever your +3.3/5V logic supply is down at ground level. You'd provide this with an isolated (5V to 5V or etc.) DC-DC converter, tying + to 600V, and - is then "595V". This is basically the same sort of thing, or, it's a hybrid I suppose between PStechPaul's solution, and a full-opto solution; physically, you're trading the need of two optos, for one opto and an output transistor (and a resistor or two to bias the gate/base), and saving the dissipation of the level-shift transistor (Paul's Q3 and R2).

One advantage of this approach is an array can be used; dual and quad phototransistor types are readily available. No worries about voltage between receivers, they're all on the common (B+) rail. Though I don't know offhand if this is economical (sometimes quad parts are much more expensive than pairs or even singles; economies of scale are weird like that).

I mentioned "photodiode" somewhat in passing, but there are kind of two ways to use them: one as a phototransistor with really shitty CTR (which is tolerable here with so little DC needed); the other by generating photovoltaic power. Granted it's a tiny voltage per diode, like half a volt -- not enough to turn on a BJT, but two in series would, but even more so, PV arrays are available -- for driving MOSFETs as AC/DC SSRs. A stack of diodes (say 10-15), all illuminated by the IR LED, generate enough voltage to turn on a MOSFET. The voltage is higher than needed for a BJT but that won't matter, it's very current-constrained, no problem "shorting out" the output here. I haven't checked these in a while though, and maybe they're too expensive compared to the other methods. (Likewise the plain-photodiode types, they're probably for analog signal application that's not very common anymore these days? I can't say I've used one myself.)

As for output transistors, note that, at low currents, 2nd breakdown isn't an issue; so it doesn't really matter if you pick BJT or MOSFET here, any will do. P type transistors are in thin supply at high voltages, but 600V I think is still within reach of available types.

  • 1
    \$\begingroup\$ I had not really considered the turn-off requirement for the capacitive load, so my circuit should include another device to discharge the capacitor. Perhaps another NPN or NMOS across the load, with its base or gate controlled by the collector of Q3 with an RC delay. \$\endgroup\$
    – PStechPaul
    Jul 9, 2022 at 23:30

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