500V unity gain buffer

Question

I am currently in the process of designing a high-voltage bias supply for use with an ionization chamber. The voltage source itself is taken care of however I need a unity buffer to drive the guard of the chamber, also raising it to 500V. A simple diagram is shown below.

In this design they mention it has to be a low-input impedance amplifier serving as the buffer because we are interested in current signals from the chamber. How would I go about picking the correct amplifier to use in this circuit? I've used op-amps to amplify very small voltage signals before, but never current signals and especially never at these voltages.

What is the input impedance of an inverting op-amp configuration? Where does the current from the source go in that topology? How else can you convert a current to voltage? (V=I**R**) — vofa, Dec 15 '16 at 15:18
Why can't you just connect the guard directly to the node between the voltage source and the current sensor, possibly with an R-C low-pass filter in between? Unless the voltage is changing rapidly, I don't see any need for an active buffer. — Dave Tweed, Dec 15 '16 at 15:34
@DaveTweed because that would defeat the purpose of having a guard, the guard is there to stop the pA of leakage between the highly sensitive current meter (Im) and ground that isn't due to ionisation events. — Sam, Dec 16 '16 at 06:19
@Sam: I'm not following you. If the guard is driven to"V", there is no leakage path from I_DUT directly to ground. — Dave Tweed, Dec 16 '16 at 12:08
@DaveTweed A lot of really sensitive current meters have really high impedances so when a tiny current is produced it can result is quite a large voltage drop across the current meter (some ionisation transimpedance probes have input impedances of teraohms), it's the drop across the current meter that will cause the guard and the 'sense' line to have different voltages (and hence some leakage and capacitive coupling) unless the guard is actively driven (think of Im as a Gohm or Tohm resistor). — Sam, Dec 16 '16 at 20:41

WhatRoughBeast · Answer 1 · 2016-12-15T18:25:11.300

This is fairly simple. Use an isolated supply to produce (let's say) +/- 15 volts, and tie its output common to your output HI. Then use any op amp you like (selected for output current), powered by this floating +/- voltage to provide your guard current. Since you can easily find AC/DC power supplies with 2 kV or more input isolation (try Digikey, for instance), this should be no problem.

EDIT - If possible, find an old-school linear supply. Modern supplies tend to be switchers, and the switching noise may in fact be objectionable. Of course, with careful attention to layout, filtering and shielding you can produce very clean outputs, but you do need to pay attention.

FURTHER EDIT - WRT the previous, you MUST use a dual supply. You cannot use a single supply with a rail-to-rail output: they just don't actually go to the rails with any current ability. And using a single supply with a virtual ground to connect to HI is not a great idea either, since it invites noise pickup on the virtual ground.

This is how I've done it on mass spectrometers +1 – Andy aka Dec 15 '16 at 15:32 — Andy aka, Dec 15 '16 at 15:32

score 1 · Answer 2 · answered Dec 15 '16 at 15:33

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If you need really low leakage to the rails you may not be able to use @WhatRough's configuration because the DC-DC will have some internal leakage.

You can use a high voltage op-amp such as Apex PA89 but it will require a power supply significantly higher than the 500V supply, maybe 550V to keep the inputs within the common mode range.

They're also just a bit pricey (more than $500 USD/ea.) so best to be careful with wiring etc.

answered Dec 15 '16 at 15:33

Spehro Pefhany

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I was thinking of an AC/DC supply, as I stated in my answer. Any leakage is then to mains. – WhatRoughBeast Dec 15 '16 at 15:53
@WhatRoughBeast Sounds good, I wasn't sure that was possible. – Spehro Pefhany Dec 15 '16 at 16:16
Dude - 3 kV for a minute is a standard safety requirement for Class II devices. Check out some of http://www.digikey.com/products/en/power-supplies-board-mount/ac-dc-converters/923 – WhatRoughBeast Dec 15 '16 at 17:02
@WhatRoughBeast Oh, I know that. It was the leakage in the exact configuration he was using I wasn't sure about. Im is supposed to be an ammeter? – Spehro Pefhany Dec 15 '16 at 17:10

score 1 · Answer 3 · answered Dec 15 '16 at 15:40

Unless I have missunderstood your diagram, you need a high input impedance, not a low input impedance. Otherwise current flows into the buffer, and Im is no longer equal to IDUT. You also need a low output impedance to drive the guard.

Probably, you'll be best off buying something from Apex, such as the PA94 or PA95. They make opamps with rail-to-rail voltages of up to 2.5kV, and currents up to 50A (though not both at once).

These opamps are expensive. It is possible to avoid shelling out for a high voltage opamp, using transitors and a low voltage opamp. It's known as "bootstrapping" and it's probably much more effort than it's worth unless you are building hundreds or thousands of units.

You will of course have to check that the input offset, input impedance, output impedance etc. are appropriate. We can't help you with that unless you give some numbers on your requirements.

WhatRoughBeast's suggestion would work out cheaper, but would require a little more setting up. If this is a one-off, then it'll be quicker to just buy the opamp. I'd also be quite cautious about introducing a switching AC/DC or DC/DC supply in close proximity to what looks like a piece of precision measurement equipment - if you go down that route than a linear AC/DC converter might save some time.

500V unity gain buffer

3 Answers3