High-value resistors and leakage

input

I tried that 40 years ago. No matter what resistance I fed through to a voltmeter, it still read line voltage. Look at a garden hose with the nozzle closed. Full pressure. You need current flow to get voltage drop.

mike

It does solve the 'parallel leakage paths' problem, because the terminal of the sense resistor that isn't at HV is at ground (within a few millivolts). So, leakage to that node is a matter of millivolts on megohms, and is negligible quantity of current, thus negligible induced error. What I mean by 'op amp current/voltage converter' is an inverting-amplifier configuration (op amp and two resistors) which holds one end of the sense resistor at pseudo-ground. Resistor stability over time is a component-choice issue

input

I guess I miscommunicated. The circuit would hold one end of the high resistance at pseudo-ground, and the other end is at HV, so the sense resistor DOES drop voltage. And sink current. The current, though, goes through the sense resistor, to the (-) amplifier node, and then through the feedback resistor to the op amp output. So, the output has a voltage Vout :== -Rf/Rsense * Vhigh

input

But how does that improve anything? Surely any leakage problem with a simple HV divider is across the resistor going to the HV, not the one going to ground? Since that one can have a low resistance. E.g. 1M and

.... Phil

Vile twit!

If you use a divider, the sense node voltage scales with the HV input. So, any on-the-printed-wiring traces make variable leakage currents to adjacent traces. Both those leakages, and the stray capacitance of the wiring, are nuisances because of the high impedance of the resistors in the divider. Replace the divider with an inverting op amp, the sense node is at pseudo-ground, so a simple grounded ring around the node is a suitable guard, and leakage currents to that guard are only driven by the small offset voltage of the op amp you use.

Additionally, the sense node of a divider cannot be ground-clamped as easily as that of an inverting op amp; this is important if the HV power and other power supplies don't have a controlled power-on sequence.

But it is only the "top" resistor that is high impedance!

pseudo-ground,

I still don't see this I'm afraid.

If you are talking about leakage from the sense node (resistor junction), this will only be at a few volts and it is low impedance (approximately that of the lower resistor in the divider). So it will not be any more significant than any other low voltage node in the circuit.

If you are talking about leakage from the HV end, you could still "guard" the sense node with an earth ring if you wanted. It would not be exactly at ground but it would only be a few volts away and would be a low impedance point.

I see no need for the sense node impedance to be particularly high. For example a 600V divider could have a 1M resistor string feeding a 1k. The sense node would be at ~0.6V, impedance would be ~1k.

OK, clamp to the supply rails instead (like with everythng else probably)?

... leakages, and the stray capacitance of the wiring, are

Oh, not necessarily. Try a 10M resistor on a 3 kV power sense application and you create a watt of heat: you can make it more accurate by lowering the dissipation (like, with 100M), but of course the pulldown resistor has to scale up at the same time. To 'unload' the HV supply, designs are frequently made with BOTH resistors high-Z.

OK, Tim mentioned only 600V IIRC.

Even with 3kV and 100M / 100k, I see no significant difference between guarding with the 3V node voltage (using an opamp) or guarding with 0V.

If the circuit board / environment is so bad that you have to worry about 3V signals at 100kohms, you have no business putting 3kV and