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I have two 10G resistors connected in series with a 3V battery. I want to determine the voltage drop across one of them, which of course is 1.5V. When I use my multimeter to check the voltage drop, it reads ~3mV, which I believe is because it has a 10M impedance so the circuit is really one 10G resistor in series with (a 10G resistor and a 10M resistor in parallel), so the voltage drop when the multimeter is part of the circuit is 2.99 mV.

schematic

simulate this circuit – Schematic created using CircuitLab

How can I measure the voltage drop? Is there something I can build so that that I can adapt the multimeter impedance to be high enough that it won't affect the circuit as much?

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    \$\begingroup\$ Look up a 1993 ED article "What's All This Femtoampere Stuff, Anyhow?" by the late Robert Pease. \$\endgroup\$ – Spehro Pefhany Mar 18 '19 at 14:00
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    \$\begingroup\$ Why would you need such a circuit if I may ask? Any load attached will casue the same effect as the multimeter. \$\endgroup\$ – Huisman Mar 18 '19 at 14:05
  • \$\begingroup\$ @Huisman for trying to build ammeters that can go to very low current. I want very low current sources, then to try to measure them. If I'm dividing the voltage down first before passing through a 10G resistor (or higher) it's especially helpful to be able to measure that the actual voltage drop is what I expect it to be. \$\endgroup\$ – John Smith Mar 18 '19 at 14:31
  • \$\begingroup\$ Could you please draw a schematic (by pressing Schematic button in editor in your original post) where the ammeter is located? I think you'd better divide the voltage by e.g. 10k pot and connect its branch with a 10G resistor to the ammeter. \$\endgroup\$ – Huisman Mar 18 '19 at 14:35
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    \$\begingroup\$ As drawn, it looks like the divider will be passing 150pA. There are definitely more things that can sneak up on you at that point to mess with your measurement. Maybe you don't need femto amp precision, but seeing what they do to ensure fA accuracy is probably a good step. \$\endgroup\$ – W5VO Mar 18 '19 at 14:57
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Do what the ancients did ==== use a Wheatstone bridge. Like this

schematic

simulate this circuit – Schematic created using CircuitLab

Rotate the 10,000 ohm potentiometer for ZERO reading.

Then measure the pot voltage (and compensate for the DVM loading)

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    \$\begingroup\$ Clever answer, so +1. To calibrate, replacing R3 with another set of 10G resistors should allow the DVM to be set to zero. \$\endgroup\$ – user105652 Mar 18 '19 at 18:20
  • \$\begingroup\$ Imbalance in impingment of external Efields, or Hfields, on the DVM leads, may be the residual error. \$\endgroup\$ – analogsystemsrf Mar 20 '19 at 2:19
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    \$\begingroup\$ In this case it is wise to stay away from the meter wires, which should be as short as possible. A metal cage would be even better. At 10G ohm, you have a very sensitive mass detector. Any charged body close by will affect the readings. \$\endgroup\$ – user105652 Mar 20 '19 at 3:07
  • \$\begingroup\$ An ancient implementation of the "bootstrapping" great idea... How interesting it is - the voltages across high resistances R1 and R2 are copied across the low resistances of R3 halfs... a unique property of the balanced bridge... BTW an op-amp follower can do the work of the man and R3 potentiometer in this arrangement. \$\endgroup\$ – Circuit fantasist May 24 '20 at 14:17
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sure, a voltage follower built with a FET op-amp that has extremely low input bias current.

https://www.mouser.co.uk/Semiconductors/Amplifier-ICs/Operational-Amplifiers-Op-Amps/_/N-4h00g?Rl=4h00gZgjdhpmZ1yvbz5oZ1yve6dbSGT

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  • \$\begingroup\$ Is it enough to use a single low input bias current follower (i.e. only use it for one of the multimeter probes with the 2nd directly probing the circuit under test) or would I need 2? \$\endgroup\$ – John Smith Mar 18 '19 at 15:08
  • \$\begingroup\$ CMOS op-amps with fempto-amp inputs are ideal for these type of devices. \$\endgroup\$ – user105652 Mar 18 '19 at 18:22
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If you can get capacitors with zero leakage **, you can hang one across each resistor. Since you are working with DC, give the circuit a few weeks to stabilize, then measure the PEAK voltage. The capacitance required would be such that the time constant RC is several seconds, where R is the load resistance of your multimeter.

** I do not know of a source for such components. If you get one, you better not touch or breathe on it. :-)

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  • \$\begingroup\$ Very original idea... So we can use a capacitor as a mobile S&H circuit to sample the voltage across components on the board and then to measure it by a stationary voltmeter:) \$\endgroup\$ – Circuit fantasist May 24 '20 at 14:25
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I was going to suggest the teflon capacitor idea, but someone beat me to it.

Whatever you do you need to be very careful of stray currents caused by surface leakage across components, you need to use teflon standoffs everywhere, and wash everything with IPA (power off obviously) and dry thoroughly, you might also want to use a guard ring too (see Bob Pease article ref in comment above).

If you can reverse the positions of R1 and R2 on the fly, you can compensate for bias currents in the measurement. Reversing the connections (on the fly) on R3 will verify you are in the middle of R3 range.

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