1
\$\begingroup\$

I know initially this question might come across stupid and obvious but all is not as it seems.

I want to measure the voltage drop over a 10M ohm resistor but my voltmeter's probes have an impedance of ~10M ohm, so if I try to measure the voltage drop the conventional way (i.e. put in parallel with load) but that would then cause the resistor value to half as it would make two 10M ohms in parallel giving ~5M ohm.

This is a similar situation if you just try and measure the voltage either side of the resistor, when you try and measure the lower potential side it will create a voltage divider and just gives me half the input voltage which is somewhat undesirable and highly inconvenient.

So my question to you folks is, how can I measure the drop over a large resistor without inadvertently putting an extra load on the circuit caused by the measurement equipment?

Some information about the circuit may be of use to you budding electroneers:

  • Voltage going into resistor = 3V DC
  • Voltage desired out of the resistor, also = 3V DC (or a little lower)
  • Resistor is going into a control pin for a 2:1 multiplexer / swtich (the TS5A3154 to be precise)

It would appear I neglected to mention the purpose of the resistor...

  • It is being used as a pull up resistor either on its own or as part of a transistor level translator (I have not yet decided on that) in order to get this pin to detect a logic 1 without drawing too much current. enter image description here

  • Note that the output of the MCU will only be ~2VDC which will cause the chip to drain more current than I would like it to, I haven't yet had time to think fully about how I will have this bit set up but this is the basic premise.

  • This question is not about the amount of current that will be drawn, just purely about measuring the drop over a large resistor which is not connected straight to ground and any attempt to use a multimeter in the normal way would cause errors in results dues to probe resistance.

\$\endgroup\$
  • \$\begingroup\$ You could make a x10 adapter for your voltmeter with a 100 Meg Ohm resistor. (100 meg in series and divide the voltage by 10.. well really 11!.. a x11 adapter (hmm Should I contact Spinal Tap?) \$\endgroup\$ – George Herold Feb 5 '15 at 16:26
  • \$\begingroup\$ The voltage drop across the resistor will depend on the current flowing through it. That will depend on the input impedance of the CTRL pin. Find that and you'll know the current, and hence the voltage. The input impedance will be very high, so the current will be next to nothing, so the voltage drop will be almost nothing. You will have more current flowing when the MCU output is active and changing the voltage at CTRL. It is that which will dictate the voltage drop, not the size of the resistor. \$\endgroup\$ – Majenko Feb 5 '15 at 16:35
  • 1
    \$\begingroup\$ In the schematic you posted, it's a 10k\$\Omega\$ resistor - so is your question right? I'd also advise against using a 10M\$\Omega\$ resistor as a pull up resistor, as they can pick up more noise than a lower valued one. In our projects we often use 100k\$\Omega\$. \$\endgroup\$ – Arsenal Feb 5 '15 at 16:52
  • \$\begingroup\$ Your schematic doesn't show any 10 megohm resistor. Can you upcate the schematic to show how a 10 megohm resitor is involved in your circuit? \$\endgroup\$ – The Photon Feb 5 '15 at 17:20
  • \$\begingroup\$ The question makes no sense at all. Is your MCU output an open-drain line? If so, why? Configure it as a standard output and just connect it to the control input. And why are you attempting independent measurement of the pin voltage, anyways? You need to provide much more information about exactly what you are trying to do. \$\endgroup\$ – WhatRoughBeast Feb 5 '15 at 19:27
4
\$\begingroup\$

Two things you can do:

  1. Measure the current flowing through the resistor, then calculate the voltage drop across it (Ohms Law).
  2. Create a Voltage Follower circuit with a much higher input impedance and measure the output voltage of that.

However, by what you're describing, I'm not sure you're using the resistor right in the first place. Please expand your question with a schematic and a full reasoning behind the use of the resistor.

\$\endgroup\$
  • \$\begingroup\$ Added an edit to better explain what the resistor is for! \$\endgroup\$ – MrPhooky Feb 5 '15 at 16:17
  • \$\begingroup\$ Oh and there is a problem measuring the current as it is absolutely tiny and I can't seem to accurately measure it with any equipment I have :/ \$\endgroup\$ – MrPhooky Feb 5 '15 at 16:23
2
\$\begingroup\$

An op-amp buffer (unity gain) with CMOS input can be used. For example an LMC6062 has fA leakage and 100uV maximum Vos. It will need about a 6V single supply minimum (9V battery would work fine) in order to measure 3V.

http://upload.wikimedia.org/wikipedia/commons/thumb/f/f7/Op-Amp_Unity-Gain_Buffer.svg/220px-Op-Amp_Unity-Gain_Buffer.svg.png

If you really want to get down to electrometer-style leakage, special construction techniques are required, but for 10M\$\Omega\$ measurements to 1% (1G\$\Omega\$) nothing special is required, just normal care, clean after soldering (and avoid no-clean solder flux).

\$\endgroup\$
1
\$\begingroup\$

JFETs have input resistances in the teraohms. Using a JFET amplifier or using a specialized DMM with a JFET input will allow you to measure such a voltage drop.

\$\endgroup\$
1
\$\begingroup\$

I usually do that with a pair of LF356 connected as unity gain or gain x2. If one end of your resistor is grounded, you need only one LF356 on the high impedance end of the resistor.

Make sure that you supply your LF356 with sufficient voltage to avoid saturation around ground and around positive supply (3 V below ground and 3 V above supply is good).

\$\endgroup\$
-1
\$\begingroup\$

I found a method shortly after posting this question which was very easy to implement and didn't involve implementing a new circuit with components I don't have.

Using an extra power supply
So it was one of my colleagues that told me this method and I thought it was a really quick and easy work around for the problem that I was having.

All you need to do is get a second power supply and a voltmeter and connect the PSU to the voltmeter and then the other lead to the 'output' of the resistor. So the original PSU is connected to where power will be and the new one is in series with the DVM connected to where the resistor goes to, in my case - the CTRL pin on the switch.

Now you have this set up, have the original power supply to whatever voltage it needs to be at (in this case 3V) and have the second starting from 0V. Increase the second one while reading the voltage 'drop' on the DVM and as you increase the PSU voltage, this difference should go down. Continue to increase the voltage until the reading on the DVM display is at 0V - this means that the voltage at the CTRL pin is the same as the voltage from the PSU.

You can then disconnect everything but keep the values the same, measure the voltages out of the PSU's (if you don't trust / have the output displays on them) and you can get the voltage drop over the large resistor by finding the difference between the two power supplies.

$$V_{PSU1}-V_{PSU2}=V_{R}$$

Below is a sketch of the set up:
enter image description here

\$\endgroup\$
  • \$\begingroup\$ This is only something you can really use on a loose resistor; this is not really usable in-circuit or in any kind of electronic design. \$\endgroup\$ – user36129 Feb 9 '15 at 14:22
  • \$\begingroup\$ @user36129 well it is for testing a prototyping so it worked for this scenario. \$\endgroup\$ – MrPhooky Feb 9 '15 at 14:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.