EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: jstarr on June 22, 2016, 03:43:56 am
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I read that to measure the voltage at a very high impedance point (say 50 MegOhm) you cannot do this accurately with a typical digital multimeter that only has a 10Megohm input as it will load down the circuit. I guess this means that the meter will carry more current than the circuit to be measured? I don't understand how that would change the voltage reading. I thought adding the DMM probe simply adds a parallel circuit node and voltage should not change at a parallel node?
tnx
john
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The meter is in series with the voltage source forming a closed loop around which the source current flows: so the combination of meter and source will act like a voltage divider. If the source has a 50 M impedance and the meter has a 10 M impedance then the meter will read one sixth of the source voltage [ = 10 / (10 + 50) ].
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I read that to measure the voltage at a very high impedance point (say 50 MegOhm) you cannot do this accurately with a typical digital multimeter that only has a 10Megohm input as it will load down the circuit.
I see what you mean, but it's not exactly correct as formulated.
If you hook up a 50M resistor to a 10V battery and then measure the voltage, you will get an accurate reading of the voltage of the battery, this is the scenario you were thinking of that doesn't make sense to you.
Now imagine you hook up the battery to something like this: https://upload.wikimedia.org/wikipedia/commons/thumb/3/31/Impedance_voltage_divider.svg/2000px-Impedance_voltage_divider.svg.png
If both resistors are 50M, then obviously where the resistors meet you would have 5V.
But if you try to measure it with a 10M meter then your circuit changes from 50M + 50M to 50M + 8.3M and you will obviously not get 5V.
So the correct phrasing would be that you can't use a meter to measure the voltage in a circuit where introducing the resistance of the meter would cause the circuit to behave differently.
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If you hook up a 50M resistor to a 10V battery and then measure the voltage, you will get an accurate reading of the voltage of the battery, this is the scenario you were thinking of that doesn't make sense to you.
However, if you put the 50 M resistor in series with the battery you will obtain a 10 V source with a 50 M source impedance. In this case if you try to measure the voltage with a 10 M meter, you will only measure 1.67 V.
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Well, if you have access to a bench DMM you might get a 'good enough' reading as many of those have high input impedance (e.g. HP 34401 >10GOhm). Or you could do this old school, use the multi meter as null meter (you will need a 2nd volt source which you can adjust), see [1]. That can give you very accurate readings even for very high impedance sources.
[1] https://en.wikipedia.org/wiki/Potentiometer_(measuring_instrument)
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I read that to measure the voltage at a very high impedance point (say 50 MegOhm) you cannot do this accurately with a typical digital multimeter that only has a 10Megohm input as it will load down the circuit. I guess this means that the meter will carry more current than the circuit to be measured? I don't understand how that would change the voltage reading. I thought adding the DMM probe simply adds a parallel circuit node and voltage should not change at a parallel node?
tnx
john
You don't have to be dealing with 50 Mohms to run into problems. Take a look at this message and read the entire thread:
https://www.eevblog.com/forum/blog/eevblog-584-what-effect-does-your-multimeter-input-impedance-have/msg396860/#msg396860 (https://www.eevblog.com/forum/blog/eevblog-584-what-effect-does-your-multimeter-input-impedance-have/msg396860/#msg396860)
If you tried to measure the voltage at the midpoint of the resistors in my (stolen) drawing with a 10 Mohm meter, you'd measure 2V857 instead of 3V0. The meter's 10 Mohm resistance is in parallel with the bottom resistor and shifts the voltage down.
You probably won't be dealing with 50 Mohm impedances very often. But 1 Mohm - now that's something you're going to see more frequently.
Ed
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I had the same problem, and ended up using a fet input opamp as a unity gain buffer in a project box with a 9v battery and a switch so I can do +/-4v or 0.5-8.5v. Put some shrouded meter sockets in it, so I can use my normal probes, and male-male shrouded banana cable to plug the meter in. This worked fine for a ~1M ohm impedance node I was measuring and didn't want to disturb the circuit as it operated. For a higher voltage, if you have a power supply with enough resolution, or a suitable divider, using the dmm as a null meter would be my go to method if the measurement only needed to be done a few times.
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This might be of interest -
http://info.tek.com/KI-Low-Level-Measurements-Handbook-LP.html (http://info.tek.com/KI-Low-Level-Measurements-Handbook-LP.html)
Regards, Dana.
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+1 to null measurement.
You basically use a variable power supply and adjust it until it equals the unknown voltage. The meter is actually connected between the unknown and the variable supply. When they are equal, the meter shows zero. In this state, there is "zero" current through the null meter so no voltage drop across the high impedance of the unknown source. Then measure the voltage of the variable supply (which is easy since it is low impedance) and you have your unknown voltage.
This is also the technique used to measure Weston cells, yesteryear's voltage standards/references. They could not tolerate even a few microamps of current at their output, so the null method was used to reduce the measurement load to near zero.
Many bench meters do "high impedence" measurements up to +-10V, others up to only 2 or 3 V. One exception is the Keithley 2001 which can do high impedance measurement up to +-21 VDC. That "high impedance" on this meter is specified as >10 GOhm (how much greater? who knows). That can still result in nearly 1% error, just from the impedance itself, without considering the input bias current which is not zero. Bias current error can be estimated by connecting a 50 Mohm resistor (your source impedance) across the terminals and noting the resulting reading. The null technique can do better, without the 21 V limit.