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.
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.