On Introductory Circuit Analysis of Boylestad I found out the following about the relation of internal resistance of a multimeter with the scale choosen on the instrument.
- Analog voltmeter: internal resistance does depend on the scale chosen
- Analog ammeter: internal resistance does depend on the scale chosen
- Digital voltmeter: internal resistance does not depend on the scale chosen
Is all of this true? How can the internal resistance of the digital voltmeter be independent on the scale chosen? And is it the same for the digital ammeter too (i.e. also in that can internal resistance does not depend on the scale choosen)?
Best Answer
It's more a function of whether the meter has an amplifier before it.
In the moving coil meter, the meter reading is directly proportional to the current flowing through it. These generally do not have an amplifier, so that current has to be provided by the thing being measured. Therefore the higher the input voltage, the higher the resistance placed in series.
My first moving coil meter was '20k/V', that is it needed a series resistance of 20kohms for every volt on the scale, which indicated that the meter read full scale for 50uA. As range switching is done by a mechanical switch, high voltage withstanding is not a problem, and input resistors are simply switched as required.
You can put an amplifier in front of a moving coil meter, but it's generally not done.
As a digital meter requires electronics and a battery, the amplifier sort of comes for free. It's most convenient there to have a high input resistance like 10Mohms, and switch the ranges by shunting the ADC input with various value resistors. This arrangement allows all the switches to stay at low voltage, even if the input voltage is 1000V. If a 10k shunt was used, then the input voltage would be 1000x the ADC voltage for full scale.