Why is it important that a digital multimeter have a high internal resistance when it’s on the voltage setting?
Posted on Mar 31, 2010 under Digital news and reports |I am really asking what does the high internal resistance prevent? And please give a little explanation, not just a flat answer, so that I can actually understand why. Much, much thanks!
March 31st, 2010 at 9:18 am
If it has a low resistance and you use it to measure the voltage across a resistor for example, the meter resistance will act like another resistor in the circuit in parallel with the one measured, increasing the total flow of current in the circuit.
Ergo, the higher the meter resistance (relative to the resistor being measured), the more accurate the reading will be.
March 31st, 2010 at 9:18 am
It minimises reading error and interference with the voltages being measured, due to its resistance being put in parallel with the circuit or part thereof.
Ideally, a voltmeter would have infinite resistance.
March 31st, 2010 at 9:18 am
When making your voltage measurement, you can model the system like below
A o————- |i|i|i —————–o B
\_____/^v^v^v^\______/
You want to measure the voltage between points A & B across the power supply and the resistive load. This puts the multimeter in parallel with the load. You now have a resisters in parallel problem.
Let:
Rl = the resistence of the load
Rm = the resistence of the meter
Rt = the effective resistence of the meter and load in parallel
The power supply can produce a given amount of current. Let’s make the assumption the current in the system is the maximum amount the supply can produce. We know that V = IR (Voltage = Current * Resistence). Since we assume I will not change when we add the multimeter into the circuit, we know Vl (voltage with only the load) and Vt (voltage across the load and meter combine).
Vl = I * Rl
Vt = I * Rt
Rearranging and substituting, we get Vt = (Vl / Rl) * Rt.
The readout on the meter is going to be Vt. We want Vt to be very close to Vl. Let’s substitute for Vt in the equation. Rt = 1 / ((1 / Rl) + (1 / Rm)).
Vt = (Vl / Rl) * (1 / ((1 / Rl) + (1 / Rm)))
Vt = (Vl / Rl) / ((1 / Rl) + (1 / Rm))
Vt = Vl / (Rl * ((1 / Rl) + (1 / Rm)))
Vt = Vl / ((Rl / Rl) + (Rl / Rm))
Vt = Vl / (1 + (Rl / Rm))
Vt / Vl = 1 / (1 + (Rl / Rm))
Vl / Vt = 1 + Rl / Rm
Our goal is to make Vl and Vt very close. Ideally they would be the same, so Vl / Vt would be 1. So, what value of Rm would be needed to make that true?
1 = 1 + Rl / Rm
0 = Rl / Rm.
There is only one exact solution, that Rl = 0. That will never happen, because the load will always have some resistence. So we can onlt control Rm. We want Rl / Rm to approach 0. Below are some experimental values for Rm.
Assume
Rm = 0.5 Rl, then Rl / Rm = 2
Rm = Rl, then Rl / Rm = 1
Rm = 5 Rl, then Rl / Rm = 0.2
Rm = 10 Rl, then Rl / Rm = 0.1
Rm = 100 Rl, then Rl / Rm = 0.01
Rm = 1000 Rl, then Rl / Rm = 0.001
Rm = 100000 Rl, then Rl / Rm = 0.00001
Rm = (1.0 * 10^10) Rl, then Rl / Rm = 1.0 * 10^-10
As you can see, increasing the resistence of the meter relative to the resistence of the load drives the measurement distortion caused by the meter to zero.
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