Making Electrical Measurements Part 2:
Loading
Meter Loading
When making a measurement with a volt-meter, an oscilloscope, or any type of
electronic measurement equipment, it is important to understand the concept of loading
if you want to be sure your readings are accurate.
For example, suppose I use a volt-meter to measure the DC voltage at the output of a
voltage-divider as shown in Figure 1, and I get a reading of 4 Volts. Assuming my
meter is working properly, am I sure it's a good reading? Well, that depends on two
things: the values of the resistors in the circuit, and the input impedance of the meter. In
order to see what's going on, I need to look at the Thevenin's Equivalent Circuit for the
voltage divider.
If you're not familiar with a Thevenin's Equivalent, it can be found in any book on
circuit analysis. Basically, it's a single voltage (Vth) in series with a single resistor (Rth)
as shown in Figure 2. The voltage (Thevenin's Voltage) is what you would measure
with a perfect volt-meter. The resistance (Thevenin's Resistance) is found from R = E/I
where E is the Thevenin's Voltage and I is the current you would get if you were to
short-circuit the output to ground.
For the voltage divider I'm trying to measure, since both resistors are equal, Vth would
be V/2 and Rth would be R/2. Figure 3 shows my meter as a resistor connected to the
Thevenin's Equivalent of the voltage divider. Note that the input impedance of the meter
looks like a resistor forming another divider. So the voltage across the leads of my
meter is not Vth as you might expect, but is a value I can calculate as:
Rin
Vm = ----------- x Vth
Rin + Rth
Now suppose that V is 12 Volts and R is 2k Ohms. Then Vth will be 6 Volts and Rth
will be 1k Ohm. Suppose that Rin of the meter is 10 Meg-Ohms. Using the above
equation I get:
10,000k
Vm = -------------- x 6 Volts = 5.9994 Volts
10,000k + 1k
Which, on a typical 3-digit meter, will read 6.00 volts. No problem since that's the