# Voltage divider output with analog voltmeter

I'm using a voltage divider to bring down 24V to 5V and attaching the output to an analog voltmeter gauge. So my question is should I be counting the resistance of the analog gauge into the value's of the resistors of the voltage divider and vary as needed to get my 5V? (see attached diagram) • Yes of course...
– user16324
Apr 20, 2016 at 22:26

Yes, you certainly should. The resistor in parallel with the coil increases the loading on the circuit under test unnecessarily. The optimum value for R2, therefore, is infinity!

Assuming that this is a fixed range meter rather than a multimeter the normal solution would be:

• Measure the coil resistance and find its full-scale deflection voltage.
• Calculate the series resistance required for full-scale voltage in your application. Since $\frac {V_{TOTAL}}{V_{METER}} = \frac {R_{TOTAL}}{R_{METER}}$ you can work out the series resistor.

The result will give you the minimum possible loading of the circuit under test by your meter. simulate this circuit – Schematic created using CircuitLab

Figure 1. Basic DC multi-voltmeter. Usually a rotary selector switch would connect the red lead to the appropriate point in the resistor chain.

If you look at the specifications of any analog multimeters you will generally find that they had a resistance of 20 kΩ/V on the DC voltage range. Changing range just switched in the appropriate series resistance. Figure 2. Simpson analog meter clearly showing 20 kΩ/V on front panel. Image source: Simpson160.com.

In contrast, most digital multimeters have a constant input impedence of 1 M on all DC voltage ranges.

The voltmeter will provide a parallel resistance to R2, changing the value of the voltage divider, hence the change in Vout. The value of the voltmeter will vary based on it's internal resistance, range, the resistance of the leads used, etc. A digital voltage meter will have a significantly higher resistance, closer to 10 Megaohms.

In theory, you should factor in the resistance of your voltmeter, but in practice the latter's resistance will be so much higher than your R1 and R2 values that it effectively has no effect and does not need to be taken into account. What you should consider though is the 'load' you will connect at the 5V point that you are trying to achieve. I am sure that you are not just creating a divider to display a voltage on a meter...