Convert Ammeter to Voltmeter

Question

I've got an old analog (0-5ma) ammeter and want to convert it to measure 0-±5 volts, as an output for my Arduino.

I'm a programmer and my knowledge of electronics is limited. I know what components do but obviously not how to use them properly.

I tried to place some 2k (1 and 2) resistors in series with the ammeter but it doesn't work as expected. This was sortof working but is wasn't a linear movement even with a capacitor attached to + and - (+ and - come from the Arduino PWM/analog output)

^{simulate this circuit – Schematic created using CircuitLab}

I know its a ammeter so it needs current and if theres nothing asking for current there will not be any so i put a very high resistor between + and - but that didn't really change anything.

So how can i change my am meter to measure 0-5v?

This is it opened, the resistance is about 2.9k (if I measured it correctly)

Could it be microAmps?

Hope some one can help, Thanks in advance.

Answer 1

If the full scale value of the ammeter is 5mA, calculate the proper value of resistor in which 5mA will flow when 5 volt is connected across the resistor. By ohms law, this will yield 1000 ohms.

Connect the resistor in series to the ammeter. Now the ammeter which converted to a voltmeter should have the full scale read out when 5 volt is connect to it.

This approach assumes the ammeter has zero internal resistance. In reality all ammeters will have some resistance. Because of this, the calculated value of the 1000 ohms series resistance must be tweaked to a finer value by introducing another small series resistance or another large parallel resistance.

Answer 2

You are missing one piece of information - the coil resistance.

Once you have that we can work out the required series resistance, \$ R_M \$.

First calculate the full-scale voltage \$ V_{FS} = {I_{FS}}{R_M} \$ (from Ohm's law).

To convert it to 5 V full scale we need to add a series resistor to drop the excess voltage at 5 mA.

$$ R1 = \frac {5 - V_{FS}}{I_{FS}} = \frac {5 - 0.005R_M}{0.005} \Omega$$

Post a photo of the meter and quote the resistance and report back.

Figure 1. For reference this series from ST have a 196 mV drop at 5 mA. (Photo is a sample from the range.)

For the meter shown in Figure 1 we can calculate the series resistance as

$$ R1 = \frac {5 - V_{FS}}{I_{FS}} = \frac {5 - 0.196}{0.005} = 960 \Omega$$

---

This is it opened, the resistance is about 2.9k (if I measured it correctly).

There's something wrong here! To get 5 mA through 2.9 kΩ you need \$ V = IR = 0.005 \times 2900 = 14.5~V \$.

You've either got an incorrect measurement or a voltmeter!