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?

enter image description here

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

enter image description here

Could it be microAmps?

Hope some one can help, Thanks in advance.

  • \$\begingroup\$ What do you mean by "it isn't working". Your question should include enough information so people don't have to ask questions, because we aren't there at your bench seeing what you can see. Please be specific electronics.stackexchange.com/help/how-to-ask \$\endgroup\$
    – Voltage Spike
    Commented Sep 1, 2016 at 19:36
  • \$\begingroup\$ I don't think it can be \$2.9k\Omega\$. That would imply a voltage drop of over \$14V\$ when at full scale. I don't know of ammeters that operate that way. \$\frac{1}{2}V\$ maybe. But not that high. \$\endgroup\$
    – jonk
    Commented Sep 1, 2016 at 19:52
  • \$\begingroup\$ It may not be an ammeter, just a meter with an amp scale (depending where it was used in a circuit). Sometimes there are additional components (along the lines of your R1) installed within the case. \$\endgroup\$ Commented Sep 1, 2016 at 20:17
  • \$\begingroup\$ These meters have never been used, even the dials are still blank \$\endgroup\$
    – Sephen
    Commented Sep 1, 2016 at 20:21
  • \$\begingroup\$ @SeanHoulihane Hmm. Then if these are truly just plain meter movements with \$5mA\$, full scale, with \$2.9k\Omega\$ resistance I'm kind of surprised a bit. It's possible. I generally imagine about \$200\Omega\$ for \$1mA\$ and perhaps something over \$3k\Omega\$ for \$50µA\$. I'd think something at \$5mA\$ would be a lot lower resistance. \$\endgroup\$
    – jonk
    Commented Sep 1, 2016 at 20:33

2 Answers 2


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.


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.

enter image description here

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!

  • \$\begingroup\$ So i only need to add resistors in series? I'm sorry if my questions are irrelevant i trying my best to get my head around it \$\endgroup\$
    – Sephen
    Commented Sep 1, 2016 at 19:51
  • \$\begingroup\$ See the update. \$\endgroup\$
    – Transistor
    Commented Sep 1, 2016 at 19:53
  • \$\begingroup\$ Added a picture... Could it be micro amps \$\endgroup\$
    – Sephen
    Commented Sep 1, 2016 at 20:05
  • \$\begingroup\$ You said it was mA. Don't ask me! Where did you get the 5 mA from? \$\endgroup\$
    – Transistor
    Commented Sep 1, 2016 at 20:07
  • \$\begingroup\$ When i bought these a time ago the guy said 5 m... something amps i assumed it was MA... I didn't mean if you could tell me if its micro amps but only if it could be possible for a ammeter to measure micro amps. \$\endgroup\$
    – Sephen
    Commented Sep 1, 2016 at 20:14

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