0-100uA analogue meter (needle) to show 0-10v

[neazoi]

Hi I have a 0-100uA analogue meter (needle) that has a marked scale of 0, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100 and subdivisions between these.
I want to be able to measure voltage of 0-10v with it, but the marked points to represent the actual voltages.
In other words, the point 10 to represent 1v, the point 50 to represent 5v, the point 90 to represent 9v etc.

Is that possible to do it and how?

I do not mind the loading of the source, as it will be a 4A capable PSU, so I won't see any voltage drop by the loading from the meter.

 

[FvM]

Already heard about ohms law? Use a series resistor. 10V/100µA = 100 kOhm.

 

[neazoi]

Already heard about ohms law? Use a series resistor. 10V/100µA = 100 kOhm.

The thing I wanted to point out was, will the meter respond linearly to the voltage changes?
I.e 1v 10% needle movement, 9v 90% movement?

 

[KlausST]

Hi,

according ohm´s law. 5V input will give 50uA, this is half of 100uA.
So this should be linear.

Nobody can tell whether your meter works linearely or not. How? You don´t give any informations about your meter.
Also you don´t give any informations about your input. Is it pure DC, AC, waveform....?
With non_DC input: Do you expect RMS reading (moving iron) or average (moving coil) reading?

Klaus

 

[neazoi]

It's pure DC. Very well regulated and cleaned from ripple. The meter is an vintage one, but one of these big ones. I have no information other than it seems well engineered, not cheap Asian.
Ok so it is linear fair enough. Thanks!

 

[std_match]

The series resistor should be less than 100k.
You must subtract the meter's internal resistance from the theoretical 100k value (10V/100uA).
In other words, the series resistor and the meter in series must have a total resistance of 100k.
You can measure the meter's internal resistance with a normal multimeter.

 

[neazoi]

The series resistor should be less than 100k.
You must subtract the meter's internal resistance from the theoretical 100k value (10V/100uA).
In other words, the series resistor and the meter in series must have a total resistance of 100k.
You can measure the meter's internal resistance with a normal multimeter.

I will put a series set trimmer after the smaller series resistor you suggest, so I can set the meter precisely at 100uA reading when 10v is applied to it.