 ### Network # Multimeter

#### juantravel

May 14, 2010
41
I have always wondered how a multimeter works(inside). I'm wondering if someone can tell me if I am on the right track. A analog multimeter uses an inductor for the meter. When you apply current to the inductor it moves the needle. By reference in how many windings the inductor has and how far the needle moved is how much current is flowing through the meter? In order to measure voltage we use the formula V=IxR? Use resistors as multipliers? Also digital multimeter uses zener diodes to measure voltage. But how does a digital multimeter measure current?

I did some more research and it looks like digital multimeter use adc to measure voltage and then use a precision resistor to apply ohms law and get the current. Is this correct?

Last edited:

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,509
First let's describe an analog multimeter in very crude terms.

Let's assume that the meter is a 50uA FSD meter. This means that 50uA is required to cause the meter to deflect fully to the right. This, in turn, means that whatever we are measuring, 50uA must flow through the meter for a full scale reading. Let's also assume the meter has a resistance of 5000 ohms.

If we want a 10V range, we need to add a resistor in series with the meter so that 50uA will pass through the meter with 10V across it.

The resistance is calculated using ohms law, V/I = R, 10/(50*10^-6) = 200k. We need to subtract the meter's resistance from this, so we need 195K in series with the meter.

If you do this calculation a few times, you'll quickly find that the total resistance is equal to 20,000 ohms for every volt in the full scale reading. This is important for a number of reasons, and you will find that analog multimeters almost always have this figure displayed somewhere on the device (in this case 20kohm/V).

For current measurements, a very similar thing applies. A resistance needs to be placed in parallel with the meter to shunt all but 50uA at the full scale current.

From the previous figures, we can see that the meter has 0.25 volts across it at FSD (V = IR). So now we need to calculate a resistance which will drop 0.25V at 50uA less than the desired current.

Let's assume the current is 2mA. This means we want to drop 0.25 V at 1950 uA. Using ohms law; V/I = R; 0.25/(1950*10^-6) = 128 ohms

The important thing here is that the meter will have 0.25 volts across it on any current range. This is a less known value for an analog meter, and one which is not typically shown on the meter itself. But let's not get into that at the moment...

Now for a digital multimeter.

A typical module may display 1999 at 199.9mV with an input impedance of 10Mohm. For simplicity, we call it 200mV, and accept that the meter can't actually read the last part of whatever range we create.

However the same sort of thing applies as previously.

For a 10V range, we could determine a series resistor that allows 200mV across the meter, but that would entail absolutely huge resistor values, ones that would be sensitive to the slightest humidity. Instead, a lower value resistor is shunted across the meter to allow more sensible series resistors.

For current ranges, we determine shunt values as for an analog meter. We need to determine the FSD current, and in this case it is V/R = I; 0.2/10,000,000 = 20nA. For many ranges, this current can be ignored.

Note that there are no zener diodes involved The module in a DMM (that I conveniently glossed over) will have a precision voltage reference and some form of A-D converter.

Note that a *real* DMM (or at least a good one) won't simply use a module, it will be rather more integrated into the design of the unit as a whole.

#### juantravel

May 14, 2010
41
Shunt

Can you point me to a tutorial on shunt's. thank you so much for the reply. Your knowledgeable. A soon as learn more I will share my knowledge with the forum.

Moderator
Jan 21, 2010
25,509

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