While researching how a multimeter worked, I figured that each part would be fairly simple and I could do one journal on the whole thing. Well, that’s just not the case. Because I like to fully understand a system from all different levels, I decided to break the multimeter down into its components to have more manageable topics. To start: the voltmeter (old school style).
Part 1: A device to react to changes in voltages.
Like so many other moving pieces that move based on a changing electric current (such as a motor), the D’Arsonval movement relies on the interactions between electricity and magnetism. It consists of a coil, permanent magnets, a spring, and a pointer (1). When electricity is run through the coil, the coil wants to turn just as inside an electric motor. However, unlike an electric motor, a spring force acts resists the turning. The result is that the coil (and the pointer that is attached to the same axle as the coil) turns until the two forces are in equilibrium. So the force from the magnetic field generated by the electrostatic magnet (the coil) acts to turn the pointer until it exactly cancels out the spring force that opposes it. The trick to measuring the current,
Wait a second… I thought we were talking about a voltmeter, not an ammeter (which measures current). Yes, we are, I’ll get to that in Part II.
Anyway, the trick to measuring the current is that the strength of the magnetic field generated is based on the voltage applied to the coil (1). So the more current you apply, the more the coil wants to turn inside the permanent magnet’s field and the further it can turn before the spring force stops it.
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| D'Arsonval Movement. Source 1. |
It turns out that a D’Arsonval movement is a very sensitive device to measure changes in current. In general, the maximum current that it can handle is 1mA.
Part II: Using that device to make a voltmeter.
Because Ohm’s law states that V (voltage) = I (current) * R (resistance), it is easy to see how a D’Arsonval movement could be used to measure voltages if you knew its resistance. However, because the internal resistance is so small (sometimes only 1kΩ), a D’Arsonval movement on its own could only measure very small voltages (1mA*1kΩ=1V) and we often want to measure larger voltages (2).
The solution: a voltage divider. By putting a big ol’ resistor in series with the D’Arsonval movement, you lower the current and burn off a lot of the voltage on the resistor (2). Importantly, since you know the current (remember that’s what the D’Arsonval movement is measuring and current is the same in series), you know how much voltage was dropped over the resistor. Given this figure:
In order to find the maximum voltage that this voltmeter can measure, assume the maximum current possible of 1mA. The total resistance is 500Ω from the D’Arsonval movement + 9.5k Ω from the resistor = 10kΩ. Therefore, the maximum voltage V = I*R = 1mA * 10k Ω = 10V.
In the same way, a different value of the resistor would allow the D’Arsonval movement to measure different voltage ranges (2). Thus, in a voltmeter that has the turn dial that lets you choose a range of voltages, the circuitry inside is moving a switch (or a turn pot but that would be less precise) to activate a certain resistor (2).
Part III: The voltmeter.
To measure voltage, you place the voltmeter in parallel with the device you want to measure voltage across. Therefore, the lower the resistance of the voltmeter, the more it disturbs the circuit. Although a small amount of current must flow through the voltmeter in order for it to be able to get a reading, the current should be negligible compared to the current through the device which the voltmeter is trying to measure voltage across. Thus, the ideal voltmeter has infinite resistance.
Looking at the figure, another way to put it is that without the voltmeter, Ir1 would be the same as Itot. The voltage drop over the resistor in question is therefore V = Itot*R1. However, when the voltmeter is connected, it draws a current equal to Iv. The current through the resistor is changed to Ir1 and the voltage drop is now equal to Ir1*R. The change in voltage drop over the resister caused by the voltmeter is:
ΔV = Itot*R1- Ir1*R
= R1*(Itot-Ir1)
= R1*Iv
Obviously you want your meter to change the circuit by a very small amount, so you want to minimize ΔV. Since you can’t change R1, you can minimize ΔV by minimizing Iv, which is equivalent to maximizing the resistance of the voltmeter because resistance and current are inversely related.
Now you see why I didn’t try to cover the ammeter and ohmmeter in the same journal.
- http://www.engineersedge.com/instrumentation/electrical_meters_measurement/darsonval_movement.htm
- http://www.allaboutcircuits.com/vol_1/chpt_8/2.html
OUCH, this one hurt my head.
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