Download ELECTRONICS 4 – Fundamentals of Electronics I

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El 04- Electronics 4
Laboratory Exercise 18 – Moving Coil Meters
OBJECTIVE: To explore the various aspects of the use and circuitry of D’Arsonval
moving coil meters.
MATERIALS: Protoboard; power supply; multimeter; meter movement; precision
potentiometer; 10K resistor.
DISCUSSION: Today’s technician regularly uses a digital multimeter that makes use of
analog-to-digital converter integrated circuit devices and precision parts to measure
voltage, current, and resistance. However, for most of the time that electronics has
existed as a science, we have had to rely on less accurate moving coil meters for
measuring circuit characteristics. While such meters are for most intents and purposes
obsolete, they can provide us with a historical perspective on electronics technology and
with practical examples of series and parallel circuit applications.
The D’Arsonval movement consists of a coil of wire suspended on a set of pivots to
which is attached a slim metal needle. The movement is allowed to turn. The movement
attempts to rotate by virtue of a current passed through its coil. This sets up a magnetic
field around the coil which interacts with a permanent field magnet nearby. The result is
motor action that makes the meter movement and its needle rotate. However, as it does
so, it is opposed by a coil spring provided for this purpose. Therefore, the amount of
meter movement is determined by the amount of current flowing through its coil – more
current provides more magnetic lines of force which in turn create more motor action.
The more that the armature wants to turn, the more it is opposed by the coil spring. So,
the needle then will reach an equilibrium where the motor action is exactly opposed by
the spring. This point is determined by just how much current the coil is carrying. The
movement is, therefore, a current measuring device.
Most meter movements are made of precision parts and are delicate. The amount of
current needed to move the needle to full scale can be anything from 1 or 2 milliamperes
to less than 200 microamperes. The meter’s coil has a resistance because of its small wire
that can range from a few ohms to over 2000 ohms. Also, there is a characteristic voltage
that appears across the meter movement when it is conducting, by virtue of Ohm’s Law.
To make practical use of a meter movement, we can extend its range by the addition of
resistors in series or parallel to the movement. The values of the resistors are typically
non-standard and of peculiar numbers. The temptation is to ignore the meter’s
characteristics, but when calculating resistors to extend the meter’s capabilities, we can’t
do that without sacrificing accuracy. Well-made instrumentation meters typically have
hand-wound resistors that are very accurate.
The first thing we need to verify is the exact internal resistance of the meter. We can’t
just simply hook an ohmmeter up to it because the ohmmeter has an internal battery that
might cause too much current to flow in the movement and damage it as a result. So, we
build a simple circuit to measure the movement’s resistance. To do this, we must know
the amount of current that is needed to drive the movement to full scale. This is usually
provided by the manufacturer and/or printed on the meter itself. To determine the internal
resistance of the device, we hook it up in parallel with a precision 10-turn potentiometer.
(A potentiometer is a variable resistor whose resistance in ohms across its terminals can
be varied by turning a shaft or knob. A radio’s volume control is an example.) We first
hook up the meter to a variable power supply with a series safety resistor and slowly
adjust the supply to obtain a full scale reading. We then install the pot across the meter
and adjust it so that the meter reads one-half scale. Since according to parallel theory, two
resistors of equal value in parallel will conduct the same current, the pot must now be
conducting the same current as the movement. We can then measure the setting of the pot
with a digital meter to see what its setting is, and thereby also know the value of the
internal resistance of the meter movement.
Once we know all the stuff about the movement, we can make it into a usable current
meter by adding resistors in "shunt", or parallel, around it. To extend the range of a 1
milliampere meter movement, for example, to 10 milliamperes, we add a resistor in
parallel with the meter so that when 1 milliampere is flowing through the meter, 9
milliamperes are flowing through the shunt. To calculate the value of the shunt resistor,
we use the value of the meter’s voltage at full scale (Imeter x Rmeter) and Ohm’s law
using the value of current through the shunt (9 milliamperes). A 1 milliampere meter with
an internal resistance of 50  will need a 5.55  resistor in shunt to provide a 10
milliampere range.
To make a voltmeter with this meter, we add multiplier resistors in series with the
movement. Using the essentials of Kirchhoff’s Voltage Law, we know that if we wish,
for example, to make 5 volt meter using the device described above, we need to place a
resistor in series with the movement that drops most of the voltage so that only about 50
millivolts is placed on the meter movement when a full 5 volts is placed on the circuit as
a whole. For the same 1 milliampere movement, we need to drop 4.95 volts across the
multiplier at 1 milliampere full scale. This requires a resistor of 4950  .
Part of the problem with this style meter is the loading affect that they have on the circuit
under test. The loading affect is a change in the circuit’s operation as a result of the
meter’s use. In the case of a current meter, the overall meter resistance made up of the
meter and the shunts in parallel will be in series with the device whose current we wish to
measure. If the resistance is fairly large in this area, the meter’s few ohms won’t make
much difference. But if the resistance of the circuit under test is itself small, then the
meter’s presence might reduce the actual current significantly and yield a reading that is
not that of the circuit without the meter.
In the case of a voltmeter, the voltmeter is placed in parallel or across the device whose
voltage we wish to measure. The overall value of the meter’s resistance can have a major
affect on the operation of the circuit and the readings we get will be wrong. Consider
what would happen if we used the meter discussed above, with an internal overall
resistance of about 5000 to measure the voltage across a 68K resistor. The 68K resistor
would effectively be replaced by the 5K resistance of the meter and the circuit would
work much differently while the meter was in place. The solution to the problem was
solved by use of the Vacuum Tube Voltmeter, or VTVM, which isolated the sensitive
meter movement from the circuit using additional circuitry and which always presented a
fixed resistance to the circuit under test, typically 11megohms.
Today’s digital meters use current IC technology to make the measurements, and have
little affect on the circuit being tested.
This exercise examines meters in three parts. In the first, we are given an unknown meter
movement and we must determine its characteristics. We want to know its full-scale
current and its internal resistance, both of which are determined by how the device is
manufactured. In the second part, we add shunts to increase its current measuring
abilities, and in the third part we add multipliers to make it into a voltmeter.
PROCEDURE:
1. Obtain a meter movement from the instructor. Inspect it to try to determine the
current it requires for a full scale reading. This can be determined usually be
examining the face of the meter and reading either its scale or a small
manufacturing notice on its face.
IFS = __________ mA
2. Set up the circuit of Figure 1 without the potentiometer in the circuit. This will
result in a simple series circuit with the meter and the 10K ballast resistor across
the power supply. DO NOT apply power yet.
3. Turn the voltage knob on the power supply all the way to zero. Then, turn on the
supply. Slowly turn the power supply knob up until the meter reads just full scale.
The current being drawn at this point should be that as determined in step 1.
4. Next, connect the potentiometer across the meter as shown. Turn its knob up or
down until the meter reads half-scale. At this point, half of the current in the
circuit should be going through the meter and the other half should be going
through the pot. What is this half current? __________ mA
5. Turn the power supply off. Carefully remove the pot from the circuit without
changing its setting. Use the multimeter to measure the resistance of the pot. This
will be the meter’s resistance.
RMETER = __________ 
6. As a convenience, we can now calculate the value of the voltage that would
appear across the meter at full scale. This is done with Ohm’s Law: VMETER =
IMETER X RMETER
VMETER = __________ mV
7. We can add a shunt resistor across the meter movement to enable the meter-shunt
combination to read more than the meter was designed to do. To do this, we must
calculate the value of the shunt.
A. Determine the scale at which the meter is to operate. For instance, if you
were given a 1mA movement and you wish to use it to measure up to
10mA, at full scale deflection the shunt would have to carry 9mA. The
formula is ISHUNT = IRANGE – IMETER.
B. Since the shunt resistor and the meter movement will be in parallel, they
will have the same voltage across them. This was calculated in step 6.
Using this, we can calculate the value of the shunt: RSHUNT = VMETER /
ISHUNT
8. Using the values from your meter movement, determine the resistance values for
shunts that would expand its ability to measure larger currents. Fill in the table
below.
Range:
ISHUNT
RSHUNT
5 mA
10 mA
100 mA
250 mA
1A
9. Using values determined in the table above, set up your movement so that it acts
like a 10mA meter. You will need to create a shunt resistor from existing resistors
but there will no doubt be some inaccuracy in this. Build a simple circuit with the
power supply, a load resistor, your meter assembly, and the multimeter in series
as shown in Figure 2.
10. Set the power supply to minimum, and then the ranges and functions of the
multimeter to make it capable of measuring 10mA. Turn the power on and slowly
advance the power supply knob so that current begins to flow. Turn the power
supply to that point where your meter assembly reads full scale.
What is the meter reading on the multimeter at this point? __________mA
How do the two meters compare?
___________________________________________________
Remove power from the circuit.
To make the meter movement perform as a voltmeter, we must place a resistor in
series with it called a multiplier. This resistor will have to drop most of the
voltage applied to the meter assembly so that only a little voltage is available for
the meter movement. Remember that the meter movement has very little voltage
across it at full scale (step 6). The formula for the multiplier is VMULT = VFS –
VMETER
12. Using the values from your meter movement, determine the resistance values for
multipliers that would expand its ability to measure larger voltages. Fill in the
table below.
11.
Range:
100 mV
500 mV
1V
5V
VMULT
RMULT
10 V
13. Using values determined in the table above, set up your movement so that it acts
like a 5 volt meter. You will need to create a multiplier resistor from existing
resistors but there will no doubt be some inaccuracy in this. Build a simple circuit
with the power supply, a load resistor, your meter assembly, and the multimeter in
series as shown in Figure 3.
14. Set the power supply to minimum, then turn the supply on. Slowly advance the
supply control until your meter assembly reads full scale. What is the reading
shown on the multimeter at this point? They should be about equal. VMULTIMETER
= __________ V
15. Now, to examine the affect of voltmeter loading, while the circuit is still setup and
running from step 14, remove a lead from your meter assembly while leaving the
multimeter in place. The multimeter should change somewhat as a result. What is
the new multimeter reading?
VMULTIMETER = __________ V
16. Remove power from the circuit and then disassemble it. Then, answer the
questions below in writing for credit.
REVIEW QUESTIONS:
1. If you were given a meter movement with no markings of any kind on it, how
could you determine what its full scale current rating would be?
2. What was the result of your test for loading affect of the voltmeter in step 15?
What was the voltage as measured on the multimeter with and with out the meter
assembly?
3. If you wanted to extend your meter to read 10 amperes, what would be the value
of the shunt?
4. If you wanted to extend your meter to read 500 volts, what would be the value of
the multiplier?
5. What value shunt resistor did you calculate in step 9? What value resistor did you
actually use? What percent inaccuracy was involved in using this value?