6.1
PMMC Instrument on AC
As discussed earlier, the PMMC instrument is polarized , that is, its terminals are identified as + and

, and it must be connected correctly for positive (on-scale) deflection to occur. When an alternating current with a very low frequency is passed through a PMMC instrument, the pointer tends to follow the instantaneous level of the ac. As the current grows positively the pointer deflection increases to a maximum at the peak of the ac. Then as the instantaneous current level falls, the pointer deflection decreases toward zero. When the ac goes negative, the pointer is deflected (off-scale) to the left of zero. This kind of pointer movement can occur only with ac having a frequency of perhaps 0.1 Hz or lower. With the normal 60 Hz or higher supply frequencies, the damping mechanism of the instrument and the inertia of the meter movement prevent the pointer from following the changing instantaneous levels. Instead, the instrument pointer settles at the average value of the current flowing through the moving coil.
The average value of purely sinusoidal ac is zero. Therefore, a PMMC instrument connected directly to measure 60 Hz ac indicates zero. It is important to note that although a PMMC instrument connected to an ac supply may be indicating zero, there can actually be a very large rms current flowing its coils.
6.2
Full-Wave Rectifier Voltmeter
Rectifier instruments use silicon or germanium diodes to convert alternating current to a series of unidirectional current pulses, which produce positive deflection when passed through a
PMMC instrument. The full-wave bridge rectifier circuit in Figure 3-15 passes the positive halfcycles of the sinusoidal input waveform and inverts the negative half-cycles. When the input is positive, diodes D
1
and D
4
conduct, causing current to flow through the meter from top to bottom, as shown. When the input goes negative, D
2
and D
3
conduct, and current again flows through the meter from the positive terminal to the negative terminal.
Figure 3-15 An ac voltmeter may be constructed of a PMMC instrument, a multiplier resistor, and a full-wave bridge rectifier. The instrument scale is correct only for pure sine waves.

The resulting current waveform is a series of positive half-cycles without any intervening spaces (see Figure 3-15). As in the case of a dc voltmeter, the rectifier voltmeter circuit in Figure
3-17 uses a series-connected multiplier resistor to limit the current flow through the PMMC instrument. The meter deflection is proportional to the average current, which is 0.637 × peak current. But the actual current (or voltage) to be indicated in ac measurements is normally the rms quantity, which is 0.707 of the peak value, or 1.11 times the average value. Since there are direct relationships between rms, peak, and average values, the meter scale can be calibrated to indicate rms volts.
A rectifier voltmeter as discussed above is for use only on pure sine-wave voltages. When other than pure sine waves are applied, the voltmeter will not indicate the rms voltage.
Example 3-6
A PMMC instrument with FSD = 100 µA and R m
= 1 k

is to be employed as an ac voltmeter with FSD =100 V (rms). Silicon diodes are used in the bridge rectifier circuit of Figure 3-15.
Calculate the multiplier resistance value required.
Solution At FSD the average current following through the PMMC instrument is
I av
= 100 µA peak current I m
=
I m
=
(
I av =
100μA

157 µA
0.637
0.637
applied peak voltage
 rectifier volt drop ) total circuit resistance rectifier volt drops = 2 V
F
(for D
1
and D
4
or D
2
and D
3
) applied peak voltage = 1.414
V rms total circuit resistance = R s
+ R m
I
R m s
=
=
1.414
V rms
1.414
V rms

R s

R m

I m
2 V
F
2 V
F

R m

1k

=
= 890.7 k

157μA
Example 3-7
Calculate the pointer indications for the voltmeter in Example 3-6, when the rms input voltage is
(a) 75 V and (b) 50 V.
Solution
(a) I av
= 0.637
I m
= 0.637×
1.414
V rms

R s

R m
2 V
F
(b)
= 0.637×
I av
= 0.637×
890.7k
1k

75µA = 0.75 FSD
890.7k
1k

50µA = 0.5 FSD
2

Example 3-8 Calculate the sensitivity of the voltmeter in Example 3-6.
Solution
I m
= 157 µA
Irms = 0.707Im = 0.707 × 157 µA

111 µA (at FSD)
Vrms = 100 V (at FSD)
100 V total R =
111μA
= 900.9k

R 900.9kΩ sensitivity = =
V 100V
= 9.009 k

/V

9 k

/V

/V
Examples 3-6 and 3-7 demonstrate that the rectifier voltmeter designed to indicate 100 V rms at full scale also indicates 0.75 FSD when 75 V rms is applied, and 0.5 FSD for 50 V rms.
Therefore, the instrument has a linear scale. At low levels of input voltage the rectifier current is also low, and this can result in errors due to variations in the diode voltage drop. The effect can be countered by using a shunt resistor across the meter, as discussed next for a half-wave rectifier voltmeter.
6.3
Half-Wave Rectifier Voltmeter
Half-wave rectification is employed in the ac voltmeter circuit shown in Figure 3-16. R
SH shunting the meter is included to cause a relatively large current to flow through diode D
1
(larger than the meter current) when the diode is forward biased. This is to ensure that the diode is biased beyond the knee and well into the linear range of its characteristics. Diode D
2 conducts during the negative half-cycles of the input. When conducting, D
2 causes a small voltage drop
( V
F
) across D
1 and the meter, thus preventing the flow of any significant reverse leakage current through the meter via D
1
. Diode D
2 also protects the meter against reverse voltages.
The waveform of voltage developed across the meter and R
SH
is a series of positive halfcycles with intervening spaces, as illustrated. In half-wave rectification, I av
= 0.5(0.637
I m
). This must be taken into account in the circuit design calculations.
Figure 3-16 Half-wave rectification may be used with a PMMC instrument and a multiplier resistor for ac voltage measurements. A shunt resistor ( R
SH
) is included to ensure a satisfactory rectifier forward current level. The additional rectifier ( D
2
) minimizes reverse leakage current through D
1
.
3

Example 3-9
A PMMC instrument with FSD = 50 µA and R m
1700

is used in the half-wave rectifier voltmeter circuit illustrated in Figure 3-16. The silicon diode ( D
1
) must have a minimum (peak) forward current of 100 µA when the measured voltage is 20% of FSD. The voltmeter is to indicate 50 V rms at full scale. Calculate the values of R
S
and R
SH
.
Solution At FSD, I av
= 50 µA.
Meter peak current,
I m
=
I av =
50μA

157µA
At 20% of FSD, diode peak current I
F
must be at least 100 µA; therefore, at 100% of FSD,
I
F (peak)
=
100%
20%
×100 µA = 500 µA
I
F (peak)
= I m
+ I
SH
I
F (peak)
= I m
+ I
SH
= 500 µA

157 µA = 343 µA
V m (peak)
= I m
R m
= 157 µA × 1700

= 266.9 mV
R
SH
=
I
V
)
=
266.9mV
343μA
= 778
)
( applied peak voltage )

V

)
I
F (peak)
=
R
S
=
1.414
V rms

V
I
)
R
S
)

V
F

V
F
= = 139.5 k

500μA
Like a dc ammeter, an ac ammeter must have a very low resistance because it is always connected in series with the circuit in which current is to be measured. This low-resistance requirement means that the voltage drop across the ammeter must be very small, typically not greater than 100 mV. However, the voltage drop across a diode is 0.3 to 0.7 V, depending on whether the diode is made from germanium or silicon. When a bridge rectifier circuit is employed, the total diode volt drop is 0.6 to 1.4 V. Clearly, a rectifier instrument is not suitable for direct application as an ac ammeter.
The use of a current transformer (Figure 3-17) gives the ammeter a low terminal resistance and low voltage drop. The transformer also steps up the input voltage (more secondary turns than primary turns) to provide sufficient voltage to operate the rectifiers, and at the same time it steps down the primary current to a level suitable for measurement by a PMMC meter.
Since the transformer is used in an ammeter circuit, the current transformation ratio I p
/ I s
= N s
/ N p is very important.
A precise load resistor ( R
L
in Figure 3-17) is connected across the secondary winding of the transformer. This is selected to take the portion of secondary current not required by the meter. For example, suppose that the PMMC instrument requires 100 µA (average) for FSD, and
4

the current transformer has N s
= 2000 and N p
= 5. If the rms primary current is 100 mA, the secondary rms current is
5
I s
=
2000
×100 mA = 250 µA or an average of
1
I s (av)
=
1.11
×250 µA = 225.2 µA
Since the meter requires 100 µA for FSD, the value of R
L
is calculated to pass the remaining 125.2 µA.
The range of the instrument can be changed by switching-in different values of load resistance. Another method of range changing involves the use of additional terminals (or taps) on the primary winding to alter the number of primary turns, as shown in Figure 3-17.
Figure 3-17 Ac ammeter circuit consisting of a current transformer, full-wave bridge rectifier, and a PMMC instrument.
Example 3-10
A rectifier ammeter with the circuit shown in Figure 3-17 is to give FSD for a primary current of
250 mA. The PMMC meter has FSD 1 mA and and N p
= 4. The diodes each have V
F
R m
=1700

. The current transformer has N
= 0.7 V, and the series resistance is R s s
= 500
= 20 k

. Calculate the required value of R
L
.
Solution
Peak meter current I m
=
I av
0.637
1mA
=
0.637
= 1.57mA
Transformer secondary peak voltage,
E m
= I m
( R
S
+ R m
) + 2 V
F
= 1.57 mA (20 k

+ 1700

) + 1.4 V

35.5 V or secondary voltage E
S
= (0.707 × 35.5 V) rms

25.1 V and rms meter current = 1.11 I av
= 1.11 mA
5

Transformer rms secondary current,
I s
= I p
N p
N s
4
= 250 mA × = 2 mA
500
I
L
= I s

I m
= 2 mA

l.11 mA = 0.89 mA
R
L
=
E
I
L s
25.1V
=
0.89mA
= 28.2 k

8.1
Reading Errors
Some sources of error in measurements made by deflection instruments are: bearing friction, improperly adjusted zero, and incorrect reading of the pointer indication. Zero and friction errors can be minimized by carefully adjusting the mechanical zero of an instrument before use and by gently tapping the meter to relieve friction when zeroing and reading. Portable instruments should normally be used lying flat on their backs. Care in deciding the exact position of the pointer on the scale will reduce reading errors.
Even with an accurately marked scale and a sharp pointer, two observers may disagree about the exact scale reading. This occurs because of parallax error : the uncertainty about the eye of the observer being directly in line with the end of the pointer. Parallax error is eliminated in good instruments by the use of a knife-edge pointer and a mirror alongside the scale. When an observer lines up the pointer and the mirror image of the pointer, the observer’s eye is exactly in the line with the pointer, and the scale can then be read accurately.
8.2
Specified Accuracy
High-quality instruments may have their accuracy specified as a percentage of the actual scale reading, or measured quantity. However, for most deflection instruments, manufacturers specify the accuracy as a percentage of FSD. This means, for example, that an instrument that gives FSD for a coil current of 100 µA, and which is specified as accurate to ±l%, has a ±l µA accuracy at all points on its scale. Thus, as demonstrated in Example 3-11, the measurement error becomes progressively greater for low scale readings.
Example 3-11
An instrument that indicates 100 µA at FSD has a specified accuracy of ±l%. Calculate the upper and lower limits of measured current and the percentage error in the measurement for (a) FSD and (b) 0.5 FSD.
Solution
(a) At FSD: indicated current = 100 µA
Error = ±1% of 100 µA = ±1 µA
actual measured current = 100 µA ±1 µA = 99 µA to 101 µA
Error = ±1% of measured current
(b) At 0.5 FSD indicated current = 0.5×100 µA = 50 µA
Error = ±1% of FSD = ±1 µA
6

actual measured current = 50 µA ±1 µA
= 49 µA to 51 µA
Error =
±1μA
50μA
×100%
= ±2% of measured current
Problems
1) An ac voltmeter uses a bridge rectifier with silicon diodes and a PMMC instrument with FSD
= 75 µA. If the meter coil resistance is 900 Ω and the multiplier resistor is 708 k Ω, calculate the applied rms voltage when the voltmeter indicates FSD. (60 V)
2) Determine the new multiplier resistance required for the voltmeter in Problem 1 to change its range to 300 V FSD.
(3.59MΩ)
3) Determine the pointer position on the voltmeter in Problem 2 when the applied rms voltage is
(a) 30 V and (b) 10 V. (9.7%, 3%)
4) A PPMC instrument with a 900 Ω coil resistance and an FSD of 75 µA is to be used with a half-wave rectifier circuit as an ac voltmeter. Silicon diodes are used, and the minimum diode forward current is to be 80 µA when the instrument indicates 0.25 FSD. Calculate the shunt and multiplier resistance values required to give 200 V FSD.
(880.9kΩ)
5) Calculate the sensitivity of the ac voltmeter in Problem 1.
(12kΩ/V)
6) Calculate the sensitivity of the ac voltmeter in Problem 4. (4.4kΩ/V)
7) A rectifier ammeter is to indicate full scale for a 1 A rms current. The PMMC instrument used has a l200 Ω coil resistance and 500 µA FSD, and the current transformer has Ns =7000 and
Np = 10. Silicon diodes are used and the meter series resistance is Rs = l50 kΩ. Determine the required secondary shunt resistance value. (97.6kΩ)
8) A rectifier ammeter has the following components: PMMC instrument with FSD = 200 µA and Rm =900 Ω; current transformer with Ns =600 and Np =5; diodes with V
F
= 0.3 V; meter series resistance Rs = 270 kΩ; transformer shunt secondary resistance R
L
=98.7 kΩ. Calculate the level of transformer primary current for instrument FSD. (100mA)
9) Calculate the sensitivity of the ac voltmeter in Example 3-9 when diode D
2
is (a) included in the circuit and (b) omitted from the circuit. (2.8 kΩ/V, 4kΩ/V)
10) A PMMC instrument with 250 µA FSD has a specified accuracy of ±2%. Calculate the measurement accuracy at currents of 200 µA and l00 µA. (±2.5%, ±5%)
11) A deflection instrument with 100 µA FSD has a ±3% specified accuracy. Calculate the possible error when the meter indication is (a) 50 µA and (b) 10 µA. (±6%, ±30%)
12) A 25 µA current is measured on an instrument with 37.5 µA FSD. If the measurement is to be accurate to within ±5%, determine the required instrument accuracy. (±3.3%)
7