Upload
muhammad-saad
View
72
Download
7
Tags:
Embed Size (px)
DESCRIPTION
non
Citation preview
ALTERNATING CURRENT METERS
2
Objectives
• Ability to know the operation & Analyzed D’Arsonval meter movement used with half wave rectification
• Abilty to know the operation & analyzed D’Arsonval meter movement used withfull wave rectification
• Types of meter movement and application for each meter movement.
3
Alternating Current Waveform
Sinusoidal wave
Square wave
Triangle wave
4
Alternating Current Waveform
Erms= E(root mean square), Ep-p= E peak-peak, Ep= E peak
5
Vavg = 0Vrms = 0.707Vp
Vavg = 0.636Vp
Vrms = 0.707Vp
Vavg = 0.318Vp
Vrms = 0.5Vp
6
Vavg = 0
Vrms = 0.707Vp
Average and RMS Value
Sine Wave
Full Wave
Vavg = 0.636Vp
Vrms = 0.707Vp
7
Vavg = 0.318Vp
Vrms = 0.5Vp
Cont..
Half Wave
8
• The PMMC instrument is polarized (terminals +ve & -ve) - it must be connected correctly for positive (on scale) deflection to occur.
• When an AC with a very low frequency is passed through a PMMC, 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
• As the instantaneous current level falls, the pointer deflection decreases toward zero. When the AC goes negative, the pointer deflected (off scale) to the left of zero
• This kind of pointer movement can occur only with AC having a frequency of perhaps 0.1Hz or lower
PMMC Instrument on AC
9
PMMC Instrument on AC
• At 50Hz 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.
•The average value of purely sinusoidal AC is zero.
• Therefore, a PMMC instrument connected directly to measure 50Hz AC indicates zero average value.
•It is important to note that although a PMMC instrument connected to an ac supply may indicating zero, there can actually be very large rms current flowing in its coils
10
1. Half wave rectification2. Full wave rectification
Two types of PMMC meter used in AC measurement :
11
D’Arsonval meter movement used with half wave rectification
To convert alternating current (AC) to unidirectional current flow, which produces positive deflection when passed through a PMMC, the diode rectifier is used. Several types of rectifiers are selected such as a copper oxide rectifier, a vacuum diode, or semiconductor or “crystal diode”.
pVV
V Prms 5.0
2
rmsrmsp
ave VVV
V 45.02
pdcave V318.0VV
12
• For example, if the output voltage from a half wave rectifier is 10Vrms so the dc voltmeter will provide an indication of approximately 4.5V dc Therefore, the pointer deflected full scale when 10V dc signal is applied.
•When we apply a 10Vrms sinusoidal AC waveform, the pointer will deflect to 4.5V This means that the AC voltmeter is not as sensitive as DC voltmeter.
•In fact, an AC voltmeter using half wave rectification is only approximately 45% as sensitive as a dc voltmeter.
Cont…
13
•Actually, the circuit would probably be designed for full-scale deflection with a 10V rms AC applied, which means the multiplier resistor would be only 45% of the value of the multiplier resistor for 10V dc voltmeter. Since we have seen that the equivalent dc voltage is equal to 45% of the rms value of the ac voltage.
Cont…
mdc
rmsm
dc
dcs R
I
E45.0R
I
ER
Sac = 0.45Sdc
14
Example 1-1
Compute the value of the multiplier resistor
for a 15Vrms ac range on the voltmeter
shown in Fig. 1.
Fig. 1: AC voltmeter using half wave rectification
RS
Ein = 15Vrms
Ifs = 1mA
Rm = 300Ω
15
Cont.
Method 1
The AC sensitivity for half wave rectifier,
Sac = 0.45Sdc = 0.45(1k) = 450/V
Rs = Sac × Rangeac – Rm
= 450 × 10 –300
= 4.2k
16
Cont.
Rs = mfs
rms RI
E45.0
300m1
1045.0
= 4.2k
=
Method 2
17
D’Arsonval meter movement used with full wave
rectification
Fig. 2: Full bridge rectifier used in an ac voltmeter circuit
During the positive half cycle, currents flows through diode D2, through the meter movement from positive to negative, and through diode D3. The polarities in circles on the transformer secondary are for the positive half cycle. Since current flows through the meter movement on both half cycles, we can expect the deflection of the pointer to be greater than with the half wave cycle, which allows current to flow only on every other half cycle; if the deflection remains the same, the instrument using full wave rectification will have a greater sensitivity.
18
Consider the circuit shown in Fig. 1-2
Fig. 1-2: AC voltmeter using full wave rectification
19
Cont.When the 10Vrms of AC signal is applied to the circuit above, where the peak value of the AC input signal is
V14.14)10(x414.1xE2E rmsp
And the average full wave output signal is
V914.14x636.0xE636.0EE pdcave
Therefore, we can see that a 10Vrms voltage is equivalent to 9Vdc for full-scale deflection.
20
Cont.
Sac = 0.9 Sdc
rmsrmspavg E9.0)xE2(636.0E636.0E
Or
This means an ac voltmeter using full wave rectification has a sensitivity equal to 90% of the dc sensitivity
21
Example 1-2Compute the value of the multiplier resistor for a 10Vrms ac range on the voltmeter in Figure 1-2.
Fig. 1-2: AC voltmeter circuit using full wave rectification
22
Solution 1-2
The dc sensitivity is
V/k1mA1
1
I
1S
fsdc
The ac sensitivity is
Sac = 0.9Sdc = 0.9 (1k) = 900 /V
23
Cont.
Therefore the multiplier resistor is
Rs = Sac x Range – Rm
= 900 x 10Vrms – 500
= 8.5k
24
Cont.
Note: Voltmeters using half wave and
full wave rectification are suitable for measuring only sinusoidal ac voltages.