Pmmc With Ac Measurement 4v561p

ALTERNATING CURRENT METERS

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. 2

Alternating Current Waveform

Sinusoidal wave

Square wave

Triangle wave 3

Alternating Current Waveform

Erms= E(root mean square), Ep-p= E peak-peak, Ep= E peak 4

Vavg = 0 Vrms = 0.707Vp

Vavg = 0.318Vp Vrms = 0.5Vp

Vavg = 0.636Vp Vrms = 0.707Vp

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Average and RMS Value Vavg = 0 Vrms = 0.707Vp Sine Wave

Vavg = 0.636Vp Vrms = 0.707Vp Full Wave

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Cont.. Vavg = 0.318Vp Vrms = 0.5Vp

Half Wave

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PMMC Instrument on AC • 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 ed 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 8

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 9

Two types of PMMC meter used in AC measurement : 1. Half wave rectification 2. Full wave rectification

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D’Arsonval meter movement used with half wave rectification To convert alternating current (AC) to unidirectional current flow, which produces positive deflection when ed 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”. VP Vrms 

2

 0.5Vp

Vave V dc  0.318Vp Vp 2  Vrms Vave    0.45Vrms   11

Cont… • 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. 12

Cont… •Actually, the circuit would probably be designed for fullscale 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.

E dc 0.45E rms Rs   Rm   Rm I dc I dc Sac = 0.45Sdc

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Example 1-1 Compute the value of the multiplier resistor for a 15Vrms ac range on the voltmeter shown in Fig. 1. R S

Ifs = 1mA Ein = 15Vrms

Rm = 300Ω

Fig. 1: AC voltmeter using half wave rectification 14

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

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Cont. Method 2

Rs

=

=

0.45E rms  Rm I fs 0.45  10  300 1m

= 4.2k

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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. 17

Consider the circuit shown in Fig. 1-2

Fig. 1-2: AC voltmeter using full wave rectification 18

Cont.

When the 10Vrms of AC signal is applied to the circuit above, where the peak value of the AC input signal is

E p  2 xE rms  1.414x (10)  14.14V And the average full wave output signal is

E ave  E dc  0.636 xE p  0.636 x14.14  9V Therefore, we can see that a 10Vrms voltage is equivalent to 9Vdc for full-scale deflection. 19

Cont. Or E avg  0.636E p  0.636( 2 xE rms )  0.9E rms This means an ac voltmeter using full wave rectification has a sensitivity equal to 90% of the dc sensitivity

Sac = 0.9 Sdc 20

Example 1-2 Compute 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 21

Solution 1-2 The dc sensitivity is

1 1 Sdc    1k / V I fs 1mA The ac sensitivity is Sac = 0.9Sdc = 0.9 (1k) = 900 /V

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Cont. Therefore the multiplier resistor is Rs

= Sac x Range – Rm = 900 x 10Vrms – 500 = 8.5k

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Cont. Note: Voltmeters using half wave and full wave rectification are suitable for measuring only sinusoidal ac voltages.

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