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CHAPTER 5 AC Meter. School of Computer and Communication Engineering, UniMAP Prepared By: Amir Razif b. Jamil Abdullah EMT 113: V-2008 1 5.0 AC Meters. 5.1 Introduction to AC Meters. 5.2 D’Arsonval Meter Movement with Half-Wave Rectification. 5.3 D’Arsonval Meter Movement with Full-Wave Rectification. 2 5.1 Introduction to AC Meters. Five principal meter movement that are commonly used in ac instruments; (i) Electrodynamometer. (ii) Iron-Vane. (iii) Electrostatic. (iv) Thermocouple. (v) D’Arsonval (PMMC) with rectifier. The d’Arsonval meter is the most frequently used meter movement, event though it cannot directly measure alternating current or voltage. In this chapter it will discuss the instruments for measuring alternating signal that use the d’Arsonval meter movement. 3 Cont’d… (a) AC Voltmeters and Ammeters AC electromechanical meter movements come in two basic arrangements: (1) Based on DC movement designs. (2) Engineered specifically for AC use. Permanent-magnet moving coil (PMMC) meter movements will not work correctly if directly connected to alternating current, because the direction of needle movement will change with each half-cycle of the AC. Permanent-magnet meter movements, like permanent-magnet motors, are devices whose motion depends on the polarity of the applied voltage, Figure 5.1. Figure 5.1: D’Arsonal Electromechanical Meter Movement. 4 Cont’d… (b) DC-style Meter Movement for AC application. If we want to use a DC-style meter movement such as the D'Arsonval design, the alternating current must be rectified into DC, Figure 5.2. This can be accomplished through the use of devices called diodes. The diodes are arranged in a bridge, four diodes will serve to steer AC through the meter movement in a constant direction throughout all portions of the AC cycle: Figure 5.2: Rectified D’Arsonal Electromechanical Meter Movement. 5 Cont’d… (c) Iron-Vane Electromechanical. The AC meter movement without the inherent polarity sensitivity of the DC types. This design avoid using the permanent magnets. The simplest design is to use a non-magnetized iron vane to move the needle against spring tension, the vane being attracted toward a stationary coil of wire energized by the AC quantity to be measured, Figure 5.3. The electrostatic meter movements are capable of measuring very high voltages without need for range resistors or other, external apparatus. Figure 5.3: Iron-Vane Electromachanical Meter Movement. 6 Cont’d… (d) AC Voltmeter with Resistive Divider. When a sensitive meter movement needs to be re-ranged to function as an AC voltmeter, series-connected "multiplier" resistors and/or resistive voltage dividers may be employed just as in DC meter design, Figure 5.4. Figure 5.4: AC Voltmeter with Resistive Divider. 7 Cont’d… (e) AC Voltmeter with Capacitive Divider. Capacitors may be used instead of resistors, though, to make voltmeter divider circuits. This strategy has the advantage of being non-dissipative; no true power consumed and no heat produced. Refer to Figure 5.5. Figure 5.5: AC Voltmeter with Capacitive Divider. 8 5.2 D’Arsonval Meter Movement with Half-Wave Rectification. In order to measure the alternating current with the d’Arsonval meter movement, we must rectify the alternating current by use of diode rectifier . Figure 5.6 is the DC voltmeter circuit modified to measure AC voltage. The diode, assume to be ideal diode, has no effect on the operation of the circuit . For example if the 10 V sine-wave input is fed as the source of the circuit, the voltage across the meter movement is just the positive half-cycle of the sine wave due to the rectifying effect of the diode. Figure 5.6: DC Voltmeter Circuit Modified to Measure AC Voltage. 9 Cont’d… The peak value of 10 Vrms sine wave is, E p 10Vrms *1.414 14.14V peak E ave E dc 0.318 * E p or E ave Ep 0.45 * E rms Edc 0.45Erms Rs Rm Rm I dc I dc If the output voltage from the half-wave rectifier is 10V only, a dc voltmeter will provide an indication of approximately 4.5 V. From the above equation, S 0.45S ac dc 10 Example 5.1: D’Arsonval Meter Half-Wave Rectifier. Compute the value of the multiplier resistor for a 10 Vrms ac range on the voltmeter shown in Figure 5.7. Figure 5.7: AC Voltmeter Using HalfWave Rectification. Solution: Find the sensitivity for a half wave rectifier. S ac 0.45S dc 0.45 * 1 450 I fs V Rs S ac * Range ac Rm . 450 10V * 300 4.2 K V 1 11 Cont’d… Commercially produced ac voltmeters that use half-wave rectification have an additional diode and shunt as shown in Figure 5.8, which is called instrument rectifier. Figure 5.8: Half-Wave Rectification Using an Instrument Rectifier and a Shunt Resistor. . 12 5.3 D’Arsonval Meter Movement with Full-Wave Rectification. The full-wave rectifier provide higher sensitivity rating compare to the half-wave rectifier. Bridge type rectifier is the most commonly used, Figure 5.9. Figure 5.9: Full Wave Bridge Rectifier Used in AC Voltmeter Circuit. 13 Cont’d… Operation; (a) During the positive half cycle (red arrow), 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. - If the deflection remains the same, the instrument using full wave rectification will have a greater sensitivity. (b) Vise-versa for the negative half cycle (blue arrow). 14 Cont’d… From the circuit in Figure 5.9, the peak value of the 10 Vrms signal with the half-wave rectifier is, E p 1.414 * E rms 14.14V peak The average dc value of the pulsating sine wave is, E ave 0.636 E p 9V Or can be compute as, Eave 0.9 * Erms 0.9 *10V 9V The AC voltmeter using full-wave rectification has a sensitivity equal to 90% of the dc sensitivity or twice the sensitivity using half-wave rectification. S ac 0.9 * S dc 15 Example 5.2: D’Arsonval Meter Full-Wave Rectifier. Each diode in the full-wave rectifier circuit in Figure 5.10 has an average forward bias resistance of 50 Ohm and is assumed to have an infinite resistance in the reverse direction. Calculate, (a) The multiplier Rs. (b) The AC sensitivity. © The equivalent DC sensitivity. Figure 5.10: AC Voltmeter Using FullWave Rectification and Shunt. Solution: (a) Calculate the current shunt and total current, I sh E m 1mA * 500 1mA Rsh 500 and . I T I sh I m 1mA 1mA 2mA 16 Cont’d… The equivalent DC voltage is, E dc 0.9 *10Vrms 0.9 *10V 9.0V RT Rs RT 2 Rd E dc 9.0V 4.5K IT 2mA Rm Rsh Rm Rsh 4500 2 * 50 (b) The ac sensitivity, (c.) The dc sensitivity, S ac 500 * 500 4.15K 500 500 RT 4500 450 / V Range 10V S dc 1 1 500 / V I T 2mA or . S dc S ac 450 / V 500 / V 0.9 0.9 17