RICHLAND COLLEGE School of Engineering Business & Technology Rev. 0 W. Slonecker Rev. 1 (8/26/2012) J. Bradbury

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1 RICHLAND COLLEGE School of Engineering Business & Technology Rev. 0 W. Slonecker Rev. 1 (8/26/2012) J. Bradbury INTC 1307 Instrumentation Test Equipment Teaching Unit 3 Alternating Current Meters Unit 3 AC Meters Objectives: Review basic AC signal concepts. Convert the D Arsonval movement into an AC ammeter and voltmeter. Describe the operation of a halfwave rectified AC meter. Describe the operation of a fullwave rectified AC meter. Determine the sensitivity of an AC voltmeter. Determine the loading effects of AC meters. AC Sinusoids The graph on the right shows one cycle of an AC sinusoid. One cycle has 360 of phase, which also equals to 2π radians. An AC signal varies as a function of time, in this case 9 ms is required for one cycle. The time of one cycle of the waveform in seconds is the period, T. The frequency in Hertz, f, is equal to the reciprocal of the period. f = 1/T. The frequency is 1/0.009 = Hertz. The peak value for the waveform is 1.0. An AC signal alternates between its positive and negative peak values. The RootMeanSquare value for a sinusoidal waveform is times the peak value. For a peak value of 10, the RMS value is 7.07 (10 divided by the square root of 2 to be exact). RMS values can be calculated for nonsinusoidal waveforms, but only sinusoids have RMS values equal to peak. The RMS value is also called the effective value. AC effective values are equivalent to the DC values needed to produce the same resistive heating effects. The signal peaktopeak value is twice the peak value. For a halfwave rectified AC signal, the average value is peak/π or peak In the graph on the right, the average value is 318mV. A rectifier eliminates polarity reversals. For a halfwave rectifier, half the signal is eliminated. Page 1 of 6

2 For fullwave rectification, half of the AC signal has its polarity reversed as shown on the right. In this case, the negative polarity is reversed. Instead of only 4 peaks in 40 ms as for the halfwave signal, there are almost 9 peaks in 40 ms for the fullwave signal. It is reasonable that the fullwave average is twice the halfwave average; 2 peak/π or peak AC Definitions Waveform Equation= Amplitude sin (2π ftθ ) f =frequency in Hertz, t=time in sec onds, θ= phase shift in radians ω=radian frequency=2πf 1 radian= 180 π =57.296, 360 =2π radians Amplitude= peak value, RMS value= Amplitude =0.707 peak value 2 Power in AC resistive circuits=i RMS V RMS Waveform Period=T =time for 1 cycle Frequency=f =Cycles per second, f = 1 T Average 1 Amplitude 2 Amplitude wave value=, Average full wave value= 2 π π AC Meters AC meters are DC meters with rectified inputs. If you understand DC meters, then all you need to do is learn the basics of rectification and you will understand AC meters. Remember; rectification converts signals with dual polarity components into signals with single polarity components. A DC signal does not change polarity if it is voltage, or does not change direction if it is current. We usually think of a DC signal as being steady and unchanging. However, the definition of DC allows fluctuations in value so long as the polarity (or direction) does not reverse. Page 2 of 6

3 Halfwave AC Voltmeter Analysis The circuit on the right is a DC voltmeter modified to measure AC voltage. Notice that the meter has a full scale current of 1 ma, while the meter resistance is relatively low, 100Ω. These values imply that the meter is probably a microammeter with a shunt resistor. To analyze the circuit, consider that sensitivity is 1/Ifsd, or 1/1mA = 1000Ω/volt. If the meter were measuring DC voltage and if the diode behaved like a short circuit, then the full scale voltage would be Ifsd R total or 1mA 10.1kΩ = 10.1 VDC. However, the halfwave rectifier means that the equivalent DC voltage applied to the meter (V pk /π) is equal to 10.1 volts. Thus V pkac = v avgac π = 10.1 π = v. Another consideration is that the diode has an intrinsic voltage drop of 0.7 volts. This adds to the V pkac voltage required for full scale. So an AC peak voltage of v pkac is needed for meter full scale deflection. Putting this voltage in terms of RMS input voltage gives = v RMS for a full scale reading. Sensitivity is (10k 100)/22.93 = ohms per volt, about 44% of the DC sensitivity. Halfwave AC Voltmeter Design Design a halfwave AC voltmeter using a movement with Ifsd = 50µA, Rm = 1500Ω, and a full scale voltage of 10 V RMS. 1. Calculate Vpk as = v pk. 2. Subtract 0.7v from Vpk to adjust for diode drop: = v pk 3. Calculate halfwave average: 13.44/π = v. This is the DC full scale value. 4. Meter sensitivity is 1/Ifsd = 20kΩ/volt. This requires total meter resistance to be kΩ = 85.56kΩ. 5. Rs = total R Rm = 85.56kΩ 1.5kΩ = 84.06kΩ The meter schematic is like the one at the top of the page only R1 = 84.06kΩ, and the meter parameters are given in the problem specification. In this case the actual AC sensitivity is total resistance divided by input RMS voltage or 8.556kΩ/volt. Therefore the halfwave sensitivity is about 0.43 times the DC sensitivity. The sensitivity factor is not a constant because the rectifier voltage drop (0.7v) is more or less of a factor depending on the magnitude of the full scale voltage specification. It should be pointed out that a rectifier produces pulsing DC. The meter movement inertia (mechanical response time) will smooth out the pulse effect on the needle if the frequency of the waveform is high enough. At lower frequencies, there will be noticeable pointer shake. R1 10kOhm Rm v pk I fs = 1 ma 10.1 v avg R m = 100Ω DC voltmeter modified to measure AC Fullwave AC Voltmeter Analysis For the meter on the right, a fullwave bridge rectifier has been added for fullwave rectification of the input waveform. The fullwave bridge changes two aspects of the analysis. First, the average voltage, v avgac, is 2v pk /π, and second, there are two diode drops in the bridge since two diodes conduct instead of one. Since the DC aspect of the meter is identical to the halfwave case, we know that 10.1v for v avgac will give a full scale reading as in the half wave case. R1 10kOhm Rm D4 D v pk I fs = 1 ma 10.1 v avg D2 R m = 100Ω Fullwave AC Voltmeter Page 3 of 6

4 Thus V pkac = v avgac π/2 = 10.1 π/2 = v, to which we add 1.4 volts for two diode drops, giving an AC input peak voltage of v pk. So for full scale deflection, the input RMS voltage is Vrms. Meter sensitivity is ohms per volt, about 83% of the DC sensitivity. Fullwave AC Voltmeter Design Design a fullwave AC voltmeter using a movement with Ifsd = 50µA, Rm = 1500Ω, and a full scale voltage of 10 V RMS. 1. Calculate Vpk as = v pk. 2. Subtract 1.4v from Vpk to adjust for diode drop: = v pk 3. Calculate fullwave average: /π = 8.11 v. This is the DC full scale value. 4. Meter sensitivity is 1/Ifsd = 20kΩ/volt. This requires total meter resistance to be kΩ/volt = 162.2kΩ. 5. Rs = total R Rm = 162.2kΩ 1.5kΩ = 160.7kΩ The meter schematic is like the one at the bottom of page 3 only R1 = 160.7kΩ, and the meter parameters are given in the problem specification. In this case the actual AC sensitivity is total resistance divided by input RMS voltage or 16.22kΩ/volt. Therefore the fullwave sensitivity is about 0.81 times the DC sensitivity. With a higher input resistance, the fullwave meter will load circuits less than the halfwave meter. As you can see, the design steps are the same as for the halfwave meter only adjusted for the bridge rectifier; i.e. different v avg and 2 diode drops. Fullwave Bridge Conduction Explanation A diode only conducts in one direction. For conventional current, the arrowhead in the symbol points in the direction of conduction. For electron current, conduction is opposite the direction of the arrowhead! The arrowhead is the anode and the line is the cathode. A diode conducts when the anode is more positive than the cathode. For the bridge circuit, when the input voltage is positive, conducts and the resulting current flow causes D2 to also conduct. When conducting, there is a 0.7 volt drop across the diode. When the input voltage is negative, the lower input lead is more positive causing D3 to conduct and also D4. The figure shows diode conduction for the two polarities of input voltage. The diodes effectively swap the conduction path like switches. Vp Vp Vp D4 D2 Vp 0.7 D2 D4 0.7 D3 Vp0.7 Vp0.7 D3 Vp0.7 V = Vp Vp0.7 V = Vp Page 4 of 6

5 You really need to understand bridge rectification before moving on. Look at it, study it, ask questions about it! AC Ammeters AC rectified ammeters are exactly like DC ammeters but with a series diode. The problem for AC ammeters is the 0.7 volt forward diode voltage. Since ideal ammeters should behave like short circuits, the forward diode voltage causes a lot of measurement error. The only AC ammeter we will consider is a halfwave ammeter for which we will ignore the forward diode voltage and only consider the effects of halfwave averaging; i avg = i pk /π. Calculations are straightforward. The AC RMS current is equal to the full scale DC current times π times i RMS = i fs π Rm ma pk I fs = 1 ma ma RMS R m = 100Ω Remember that this simple analysis ignores the forward diode voltage! AC Meter Loading Meter loading for AC meters works just like it did for DC meters, except that sensitivity must be adjusted for signal rectification. This works out to roughly 44% sensitivity for halfwave meters and 81% for fullwave meters. In the circuit on the right, Vth = 15 v RMS, Rth Rth = 10kΩ, Rx = 10kΩ. Using voltage division it is easy to see that the voltage across Rx is 7.5 v RMS. Vth Rx The voltmeter is a halfwave rectified DC meter with a sensitivity of 10kΩ/v on a 20 volt scale. What does the meter read? Using a sensitivity of 44% of DC sensitivity gives a meter resistance of kΩ/v = 88kΩ. This parallels the 10kΩ load to give a total load resistance of 8.98kΩ. Using voltage division, the meter reads kΩ/18.98kΩ = v RMS, for v error or 5.37% error V Using the same meter with a fullwave rectifier gives a meter resistance of kΩ/v = 162kΩ. This changes the Rx load from 10k to 9.42k, for a meter voltage of kΩ/19.42kΩ = v RMS, for v error or 3% error. Page 5 of 6

6 Other AC Meters Electrodynamometer movement is the most fundamental meter used prior to electronic digital instruments. This meter is a currentcontrolled device. The pointer deflects because of current flow through a moving coil wound in series with two fixed coils. It can measure DC or AC current and voltage. The ironvane movement consists of a fixed coil of many turns and two iron vanes placed inside the fixed coil. As current passes through the fixed coil, the resulting magnetic field magnetizes the iron vanes with the same magnetic polarity, causing them to repel each other. One vane is fixed while the other is free to rotate, moving the pointer. The thermocouple meter (also called a bolometer) works by the measured heating in a thermocouple junction. The heating causes a current to flow through a meter movement. Page 6 of 6