Explain and predict using a block diagram, how a true RMS meter processes an input signal (sine and complex waves).

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OBJECIVE HREE When you complete this objective you will be able to Explain and predict using a block diagram, how a true RMS meter processes an input signal (sine and complex waves). Learning Activity Complete each of the Learning Activities listed below. Read the learning material. Learning Material rue RMS responding voltmeter Complex waveforms are most accurately measured with a true RMS responding voltmeter. his instrument produces a meter indication by sensing waveform heating power, which is proportional to the square of the RMS value of the voltage. his heating power can be measured by feeding an amplified version of the input waveform to the heater element of a thermocouple whose output voltage is then proportional to E RMS. One difficulty with this technique is that the thermocouple is often non-linear in its behavior. his difficulty is overcome in some instruments by placing two thermocouples in the same thermal environment, as shown in the block diagram of the true RMS responding voltmeter of Figure 5-. he effect of the nonlinear behavior of the couple in the input circuit (the measuring thermocouple) is cancelled by similar non-linear effects of the couple in the feedback circuit (the balancing thermocouple). he two couple elements form part of a bridge in the input circuit of a DC amplifier. he unknown AC input voltage is amplified and applied to the heating element of the measuring thermocouple. he application of heat produces an output voltage that upsets the balance of the bridge. he unbalance voltage is amplified by the DC amplifier and fed back to the heating element of the balancing thermocouple. Bridge balance will be re-established when the feedback current delivers sufficient heat to the balancing thermocouple, so that the voltage outputs of both couples are the same. At this point the DC current in the heating element of the feedback couple is equal to the AC current in the input couple. his DC current is therefore directly proportional to the effective, or RMS, value of the input voltage and is indicated on the meter movement in the output circuit of the DC amplifier. he true RMS value is measured independently of the waveform of the AC signal, provided that the peak excursions of the waveform do not exceed the dynamic range of the AC amplifier. 5-3

Measuring hermocouple Input Voltage ac Amplifier + dc Amplifier - + Balancing hermocouple Indicating Meter Feedback Current Figure 5-: Block diagram of a true RMS reading voltmeter. he measuring and balancing thermocouples are located in the same thermal environment. A typical laboratory-type RMS responding voltmeter provides accurate RMS readings of complex waveform having a crest factor (ratio of peak value to RMS value) of /. At per cent of full-scale meter deflection, where there is less chance of amplifier saturation, waveforms with crest factors as high as / could be accommodated. Voltages throughout a range of µv to 3 V within a frequency range of Hz to MHz may be measured with most good instruments. Figure 5-3 shows an IC that converts an input signal to an equivalent DC voltage. +V Error Amplifier RMS Sensor Chip + - V in V out -V Figure 5-3: Simplified RMS circuit 5-4

hermocouple-based thermal elements suffer from the following drawbacks: long settling time (the time required for Vout to settle when Vin changes), sensitivity to external temperatures, low output voltages, susceptibility to damage from overload he following information refers to the fluke 79A RMS circuit in Figure 5-3: fast settling time, 3 to 6 sec., large output voltage Vdc compared to thermocouples mv s, temperature stability to 35 degrees C, input voltages from mv to Vac, frequency response Hz to MHz, Hi accuracy PPM to 5 PPM depending on voltage range and frequency selected Explain below how Figure 5-3 operates. rue RMS Meters Only meters that are specified as true RMS can determine the true RMS voltage of waveforms other than sine waves. rue RMS Meter Meter will disply the true RMS voltage of the input signal. Signal must fall between the meters limits of frequency, crest factor and voltage. Figure 5-4 On some true RMS voltmeters (fluke 8A) the true RMS value must be determined with two measurements.. Set voltmeter to volts DC and record the value.. Set voltmeter to volts AC and record the value Vac RMS Calculate: V RMS ( Vac ) ( Vdc) RMS + Some true RMS meters (fluke scope meter, fluke 45) will display the true RMS value directly. he meter measures Vac RMS and Vdc then completes the calculation below and displays the true RMS value. 5-5

Average Responding Meters Average responding RMS indicating for sine wave only meter Meter will display incorrect values for any waveform other than a sine wave. Figure 5-5 Determine the true RMS voltage for the ½ wave rectified signal shown below. Also determine what a true RMS meter would display when set to volts DC and volts AC. +5 Vp / / Wave V RMS ( f ( t) dt t Figure 5-6(a) V RMS (Vp sin ωt)dt Vp sin ( ωt)dt 5-6

Calculate Vdc for ½ wave rectifier sine wave shown in Figure 5-6(b). +5 / Wave / Figure 5-6(b) Vdc V p sin ωt dt Vp sin( ωt)dt - Vp cos( ωt) ω + - π Figure 5-7 Vdc.59 V Recall V RMS ( Vac RMS ) + (Vdc) Vac RMS (V RMS ) - (Vdc) herefore: Vac RMS when set to volts AC. (.5) - (.59).93 V. rue RMS meter display 5-7

Determine the V RMS voltage for the signal shown below. Also determine what a true RMS meter would indicate when set to volts AC and volts DC. +5-5 / Figure 5-8(a) rue RMS V RMS t (f(t)) dt f(t) + 5; 5; ( t < ( ) t < ) V RMS (5V) dt + (-5V) dt 5t + 5t Vdc for Figure 5-8(b) volts. 5-8

What would a true RMS meter indicate on volts AC range, using Figure 5-8(b)? +5-5 / Figure 5-8(b) Recall V RMS V RMS ( Vac RMS ) + (Vtrue RMS) ( 5V) - (Vdc) - (Vdc) 5 volts An average responding RMS indicating meter would display 5.555 volts when set to volts AC. he average responding meter would therefore have a 5-5.555 x -. % error when measuring the voltage of Figure 5-8(a). 5 Determine V RMS for signal shown in Figure 5-9. 5 V V /4 Figure 5-9 Duty Cycle 5% f(t) 5V; t < 4 V; t < 4 4 4 5 V RMS (5v) dt (v) t 5 (v) 4 5(v) 4.5 V V RMS.5 V 5-9

Vdc Vac RMS Another method to predict what the meter will read on volts AC function, is shown using Figure 5-9 as the input. 5 V V 3.75 V Ideal Rectifier V.5 V 3.75 V /4 Figure 5-3.5 V 5-3

Vac RMS 4 (3.75) dt + (.5) dt 4 4 4.65 t.565 t 4 Determine the V RMS voltage of the signal shown in Figure 5-3. Also determine what a true RMS meter would indicate when set to volts DC and volts AC function. Frequency - khz Slope v.5 ms v s V.5 ms.5 ms Figure 5-3 V RMS (f(t) dt t f(t) v s t < t.5 ms.5 ms < t. ms V RMS 5-3

Determine V RMS voltage of the signal shown below. Also determine what a true RMS meter would indicate when set to volts DC and volts AC function. 8 V 6 ms Period 8 ms V V t Figure 5-3 V RMS (f(t) dt t f(t) 8 v v < t ms ms < t 8 ms V RMS 4.36 volts Vdc area method 5-3

Vdc calculus method: Vdc Vac RMS ( V ) - (Vdc) RMS ( 4.36) - (3.5).6 volts.v/div 5 µs/div V max 6 v 36.8% t 33 µs Figure 5-33 Measured Vdc for the signal in Figure 5-33.3 volts. Measured V RMS.76 volts. 5-33

Determine the equation for the signal in Figure 5-33. he wave form is an exponential fall. Recall time constant e -.3679 36.8% of V max. From Figure 5-33, determine the time for the waveform to get to 36.8% of V max. τ 33 µs Equation v(t) 6 e -t/τ for waveform in Figure 5-33. Verify the above equation let t5 µs, v(t).3 volts Answer:.3 v 6 e 5 µ s 33 µ s Determine Vdc, V RMS and what a true RMS voltmeter would indicate on volts AC (Vac RMS ) for the signal in Figure 5-33. he exponential fall equation Ve -x requires five time constants τ, for the waveform to decay to approximately. Determine V RMS of Figure 5-33: v(t) 6e -t/τ τ 33 µs V RMS (f(t) dt (6e t - τ ) dt 5-34

Note: he measured and calculated voltage values are slightly different. One reason for this is the approximation of τ. 5-35

Determine Vdc of Figure 5-33: v(t) 6e -t/τ τ 33 µs Vdc (f(t) dt ms ms 6e -t τ dt Note: he measured and calculated Vdc voltage values are slightly different. One reason for this is the approximation of τ. Vdc (recall Vdc is what a true RMS or average responding voltmeter indicates when set to volts DC.) V RMS V RMS (Vdc) + (Vac ) RMS Solve for; Vac RMS 5-36