• Lecture 8: RMS ECEN 1400 Introduction to Analog and Digital Electronics Lecture 8 Root mean square • • • • Concept RMS of a sinusoidal signal Examples Application to rectified signals Robert R. McLeod, University of Colorado http://www.sfu.ca/sonic-studio/handbook/Root_Mean_Square.html 91 • Lecture 8: RMS ECEN 1400 Introduction to Analog and Digital Electronics A useful number for describing time-varying waveforms Although peak or peak-to-peak amplitude might seem like the most obvious descriptions of a time-varying waveform, there are two reasons to want something different. 1. What we often care most about is the average power we can extract, not instantaneous voltage or current. 2. We would like a way to compare waveforms of different shape on this basis (average power). Consider a square wave voltage with a 25% duty cycle (on for only 25% of the period). What average power would it deliver to a resistor? Paverage = T 4 (V 2 peak ) @ D V V 1.0 0.8 0.6 0.4 0.2 R + 34T (0 R ) T 0.01 0.02 0.03 0.04 0.05 @ D t s 2 1 V peak (V peak 2 ) = = 4 R R 2 So we observe that this AC voltage would deliver the same power as a DC voltage of Vpeak/2. This is the RMS voltage description of this waveform. Robert R. McLeod, University of Colorado 92 • Lecture 8: RMS ECEN 1400 Introduction to Analog and Digital Electronics Root mean square voltage of a sinusoidal function Q: What DC voltage would deliver the same average power to a load as a sinusoidal AC voltage Vp sin(ωt)? AC DC V (t ) = V p sin (2π t T ) V = VEquiv P(t ) = V 2 (t ) R 2 PAverage = VEquiv R T PAverage 1 = ∫ P(t )dt T0 T Setting them equal [ ] VP2 2 = VEquiv R 2R 1 2 = ∫ V p sin (2π t T ) R dt T0 1 VP2 T = T R 2 VP2 = 2R VEquiv VP2 VP = = 2 2 VRMS V (t ) VP = 2 For a sine wave! V 2 (t ) 1.0 1.0 0.8 0.5 t 0.5 1.0 1.5 ! 0.5 ! 1.0 Robert R. McLeod, University of Colorado VP2 2 0.6 2.0 0.4 0.2 0.5 1.0 1.5 2.0 t 93 • Lecture 8: RMS ECEN 1400 Introduction to Analog and Digital Electronics In general, then… RMS stands for “root mean square” because to calculate it, you take the square Root of the Mean of the Square of a signal. Why? 2 VRMS Mean[V 2 (t )] P = = ∴ V = Mean[V 2 (t )] Average R RMS R This works the same way for current 2 PAverage = I RMS R = Mean[I 2 (t )]R ∴ I RMS = Mean[I 2 (t )] So you can use RMS quantities as if they were DC: V =I R RMS RMS 2 2 Paverage = VRMS I RMS = VRMS R = I RMS R Neat! This is why wall voltage of 120 V is actually 120 V RMS = 169.7 V peak Robert R. McLeod, University of Colorado 94 • Lecture 8: RMS ECEN 1400 Introduction to Analog and Digital Electronics RMS voltage example VP = 169.7 V The multimeter reads RMS voltage of 120 V. The oscilloscope reads VP = 169.7 V = 120 2 The average power delivered to the load is PAverage ( 2 VRMS VP2 VRMS 2 = = = R 2R 2R Robert R. McLeod, University of Colorado ) 2 1202 = = 14.4 [W ] 1K 95 • Lecture 8: RMS ECEN 1400 Introduction to Analog and Digital Electronics RMS applied to rectifiers Half-wave VDC T 2 T VDC 1 1 1 T V = ∫ V (t )dt = ∫ VP sin (2π t T )dt = VP = P T0 T 0 T π π T VRMS 1 1 2 ( ) = V t dt = T ∫0 T T 2 2 ( ) [ ] V sin 2 π t T dt = ∫ P 0 1 2 T VP VP = T 4 2 Full-wave VDC T VDC T 1 1 1 2T 2 = ∫ V (t )dt = ∫ VP sin (2π t T ) dt = VP = V T0 T0 T π π P T VRMS T 1 1 1 2 T VP 2 2 ( ) ( ) [ ] = V t dt = V sin 2 π t T dt = VP = P T ∫0 T ∫0 T 2 2 VDC of full-wave rectifier is twice that of half-wave rectifier VRMS of full-wave rectified output is same as input, ignoring diode forward bias drop. Robert R. McLeod, University of Colorado http://en.wikipedia.org/wiki/Rectifier 96 • Lecture 8: RMS ECEN 1400 Introduction to Analog and Digital Electronics Quiz 8.1 Q: Why is the RMS voltage of a rectified waveform the same as the RMS voltage of the original, bipolar waveform (assuming the same peak voltage in each case)? A: The instantaneous power delivered by the bipolar waveform does not depend on the sign of the voltage, so the same power is delivered in the positive cycle as the negative cycle. Changing the sign of the voltage does not change the power, so the RMS voltages are also the same. Robert R. McLeod, University of Colorado 97 • Lecture 8: RMS ECEN 1400 Introduction to Analog and Digital Electronics Quiz 8.2 Q: Line voltage of 120 V RMS is connected directly to a 5 Kohm resistor. The average power dissipated and peak current are? A: Average P = (Vrms)2/R=2.88 W Irms = Vrms/R = 24 mA Ipeak = Sqrt(2) Irms = 33.9 mA Robert R. McLeod, University of Colorado 98 • Lecture 8: RMS ECEN 1400 Introduction to Analog and Digital Electronics D2L 8.3 V 0.5 0.4 0.3 0.2 0.1 0.2 0.4 0.6 0.8 1.0 t tT T Q: The plot shows one period of a triangle wave with a peak voltage of 0.5 V. What is the RMS voltage of this waveform? A: Since the waveform is symmetric, the average of the first half of the period is the same as the second half, which simplifies the calculation. The equation for this time is V=t, so we need the average of the function t^2 over the period t=0 to .5. The integral of t^2 is t^3/3, so the average = (.5^3/3-0)/.5 = 1/12. Finally, taking the square root, we get 1/(2 sqrt(3)) or about 0.3 V Robert R. McLeod, University of Colorado 99