Chapter 5 - Experiment 3 AC Circuit Analysis In this lab you will use the oscilloscope to measure the amplitude and phase angle of AC voltages in a circuit. You will verify KVL and KCL for the circuit, using phasors to represent the AC voltages and currents. You will find the Thevenin Equivalent of the circuit by measuring the amplitude and phase of the open-circuit voltage and of the current in a one-ohm load resistor approximating a short circuit. In the second part of this experiment, you will measure the response of a low pass filter, a high pass filter and a band pass filter. 5.1 AC Circuit Analysis AC circuit analysis is presented in detail in the lecture segment of the course and in Part 2 of reference [10], the course textbook. The following summarizes some important concepts that are used in this experiment. In AC circuit analysis the generator is a cosine function cos wt at radian frequency w = 2πf where f is the frequency in Hz. All the voltages and currents in the circuit are cosine functions at the frequency of the generator. Any voltage in the circuit can be written as v(t) = Vm cos(wt + 0) where Vm is the amplitude of the voltage and 0 is the phase angle of the voltage. It is convenient to do calculations with voltages and currents represented by complex numbers called phasors. The phasor for Vm cos(wt + 0) is the complex number V, which can be expanded in polar form as V = Vm20. The magnitude of the phasor is the amplitude of the cosine Vm and the angle of the phasor is the phase angle of the cosine, 0. Similarly, we can represent a current i(t) = Im cos(wt + ) with the phasor I = Im24. Ꭱ The impedance of a component is defined as the ratio of the phasor voltage V across the component to the phasor current I flowing through the component, Z = ✗. The impedance of a resistor is ZR = R; of an inductor is Z = jwL; and of a capacitor is Zc where j = √1. Impedances obey V = ZI which is like Ohm's Law for resistors. We can combine impedances like resistors. Impedances Z₁ and Z2 in series add to give an equivalent impedance of 1 == jwC' Impedances in parallel combine like resistors in parallel, using the product over sum rule, Zeq = Z₁ + Z2 Zea = Z₁ Z₂ Z₁ + Z₂ Admittance is defined as the reciprocal of impedance, Y = 1/Z. Admittances in parallel add, but admittances in series combine using the product over sum rule. We can measure impedance by measuring the amplitude and phase of the voltage V across the impedance, and measuring the amplitude and phase of the current I flowing through the impedance, and evaluating Z = However, you cannot measure current directly with the oscilloscope. You can insert a small resistor of value R in series with the impedance and measure the voltage VR across the resistor. Then the current is I = VR R 51 We can solve AC circuits by converting the generator to a phasor, and each of the components to an impedance. The R, L and C components in the circuit are called branches. The phasor voltages obey Kirchhoff's Voltage Law (KVL). The phasor currents at the junctions obey Kirchhoff's Current Law (KCL). Node voltages are voltages measured relative to a ground node or datum node. We can use node analysis with phasors to solve AC circuits efficiently for the values of the phasors representing the node voltages. Or we can use mesh analysis to find phasor values for circulating mesh currents, and then find the branch currents from the mesh currents. To illustrate the use of phasors and impedance, let's solve the low pass circuit of Figure. 5.1. The generator voltage is Vs cos wt, with amplitude Vs and phase zero degrees. Resistance Rs is the internal resistance of the generator. There are three branches: Rs, L and R. There are two node voltages, v₁(t) = V₁m cos(wt + 0₁) and v₂ (t) = V2m cos(wt + 02). By “solve the circuit” is meant finding the amplitude V1m and phase 0₁ of v₁(t), and the amplitude V2m and phase 02 of v₂ (t). We will use phasor analysis to solve the circuit. i₁(t) ш Rs эт L + + Vs cos wt V₁(t) R v2(t) Figure 5.1 Low pass filter circuit in the time domain. Figure 5.2 shows the circuit transformed into the phasor domain. The generator voltage is represented by phasor V≤0°. The three branches are represented by their impedances. The impedance of the inductance is jwL. Node voltage v₁ (t) = V1m cos(wt + 0₁) is represented by phasor V₁ = √1m²01, and v2(t) becomes phasor V2. Since Rs, jwL and R are in series the impedance at the generator terminals is The current phasor is The voltage across R is Zin = Rs + jwL + R 11 = V₂ = RI₁ Vs Zin = RVS Zin Voltage V₁ is the voltage across the series combination of jwL and R and is (R+ jwL)Vs V₁ = (R + jwL)I₁ = Zin We will use this solution to the low pass filter circuit below. 52 Vs20° Rs +51 V₁ 11 m jwL R +1 Figure 5.2 Low pass filter circuit in the phasor domain. 5.2 Filter Circuits A filter is a circuit that separates signals by frequency[10]. Suppose the input to the filter contains signals at two frequencies, namely a low frequency and a high frequency. A low pass filter is a circuit that passes the low frequency through the filter with little reduction in amplitude, but rejects the high frequency, so that the amplitude of the high frequency signal is much less at the output of the filter than it is at the input. A high pass filter rejects a low frequency signal but passes a high frequency signal. A band pass filter passes signals in a band of frequencies, and rejects signals below the band and above the band. We can build simple low pass, high pass and band pass filters as described in the following. 5.2.1 Low Pass Filter Figure 5.2 shows a circuit for a low pass filter. The generator is an AC voltage. Vs(t) = V cos wt. Resistor Rs is the internal resistance of the function generator, Rs = 50. We will use the oscilloscope to measure the input voltage to the filter V₁, and the output voltage of the filter V₂. From the analysis done above, the ratio of V₂ to V₁ is V2 = ᎡᏙ Zin V₁¯¯ (R + jwL)Vs Zin R R + jwL The output voltage phasor is related to the input voltage phasor by 1 V2 = V₁ 1+ jwL R Since the amplitude of the voltage is the magnitude of the phasor, the output amplitude is given by 1 2 [V2] :|V₁| 1 + (2) ~ At a low frequency the impedance of the inductor wL is much smaller than that of the resistor, wL << R, and << 1 and 1+ WL R 2 WL ≈ 1. The inductor is like a short circuit compared to the resistor, and the output amplitude [V2] is almost the same as the input amplitude [V₁]. But at a high frequency the inductor has a much larger impedance than the resistor, wL » R, so the inductor is almost an open circuit, and the output voltage is small. The filter passes low frequencies but blocks high frequencies and so is called a low pass filter. 53 The half-power frequency is defined as the frequency were the power in the output voltage is half of the power in the input voltage. In decibels, a reduction of one half in the power is 10log 0.5=-3 dB, so the power is reduced by 3 dB and the half power frequency is also called the 3 dB frequency, fзdв· Since power is proportional to the square of the voltage, at the half power frequency the voltage is reduced by 1/√√2. We can find the half-power frequency fзd by changing the frequency until the output voltage amplitude is 1/√√2 of the input voltage. At half-power frequency WзdB = 2лƒ³d, the output voltage is |V2| = |V₁| and 1 1 1 2 √1+ (wy 2) W3d R Solve this equation to find 1 R f3dB = 2πΙ We expect a signal at a frequency much less than ƒ³ã to pass through the filter, but a signal at a frequency much higher than fзd will be rejected by the filter, with a substantial reduction in amplitude. 5.2.2 High Pass Filter Figure 5.3 shows a circuit for a high pass filter. Solve the circuit using phasor analysis to show that 1 V2 = V₁ 1 1 + jwRC Since the amplitude of the voltage is the magnitude of the phasor, the output amplitude ratio is given by 1 [V2] = |V₁| 2 1 + (WRC) 1 WC 1 ως At a low frequency the impedance of the capacitor is much bigger than that of the resistor, >> R, and so the capacitor is like an open circuit compared to the resistor, and the output amplitude [V2] is small compared to the input amplitude. But at a high frequency the capacitor has a much smaller impedance than the resistor, << R, so the capacitor is like a short circuit, and the output voltage is almost equal to the input voltage. We can show that the half-power frequency 1 ως f3d = 1 1 2π RC We expect an input frequency much less than ƒ³ã³ to be rejected the filter with a substantial reduction in amplitude, but an input frequency much higher than fзd will be passed by the filter. 54 ..... Vs20° RS +51 C Figure 5.3 High Pass Filter R V2 +1 5.2.3 Band Pass Filter 1 ως Figure 5.4 shows a circuit for a band pass filter. At low frequencies in the capacitor impedance is large and the capacitor behaves like an open circuit. At high frequencies, the inductor impedance wL is large and the inductor behaves like an open circuit. At the center frequency wc = 2πfc, the series impedance of the inductor and capacitor is zero jwcL j WcC = 0 so the inductor and capacitor in series behave as a short circuit and the output voltage is the same as the input voltage. Solve this equation to find that the center frequency is fc 1 1 2π √LC Using the voltage divider relationship the output voltage is R V2 = V₁ 1 R+jwL + jwC The amplitude of the output voltage is 1 |V2| 1 + (CL R WRC At the half power frequencies 2 WL 1 1 + R WRC = √2 Accounting for the possibility of a positive or a negative square root we have WL 1 R = ±1 WRC hence Using the quadratic formula LCw² WRC-1 = 0 W1,2 ±RC±√R²C² + 4LC 2LC Since √R2C2 + 4LC > RC, take the positive values for the half power frequencies, 55