ELEC273 Lecture Notes Set C Circuit Theorems The course web site is: http://users.encs.concordia.ca/~trueman/web_page_273.htm Final Exam (confirmed): Friday December 5, 207 from 9:00 to 2:00 (confirmed) Homework on phasors and impedance. The final exam in ELEC 273 is December 4, 208 from 2:00 to 5:00.
Network Theorems: Chapter 0 Sinusoidal Steady State nalysis Superposition Thevenin s Theorem Norton s Theorem Chapter C Power nalysis The Maximum Power Transfer Theorem
Thevenin s Theorem for C Circuits ZZ TT Two Terminal Circuit VV TT Thevenin s Theorem ny two-terminal network consisting of voltage sources, current sources, dependent sources, inductors, capacitors and resistors is equivalent to a voltage source VV TT in series with an impedance ZZ TT. Procedure to find the Thevenin Equivalent Circuit:.Open-circuit test: terminate the circuit with an open circuit and find the open-circuit voltage, VV oooo. 2.Short-circuit test: terminate the circuit with a short circuit and find the short-circuit current, II ssss. 3.The Thevenin equivalent voltage source is VV TT = VV oooo. 4.The Thevenin equivalent impedance is ZZ TT = VV oooo. II ssss Note that ZZ TT can also be found as the impedance of the dead circuit with the independent sources set equal to zero.
Example: Thevenin s Theorem 2 Ω 0 sin 3tt 2 H )Find the Thevenin Equivalent Circuit at a frequency of ωω=3 rad/sec. Use phasors relative to sine. 2)Find the voltage across a load consisting of a ohm resistor in series with a capacitor of value C=0.6667 F, using: The Thevenin equivalent circuit The original circuit The answer must be the same in both cases!
Find the Thevenin Equivalent Circuit. 0 sin 3tt 2 Ω 2 H In this example I have written my phasors relative to sine. Hence 0 sin 3t becomes phasor 0 angle zero. However, we must remember to convert phasors back to sine. 0 2 Ω jjj.5 Ω VV oooo Thus phasor VV = ee jjθθ converts to time function vv tt = sin(ωωωω θθ). Convert the circuit to phasors and impedance at frequency ωω = 3 rad/sec. The impedance of the inductance is jjωωωω = j3x = jjj.5 ohms 2 ) Find the open-circuit voltage. 2) Find the short-circuit current.
0 2 Ω jjj.5 Ω VV oooo Find the open-circuit voltage 0 V 2 oc Voc Voc = 0 j.5 Solve for VV oooo : 0 Voc Voc Voc 6 j 6 j 6 j = 0 2 j.5 3 j(0 V ) 4V 6 jv = oc oc oc 0 30 j 3 jv 4V 6 jv = oc oc oc 0 4 Voc 9 jvoc = 30 j V oc 30 j 30.57 = = = 3.046 0.482 = 3.046 24 4 9 j 9.849.53 V T = V oc = 3.046 24
2)Find the short-circuit current 0 2 Ω jjj.5 Ω II ssss I sc 0 = = 5 0 2
3)Find the impedance V oc I sc = 3.046 24 = 5 0 Voc 3.046 24.0 ZT = = = 0.6092 24.0 = 0.5565 I 5 0 sc V T = V oc = 3.046 24 j0.2477 This phasor is written relative to sine so the time function is 3.046 sin(ωωωω 24 ) ZZ TT = 0.5565 jjj.2477 Ω 0.5565 Ω 0.08257 H VV TT = 3.046 24 VV TT = 3.046 24 The reactance of 0.2477 ohms is equivalent to an inductor of value ωωll = 0.2477 so LL = 0.2477 = 0.2477 ωω 3 = 0.08257 H
2)Find the voltage across a load consisting of a ohm resistor in series with a capacitor of value C=0.6667 F, using: The original circuit The Thevenin equivalent circuit Using the original circuit: 0 sin 3tt 2 Ω 2 H 0.6667 F The impedance of the capacitor is ZZ CC = JJωωωω = jjjjjj.6667 = jjj ohms Draw the circuit in the frequency domain using phasors and impedances. 0 2 Ω jjj.5 Ω jjj Ω
Using the original circuit, find VV LL : VV LL 0 2 Ω jjj.5 Ω jjj Ω Write a node equation: 0 V 2 V 2 L V L L VL VL j.5 VL VL j.5 j.5 V V L j2 L = j2 0 2 ( j2) 2( j2) 2( j.5) ( j2) ( j2) = 0 2( j.5) = 5 V L V L j3.5 6 j3 = 5 6 j3 = 5 = 2.906 8. 9 j3.5 Do we get the same answer using the Thevenin Equivalent Circuit?
Find the load voltage using the Thevenin Equivalent Circuit. Define ZZ LL = jjj Ω VV LL ZZ TT VV TT jjj Ω V T = V oc = 3.046 24 Z T = 0.6092 24.0 = 0.5565 j0.2477 Z L = j2 V T V Z T L V Z L L = 0 V L = ( j2)(3.046 24.0) 0.5565 j0.2477 ( j2) V L = ZT Z L Z Z Z LV T = Z Z T L V L = ZT Z L V L T L V Z T T V Z Note that this equation is simply a voltage divider between ZZ TT and ZZ LL. T T V L V L V L (2.236 63.4)(3.046 24.0) =.5565 j.7523 (2.236 63.4)(3.046 24.0) = 2.343 48.8 = 2.906 8. 95 This agrees with the answer obtained above of V L = 2.906 8. 9
Norton s Theorem for C Circuits. Two Terminal Circuit II NN YY NN Norton s Theorem ny two-terminal network consisting of voltage sources, current sources, dependent sources, capacitors, inductors and resistors is equivalent to a current source II NN in parallel with an admittance YY NN. Procedure to find the Norton Equivalent Circuit:.Open-circuit test: terminate the circuit with an open circuit and find the open-circuit voltage, VV oooo. 2.Short-circuit test: terminate the circuit with a short circuit and find the short-circuit current, II ssss. 3.The Norton equivalent current source is II NN = II sscc. 4.The Norton equivalent susceptance is YY NN = II sscc. VV oocc Note that YY NN can also be found as the admittance of the circuit with the independent sources set equal to zero.
Norton Equivalent from the Thevenin Equivalent Find the open-circuit voltage VV oooo and the short-circuit current II ssss. ZZ TT VV TT Thevenin: VV TT = VV oooo ZZ TT = VV oooo II ssss II NN YY NN Norton: II NN = II SSSS YY NN = II ssss VV oooo ZZ TT VV TT II SSSS = VV TT ZZ TT We can change The Thevenin Equivalent Circuit to the Norton Equivalent Circuits using: YY NN = ZZ TT and II NN = II SSSS = VV TT ZZ TT
Example: Find the Norton Equivalent Circuit 0 sin 3tt 2 Ω 2 H We found the Thevenin Equivalent Circuit earlier in this set of notes. Find the Norton Equivalent Circuit at a frequency of ωω=3 rad/sec. Method:.Find the open-circuit voltage. 2.Find the short-circuit current. 3.Find the Norton Equivalent Circuit: II NN = II SSSS YY NN = II ssss VV oooo 0 sin 3tt 0 sin 3tt 2 Ω 2 Ω 2 H 2 H VV oooo II ssss Open-circuit load. Short-circuit load.
Find the open-circuit voltage: 0 sin 3tt 2 Ω 2 H We solved for the open-circuit voltage and the short-circuit current when we found the Thevenin equivalent earlier in these notes. 0 2 Ω jjj.5 Ω VV oooo Convert the circuit to phasors and impedance. The impedance of the inductance is jjωωωω = j3x = jjj.5 ohms 2 ) Find the open-circuit voltage 0 V 2 oc Voc Voc j.5 = 0 V T = V oc = 3.046 24
Find the short-circuit current: 0 2 Ω jjj.5 Ω II ssss I sc 0 = = 5 0 2
3)Find the admittance: V oc I sc Y N = 3.046 24 = 5 0 I sc 5 0 = = =.64 24. 0 V 3.046 24.0 oc I N = I sc = 5 0 2 Ω 0 jjj.5 Ω II NN YY NN
Norton Equivalent, example #2: 5.3 mh 0 cos ωωωω 0.62 mh 330 μμf Find the Norton Equivalent Circuit at 60 Hz. Convert the Norton Equivalent to the Thevenin Equivalent Circuit. Procedure to find the Norton Equivalent Circuit:.Open-circuit test: terminate the circuit with an open circuit and find the open-circuit voltage, VV oooo. 2.Short-circuit test: terminate the circuit with a short circuit and find the short-circuit current, II ssss. 3.The Norton equivalent current source is II NN = II ssss. 4.The Norton equivalent susceptance is YY NN = II ssss. VV oooo
Convert to phasors and impedances: 5.3 mh 0 cos ωωωω 0.62 mh 330 μμf Voltage source: vv ss tt = 0 cos ωωωω becomes phasor VV ss = 0 volts t ff =60 Hz the radian frequency is ωω = 2ππππ = 377 r/s jjωωωω = jjjjjjjjjj.3xx0 3 = jjj ohms jjωωωω = jjjjjjjjj0.62xx0 3 = jj4 ohms = jj = jjωωωω ωωcc jj 377xxxxxxxx0 6 = jjj ohms
.Find the open-circuit voltage: VV oooo VV oooo - 0 VV oooo jjj VV oooo jjj VV oooo jjj = 0 VV oooo jjj jjj jjj = 0 jjj jjj jjj jj2 jjj ( jj2)( jj4) VV oooo ( jjj)( jjj)(jjj) jjj jjj jjj jjj ( jjj)( jjj) VV oooo ( jjj)(jjj) 2jj 8 2jj 4 jjj jjj 8 VV oooo = 0 VV oooo 5 jjj 8 jj2 = 0 jjj 8 = 0 jjj = 0 VV oooo 5 jjj 8 jj2 = 0 VV oooo = 0 VV oooo = 0 8 jjj 5 2jj 8.25 4.0 5.39 2.8 VV oooo = 68.4 35.8
2.Find the short-circuit current: II ssss II ssss = 0 jjj = 22 j44 = 49.9 63.4
3 and 4.Find the Norton Equivalent Circuit: VV oooo = 68.4 35.8 II ssss = 0 = 22 j44 = 49.9 63.4 jjj Y N = I V sc oc = 49.9 63.4 68.4 35.8 = 0.292 27.6 Siemens I N = I sc = 49.9 63. 4 mps II NN YY NN
Check by finding the dead-circuit impedance: jjj jjj = jjj jjj jjjjjj = 0.8 jjj.6 ohms ZZ TT ZZ TT = jj2 0.8 jj3.6 = jj2 0.8 jj3.6 jj2 0.8 jj3.6 ZZ TT = 3.034 jj.586 = 3.423 27.6 ohms YY NN = ZZ TT = 3.423 27.6 = 0.2920 27.6 Siemens This agrees with the value found on the previous slide. ZZ TT
Convert the Norton Equivalent to the Thevenin Equivalent Circuit: Y N I N = 0.292 27. 6 = 49.9 63. 4 VV oooo - VV oooo = II NN YY NN = 49.9 63.4 0.292 27.6 = 36.6 jjjj.5 = 68.4 35.8 volts VV TT = VV oooo = 98.5 = 68.4 35.8 ZZ TT = = = 3.423 27.6 ohms YY NN 0.292 27.6
The Superposition Theorem for C Circuits VV oo VV ooo VV ooo VV sss Linear Circuit VV sss VV sss Linear Circuit Linear Circuit VV sss Two sources acting together. Source # acting alone. Source #2 acting alone. The Superposition Theorem is exactly the same for C circuits as for DC circuits. The response of a circuit VV 0 with two sources VV sss and VV sss acting together is equal to the sum of: the response VV 0 with source VV sss acting alone and VV sss =0, plus the response VV 02 with source VV ss2 acting alone and VV ss =0, VV 0 = VV 0 VV 02.
Superposition: DC Sources and C Source Question from an old final exam! In the circuit below there are two DC sources, VV 2 = 3 volts and II = amp. There is also an C source, vv tt = 0. cos 000tt volts. Use Superposition to find the voltage across the current generator vv tt. The component values are RR =, RR 2 = 2 Ω, CC = 500 microfarads, CC 2 = 40 microfarads, LL = mh. CC = 500 μμf CC 2 = 40 μμf RR 2 = 2 Ω LL = mh vv(tt) 0. cos 000tt vv xx 5vv xx RR = vv II = amp VV 2 = 3 volts Use Superposition to separate this problem into: )The DC response vv DDDD to the DC sources VV 2 = 3 volts and II = amp 2)The C response vv CC (t) to the C source vv tt = 0. cos(000tt) volts at ωω=000 r/s Then the response of the circuit is vv tt = vv DDDD vv (t)
Solve with the DC sources: CC. cos 000tt vv xx vv xx RR CC 2 RR 2 LL 5vv xx vv(tt) vv vv DDDD amp 2 Ω 5vv xx 3 volts 3 Draw the DC circuit: C source becomes zero volts = short circuit Capacitor: ii = CC dddd = 0, C s become open dddd circuits at DC. Inductor: vv = LL dddd = 0, L s become short dddd circuits at DC. Note that vv xx = 0 at DC. Node equation: vv DDDD 3 = 0 2 3 vv DDDD 2 = 0 vv DDDD = volt at DC The C source is set to zero. The capacitors become open circuits. The inductor becomes a short circuit. Since CC and CC 2 are open-circuits at DC, there is zero DC current in RR and so vv xx = 0 Then the dependent source is zero: 5vv xx = 0
Solve with the C source: use phasors and impedance. Use Superposition to separate this problem into: )The DC response vv DDDD to the DC sources VV 2 = 3 volts and II = amp 2)The C response vv CC (tt) to the C source vv tt = 0. cos(000tt) volts at ωω=000 r/s CC = 500 μμf CC 2 = 40 μμf RR 2 = 2 Ω LL = mh vv CC (tt) 0. cos 000tt vv xx 5vv xx RR = vv II = amp VV 2 = 3 volts Draw the C circuit. CC = 500 μμf CC 2 = 40 μμf 2 Ω mh 0. cos 000tt vv xx 5vv xx vv CC (t)
Use phasors and impedance: 0. cos 000tt CC = 500 μμf jjj vv xx VV xx CC 2 = 40 μμf 2 Ω mh 5vv xx jjjj VV 2 Ω jjj Draw the C circuit: The DC sources are set to zero. vv tt = 0. cos(000tt) becomes phasor VV = 0. The frequency is ωω = 000 r/s. = jjωωcc jjjjjjjjjjjjjjj06 = jjj = jjωωcc 2 jjjjjjjjjj40xx06 = jjjj jjωωωω = jjjjjjjjjjjjj0 3 = jjj Node equations: 0. VV xx 5VV xx VV 0. VV xx jjj VV xx VV xx VV jjjj = 0 VV xx VV jjjj 5VV xx VV 2 jj = 0 Solve the equations to find VV = 0.4536 8.
Superposition: add the DC solution to the C solution. vv xx 2 Ω 5vv xx vv 3 DC Solution: vv DDDD = volt VV xx VV 0. jjj VV xx jjjj 2 Ω jjj 5VV xx VV C Solution: VV = 0.4536 8. vv tt = 0.4536 cos 000tt 8. volts CC. cos 000tt vv xx RR CC 2 RR 2 LL 5vv xx vv(tt) vv amp 3 volts With both the DC sources and the C source active: vv tt = vv DDDD vv tt So vv(tt) = 0.4536 cos 000tt 8. volts
Superposition with two frequencies: vv sss tt = 0. cos ωω 2 tt 50 Ω vv 50 Ω vv 2 CC LL RR vv 3 RR vv 4 vv sss tt = 0. cos ωω tt vv 4 (tt) n antenna receives two signals at the same time, one at frequency, ff = GHz and the other at frequency ff 2 = 2 GHz. The antenna is modelled with two voltage sources in series. The circuit uses two ideal op-amps. The component values are RR = ohm, RR = ohm, CC = pf and LL = 6.333 nh. Use Superposition to find the output voltage vv 4 (tt).
nalysis of the first stage: vv 50 Ω 0 vv sss vv ss2 50 Ω vv sss tt 0 vv sss tt vv 2 vv sss vv sss vv 50 vv vv 2 50 = 0 The op-amps are ideal with infinite gain and very high input impedance. Consider vv to be a virtual ground, so vv = 0. Then vv sss vv sss 50 vv 2 50 = 0 vv 2 = vv sss vv sss The output of the first stage is the sum of the two voltages.
Use phasors to analyze the second stage at some frequency ωω: vv ss tt = cos ωωtt phasor VV ss = 50 Ω 50 Ω VV 0 VV 2 = VV ss VV 3 0 0 0 jjωωll RR jjωωωω RR VV 4 VV 4 Drive the circuit with a generator at frequency ωω: vv ss tt = cos ωωtt The phasor is VV ss = ee jjj Write a node equation at VV 3 : VV 2 VV 3 RR jj ωωωω ωωωω VV 3 VV 4 RR = 0 mplifier input VV 3 is a virtual ground, VV 3 = 0 VV 4 = RR VV ss RR jj ωωωω ωωωω The output of the first stage is VV 2 = VV ss VV 2 RR jj ωωωω ωωωω VV 4 RR = 0 VV 4 = RR VV ss RR jj ωωωω ωωωω
t frequency ff = GHz, ωω = 2ππff = 6.283xx0 9 : vv ss tt = 0. cos ωω tt VV ss 50 Ω VV 0 50 Ω VV 2 VV 3 0 0 0 jjωω LL RR jjωω CC RR VV 4 VV 4 = RR VV ss RR jj ωω LL ωω CC RR = ohm, RR = ohm, CC = pf and LL = 6.333 nh, VV ss = 0. ωω LL ωω CC = 6.283xx09 xxx.333xx0 9 6.283xx0 9 = 39.79 59.5 = 9.36 xxxxx02 VV 4 = RR VV ss RRjj ωω LL ωωcc = 0. jj9.36 = 8.378xx04 89.5 degrees vv 4 tt = 8.378xx0 4 cos ωω tt 89.5 The output voltage is small at GHz.
t frequency ff 2 = 2 GHz, ωω 2 = 2ππff 2 =.257xx0 0 : vv ss tt = 0. cos ωω 2 tt VV ss 50 Ω VV 0 50 Ω VV 2 VV 3 0 0 0 jjωω 2 LL RR jjωω 2 CC RR VV 4 VV 4 = RR VV ss RR jj ωω 2 LL ωω 2 CC RR = ohm, RR = ohm, CC = pf and LL = 6.333 nh, VV ss = 0. ωω 2 LL ωω 2 CC =.257xx00 xxx.333xx0 9.257xx0 0 = 79.54 79.55 = 0.0 0 xxxxx02 VV 4 = RR VV ss RR jj ωω 2 LL ωω 2 CC = 0. jjj = 0. 0 degrees vv 4 tt = 0. cos ωω 2 tt
Superposition with two frequencies: vv sss tt = 0. cos ωω 2 tt 50 Ω vv 50 Ω vv 2 CC LL RR vv 3 RR vv 4 vv sss tt = 0. cos ωω tt vv 4 (tt) y superposition, with both sources active: vv 4 tt = 8.378xx0 4 cos ωω tt 89.5 0. cos ωω 2 tt The design of the circuit makes ωωωω ωωωω =0 at 2 GHz. Then the circuit passes 2 GHz but rejects GHz. The circuit behaves as a filter to recover the 2 GHz signal from the two signals received by the antenna.
To master this set of lecture notes, solve the problems yourself: 2 Ω 0 sin 3tt 2 H )Find the Thevenin Equivalent Circuit at a frequency of ωω=3 rad/sec. Use phasors relative to sine. 2)Find the voltage across a load consisting of a ohm resistor in series with a capacitor of value C=0.6667 F, using: The Thevenin equivalent circuit The original circuit The answer must be the same in both cases!
Find the Norton Equivalent Circuit 0 sin 3tt 2 Ω 2 H Find the Norton Equivalent Circuit at a frequency of ωω=3 rad/sec.
5.3 mh 0 cos ωωωω 0.62 mh 330 μμf ) Find the Norton Equivalent Circuit at 60 Hz. 2)Convert the Norton Equivalent to the Thevenin Equivalent Circuit.
Superposition In the circuit below there are two DC sources, VV 2 = 3 volts and II = amp. There is also an C source, vv tt = 0. cos 000tt volts. Use Superposition to find the voltage across the current generator vv tt. The component values are RR =, RR 2 = 2 Ω, CC = 500 microfarads, CC 2 = 40 microfarads, LL = mh. CC = 500 μμf CC 2 = 40 μμf RR 2 = 2 Ω LL = mh vv(tt) 0. cos 000tt vv xx 5vv xx RR = vv II = amp VV 2 = 3 volts
Superposition: vv sss tt = 0. cos ωω 2 tt 50 Ω vv 50 Ω vv 2 CC LL RR vv 3 RR vv 4 vv sss tt = 0. cos ωω tt vv 4 (tt) n antenna receives two signals at the same time, one at frequency, ff = GHz and the other at frequency ff 2 = 2 GHz. The antenna is modelled with two voltage sources in series. The circuit uses two ideal op-amps. The component values are RR = ohm, RR = ohm, CC = pf and LL = 6.333 nh. Use Superposition to find the output voltage vv 4 (tt).