Hart Chapter 5 solutions

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CHAPTER 5 SOLUTIONS
3/9/10
5-1)
	
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5-2)
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5-3)
	
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5-4)
	
	
	
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5-5) 
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5-6) 
	
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5-7)
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5-8)
	
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5-9) S1 is on from α to π, and D2 is on from π to 2π.
	
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5-10) 
	
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5-11) a) Using Eq. 5-9,
	
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5-12) Using Eq. 5-9,
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5-13) Using Eq. 5-9,
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5-14) Using Eq. 5-9,
PSpice: P = AVG(W(R)) in Probe gives 523 W (read at the end of the trace). The difference between PSpice and the theoretical output is because of the nonideal SCR model in PSpice. The PSpice result will be more realistic. The THD is 22.4% from the PSpice output file using Fourier terms through n = 9.
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5-15) Use the PSpice circuit of Example 5-3. The .STEP PARAM command is quite useful for determining α. (a) α ≈ 81° for 400 W. (b) α ≈ 46° for 700 W.
SINGLE-PHASE VOLTAGE CONTROLLER (voltcont.cir)
*** OUTPUT VOLTAGE IS V(3), OUTPUT CURRENT IS I(R) ***
**************** INPUT PARAMETERS *********************
.PARAM VS = 120 ; source rms voltage
.PARAM ALPHA = 81 ; delay angle in degrees
.STEP PARAM ALPHA 10 90 20 ; try several values of alpha. Modify the range for more precision
.PARAM R = 15 ; load resistance
.PARAM L = 15mH ; load inductance
.PARAM F = 60 ; frequency
.PARAM TALPHA = {ALPHA/(360*F)} ; converts angle to time delay
.PARAM PW = {0.5/F} ; pulse width for switch control
***************** CIRCUIT DESCRIPTION *********************
VS 1 0 SIN(0 {VS*SQRT(2)} {F})
S1 1 2 11 0 SMOD
D1 2 3 DMOD ; forward SCR
S2 3 5 0 11 SMOD
D2 5 1 DMOD ; reverse SCR
R 3 4 {R}
L 4 0 {L}
**************** MODELS AND COMMANDS ********************
.MODEL DMOD D(n=0.01)
.MODEL SMOD VSWITCH (RON=.01)
VCONTROL 11 0 PULSE(-10 10 {TALPHA} 0 0 {PW} {1/F}) ;control for both switches
.TRAN .1MS 50MS 0MS 1u UIC ; one period of output
.FOUR 60 I(R) ; Fourier Analysis to get THD
.PROBE
.END
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5-16) Modify the PSpice circuit file of Example 5-3. Use the .STEP PARAM command (see Prob. 5-15) for determining α. (a) α ≈ 80° for 600 W. (b) α ≈ 57° for 1000 W.
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5-17) The single-phase voltage controller of Fig. 5-4a is suitable for this application. Equation (5-9) applies for each half-period of the input sine wave. For 250 W delivered to the load, each half period must deliver 125 W. Therefore, the rms value of the current in Eq. (5-9) must be 2.28 A, found by using I2R = 125. A closed-form solution is not possible, but trial-and-error numerical techniques give α ≈ 74°. A similar but perhaps easier method is to use PSpice simulations using the PSpice A/D circuit file in Example 5-3. Modifying the diode model to .MODEL DMOD D(n=.01) to represent an ideal diode, and with trial-and-error values of α, gives α ≈ 74°.
The average and rms currents are determined from a numerical integration of the current expression from Eq. (5-9) or from a PSpice simulation. ISCR,avg = 1.3 A, ISCR,rms = 2.3 A. The maximum voltage across the switches is 120√2sin(74°) = 163 V.
5-18) The PSpice circuit file is shown below. The total average load power is three times the power in one of the phase resistors. Enter 3*AVG(W(RA)) in Probe. The results are (a) 6.45 kW for 20°, (b) 2.79 kW for 80°, and (c) 433 W for 115°. Note that the .STEP PARAM command can be used to run the three simulations at once.
THREE-PHASE VOLTAGE CONTROLLER -- R-L LOAD (3phvc.cir)
*SOURCE AND LOAD ARE Y-CONNECTED (UNGROUNDED)
********************** INPUT PARAMETERS ****************************
.PARAM Vs=480 ; rms line-to-line voltage
.PARAM ALPHA=20 ; delay angle in degrees
.STEP PARAM ALPHA LIST 20 80 115
.PARAM R=35 ; load resistance (y-connected)
.PARAM L = 1p ; load inductance
.PARAM F=60 ; source frequency
********************** COMPUTED PARAMETERS **************************
.PARAM Vm={Vs*SQRT(2)/SQRT(3)} ; convert to peak line-neutral volts
.PARAM DLAY={1/(6*F)} ; switching interval is 1/6 period
.PARAM PW={.5/F} TALPHA={ALPHA/(F*360)}
.PARAM TRF=10US ; rise and fall time for pulse switch control
*********************** THREE-PHASE SOURCE **************************
VAN 1 0 SIN(0 {VM} 60) 
VBN 2 0 SIN(0 {VM} 60 0 0 -120)
VCN 3 0 SIN(0 {VM} 60 0 0 -240)
***************************** SWITCHES ********************************
S1 1 8 18 0 SMOD ; A-phase
D1 8 4 DMOD
S4 4 9 19 0 SMOD
D4 9 1 DMOD
S3 2 10 20 0 SMOD ; B-phase
D3 10 5 DMOD
S6 5 11 21 0 SMOD
D6 11 2 DMOD
S5 3 12 22 0 SMOD ; C-phase
D5 12 6 DMOD
S2 6 13 23 0 SMOD
D2 13 3 DMOD
***************************** LOAD **********************************
RA 4 4A {R} ; van = v(4,7)
LA 4A 7 {L}
RB 5 5A {R} ; vbn = v(5,7)
LB 5A 7 {L}
RC 6 6A {R} ; vcn = v(6,7)
LC 6A 7 {L}
************************* SWITCH CONTROL *****************************
V1 18 0 PULSE(-10 10 {TALPHA} {TRF} {TRF} {PW} {1/F}) 
V4 19 0 PULSE(-10 10 {TALPHA+3*DLAY} {TRF} {TRF} {PW} {1/F}) 
V3 20 0 PULSE(-10 10 {TALPHA+2*DLAY} {TRF} {TRF} {PW} {1/F})
V6 21 0 PULSE(-10 10 {TALPHA+5*DLAY} {TRF} {TRF} {PW} {1/F})
V5 22 0 PULSE(-10 10 {TALPHA+4*DLAY} {TRF} {TRF} {PW} {1/F})
V2 23 0 PULSE(-10 10 {TALPHA+DLAY} {TRF} {TRF} {PW} {1/F})
************************ MODELS AND COMMANDS *************************
.MODEL SMOD VSWITCH(RON=0.01)
.MODEL DMOD D
.TRAN .1MS 50MS 16.67ms 10US UIC
.FOUR 60 I(RA) ; Fourier analysis of line current
.PROBE
.OPTIONS NOPAGE ITL5=0
.END
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5-19) The PSpice input file from Example 5-4 is used for this simulation. In Probe, enter the expression 3*AVG(W(RA)) to get the total three-phase average power in the load, resulting in 368 W. Switch S1 conducts when the current in phase A is positive, and S4 conducts when the current is negative.
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5-20) The smallest value of α is 120°. The conduction angel must be less thanfor equal to 60°. The extinction angle is 180°, so α is 120° or greater.
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5-21) 
THREE-PHASE VOLTAGE CONTROLLER -- R-L LOAD 
*MODIFIED FOR A DELTA-CONNECTED LOAD
*SOURCE IS Y-CONNECTED (UNGROUNDED)
********************** INPUT PARAMETERS ****************************
.PARAM Vs=480 ; rms line-to-line voltage
.PARAM ALPHA=45 ; delay angle in degrees
.PARAM R=25 ; load resistance (y-connected)
.PARAM L = 1p ; load inductance
.PARAM F=60 ; source frequency
********************** COMPUTED PARAMETERS **************************
.PARAM Vm={Vs*SQRT(2)/SQRT(3)} ; convert to peak line-neutral volts
.PARAM DLAY={1/(6*F)} ; switching interval is 1/6 period
.PARAM PW={.5/F} TALPHA={ALPHA/(F*360)}
.PARAM TRF=10US ; rise and fall time for pulse switch control
*********************** THREE-PHASE SOURCE **************************
VAN 1 0 SIN(0 {VM} 60) 
VBN 2 0 SIN(0 {VM} 60 0 0 -120)
VCN 3 0 SIN(0 {VM} 60 0 0 -240)
***************************** SWITCHES ********************************
S1 1 8 18 0 SMOD ; A-phase
D1 8 4 DMOD
S4 4 9 19 0 SMOD
D4 9 1 DMOD
S3 2 10 20 0 SMOD ; B-phase
D3 10 5 DMOD
S6 5 11 21 0 SMOD
D6 11 2 DMOD
S5 3 12 22 0 SMOD ; C-phase
D5 12 6 DMOD
S2 6 13 23 0 SMOD
D2 13 3 DMOD
***************************** LOAD **********************************
RA 4 4A {R} ; 
LA 4A 2 {L}
RB 5 5A {R} ;
LB 5A 3 {L}
RC 6 6A {R} ; 
LC 6A 1 {L}
************************* SWITCH CONTROL *****************************
V1 18 0 PULSE(-10 10 {TALPHA} {TRF} {TRF} {PW} {1/F}) 
V4 19 0 PULSE(-10 10 {TALPHA+3*DLAY} {TRF} {TRF} {PW} {1/F}) 
V3 20 0 PULSE(-10 10 {TALPHA+2*DLAY} {TRF} {TRF} {PW} {1/F})
V6 21 0 PULSE(-10 10 {TALPHA+5*DLAY} {TRF} {TRF} {PW} {1/F})
V5 22 0 PULSE(-10 10 {TALPHA+4*DLAY} {TRF} {TRF} {PW} {1/F})
V2 23 0 PULSE(-10 10 {TALPHA+DLAY} {TRF} {TRF} {PW} {1/F})
************************ MODELS AND COMMANDS *************************
.MODEL SMOD VSWITCH(RON=0.01)
.MODEL DMOD D
.TRAN .1MS 50MS 16.67ms 10US UIC
.FOUR 60 I(RA) ; Fourier analysis of line current
.PROBE
.OPTIONS NOPAGE ITL5=0
.END
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5-22) The PSpice circuit file modification must include a very large resistor (e.g., one megaohm) connected between the neutral of the load to ground to prevent a “floating node” error because of the series capacitor. The steady-state phase A current has two pulses for each of the switches, assuming that the gate signal to the SCRs is continuously applied during the conduction interval. The rms current is approximately 5.52 A. The total average power for all three phases is approximately 1.28 kW. The THD for the load current is computed as 140% for harmonics through n = 9 in the .FOUR command. However, the current waveform is rich in higher-order harmonics and the THD is approximately 300% for n = 100. It should be noted that this load is not conducive for use with the voltage controller because the load voltage will get extremely large (over 5 kV) because of stored charge on the capacitor.
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5-23) With the S1-S4 switch path open, the equivalent circuit is as shown. The current in phase A is zero, so the voltage across the phase-A resistor is zero. The voltage at the negative of V14 is then Vn, and the voltage at the positive of V14 is Va. The voltage across the phase B resistor is half of the voltage from phase B to phase C, resulting in
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