display barcode in ssrs report (a) Solve (10.9) for D to nd D V2 15 0:5555 V1 V2 12 15 in Software

Printer Code39 in Software (a) Solve (10.9) for D to nd D V2 15 0:5555 V1 V2 12 15

(a) Solve (10.9) for D to nd D V2 15 0:5555 V1 V2 12 15
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2 V2 15 2 22:5 W RL 10
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(b) Based on (1.23) for this lossless converter, Pin Po
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[CHAP. 10
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The average value of input current follows as I1 Pin 22:5 1:875 A 12 V1
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10.14 If Rx is the inherent resistance of the inductor L for the buck-boost converter of Fig. 10-6, derive 0 an expression for the actual voltage gain GV V2 =V1 that is valid for continuous inductor current. Assume that iL is described by straight line segments.
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The circuit of Fig. 10-14(a) represents the circuit of Fig. 10-6 with Q ON and D OFF. di L L Rx IL V1 dt 0 t DTs By KVL, 1
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Fig. 10-14 The circuit of Fig. 10-14(b) is valid for Q OFF and D ON. L diL Rx iL v2 dt Whence, t Ts 2
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In similar manner to the procedure of Problem 10.11, integrate (1) and (2), add the results, and divide by Ts to nd L iL Ts 1 Ts V DTs 1 Ts diL Rx iL dt 1 dt v dt 3 Ts 0 Ts iL 0 Ts 0 Ts DTs s For a periodic iL , the rst term of (3) must be zero. Recognize the average values of iL and v2 , respectively, in the second term on each side of the equation to give Rx IL DV1 1 D V2 From Fig. 10-14, C dv2 v 2 dt RL dv v C 2 iL 2 dt RL 0 t DTs DTs t Ts 5 6 4
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Integrate, add, and divide by Ts for (5) and (6). C v2 Ts 1 Ts 1 1 Ts dv2 iL dt v dt Ts v2 0 Ts DTs R Ts 0 2
The rst term of (7) must be zero for periodic v2 . Owing to the straight-line segment description of iL , the rst term on the right-hand side of (7) can be written as 1 D IL . Recognize the average value of v2 in the last term. Thus, (7) becomes 0 1 D IL V2 RL 8
CHAP. 10]
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Solve (8) for IL , substitute the result into (4), and rearrange to yield
0 GV
V2 D 1 D RL V1 Rx 1 D 2 RL
10.15 By SPICE methods, model the buck-boost converter of Fig. 10-6 with fs 30 kHz, D 0:4, L 70 H; C 100 F; and RL 10 . Use the model to generate a set of waveforms analogous to Fig. 10-7.
The netlist code is shown below where the initial conditions on inductor current and capacitor voltage were determined after running a large integer number of cycles to nd the repetitive values.
Prb10_15.CIR * BUCK-BOOST CONVERTER * D=DUTY CYCLE, fs=SWITCHING FREQUENCY .PARAM D=0.4 fs=30e3Hz V1 1 0 DC 15V SW 1 2 4 2 VCS VSW 4 2 PULSE(0V 1V 0s 5ns 5ns {D/fs} {1/fs}) L 2 0 70uH IC=0.229A D 3 2 DMOD C 0 3 100uF IC=10.02V RL 0 3 10ohm .MODEL DMOD D(N=0.01) .MODEL VCS VSWITCH(RON=1e-6ohm) .TRAN 1us 0.166667ms 0s 100ns UIC .PROBE .END
Execute hPrb10_15.CIRi and use the Probe feature of PSpice to plot the waveforms of Fig. 10-15.
Fig. 10-15
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[CHAP. 10
Supplementary Problems
10.16 Determine the smallest value of inductance that could have been used for the buck converter of Example 10.5 and the inductor current remain continuous. Ans: L Lc 50 H
Find the values of Imax and Imin for the buck converter of Example 10.5 if the value of L Lc 50 H, as determined in Problem 10.16. Ans: Imax 2:4 A; Imin 0
Use the procedure of Problem 10.3 to nd an expression for the current gain GI I2 =I1 for the ideal boost converter of Fig. 10-4. Ans: GI 1 D
Use Problem 10.6 as a guideline to derive expressions for Imax and Imin shown on the boost converter waveforms of Fig. 10-5. Ans: Imax V2 V D 1 ; 1 D RL 2 fs L Imin V2 V D 1 1 D RL 2 fs L
0 The actual gain GV for the boost converter with inherent inductor resistance was determined in Problem 0 10.11. Determine cycle D Dp for which GV has a maximum value. p the duty Ans: Dp 1 Rx =RL
Find an expression for the peak-to-peak ripple voltage of a boost converter. Ans: v2 DV2 = fs RL C
Execute hPrb10_12.CIRi of Problem 10.12 and plot v2 V 3 . From the plot, determine the peak-to-peak ripple voltage v2 . Ans: v2 0:383 V
Determine the smallest value of inductance that could have been used for the buck-boost converter of Problem 10.15 and the inductor current remain continuous. Ans: L Lc 24 H
0 The actual gain GV for the buck-boost converter with inherent inductor resistance was determined in 0 Problem 10.14. Determine the duty cycle D Dp for which GV has a maximum value.
Ans:
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