display barcode in ssrs report Since v2 QC =C, the total increment in v2 is v2 QC =C 1 in Software

Generation Code-39 in Software Since v2 QC =C, the total increment in v2 is v2 QC =C 1

Since v2 QC =C, the total increment in v2 is v2 QC =C 1
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The total increment in charge QC is given by the half-period duration amp-second, triangle-shaped area of iL above I2 IL in Fig. 10-3. 1 T QC Imax I2 s 2 2 2
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Use I2 V2 =RL and (5) of Problem 10.6 in (2) and substitute the result into (1) to yield the peak-to-peak ripple voltage.   1 1 V1 V2 D Ts v2 3 C 2 2 fs L 2 From (10.5), V1 V2 =D. Substitute into (3), use Ts 1=fs , and rearrange to nd v2 1 D V2 8 fs2 LC 4
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For the buck converter of Example 10.5, (a) calculate the percent voltage ripple by (4) of Problem 10.8 and (b) formulate a SPICE simulation to numerically determine the percentage ripple.
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(a) By (4) of Problem 10.8 with the values of Example 10.5 and using (10.5), v2 1 D 1 0:5 100% 100% 2% V2 8 fs2 LC 8 25 103 2 100 10 6 50 10 6 (b) Execute hEx10_5.CIRi and use the Probe feature of PSpice to plot v2 V 3 with marked values shown in Fig. 10-11. Then, 6:064 5:943 6:0035 2 v2 6:064 5:943 0:121 v2 0:121 100% 2:01% 6:0035 V2 V2 The error in the two methods is much less than 1 percent.
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10.10 A boost converter with continuous inductor current is fed from a 12-V source with a 60 percent duty cycle while supplying a power of 60 W to the connected load. Determine (a) the output voltage, (b) the load resistance, and (c) the load current.
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Fig. 10-11 (a) By (10.7), V2 (b) Based on (1.23), RL (c) By Ohm s law, I2 V2 30 2A RL 15
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2 V2 30 2 15  Po 60
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V1 12 30 V 1 D 1:06
10.11 Let Rx be the inherent resistance of the inductor L for the boost converter of Fig. 10-4 and derive 0 an expression for the actual voltage gain GV V2 =V1 that is valid for continuous inductor current. Treat V2 as constant in value. Assume that iL can be described by straight line segments.
Figure 10-12(a) represents the circuit of Fig. 10-4 with Q ON and D OFF from which KVL gives L diL R x iL V 1 dt 0 t DTs 1
Fig. 10-12
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[CHAP. 10
The equivalent circuit of Fig. 10-12(b) is valid for Q OFF and D ON, yielding diL Rx IL V1 V2 dt 2
Integrate both (1) and (2) over their time regions of validity to give iL DTs L iL Ts L
iL DTs iL 0
diL Rx diL Rx
DTs Ts
DTs 0
dt V1 dt V1
DTs dt Ts
DTs 0
3 Ts dt
dt V2
Add (3) and (4) and divide by Ts to nd L Ts iL Ts
iL 0
diL Rx
1 Ts
Ts
iL dt
V1 Ts
DTs
dt
V2 Ts
Ts dt
If iL is periodic, iL 0 iL Ts . Hence, the rst term of (5) has a value of zero. Rx hiL i Rx IL . Thus, (5) can be written as Rx IL V1 1 D V2 From the waveform sketch of Fig. 10-5, I2 1 Ts Ts iL dt
The second term is
Since iL is described by straight line segments, it follows that I2 Ts IL 1 D Ts or IL I2 V2 1 D RL 1 D 7
Substitute (7) into (6) and rearrange to yield
0 GV
V2 1 D RL V1 Rx RL 1 D 2
10.12 Use SPICE methods to model the boost converter of Fig. 10-4 with fs 20 kHz, D 0:25, L 50 H, C 100 F, and RL 7:5 . From the model, generate a set of waveforms analogous to Fig. 10-5.
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.
CHAP. 10]
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Prb10_12.CIR * BOOST CONVERTER * D=DUTY CYCLE, fs=SWITCHING FREQUENCY .PARAM D=0.25 fs=20e3Hz V1 1 0 DC 15V SW 2 0 4 0 VCS VSW 4 0 PULSE(0V 1V 0s 5ns 5ns {D/fs} {1/fs}) L 1 2 50uH IC=1.657A D 2 3 DMOD C 3 0 100uF IC=20.05V RL 3 0 7.5ohm .MODEL DMOD D(N=0.01) .MODEL VCS VSWITCH(RON=1e-6ohm) .TRAN 1us 0.25ms 0s 100ns UIC .PROBE .END
Execute hPrb10_12.CIRi and use the Probe feature of PSpice to plot the waveforms shown in Fig. 10-13.
Fig. 10-13
10.13 A lossless buck-boost converter with continuous inductor current supplies a 10  load with a regulated output voltage of 15 V. The input voltage is 12 V. Determine the value of (a) duty cycle, (b) input power, and (c) average value of input current.
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