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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.

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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Clear the denominators by multiplying both sides by (x + 1)(x 1), (x + 1)(x 1)

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

If a check is lost or stolen, the de positors can request the bank not to pay (a stop order).

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

CHAP. 40]

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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

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

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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

and with x = 1, 3 2 With all constants known, the antiderivative of f (x) will be the sum of terms of the form 3 = 0 + A2 (2) or A2 = A dx = A ln |x a| x a

CHAP. 10]

Endorsements When a check is given to the bank for deposit, the depositor signs the check to show that he or she accepts responsibility for the amount of that check. The depositor s signature is known as an endorsement. This en dorsement transfers the ownership of the check and guarantees to the in dividual that the depositor will guarantee its payment. Different kinds of endorsements serve different needs. 1. Blank endorsements. A blank endorsement consists only of the name of the endorser. Its disadvantage lies in the fact that a lost or stolen check with a blank endorsement may be cashed by the nder or thief.

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.

Case 2: D(x) is a product of linear factors, at least one of which is repeated. This is treated in the same manner as in Case 1, except that a repeated factor (x a)k gives rise to a sum of the form A2 Ak A1 + + + 2 x a (x a) (x a)k EXAMPLE

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

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