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Fig. 10-3 Buck converter waveform
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For this triangular waveform, under the assumption that the ac component of iL ows through C and the dc component of iL ows through RL , I2 1 V 1 D Ts I 2 2 max 2Lc
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However, I2 V2 =RL which can be equated to the above expression for I2 . Rearrangement of the result gives Lc RL 1 D Ts RL 1 D 2fs 2 10:6
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Example 10.2. A buck converter having a switching frequency of 25 kHz is to be operated with a duty cycle such that 0:1 D 1. The load is described by RL 5 . Determine the value of critical inductance L Lc so that current iL is continuous. The critical inductance must be determined for the minimum value of duty cycle. By (10.6), Lc RL 1 D 5 1 0:1 90 H 2fs 2 25 103
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BOOST CONVERTER
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The boost converter SMPS circuit of Fig. 10-4 produces an average value output voltage V2 hv2 i > V1 . When the ideal transistor Q is ON, vQ 0. Conversely, when Q is OFF continuity of current through inductor L requires that ideal diode D be in the forward conducting state. With vD 0; vQ V2 . Thus, vQ is a rectangular pulse with a delay of DTs and duration 1 D Ts . If capacitor C is large, reasonable approximations are that the time-varying component of iD ows through C and that the voltage across the load resistor RL is constant. Due to periodic switching of Q, voltage vL and current iC are periodic once initial transients die out.
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Fig. 10-4 Boost converter
As a consequence of the above approximations, vL and iL can be appropriately determined by (10.1) and (10.2) when Q is OFF and by (1) and (2) of Example 10.1 when Q is ON (replace VC with V1 ). Diode current iD must be equal to iL when Q is OFF. Since load current I2 is the average value of iD , iC iD I2 . Figure 10-5 displays the resulting waveforms for vQ ; iL ; vL ; iD ; and iC : Based on (10.3) and the vL waveform of Fig. 10-5, V1 DTs V2 V1 1 D Ts
Fig. 10-5 Boost converter waveform
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Rearrangement nds the ideal boost converter voltage gain as GV V2 1 V1 1 D 10:7
Unlike the buck converter, the boost converter gain is not a linear function of D. From (10.7), the ideal gain approaches in nity as D approaches 1. When parasitic resistances of the inductor and capacitor 0 are considered, the actual gain GV departs signi cantly from the ideal gain for values of D > 0:8. (See Problem 10.11.) The common case of continuous current iL exists only if the value of L ! Lc (critical inductance) that results in marginally continuous conduction for iL . For this case, Imin 0 in Fig. 10-5 and iL 0 0. By application of (10.2), iL t Evaluate for t DTs and use (10.7) to nd iL DTs iD DTs Imax For the triangular iD waveform, I2 hiD i 1 1 D Ts V I 2 D 1 D 2 Ts 2 max Ts 2Lc After rearrangement, 10:8
V1 t Lc
V1 V 1 D DTs 2 DTs Lc Lc
But I2 V2 =RL , which can be equated to the above expression for I2 . Lc 1 D DTs RL 1 D DRL 2 2fs
Example 10.3. A boost converter with a 20-kHz switching frequency is operating with a 50 percent duty cycle. The connected load is 7 . Determine the value of critical inductance so that current iL is continuous. By (10.8), Lc 1 D 2 DRL 1 0:5 2 0:5 7 21:9 H 2fs 2 20 103
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