If Rx is the inherent resistance of the inductor L for the buck-boost converter of Fig. 10-6, derive 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.
The circuit of Fig. 10-14(a) represents the circuit of Fig. 10-6 with Q ON and D OFF. By KVL,
LdtdiL+RxIL=V10≤t≤DTs (1)
The circuit of Fig. 10-14(b) is valid for Q OFF and D ON. Whence,
LdtdiL+RxiL=−v2DTs≤t≤Ts (2)
In similar manner to the procedure of Problem 10.11, integrate (1) and (2), add the results, and divide by Ts to find
TsL∫iL(0)iL(t)diL+RxTs1∫0TsiLdt=TsV1∫0DTsdt−Ts1∫DTstsvsdt (3)
For a periodic iL, the first 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
RxIL=DV1 – (1 – D)V2 (4)
From Fig. 10-14,
Cdtdv2=−RLv20≤t≤DTs (5)
Cdtdv2=iL – RLv2DTs≤t≤Ts (6)
Integrate, add, and divide by Ts for (5) and (6).
TsC∫v2(0)v2(Ts)dv2=Ts1∫DTsTsiLdt – R1Ts1∫0Tsv2dt (7)
The first term of (7) must be zero for periodic v2. Owing to the straight-line segment description of iL, the first 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 – RLV2 (8)
Solve (8) for IL, substitute the result into (4), and rearrange to yield
GV′=V1V2=Rx + (1 – D)2RLD(1 – D)RL (9)