Let Rx be the inherent resistance of the inductor L for the boost converter of Fig. 10-4 and derive 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
LdtdiL+RxiL=V10≤t≤DTs (1)
The equivalent circuit of Fig. 10-12(b) is valid for Q OFF and D ON, yielding
LdtdiL+RxIL=V1 – V2DTs≤t≤Ts (2)
Integrate both (1) and (2) over their time regions of validity to give
L∫iL(0)iL(DTs)diL+Rx∫0DTsdt=V1∫0DTsdt (3)
L∫iL(DTs)iL(Ts)diL+Rx∫DTsTsdt=V1∫DTsTsdt – V2∫DTsTsdt (4)
Add (3) and (4) and divide by Ts to find
TsL∫iL(0)iL(Ts)diL+RxTs1∫0TsiLdt=TsV1∫0DTsdt – TsV2∫DTsTsdt (5)
If iL is periodic, iL(0)=iL(Ts). Hence, the first term of (5) has a value of zero. The second term is Rx〈iL〉=RxIL. Thus, (5) can be written as
RxIL=V1 – (1 – D)V2 (6)
From the waveform sketch of Fig. 10-5,
I2=Ts1∫DTsTsiLdt
Since iL is described by straight line segments, it follows that
I2Ts=IL(1 – D)Tsor
IL=1 – DI2=RL(1 – D)V2 (7)
Substitute (7) into (6) and rearrange to yield
GV′=V1V2=Rx + RL(1 – D)2(1 – D)RL (8)