Question 10.SP.3: Assume the buck converter of Fig. 10-2 is lossless so that t......
Assume the buck converter of Fig. 10-2 is lossless so that the input power (P_{in}) is equal to the output power (P_o). Derive an expression for the current gain G_I = I_2/I_1.
The input power and output power are found by use of (1.20).
F_{0} = {\frac{1}{T}}\int_{t_{0}}^{t_{0} + T}f(t)\,d t (1.20)
P_{in} = {\frac{1}{T_{s}}}\int_{0}^{T_{s}}\,V_{1}i_{1}\,d t = V_{1}\,{\frac{1}{T_{s}}}\int_{0}^{T_{s}}\,i_{1}\,d t = V_{1}I_{1} (1)
P_{o} = {\frac{1}{T_{s}}}\int_{0}^{T_{s}}v_{2}i_{2}\,d t = V_{2}I_{2}\,{\frac{1}{T_{s}}}\int_{o}^{T_{s}}\,d t = V_{2}I_{2} (2)
Constant values for v_2 and i_2 were assumed in (2). Equate (1) and (2). Rearrange the result and use (10.5) to find
G_{V} = {\frac{V_{2}}{V_{1}}} = D (10.5)
G_{I} = {\frac{I_{2}}{I_{1}}} = {\frac{V_{1}}{V_{2}}} = {\frac{1}{D}}