419 and 420 we obtain Hence the current gain is given by I DVin o R D2Vin Jin

# 419 and 420 we obtain hence the current gain is given

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Substituting these equations in Eqs. ( 4.19) and ( 4.20), we obtain Hence, the current gain is given by I = DVin o R D2Vin Jin=~ (4.21) (4.22) (4.23) This relation can be obtained by equating the average input and output power, to yield From Eqs. (4.15) and (4.23), it is clear that the current and voltage relations for the converter are equivalent to a de transformer model with a ratio of D, as shown in Fig. 4.13. The sinusoidal curve and straight line drawn across the transformer wind- ings indicate that the transformer is capable of transferring ac and de, respectively. Critical Inductor Value It is clear that for I Lmin ;;t:. 0 , the converter will operate in the continuous conduction mode (cern). To find the minimum inductor value that is needed to keep the converter in the cern, we set JLmin to zero and solve for L: I :D + _ (1 (1-D)TJ _ ]Lmin - D vin R- 2L - 0 /() ( 1-D) Lcrit = -2- TR (4.24) + Figure 4.13 Equivalent circuit representation for the buck converter, referred to as a de-de transformer.
142 Chapter 4 Nonisolated Switch-Mode de-de Converters where Lcrit is the critical inductance minimum value for a given D, T, and R converter enters the discontinuous conduction mode (dcm) of operation. Output Voltage Ripple Since we have assumed that the output voltage has no ripple, the entire ac current from the inductor passes through the parallel capacitor, and only de rent is delivered to the load resistor. In practice, the value of the output is an important design parameter since it influences the overall size of the de converter and how much of the switching frequency ripple is being Having said that, it is design practice to choose a larger output capacitor in to limit the ac ripple across V 0 Theoretically speaking, if C ~ oo, the acts like a short circuit to the ac ripple, resulting in zero output voltage we assume Cis finite, then there exists a voltage ripple superimposed on the age output voltage. In order to derive an expression for the capacitor ripple age, we first obtain an expression for the capacitor current, which is given by following relation: As a result, the initial capacitor current at t = 0 is given by and at t = DT, Jc(DT) = JL(DT) -1 0 = + (/Lmax; /Lmin) The resultant capacitor current and voltage are shown in Fig. 4.14. DT T DT l+DT 2 T Figure 4.14 Capacitor current and voltage waveforms.
4.2 Continuous Conduction Mode 143 The instantaneous capacitor current can be expressed in terms of /).I from Eqs. ( 4.21) and ( 4.22), as shown in the following equations: i {t) = ILmax- ILmin! _ ILmax- ILmin = /).I! _/).I e DT 2 DT 2 0 \$ t < DT (4.25a) I -I . IL -IL · i (!} = _ Lmax Lmm{t-DT) + max mm e (I -D)T 2 = -/).I (t-DT)+/).I (I -D)T 2 0\$ t < DT (4.25b) where /).I= ( Vin(I- D)TD)/ L. From the capacitor voltage-current relation, ie = C(dv/ dt), the capacitor volt- age, vel(!), can be expressed by the following integral fort 2::0: Vel (t) = .!.Jt ie dt + Ve(O) c 0 where Ve(O) is the initial capacitor voltage at t = 0. Substituting for ie(t) from Eq. (4.25a), we obtain the following equation: Vel(!)= bf: (~~~-~I) dt+ Ve(O) Evaluating this integral yields 1 /).I t 2 /).I v l(t) = ---- -t+ v (0) e CDT2 2C e 0\$ t < DT Similarly, for t 2:: D T, the capacitor voltage is given by vc 2 (t) = .!.Jt ie dt + Ve(DT) C DT (4.26) where Ve(DT) is the initial capacitor voltage when the switch is turned off at t = DT. From Eq. (4.27) vc 2 (t) is given by v (t) = -/).I (t-DT) 2 + /).I(t-DT)+ V (DT) DT\$t < T e 2 C(l -D)T 2 2C e (4.27) Since the capacitor voltage is in the steady state, we have v e 1 (t = D T) = ve 2 (t = DT) and vel (0) = ve 2 (T),