Question

Problem 2 A 1.5 volt energy source is to used to power a 5 volt, 1 amp load. A buck-boost converter is used for this application. A suitable transistor is found with an on-resistance of 35 mQ, and a Schottky diode is found with a forward drop of 0.5 volts. The on resistance of the Schottky diode may be ignored. The circuit is shown in the following figure. 1 A 100 HH 5 V Load 1.5 V DT T f, 40 kHaz a) Derive the equivalent DC transformer circuit that models the DC properties of this converter. include conduction losses and inductor copper voltage losses. (0%) b) If we desire an efficiency of at least 70% under normal conditions. How large can the inductor winding resistance be (096) c) What should be the duty cycle of the converter. (096) d) Compute the power loss in each element. (20%)

Answer 1

(a) Condition for Boost convertor 0.5 < D< Dmax Condition for Buck convertor 0 D<0.5 Now V 1.5 V V5 V The load voltage is V. S 1-D 1.5 5 1- D 1.5 1 -D 5 1 D 0.3 D 1-0.3 D 0.7

(b) The output power is Out =(5)(1) -5 W Efficiency is, Pout Paut+Posses 5 0.7 5+ Plosses 0.7(5 Pses 5 loss 3.5 0.7P 5 losses 0.7Posses5-3.5 1.5 Plosses 0.7 Posses 2.143 W The output current is, I 1, (1-D) 1 1, (1-0.7) S 1 I 0.3 I 3.33 A

The power loss is, losses 2.14 (3.33) R 2.14 R = (3.33) R 193 m2 (c) In Schottky diode, power loss is, sch on o =(0.5)(1) 0.5 W In inductor, power loss is, osses IR -(3.33) (193x10 -2.143 W In transistor, power loss is, trans on -(3.33) (35x10) =0.388 W

(d) The plot of output voltage (V) v/s duty cycle (D) is as follows: Vo Vo max Dmax 0.5 0 P 0 1 0 V = 0 At D 0 Unstable

V S At D 0.5 2 x100 P °Г'А x100 VI 0 0 S S (0.5V,)I x 100 I V 1 D - 25% At D 1 V 0 P0 n 0 At D Dmax V Vomax 1=1max max 0.25n max D 1 0 0.5 Dmax Po