Question

In the circuit shown below,



If the switching frequency is $5kHz$, duty ratio is $0.4$ and circuit reached to steady state, then the average value of inductor current is ____ $A$.
(Assuming the switch and diode to be ideal and discontinuous mode of current conduction)

• A. 0.24

Answer 1

The correct option is A 0.24

During $T_{on}$ the circuit behaves as,
$V_s=L\frac{dI}{dt}$
$dI=\frac{V_o}{L}dt$
Integrating on both sides we get
$I_P=\frac{V_o}{L}{T}_{on}$
$T_{on}=\alpha T=\frac{\alpha }{f}=\frac{0.4}{5\times {10}^3}=80\times {10}^{-6}$
$I_P=\frac{50}{5\times {10}^{-3}}\times (80\times {10}^{-6})=0.8A$
During $T_{off}$ it is $T_{on}\le t\le \beta T$
Applying KVL in the circuit,
$V_L=V_s-V_0$
$(V_L{)}_{avg}=0$
$V_s{T}_{on}+(V_s-V_0)(\beta T-T_{on})=0$
$V_s\beta T=V_0(\beta T-T_{on})$
$\frac{V_0}{V_s}=\frac{\beta }{\beta -\alpha }$
$\beta =0.6$
From the graph of $I_L$


$I_{L\left(avg\right)}=\frac{\frac{1}{2}\times b\times h}{T}$
$=\frac{\frac{1}{2}\times \beta T\times I_P}{T}=\frac{1}{2}\times 0.6\times 0.8=0.24A$