Question

The output of a buck-boost converter is 80 V with input voltage 100 V. If the frequency of switch is 10 kHz, then the turn off time of the switch will be

• A. 112 $\mu$sec
• B. 45 $\mu$sec
• C. 56 $\mu$sec
• D. 90 $\mu$sec

Answer 1

The correct option is C 56 $\mu$sec

$V_0=80V$

$V_{in}=100V$
For buck-boost converter,

$V_0=\frac{V_{in}D}{1-D}$

or $80=\frac{100D}{1-D}$

or $D=0.44$

$\frac{T_{on}}{T}=0.44$

$T=\frac{1}{f}=\frac{1}{10\times {10}^3}=100\mu s$

$T_{on}=0.44\times 100=44\mu s$

$T_{off}=T-T_{on}$

$T_{off}=100-44=56\mu s$