A 5.0 , V stabilized power supply is required | vedclass

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(B) The zener diode operates at a constant voltage $V_{Z} = 5.0 \, V$. Given the input voltage $V_{S} = 12 \, V$ and the maximum power rating $P_{Z} =

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$17$

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(B) The zener diode operates at a constant voltage $V_{Z} = 5.0 \, V$. Given the input voltage $V_{S} = 12 \, V$ and the maximum power rating $P_{Z} = 2.0 \, W$. The maximum current $I_{Z_{\max}}$ that the zener diode can handle is calculated as: $I_{Z_{\max}} = \frac{P_{Z}}{V_{Z}} = \frac{2.0 \, W}{5.0 \, V} = 0.4 \, A = 400 \, mA$. The series resistance $R_{S}$ is used to drop the excess voltage $(V_{S} - V_{Z})$ and limit the current to $I_{Z_{\max}}$. $R_{S} = \frac{V_{S} - V_{Z}}{I_{Z_{\max}}} = \frac{12 \, V - 5.0 \, V}{0.4 \, A} = \frac{7}{0.4} \, \Omega = 17.5 \, \Omega$.

$A$ $5.0 \, V$ stabilized power supply is required to be designed using a $12 \, V$ $DC$ power supply as an input source. The maximum power rating of the zener diode is $2.0 \, W$. The minimum value of resistance $R_{s}$ in $\Omega$ connected in series with the zener diode will be (in $.5$)

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