A Zener diode of power rating 2 , W is to be | vedclass

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(B) The power rating of the Zener diode is $P_z = 2 \, W$ and its breakdown voltage is $V_z = 10 \, V$.The maximum current that can flow through the Z

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$20$

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(B) The power rating of the Zener diode is $P_z = 2 \, W$ and its breakdown voltage is $V_z = 10 \, V$.The maximum current that can flow through the Zener diode is $I_{z,max} = \frac{P_z}{V_z} = \frac{2 \, W}{10 \, V} = 0.2 \, A$.To ensure safe operation, the Zener diode must handle the maximum input voltage of $V_{in,max} = 14 \, V$ without exceeding its power rating.When $V_{in} = 14 \, V$, the voltage drop across the series resistor $R_s$ is $\Delta V_{Rs} = V_{in,max} - V_z = 14 \, V - 10 \, V = 4 \, V$.Assuming the load current is negligible for the worst-case calculation (or that the Zener must handle the full current when no load is connected), the current through $R_s$ is $I = I_{z,max} = 0.2 \, A$.Using Ohm's law, $R_s = \frac{\Delta V_{Rs}}{I} = \frac{4 \, V}{0.2 \, A} = 20 \, \Omega$.

$A$ Zener diode of power rating $2 \, W$ is to be used as a voltage regulator. If the Zener diode has a breakdown voltage of $10 \, V$ and it has to regulate an unregulated input voltage that fluctuates between $6 \, V$ and $14 \, V$, the value of $R_s$ for safe operation should be $.... \Omega$.

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