A Zener diode having breakdown voltage V_z = | vedclass

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(B) The voltage across the series resistor $R_s$ is given by $V_{R_s} = V_{in} - V_z = 10 \, V - 6 \, V = 4 \, V$.For the Zener diode to act as a volt

Vedclass · Accepted Answer

$400$

Vedclass · Answer

(B) The voltage across the series resistor $R_s$ is given by $V_{R_s} = V_{in} - V_z = 10 \, V - 6 \, V = 4 \, V$.For the Zener diode to act as a voltage regulator, it must maintain a minimum current $I_{z,min} = 10 \, mA$ even when the load current $I_L$ is zero.The total current $I$ flowing through $R_s$ is $I = I_z + I_L$.To find the maximum value of $R_s$, we consider the condition where the load current $I_L$ is minimum (i.e., $I_L = 0$).Thus, $I = I_{z,min} = 10 \, mA$.Using Ohm's law, $R_s = \frac{V_{R_s}}{I} = \frac{4 \, V}{10 \, mA} = \frac{4 \, V}{10 	imes 10^{-3} \, A} = 400 \, \Omega$.Therefore, the maximum value of $R_s$ is $400 \, \Omega$.

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