A Zener diode of breakdown voltage V_{Z} = 8, | vedclass

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(B) The total current $I$ flowing through the series resistor $R$ is given by:$I = \frac{V_{i} - V_{Z}}{R} = \frac{10\,V - 8\,V}{100\,\Omega} = \frac{

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$2$

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(B) The total current $I$ flowing through the series resistor $R$ is given by:$I = \frac{V_{i} - V_{Z}}{R} = \frac{10\,V - 8\,V}{100\,\Omega} = \frac{2\,V}{100\,\Omega} = 20\,mA$.For the Zener diode to operate in the breakdown region,the load current $I_{L}$ must satisfy $I = I_{Z} + I_{L}$,where $0 \le I_{Z} \le I_{ZM}$.$1$. To find $R_{L, \max}$,we need the minimum load current $I_{L, \min}$. This occurs when the Zener current is at its maximum $(I_{Z} = I_{ZM} = 10\,mA)$.$I_{L, \min} = I - I_{ZM} = 20\,mA - 10\,mA = 10\,mA$.$R_{L, \max} = \frac{V_{Z}}{I_{L, \min}} = \frac{8\,V}{10\,mA} = 800\,\Omega$.$2$. To find $R_{L, \min}$,we need the maximum load current $I_{L, \max}$. This occurs when the Zener current is at its minimum $(I_{Z} = 0)$.$I_{L, \max} = I = 20\,mA$.$R_{L, \min} = \frac{V_{Z}}{I_{L, \max}} = \frac{8\,V}{20\,mA} = 400\,\Omega$.The ratio of the maximum and minimum value of $R_{L}$ is:$\frac{R_{L, \max}}{R_{L, \min}} = \frac{800\,\Omega}{400\,\Omega} = 2$.

Similar Questions

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