An 8,V Zener diode along with a series resist | vedclass

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Question

(A) The voltage across the series resistor $R$ is given by the difference between the supply voltage $V_{S}$ and the Zener breakdown voltage $V_{Z}$.$

Vedclass · Accepted Answer

$480$

Vedclass · Answer

(A) The voltage across the series resistor $R$ is given by the difference between the supply voltage $V_{S}$ and the Zener breakdown voltage $V_{Z}$.$V_{R} = V_{S} - V_{Z} = 20\,V - 8\,V = 12\,V$.To find the minimum value of the series resistance $R$, we must use the maximum Zener current $I_{Z,max} = 25\,mA = 25 	imes 10^{-3}\,A$ (assuming load current is negligible).Using Ohm's law, $V_{R} = I_{Z,max} 	imes R$.$12\,V = (25 	imes 10^{-3}\,A) 	imes R$.$R = \frac{12}{25 	imes 10^{-3}}\,\Omega = \frac{12000}{25}\,\Omega = 480\,\Omega$.

An $8\,V$ Zener diode along with a series resistance $R$ is connected across a $20\,V$ supply as shown in the figure. If the maximum Zener current is $25\,mA$ and the load current is negligible, then the minimum value of $R$ will be $\Omega$.

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