A Zener diode, having a breakdown voltage of | vedclass

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(B) The Zener diode is in parallel with the load resistor of $1 	ext{ k}\Omega$. Since the Zener breakdown voltage is $15 	ext{ V}$, the voltage acr

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$5$

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(B) The Zener diode is in parallel with the load resistor of $1 	ext{ k}\Omega$. Since the Zener breakdown voltage is $15 	ext{ V}$, the voltage across the load resistor is $V_L = 15 	ext{ V}$.The current through the load resistor is $I_L = \frac{V_L}{R_L} = \frac{15 	ext{ V}}{1000 \Omega} = 15 	imes 10^{-3} 	ext{ A} = 15 	ext{ mA}$.The voltage drop across the series resistor $R_s = 250 \Omega$ is $V_s = V_{in} - V_L = 20 	ext{ V} - 15 	ext{ V} = 5 	ext{ V}$.The total current supplied by the source is $I = \frac{V_s}{R_s} = \frac{5 	ext{ V}}{250 \Omega} = 0.02 	ext{ A} = 20 	ext{ mA}$.Applying Kirchhoff's Current Law at the junction, the current through the Zener diode $I_z$ is $I_z = I - I_L = 20 	ext{ mA} - 15 	ext{ mA} = 5 	ext{ mA}$.

$A$ Zener diode, having a breakdown voltage of $15 \text{ V}$, is used in a voltage regulator circuit as shown. The current through the Zener diode is (in $\text{ mA}$)

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