In a Zener regulated power supply, a Zener di | vedclass

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(B) Given: Zener voltage $V_z = 6 \, V$, load current $I_L = 4 \, mA$, and input voltage $V_{in} = 10 \, V$.We are given that the Zener current $I_Z$

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$167 \, \Omega$

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(B) Given: Zener voltage $V_z = 6 \, V$, load current $I_L = 4 \, mA$, and input voltage $V_{in} = 10 \, V$.We are given that the Zener current $I_Z$ is five times the load current $I_L$.So, $I_Z = 5 	imes I_L = 5 	imes 4 \, mA = 20 \, mA$.The total current $I_S$ flowing through the series resistor $R_S$ is the sum of the Zener current and the load current:$I_S = I_Z + I_L = 20 \, mA + 4 \, mA = 24 \, mA = 24 	imes 10^{-3} \, A$.The potential drop across the series resistor $R_S$ is the difference between the input voltage and the Zener voltage:$V_S = V_{in} - V_z = 10 \, V - 6 \, V = 4 \, V$.Using Ohm's law, the value of the series resistor $R_S$ is:$R_S = \frac{V_S}{I_S} = \frac{4 \, V}{24 	imes 10^{-3} \, A} = \frac{4000}{24} \, \Omega \approx 166.67 \, \Omega$.Rounding to the nearest value, we get $R_S \approx 167 \, \Omega$.

In a Zener regulated power supply, a Zener diode with $V_z = 6 \, V$ is used for regulation. The load current is $4 \, mA$ and the unregulated input voltage is $10 \, V$. To get a Zener current five times the load current, the value of the series resistor $R_S$ is nearly:

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