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(NONE) The voltage drop across the series resistor $R_s$ is given by:
$V_s = V_{in} - V_z = 8 \, V - 5 \, V = 3 \, V$
The current through the load r

Vedclass · Accepted Answer

$0.6 \, k \Omega$

Vedclass · Answer

(NONE) The voltage drop across the series resistor $R_s$ is given by:
$V_s = V_{in} - V_z = 8 \, V - 5 \, V = 3 \, V$
The current through the load resistor $R_L$ is:
$I_L = \frac{V_z}{R_L} = \frac{5 \, V}{1 	imes 10^3 \, \Omega} = 5 \, mA$
The maximum current through the Zener diode is determined by its power rating $P_z = 10 \, mW$:
$I_{z,max} = \frac{P_z}{V_z} = \frac{10 \, mW}{5 \, V} = 2 \, mA$
For the Zener diode to regulate, the current through the series resistor $I_s$ must be sufficient to supply the load current and the Zener current. The total current $I_s$ is $I_L + I_z$.
To find the required $R_s$, we use the condition $I_{s} = I_L + I_z$.
For the circuit to function as a regulator, the current through the Zener diode must be between $0$ and $I_{z,max}$.
Thus, $I_{s,min} = I_L + 0 = 5 \, mA$ and $I_{s,max} = I_L + I_{z,max} = 5 \, mA + 2 \, mA = 7 \, mA$.
The resistance $R_s$ must satisfy:
$R_{s,min} = \frac{V_s}{I_{s,max}} = \frac{3 \, V}{7 \, mA} = \frac{3}{7} \, k \Omega \approx 0.43 \, k \Omega$
$R_{s,max} = \frac{V_s}{I_{s,min}} = \frac{3 \, V}{5 \, mA} = 0.6 \, k \Omega$
Given the options, none match the calculated range exactly, but $R_s$ must be chosen to ensure the Zener diode operates within its limits. Typically, such questions look for a value within the range or a specific design constraint.

In the given circuit, if the power rating of the Zener diode is $10 \, mW$, the value of series resistance $R_s$ to regulate the input unregulated supply is:

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