A Zener diode with 5 V Zener voltage is used | vedclass

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(D) Let the load current be $I_L = i$. According to the problem, the Zener current $I_Z = 4I_L = 4i$. The total current flowing through the series res

Vedclass · Accepted Answer

$I_L = 10\ mA; R_L = 500\ \Omega$

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(D) Let the load current be $I_L = i$. According to the problem, the Zener current $I_Z = 4I_L = 4i$. The total current flowing through the series resistor is $I = I_Z + I_L = 4i + i = 5i$. The voltage across the series resistor is $V_R = V_{in} - V_Z = 25\ V - 5\ V = 20\ V$. Using Ohm's law for the series resistor: $V_R = I 	imes R_s$ $20\ V = (5i) 	imes 400\ \Omega$ $20 = 2000i$ $i = \frac{20}{2000} = 0.01\ A = 10\ mA$. Thus, the load current $I_L = 10\ mA$. The load resistance $R_L$ is given by $R_L = \frac{V_L}{I_L} = \frac{5\ V}{10 	imes 10^{-3}\ A} = 500\ \Omega$.

$A$ Zener diode with $5\ V$ Zener voltage is used to regulate an unregulated $DC$ voltage input of $25\ V$. For a $400\ \Omega$ resistor connected in series, the Zener current is found to be $4$ times the load current. The load current $(I_L)$ and load resistance $(R_L)$ are:

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