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(B) The Zener diode acts as a voltage regulator, maintaining a constant voltage of $V_Z = 6\,V$ across the load resistance $R_L = 4\,k\Omega$ when it

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(B) The Zener diode acts as a voltage regulator, maintaining a constant voltage of $V_Z = 6\,V$ across the load resistance $R_L = 4\,k\Omega$ when it is in the breakdown region.The load current $I_L$ is constant and is given by:$I_L = \frac{V_Z}{R_L} = \frac{6\,V}{4\,k\Omega} = 1.5\,mA$The total source current $I_S$ is given by:$I_S = \frac{V_{in} - V_Z}{R_S}$The Zener current $I_Z$ is given by:$I_Z = I_S - I_L = \frac{V_{in} - 6}{2\,k\Omega} - 1.5\,mA$To maximize $I_Z$, we must use the maximum input voltage $V_{in} = 16\,V$:$I_{S,max} = \frac{16\,V - 6\,V}{2\,k\Omega} = \frac{10\,V}{2\,k\Omega} = 5\,mA$Therefore, the maximum Zener current is:$I_{Z,max} = I_{S,max} - I_L = 5\,mA - 1.5\,mA = 3.5\,mA$

The figure shows a $DC$ voltage regulator circuit with a Zener diode of breakdown voltage $= 6\,V$. If the unregulated input voltage varies between $10\,V$ to $16\,V$, then what is the maximum Zener current in $mA$ (in $.5$)?

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