The 6 V Zener diode shown in figure has negligible resistance and a knee current of 5 mA. The minimum value of $R$ (in $\Omega$) so that the voltage across it does not fall below 6 V is
Show Hint
In Zener regulator problems, the "minimum load resistance" always corresponds to the "maximum load current." This occurs when the Zener diode is drawing its minimum required current (the knee current).
Concept:
A Zener diode maintains its rated voltage ($V_z = 6$ V) as long as the current through it ($I_z$) is at least the knee current ($I_k = 5$ mA). To find the minimum resistance $R$, we must consider the scenario where the Zener diode is just barely "on" (carrying exactly 5 mA), allowing the maximum possible current to flow to the load $R$.
Step 1: Find the total current ($I_t$).
The total current flows through the $50\Omega$ series resistor. The voltage drop across this resistor is the source voltage minus the Zener voltage: $10\text{ V} - 6\text{ V} = 4\text{ V}$.
\[ I_t = \frac{V_{series}}{R_{series}} = \frac{4\text{ V}}{50\Omega} = 0.08\text{ A} = 80\text{ mA} \]
Step 2: Find the maximum load current ($I_L$).
By Kirchhoff's Current Law, $I_t = I_z + I_L$. To keep the Zener at the breakdown voltage, $I_z$ must be at least $5\text{ mA}$.
\[ I_{L(max)} = I_t - I_{z(min)} = 80\text{ mA} - 5\text{ mA} = 75\text{ mA} \]
Step 3: Calculate minimum resistance $R$.
Since the load $R$ is in parallel with the Zener, the voltage across it is $6\text{ V}$.
\[ R_{min} = \frac{V_{z}}{I_{L(max)}} = \frac{6\text{ V}}{0.075\text{ A}} = 80\Omega \]
