In the circuit shown below, maximum Zener diode current will be ______ mA.

The Zener diode, the load resistor $$R_L$$ and the series resistor $$R_S$$ are connected across a dc source of $$V_S$$ volts.

The Zener voltage is $$V_Z = 6\text{ V}$$ and the series resistance is $$R_S = 1\text{ k}\Omega$$ (as shown in the given circuit diagram).

Step 1 - Identify the condition for the maximum Zener current.
  • The total current flowing through $$R_S$$ splits into the load current $$I_L$$ and the Zener current $$I_Z$$:
    $$I_S = I_L + I_Z$$
  • To make $$I_Z$$ as large as possible we must make $$I_L = 0$$. This happens when the load is disconnected (open-circuit) or its resistance is very large.
  • Therefore, at $$I_Z(\max)$$ the entire current $$I_S$$ flows through the Zener diode.

Step 2 - Find the current through the series resistor $$R_S$$ under this condition.
  • The voltage across $$R_S$$ equals the difference between the source voltage and the regulated Zener voltage: $$V_S - V_Z$$.
  • Ohm’s law gives
    $$I_S = \frac{V_S - V_Z}{R_S}$$.

Step 3 - Insert the numerical values.
  • Here $$V_S = 15\text{ V}$$, $$V_Z = 6\text{ V}$$ and $$R_S = 1\text{ k}\Omega$$.
  • Hence
    $$I_S = \frac{15 - 6}{1\,\text{k}\Omega} = \frac{9\text{ V}}{1000\ \Omega} = 0.009\text{ A}$$.

Step 4 - Convert to milli-amperes and state the answer.
  • $$I_S = 0.009\text{ A} = 9\text{ mA}$$.
  • Because $$I_L = 0$$ in this situation, $$I_Z(\max) = I_S = 9\text{ mA}$$.

Therefore, the maximum Zener diode current is 9 mA.