Animated Solution for Physics - Semiconductor Electronics: For the circuit shown below, calculate the value of $I_z$.

Welcome to the fascinating world of voltage regulation! In this problem, we are tasked with finding the current flowing through a Zener diode, denoted as $I_z$. Let's first understand the landscape of our circuit. We have a primary DC voltage source providing an input voltage of $V_i=100\text{V}$. This source pushes current through a series resistor $R_S=1000\Omega$.

Beyond this resistor, the circuit splits into two parallel branches: one containing our Zener diode with a breakdown voltage of $V_z=50\text{V}$, and the other containing a load resistor $R=2000\Omega$. Because the input voltage ($100\text{V}$) is significantly higher than the Zener breakdown voltage ($50\text{V}$), the Zener diode is firmly in its active breakdown region. This means it acts like a strict traffic cop, clamping the voltage across itself—and consequently across the parallel load resistor—to exactly $50\text{V}$.

To unravel the currents in this circuit, we must start from the source. The total current $I$ flows out of the $100\text{V}$ supply and passes entirely through the series resistor $R_S$.

What is the potential difference across this resistor? On its left side, it feels the full force of the $100\text{V}$ input. On its right side, the Zener diode has clamped the voltage down to $50\text{V}$. Therefore, the voltage drop across $R_S$ is simply the difference between the two:

Now, armed with the voltage across the resistor and its resistance, we can invoke Ohm's Law to find the total current $I$:

Converting this to a more convenient unit, we get $I=50\text{ mA}$. This is the total current entering the main junction of our circuit.

Next, let's shift our focus to the load resistor $R$. Because it is wired in parallel with the Zener diode, it experiences the exact same clamped voltage of $50\text{V}$.

Using Ohm's Law once again, we can determine the current $I_1$ flowing specifically through this load branch:

Converting this to milliamperes gives us $I_1=0.025\text{ A}=25\text{ mA}$. So, the load is drawing exactly half of the total current supplied by the source.

Finally, we arrive at the climax of our problem. We need to find $I_z$, the current flowing through the Zener diode. To do this, we apply Kirchhoff's Current Law (KCL) at the node just above the Zener diode.

KCL states that the total current entering a junction must equal the total current leaving it. In our circuit, the total current $I$ enters the node, and it splits into two exiting paths: $I_z$ (through the diode) and $I_1$ (through the load). Mathematically, this is expressed as:

We want to isolate $I_z$, so we rearrange the equation:

Now, we simply substitute the values we worked so hard to calculate:

And there is our answer! The Zener diode is conducting $25\text{ mA}$ of current to maintain the voltage regulation. This perfectly matches option (a).

A Quick Thought Experiment: What makes the Zener diode so special? Imagine if we suddenly swapped our $2000\Omega$ load resistor for a smaller $1000\Omega$ one. The load would now demand $50\text{ mA}$ of current. Because the total current $I$ is fixed at $50\text{ mA}$ by the series resistor, the Zener diode would gracefully reduce its own current $I_z$ to $0\text{ mA}$ to compensate, ensuring the load still gets its required $50\text{V}$. It acts as a dynamic buffer, absorbing whatever current the load doesn't need!