For the circuit shown below, the current through the Zener diode is
(a) 14 mA
(b) zero
(c) 5 mA
(d)

Q: Why do we assume the Zener diode is an open circuit initially?

To check if the potential difference across its terminals is sufficient to push it into the breakdown region. If the voltage is less than the Zener rating, it won't conduct.

Q: What happens if the open-circuit voltage is less than 50V?

The Zener diode will not conduct (it remains reverse-biased but not in breakdown), acting as an open circuit, and the current through it will be zero.

Q: Why is the voltage across the 10kΩ resistor exactly 50V?

Because it is connected in parallel with the Zener diode, which maintains a constant voltage of $50\text{ V}$ when operating in its breakdown region.

Q: How did we get 70V across the 5kΩ resistor?

By applying Kirchhoff's Voltage Law. The source is $120\text{ V}$, and the parallel combination takes $50\text{ V}$, leaving $120 - 50 = 70\text{ V}$ for the series resistor.

Q: Can the Zener current be negative?

In this context, we define the Zener current in the direction of reverse bias. If the math yielded a negative value, it would mean our assumption of breakdown was wrong.

Question

For the circuit shown below, the current through the Zener diode is

Options:

(A) 14 mA

(B) zero

(C) 5 mA

(D) 9 mA

Sigma Percentile · Accepted Answer

The Zener diode is one of the most fascinating components in semiconductor electronics. It acts as a reliable voltage regulator, but only if it is pushed into its breakdown region. Let's embark on a thrilling journey to decode this circuit and find the exact current flowing through our Zener diode!

### Analyzing the Setup

Imagine you are looking at a water pipe system. Our voltage source is a powerful pump providing a constant pressure of $120	ext{ V}$. The current flows through a main pipe, which has a resistance of $5	ext{ k}\Omega$, before splitting into two parallel branches.

One branch contains our Zener diode, which is rated at $50	ext{ V}$. The other branch is a load resistor of $10	ext{ k}\Omega$. Our ultimate goal is to find the current flowing through the Zener diode branch.

### The Breakdown Check: A Crucial First Step

Before we dive into calculations, we must ask a critical question: **Is the Zener diode actually working in the breakdown region?**

To answer this, we perform a thought experiment. Imagine removing the Zener diode completely. The circuit now becomes a simple series circuit with the $120	ext{ V}$ source, the $5	ext{ k}\Omega$ resistor, and the $10	ext{ k}\Omega$ resistor.

Using the voltage divider rule, the potential difference across the terminals where the Zener diode was connected (let's call them A and B) would be:

$$V_{AB} = \frac{120 	imes 10}{5 + 10} = \frac{1200}{15} = 80	ext{ V}$$

Since $80	ext{ V}$ is significantly greater than the Zener's breakdown voltage of $50	ext{ V}$, the diode will indeed enter the breakdown region. This means it will lock the voltage across the parallel branches to exactly $50	ext{ V}$.

### The Master Equation: Kirchhoff's Current Law

Now that we know the voltage across the parallel section is fixed at $50	ext{ V}$, we can apply **Kirchhoff's Current Law (KCL)** at the junction (Node A).

The total current arriving from the source ($I_1$) must split into the Zener current ($I_2$) and the load current ($I_3$). Mathematically, this is expressed as:

$$I_1 = I_2 + I_3$$

To find the Zener current ($I_2$), we need to calculate $I_1$ and $I_3$ first.

### Final Calculation: Unveiling the Zener Current

Let's find the total current $I_1$. The voltage drop across the $5	ext{ k}\Omega$ series resistor is the difference between the source voltage and the Zener voltage:

$$V_{5	ext{k}\Omega} = 120	ext{ V} - 50	ext{ V} = 70	ext{ V}$$

Using Ohm's law, the total current is:

$$I_1 = \frac{70	ext{ V}}{5	ext{ k}\Omega} = 14	ext{ mA}$$

Next, we calculate the load current $I_3$. The voltage across the $10	ext{ k}\Omega$ load resistor is exactly the Zener voltage, $50	ext{ V}$.

$$I_3 = \frac{50	ext{ V}}{10	ext{ k}\Omega} = 5	ext{ mA}$$

Finally, we substitute these values back into our master equation to find the Zener current:

$$I_2 = I_1 - I_3$$

$$I_2 = 14	ext{ mA} - 5	ext{ mA} = 9	ext{ mA}$$

And there we have it! The current flowing through the Zener diode is a perfect **$9	ext{ mA}$**. Always remember to check the breakdown condition first, as it dictates the entire behavior of the circuit!

Animated Solution for Physics - Semiconductors: For the circuit shown below, the current through the Zener diode is

Select Answer:

Visualized Solution

Circuit Analysis

Checking Zener Breakdown

Kirchhoff's Current Law

Calculating Total Current $I_{1}$

Calculating Load Current $I_{3}$

Calculating Zener Current $I_{2}$

The Way Forward

The Sigma Insight: P-N Junction Diode

Analyzing the Setup

The Breakdown Check

The Master Equation

Final Calculation

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