The Zener diode has a V_z = 30 V. The current passing through the diode for the following circuit is

Q: How do we know the Zener diode is actually in the breakdown region?

We can check by assuming the Zener is removed and calculating the voltage across the load. If it's greater than $V_z$, the Zener will enter breakdown. Here, the open-circuit voltage would be $90 \times (5/9) = 50\text{ V}$, which is strictly greater than $30\text{ V}$. Thus, breakdown is confirmed.

Q: Why is the voltage across the $5\text{ k}\Omega$ resistor exactly $30\text{ V}$?

Because it is connected in parallel with the Zener diode. In a parallel circuit, the voltage across all branches is identical by definition.

Q: What would happen if the supply voltage was only $20\text{ V}$?

The Zener diode would not reach its breakdown voltage. It would act like an open circuit, and the circuit would just be a simple series combination of the $4\text{ k}\Omega$ and $5\text{ k}\Omega$ resistors.

Q: Why do we subtract $V_z$ from the supply voltage to find the series resistor's voltage?

This is a direct application of Kirchhoff's Voltage Law (KVL). The sum of voltage drops in a closed loop must equal the supply voltage. So, $V_{supply} = V_{series} + V_{parallel}$.

Q: Can the Zener current be negative?

In this context, we define the Zener current $I_z$ as flowing from cathode to anode (the reverse direction). If the calculated $I_z$ was negative, it would mean the diode isn't in breakdown, but here it's firmly positive.

Question

The Zener diode has a $V_z = 30	ext{ V}$. The current passing through the diode for the following circuit is ......... mA.

Sigma Percentile · Accepted Answer

### The Setup: Understanding the Circuit

Let's carefully analyze the given circuit. We have a $90	ext{ V}$ DC power supply connected to a $4	ext{ k}\Omega$ series resistor. This is then connected to a parallel combination of a Zener diode and a $5	ext{ k}\Omega$ load resistor.

The Zener diode is reverse biased here. Its breakdown voltage is given as $30	ext{ V}$. This means it will lock the voltage across the parallel branch to exactly $30	ext{ V}$, provided the supply is sufficient to push it into the breakdown region.

### The Zener Lock: Fixing the Voltage

Since the $5	ext{ k}\Omega$ resistor is in parallel with the Zener diode, the voltage across it is also forced to be $30	ext{ V}$. Using Ohm's law, we can find the current flowing through this load resistor. Let's call it $I_1$.

Dividing $30	ext{ V}$ by $5000\,\Omega$ gives us $0.006	ext{ A}$, which is exactly $6	ext{ mA}$.

$$I_1 = \frac{30}{5000} = 6	ext{ mA}$$

So, $6	ext{ mA}$ of current flows through the load resistor.

### Tracing the Currents: Ohm's Law in Action

Now, let's look at the $4	ext{ k}\Omega$ series resistor. By Kirchhoff's Voltage Law, the voltage drop across it will be the total supply voltage minus the Zener voltage.

Subtracting $30	ext{ V}$ from the $90	ext{ V}$ supply, we get $60	ext{ V}$. This $60	ext{ V}$ is the potential difference across the $4	ext{ k}\Omega$ resistor.

$$V' = 90 - 30 = 60	ext{ V}$$

With $60	ext{ V}$ across $4000\,\Omega$, we can again use Ohm's law to find the total current, $I$, drawn from the battery. Dividing $60$ by $4000$ gives $0.015	ext{ A}$, or $15	ext{ mA}$. This is the total current entering the parallel junction.

$$I = \frac{60}{4000} = 15	ext{ mA}$$

### The Junction Split: Kirchhoff's Current Law

At the junction, this total current splits into two paths: one part goes through the Zener diode, and the rest goes through the load resistor. According to Kirchhoff's Current Law, the total current equals the sum of the Zener current and the load current.

$$I = I_z + I_1$$

To find the Zener current, we simply subtract the load current from the total current. $15	ext{ mA}$ minus $6	ext{ mA}$ gives us **$9	ext{ mA}$**.

$$I_z = 15 - 6 = 9	ext{ mA}$$

And that is our final answer!

### Beyond the Problem: Power Limits

What if the load resistance was removed? All $15	ext{ mA}$ would flow through the Zener diode. It's crucial to ensure this doesn't exceed the diode's maximum power rating. Always think about the limits of your components!

Animated Solution for Physics - Semiconductors: The Zener diode has a $V_{z} = 30 V$ . The current passing through the diode for the following circuit is ......... mA.

Enter Numerical Value:

Visualized Solution

Circuit Analysis

Zener Breakdown Voltage

Current through Load Resistor

Calculating $I_{1}$

Voltage across Series Resistor

Calculating $V^{'}$

Total Source Current

Calculating $I$

Kirchhoff's Current Law

Final Zener Current

The Way Forward

The Sigma Insight: P-N Junction Diode

The Setup

The Zener Lock

Tracing the Currents

The Junction Split

Beyond the Problem

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Animated Solution for Physics - Semiconductors: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Zener diode has a Vz​=30 V. The current passing through the diode for the following circuit is ......... mA.