The value of power dissipated across the Zener diode (Vz = 15V) connected in the circuit as shown in

Q: Why is the voltage across $R_L$ exactly $15\ \text{V}$?

Because $R_L$ is connected in parallel with the Zener diode. In a parallel circuit, the voltage across all branches is identical. Since the Zener diode is in breakdown, it locks the voltage at $15\ \text{V}$.

Q: What would happen if the battery was only $10\ \text{V}$?

If the supply voltage is less than the Zener breakdown voltage ($15\ \text{V}$), the diode will not enter the breakdown region. It will act like an open switch, and the circuit will behave as a simple series circuit with $R_s$ and $R_L$.

Q: Why do we subtract $I_L$ from $I$ to find $I_z$?

This is an application of Kirchhoff's Current Law (KCL). The total current $I$ arrives at the junction and splits into two paths: one through the Zener diode ($I_z$) and one through the load ($I_L$). Therefore, $I = I_z + I_L$.

Q: Can the Zener diode burn out in this circuit?

Yes! Every Zener diode has a maximum power rating. If the power dissipated ($0.5\ \text{W}$ in this case) exceeds its maximum rating, the diode will overheat and get destroyed.

Q: What is the purpose of the series resistor $R_s$?

The series resistor $R_s$ is crucial. It absorbs the excess voltage from the supply ($7\ \text{V}$ here) and limits the total current flowing into the circuit, protecting the Zener diode from receiving too much current and burning out.

Question

The value of power dissipated across the Zener diode ($V_z = 15	ext{V}$) connected in the circuit as shown in the figure is $x 	imes 10^{-1}\ 	ext{W}$.

The value of $x$, to the nearest integer, is ……… .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome to the fascinating world of Zener diodes! Imagine you are an electrical engineer tasked with ensuring a delicate component receives exactly $15\ 	ext{V}$, no matter what.

That is exactly what this circuit does. We have a $22\ 	ext{V}$ main power supply, but our load resistor $R_L$ only needs $15\ 	ext{V}$.

To achieve this, we place a Zener diode in parallel with the load. The Zener diode acts like a strict bouncer, refusing to let the voltage across it rise above its breakdown voltage of $15\ 	ext{V}$.

### The Master Equation

First, we must confirm the bouncer is awake. Since our supply is $22\ 	ext{V}$, which is greater than $15\ 	ext{V}$, the Zener diode is actively in its breakdown region.

This means the voltage across the parallel load resistor $R_L$ is locked at $15\ 	ext{V}$. But where does the extra voltage go?

It is dropped across the series resistor $R_s$. The voltage across $R_s$ is simply the difference: $V_s = 22 - 15 = 7\ 	ext{V}$.

Now, let's track the flow of electrons. Using Ohm's law, the total current leaving the battery is $I = \frac{V_s}{R_s} = \frac{7}{35} = \frac{1}{5}\ 	ext{A}$.

This total current reaches the junction and splits. A portion goes through the load, and the rest goes through the Zener diode.

The current through the load is $I_L = \frac{V_z}{R_L} = \frac{15}{90} = \frac{1}{6}\ 	ext{A}$.

### Final Calculation

To find the current forced through the Zener diode, we apply Kirchhoff's Current Law. We subtract the load current from the total current.

This gives us $I_z = I - I_L = \frac{1}{5} - \frac{1}{6} = \frac{1}{30}\ 	ext{A}$.

Finally, we need the power dissipated by the Zener diode. Power is the product of voltage and current, so $P = V_z 	imes I_z = 15 	imes \frac{1}{30} = 0.5\ 	ext{W}$.

The question asks for the answer in the format $x 	imes 10^{-1}\ 	ext{W}$. We can rewrite $0.5\ 	ext{W}$ as **$5 	imes 10^{-1}\ 	ext{W}$**.

Therefore, our final integer value is **$x = 5$**.

Animated Solution for Physics - Semiconductors: The value of power dissipated across the Zener diode ( $V_{z} = 15 V$ ) connected in the circuit as shown in the figure is $x \times 1 0^{- 1} W$ . The value of $x$ , to the nearest integer, is ……… .

Enter Numerical Value:

Visualized Solution

Circuit Analysis

Zener State

Voltage across $R_{s}$

Total Current $I$

Load Current $I_{L}$

Zener Current $I_{z}$

Power Dissipation

Final Value of $x$

Conceptual Check

The Sigma Insight: P-N Junction Diode

Analyzing the Setup

The Master Equation

Final Calculation

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