For the given circuit, the power across Zener diode is ............ mW.

Q: Why do we assume the voltage across the load is $10\text{ V}$?

Because the Zener diode is connected in parallel with the load. When the unregulated voltage is high enough to cause Zener breakdown, the diode maintains a constant voltage of $10\text{ V}$ across its terminals, and thus across the load.

Q: What if the calculated voltage without the Zener was less than $10\text{ V}$?

If the voltage without the Zener was less than $10\text{ V}$, the Zener diode would not enter the breakdown region. It would act like an open circuit, and no current would flow through it. The power dissipated by it would be zero.

Q: Why did we subtract $10\text{ V}$ from $24\text{ V}$ for the series resistor?

According to Kirchhoff's Voltage Law, the sum of voltage drops in a closed loop must equal the source voltage. The source is $24\text{ V}$, and the parallel combination of the Zener and load takes $10\text{ V}$. The remaining $14\text{ V}$ must drop across the series resistor.

Q: Can we use the power formula $P = I^2 R$ for the Zener diode?

No, a Zener diode is a non-linear device and doesn't have a constant resistance. You must use the fundamental power formula $P = V \times I$.

Q: Why are the currents in milli-amperes?

The voltages are in volts and the resistances are in kilo-ohms ($10^3\ \Omega$). According to Ohm's law, $I = V/R$, so dividing volts by kilo-ohms naturally gives current in milli-amperes ($10^{-3}\text{ A}$).

Question

For the given circuit, the power across Zener diode is ............ mW.

Sigma Percentile · Accepted Answer

The problem asks us to find the power dissipated by a Zener diode in a voltage regulator circuit. This is a classic application of Zener diodes, and it tests our understanding of circuit analysis and the breakdown region of a diode.

### Analyzing the Setup

Imagine you are looking at a water pipe system where the Zener diode acts as a pressure relief valve. We have a $24	ext{ V}$ unregulated DC source, a series resistor of $1	ext{ k}\Omega$, and a load resistor of $5	ext{ k}\Omega$. The Zener diode is connected in parallel with the load and has a breakdown voltage of $V_Z = 10	ext{ V}$.

Our ultimate goal is to find the power consumed by this Zener diode. To do that, we need to know two things: the voltage across it and the current flowing through it.

### The Master Equation

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

If the voltage across the load (without the Zener) is less than $10	ext{ V}$, the Zener diode won't even turn on! Let's check this by using the voltage divider rule. If we remove the Zener diode, the voltage across the load would be:
$$V = 24	ext{ V} 	imes \frac{5	ext{ k}\Omega}{1	ext{ k}\Omega + 5	ext{ k}\Omega} = 20	ext{ V}$$

Since $20	ext{ V}$ is greater than the Zener breakdown voltage of $10	ext{ V}$, the diode will indeed enter the breakdown region. It will act like a strict traffic cop, clamping the voltage across the load to exactly $10	ext{ V}$.

### Calculating the Currents

Now that we know the voltage across the load is fixed at $10	ext{ V}$, we can easily find the load current $I_L$ using Ohm's law:
$$I_L = \frac{V_Z}{R_L} = \frac{10	ext{ V}}{5	ext{ k}\Omega} = 2	ext{ mA}$$

Next, let's look at the series resistor. The battery provides $24	ext{ V}$, and $10	ext{ V}$ is dropped across the parallel combination of the Zener and the load. According to Kirchhoff's Voltage Law, the remaining voltage must drop across the series resistor:
$$V_S = 24	ext{ V} - 10	ext{ V} = 14	ext{ V}$$

With the voltage across the series resistor known, we can find the total current $I$ flowing from the source:
$$I = \frac{V_S}{R_S} = \frac{14	ext{ V}}{1	ext{ k}\Omega} = 14	ext{ mA}$$

Now, focus on the junction where the current splits. According to **Kirchhoff's Current Law (KCL)**, the total current $I$ must equal the sum of the Zener current $I_Z$ and the load current $I_L$:
$$I = I_Z + I_L$$

Rearranging this to solve for the Zener current:
$$I_Z = I - I_L = 14	ext{ mA} - 2	ext{ mA} = 12	ext{ mA}$$

### Final Calculation

We finally have everything we need! The power dissipated by any component is the product of the voltage across it and the current through it. For our Zener diode:
$$P_Z = V_Z 	imes I_Z$$

Substituting our values:
$$P_Z = 10	ext{ V} 	imes 12	ext{ mA} = 120	ext{ mW}$$

The power across the Zener diode is **$120	ext{ mW}$**. Always remember to verify the Zener state before diving into the currents. It's a small step that saves you from big mistakes!

Animated Solution for Physics - Semiconductors: For the given circuit, the power across Zener diode is ............ mW.

Enter Numerical Value:

Visualized Solution

$C i r c u i t A na l y s i s$

$Z e n er B r e ak d o w n C h ec k$

$L o a d C u r r e n t$

$S er i es R es i s t or V o l t a g e$

$T o t a l C u r r e n t$

$Z e n er C u r r e n t$

$P o w er D i ss i p a t i o n$

$C o n c l u s i o n$

The Sigma Insight: P-N Junction Diode

Analyzing the Setup

The Master Equation

Calculating the Currents

Final Calculation

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