The figure represents a voltage regulator circuit using a Zener diode. The breakdown voltage of the

Q: Why is the voltage across $R_L$ constant?

Because the load resistor $R_L$ is connected in parallel with the Zener diode. As long as the Zener diode is in its breakdown region, it maintains a constant voltage ($V_Z$) across its terminals, forcing the load to have the exact same voltage.

Q: What if $V_B$ was less than $6\text{ V}$?

If the battery voltage drops below the Zener breakdown voltage ($6\text{ V}$), the Zener diode will not enter the breakdown region. It will act like an open circuit, and the voltage across the load will simply be determined by the voltage divider rule between $R_i$ and $R_L$.

Q: Why do we need the series resistor $R_i$?

The series resistor $R_i$ is crucial. It absorbs the excess voltage from the battery ($V_B - V_Z$) and limits the total current flowing into the circuit. Without it, the Zener diode would draw an immense amount of current and instantly burn out.

Q: Does the load current ever change in this setup?

No, as long as the Zener diode remains in breakdown, the voltage across the load is locked at $6\text{ V}$. Since $R_L$ is a fixed resistor, Ohm's law ($i = V/R$) dictates that the load current must remain perfectly constant.

Q: How does the Zener diode regulate voltage?

The Zener diode regulates voltage by dynamically varying its internal resistance. When the input voltage increases, it draws more current (absorbing the excess), and when the input voltage decreases, it draws less current, all while keeping the voltage across its terminals fixed.

Question

The figure represents a voltage regulator circuit using a Zener diode. The breakdown voltage of the Zener diode is $6	ext{ V}$ and the load resistance is $R_L = 4	ext{ k}\Omega$. The series resistance of the circuit is $R_i = 1	ext{ k}\Omega$. If the battery voltage $V_B$ varies from $8	ext{ V}$ to $16	ext{ V}$, what are the minimum and maximum values of the current through Zener diode?

Options:

(A) $1.5	ext{ mA}, 8.5	ext{ mA}$

(B) $1	ext{ mA}, 8.5	ext{ mA}$

(C) $0.5	ext{ mA}, 8.5	ext{ mA}$

(D) $0.5	ext{ mA}, 6	ext{ mA}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are designing a power supply for a sensitive electronic device. Your raw power source is a battery, but its voltage isn't stable—it fluctuates between $8	ext{ V}$ and $16	ext{ V}$. If you feed this directly to your device, it might fry! This is where our hero, the **Zener diode**, steps in to save the day.

In our circuit, the Zener diode is placed in parallel with the load (your sensitive device, represented by $R_L$). A series resistor $R_i$ acts as a buffer between the fluctuating battery and the stable load.

### The Master Equation: Locking the Voltage

The defining characteristic of a Zener diode is its ability to maintain a constant voltage across its terminals when operated in its breakdown region. Here, the breakdown voltage is given as $V_Z = 6	ext{ V}$.

Because the load resistor $R_L$ is connected perfectly in parallel with the Zener diode, the voltage across the load is forcibly locked to this same value.

$$V_L = V_Z = 6	ext{ V}$$

This is the magic of voltage regulation! No matter what the battery does, the load sees exactly $6	ext{ V}$.

### Calculating the Constant Load Current

Since the voltage across the load is fixed, the current flowing through it must also be perfectly constant. We can find this using Ohm's Law:

$$i_L = \frac{V_L}{R_L}$$

Substituting our known values:

$$i_L = \frac{6}{4 	imes 10^3} = 1.5 	imes 10^{-3}	ext{ A} = 1.5	ext{ mA}$$

So, the load constantly draws $1.5	ext{ mA}$ of current.

### The Role of the Series Resistor

Now, let's look at the series resistor $R_i$. It has to absorb the excess voltage from the battery. By Kirchhoff's Voltage Law, the voltage drop across $R_i$ is the difference between the battery voltage and the regulated Zener voltage:

$$V_{Ri} = V_B - V_Z$$

The total current $i_i$ flowing out of the battery passes through this resistor. At the junction above the Zener diode, this total current splits. One part goes to the load ($i_L$), and the rest is "swallowed" by the Zener diode ($i_Z$).

$$i_i = i_Z + i_L$$

Rearranging this, we find the current through the Zener diode:

$$i_Z = i_i - i_L$$

### Final Calculation: The Two Extremes

Let's test our circuit at the two extreme battery voltages.

**Case 1: Minimum Battery Voltage ($V_B = 8	ext{ V}$)**

When the battery is at its weakest, the voltage across the series resistor is:

$$V_{Ri} = 8 - 6 = 2	ext{ V}$$

The total current is:

$$i_{i(	ext{min})} = \frac{2}{1 	imes 10^3} = 2	ext{ mA}$$

Subtracting the constant load current, the minimum Zener current is:

$$i_{Z(	ext{min})} = 2	ext{ mA} - 1.5	ext{ mA} = 0.5	ext{ mA}$$

**Case 2: Maximum Battery Voltage ($V_B = 16	ext{ V}$)**

When the battery surges to its maximum, the series resistor takes the hit:

$$V_{Ri} = 16 - 6 = 10	ext{ V}$$

The total current spikes to:

$$i_{i(	ext{max})} = \frac{10}{1 	imes 10^3} = 10	ext{ mA}$$

Again, the load only takes its required $1.5	ext{ mA}$. The Zener diode absorbs all the excess:

$$i_{Z(	ext{max})} = 10	ext{ mA} - 1.5	ext{ mA} = 8.5	ext{ mA}$$

The Zener diode dynamically adjusts its current between **$0.5	ext{ mA}$** and **$8.5	ext{ mA}$** to ensure the load remains perfectly protected.

Select Answer:

Visualized Solution

Analyzing the Voltage Regulator Circuit

Voltage Across the Load

Calculating Load Current ( $i_{L}$ )

Voltage Across Series Resistor ( $R_{i}$ )

Kirchhoff's Current Law (KCL)

Case 1: Minimum Battery Voltage ( $V_{B} = 8 V$ )

Minimum Zener Current

Case 2: Maximum Battery Voltage ( $V_{B} = 16 V$ )

Maximum Zener Current

Conclusion

Power Dissipation Consideration

The Sigma Insight: Semiconductor and p-n Junction Diode

Analyzing the Setup

The Master Equation

Calculating the Constant Load Current

The Role of the Series Resistor

Final Calculation

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