The circuit shown below is working as a 8 V DC regulated voltage source. When 12 V is used as input,

Q: Why do we subtract the output voltage from the input voltage to find the drop across the resistors?

The input voltage is the total potential provided to the circuit. The output voltage is the potential remaining after the drop across the series resistors. Therefore, the difference between them must be the voltage consumed by those resistors.

Q: Why is the current the same through the resistors and the Zener diodes?

In this specific circuit, the load is not drawing any current (it's an open circuit at the output terminals). Thus, all the current flowing through the series resistors must continue its path through the Zener diodes to the ground.

Q: What if the Zener diodes were not identical?

If they weren't identical, they would have different Zener breakdown voltages. The total output voltage would be the sum of their individual breakdown voltages, and the power dissipated by each would be different, calculated using their specific $V_Z$.

Q: Why are the Zener diodes connected in reverse bias?

Zener diodes are designed to operate in their breakdown region to regulate voltage. This breakdown only occurs when they are reverse-biased. If forward-biased, they would act like normal diodes with a small voltage drop (around 0.7V).

Q: Can we use a single 8V Zener diode instead of two 4V ones?

Yes, functionally it would provide the same 8V regulation. However, using two diodes distributes the power dissipation. A single diode would have to dissipate the entire 80 mW of power, which might exceed its power rating and cause it to burn out.

Question

The circuit shown below is working as a 8 V DC regulated voltage source. When 12 V is used as input, the power dissipated (in mW) in each diode is; (considering both Zener diodes are identical) ......... .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are looking at a water pipe system where the pressure drops as water flows through different sections. Our circuit is very similar! We start with an input voltage of $12 	ext{ V}$ and end up with a regulated output of $8 	ext{ V}$.

In between, the current must pass through two $200 \ \Omega$ resistors connected in series. After these resistors, the current flows through two identical Zener diodes, also connected in series, before returning to the ground.

Our goal is to find out exactly how much power is being dissipated as heat by each of these Zener diodes.

### The Master Equation

First, we need to figure out what happens across those two resistors. If we start with $12 	ext{ V}$ and end up with $8 	ext{ V}$ right after the resistors, the difference must have been "dropped" across them.

We can write this as:

$$ \Delta V = V_{in} - V_{out} $$

Substituting our values, we get:

$$ \Delta V = 12 	ext{ V} - 8 	ext{ V} = 4 	ext{ V} $$

So, a total of $4 	ext{ V}$ is dropped across the combined resistance of the two resistors. Since they are in series, their equivalent resistance is simply $200 \ \Omega + 200 \ \Omega = 400 \ \Omega$.

Now, we can use Ohm's Law to find the total current flowing through the circuit.

$$ I = \frac{\Delta V}{R_{eq}} $$

$$ I = \frac{4 	ext{ V}}{400 \ \Omega} = 0.01 	ext{ A} $$

This $0.01 	ext{ A}$ (or $10 	ext{ mA}$) is the current flowing through the entire series circuit, including our Zener diodes.

### Final Calculation

Now, let's shift our focus to the Zener diodes. We know they are connected in series and the total voltage across both of them is our output voltage, $8 	ext{ V}$.

Because the problem states that both Zener diodes are identical, they will share this voltage drop equally.

$$ V_Z = \frac{8 	ext{ V}}{2} = 4 	ext{ V} $$

Each Zener diode has exactly $4 	ext{ V}$ across it. To find the power dissipated by a single diode, we use the electrical power formula:

$$ P = V_Z 	imes I $$

Substituting the values we found:

$$ P = 4 	ext{ V} 	imes 0.01 	ext{ A} = 0.04 	ext{ W} $$

To convert this to milliwatts, we multiply by $1000$.

$$ P = 40 	ext{ mW} $$

And there we have it! Each Zener diode is dissipating **$40 	ext{ mW}$** of power to keep the voltage regulated.

Animated Solution for Physics - Semiconductors: The circuit shown below is working as a 8 V DC regulated voltage source. When 12 V is used as input, the power dissipated (in mW) in each diode is; (considering both Zener diodes are identical) ......... .

Enter Numerical Value:

Visualized Solution

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

$V o l t a g e D r o p A cr oss R es i s t or s$

$C a l c u l a t in g V o l t a g e D r o p$

$O h m^{'} s L a w f or C u r r e n t$

$C a l c u l a t in g C u r r e n t$

$V o l t a g e A cr oss Z e n er D i o d es$

$V o l t a g e p er D i o d e$

$P o w er D i ss i p a t i o n F or m u l a$

$C a l c u l a t in g P o w er$

$F ina l A n s w er$

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

The Sigma Insight: P-N Junction Diode

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

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Animated Solution for Physics - Semiconductors: The circuit shown below is working as a 8 V DC regulated voltage source. When 12 V is used as input, the power dissipated (in mW) in each diode is; (considering both Zener diodes are identical) ......... .

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Circuit Analysis

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Voltage Drop Across Resistors

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Voltage Drop

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ohm′s Law for Current

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Current

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Voltage Across Zener Diodes

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Voltage per Diode

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Power Dissipation Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Power

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion