When a diode is forward biased, it has a voltage drop of 0.5 V. The safe limit of current through th

Q: Why do we subtract the diode's voltage instead of adding it?

In forward bias, the diode acts as a consumer of electrical energy to overcome its internal barrier. It drops the potential, just like a resistor, which is why it gets a negative sign in our KVL loop.

Q: What would happen if we didn't connect this resistor $R$?

Without the resistor, the only voltage drop would be the diode's $0.5\text{ V}$. The remaining $1.0\text{ V}$ from the battery would force an enormous, practically infinite current through the diode, instantly burning it out. The resistor acts as a crucial current limiter.

Q: Why did we convert $10\text{ mA}$ to $10 \times 10^{-3}\text{ A}$?

Physics equations require standard SI units to yield correct results. Since voltage is in Volts and resistance is in Ohms, current must be in Amperes. Mixing milliamperes with Volts would give you a resistance in kilo-ohms, leading to a silly mistake!

Q: Does the diode's voltage drop change if the current changes?

For an ideal practical diode model, we assume the forward voltage drop remains relatively constant (like $0.5\text{ V}$ or $0.7\text{ V}$) once it's fully conducting, regardless of small changes in current. It's a simplification that works beautifully for circuit design.

Q: What if the battery was connected the other way around?

That would put the diode in reverse bias. The depletion region would widen, acting like an open switch. The current would be practically zero, and the entire $1.5\text{ V}$ would appear across the diode, with no voltage drop across the resistor.

Question

When a diode is forward biased, it has a voltage drop of $0.5	ext{ V}$. The safe limit of current through the diode is $10	ext{ mA}$. If a battery of emf $1.5	ext{ V}$ is used in the circuit, the value of minimum resistance to be connected in series with the diode, so that the current does not exceed the safe limit is

Options:

(A) $300\ \Omega$

(B) $50\ \Omega$

(C) $100\ \Omega$

(D) $200\ \Omega$

Sigma Percentile · Accepted Answer

### The Delicate Nature of Diodes

Imagine a semiconductor diode as a highly sensitive, one-way toll booth on an electrical highway. When it is **forward-biased**, it allows traffic (current) to flow, but it demands a small, fixed payment in the form of a voltage drop. In our problem, this toll is exactly $0.5	ext{ V}$.

However, this toll booth is fragile. If too many cars rush through it at once, it will overheat and be destroyed. The problem states that the **safe limit of current** is $10	ext{ mA}$. To prevent a catastrophic traffic jam, we must install a traffic controller—a resistor $R$—in series with the diode.

### Setting Up the Circuit

Let's visualize the entire system. We have a battery acting as our main pump, providing an electromotive force (EMF) of $1.5	ext{ V}$. This battery pushes current through a single, closed loop containing our protective resistor $R$ and the forward-biased diode.

Because all these components are in series, the exact same current $I = 10	ext{ mA}$ flows through every single part of the circuit. Our goal is to find the exact value of $R$ that chokes the current down to this safe maximum limit.

### The Master Equation: Kirchhoff's Voltage Law

To solve this, we invoke one of the most powerful tools in circuit analysis: **Kirchhoff's Voltage Law (KVL)**. KVL states that the total energy supplied by the battery must be exactly consumed by the components in the loop. Mathematically, the sum of all voltage changes around a closed loop is zero: $\sum \Delta V = 0$.

Let's take a walk around our circuit loop, starting from the negative terminal of the battery:
1. We move through the battery from negative to positive, gaining potential: $+1.5	ext{ V}$.
2. We pass through the diode. Since it's forward-biased, it consumes energy to push electrons across its depletion region. This is a voltage drop: $-0.5	ext{ V}$.
3. Finally, we pass through the resistor. According to Ohm's Law, the voltage drop across a resistor is the product of current and resistance: $-I \cdot R$.

Putting it all together into a single, elegant equation:
$$1.5 - 0.5 - I \cdot R = 0$$

### The Final Calculation

Before we plug in the numbers, we must heed a critical warning: **Always use standard SI units!** The current is given as $10	ext{ mA}$. If we plug $10$ directly into our equation, our resistance will be off by a factor of a thousand. We must convert milliamperes to Amperes:
$$I = 10 	imes 10^{-3}	ext{ A} = 10^{-2}	ext{ A}$$

Now, let's substitute this into our KVL equation:
$$1.5 - 0.5 - (10^{-2}) \cdot R = 0$$

Subtracting the diode's voltage drop from the battery's supply leaves us with the net voltage available for the resistor:
$$1.0 - 10^{-2} \cdot R = 0$$

Moving the resistor's voltage drop to the right side:
$$10^{-2} \cdot R = 1.0$$

Finally, isolating $R$:
$$R = \frac{1.0}{10^{-2}}$$
$$R = 100\ \Omega$$

And there we have it! A $100\ \Omega$ resistor is the perfect traffic controller. It absorbs exactly $1.0	ext{ V}$ of potential while allowing exactly $10	ext{ mA}$ of current to flow, keeping our delicate diode perfectly safe.

Select Answer:

Visualized Solution

Visualizing the Circuit

Applying Kirchhoff's Voltage Law

Substituting the Values

Simplifying the Equation

Solving for Resistance

Final Answer

The Way Forward

The Sigma Insight: P-N Junction Diode

The Delicate Nature of Diodes

Setting Up the Circuit

The Master Equation

The Final Calculation

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