Animated Solution for Physics - Semiconductors: The reading of the ammeter for a silicon diode in the given circuit is

Imagine you are looking at a simple electronic circuit. We have a $3\text{ V}$ battery acting as our power source, connected in series with a silicon diode, a $200\Omega$ resistor, and an ammeter. Our ultimate goal is to determine the reading on the ammeter, which simply means finding the current flowing through this series circuit.

The very first thing we must do when dealing with diodes is to check their biasing. Look closely at the battery's orientation. The long line represents the positive terminal, and it is connected directly to the p-side (the flat triangle side) of the silicon diode. Because the positive terminal is connected to the p-side, the diode is forward biased. This means it will act like a closed switch and allow current to flow through the circuit.

Now, here is where a lot of students make a silly mistake. They assume the diode is an ideal, perfect conductor. But this is a practical silicon diode!

When a silicon diode is forward biased, it doesn't just let current pass for free. It requires a small amount of energy to overcome its internal depletion region. This is known as the barrier potential. For a standard silicon diode, this barrier potential is approximately $0.7\text{ V}$.

Think of it like a toll booth on a highway. The battery provides $3\text{ V}$ of "energy," but the diode takes a $0.7\text{ V}$ "toll" right at the start.

To find out how much voltage is left for the rest of the circuit, we apply Kirchhoff's Voltage Law (KVL). KVL states that the total voltage supplied by the battery must equal the sum of the voltage drops across all the components in the loop.

Since the diode consumes $0.7\text{ V}$, the remaining voltage must be dropped entirely across the $200\Omega$ resistor. We can write this mathematically as:

Let's substitute our known values into this equation:

So, exactly $2.3\text{ V}$ is available to push current through the resistor.

Now that we know the voltage across the resistor and its resistance, finding the current is a straightforward application of Ohm's Law.

Substituting our values:

Let's do the math carefully. Dividing $2.3$ by $200$ gives us:

However, if we look at our options, they are all given in milliamperes ($\text{mA}$). To convert Amperes to milliamperes, we simply multiply by $1000$:

And there we have it! The ammeter will read $11.5\text{ mA}$, which perfectly matches option (c).

Always remember to check the biasing first, and never forget the $0.7\text{ V}$ drop for a silicon diode!