Animated Solution for Physics - Semiconductors: In the forward bias arrangement of a $p-n$ junction rectifier, the $p$ end is connected to the ....... terminal of the battery and the direction of the current is from ....... to ...... in the rectifier.

Imagine a bustling city divided into two distinct neighborhoods. On one side, you have the $p$-type region, which is teeming with 'holes' (positive charge carriers) looking for electrons. On the other side, you have the $n$-type region, packed with free electrons (negative charge carriers) eager to find a home.

When these two neighborhoods are joined to form a $p-n$ junction, a fascinating thing happens at the border. A few electrons wander over to the $p$-side, and a few holes wander over to the $n$-side. This creates a 'no man's land' called the depletion region, which acts as a natural barrier, preventing any further free movement of charges.

Forward Bias

To make this junction useful as a rectifier (a device that lets current flow in only one direction), we need to give these charge carriers a push. This is where forward biasing comes into play.

When we connect a battery to the $p-n$ junction, we have a choice. To forward bias the diode, we connect the $p$-side to the positive terminal of the battery and the $n$-side to the negative terminal.

Why does this work? The positive terminal of the battery repels the positive holes in the $p$-region, pushing them toward the junction. Simultaneously, the negative terminal repels the negative electrons in the $n$-region, also pushing them toward the junction. This combined push squeezes the depletion region, making it narrower and narrower until it practically disappears.

Once the barrier is overcome, a significant number of charge carriers can cross the junction. Electrons flood from the $n$-side to the $p$-side, and holes flood from the $p$-side to the $n$-side.

However, when we talk about electrical current in standard physics, we always refer to conventional current, which is defined as the direction in which positive charges would move. Since holes (positive charges) are moving from the $p$-side to the $n$-side, the conventional current inside the rectifier flows exactly in that direction: from the $p$-side to the $n$-side.

So, to fill in the blanks of our question: In a forward bias arrangement, the $p$ end is connected to the positive terminal, and the direction of the current is from the $p$-side to the $n$-side.