Question Video: Identifying the Junction Depletion Region Corresponding to Reverse Bias Physics • 9th Grade

Each of the following diagrams shows a p-n junction. The charged sides of the junction’s depletion region are shown in red and blue. The junction is shown under conditions of forward bias, reverse bias, and zero bias. Which bias shown correctly represents reverse bias? [A] Option A [B] Option B [C] Option C

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Video Transcript

Each of the following diagrams shows a p-n junction. The charged sides of the junction’s depletion region are shown in red and blue. The junction is shown under conditions of forward bias, reverse bias, and zero bias. Which bias shown correctly represents reverse bias?

Our three answer options (A), (B), and (C) show us candidates for representations of reverse bias across a semiconductor diode. The red and blue regions represent differently charged sides of the depletion region. To get started, say that we have a p-type and an n-type semiconductor material joined together like this. This means that on the right side of our diode the negatively charged electrons are the charge carriers, while on the left the mobile charge carriers are positively charged holes. The free electrons near the junction between these two material types will move across that barrier and cause the destruction, we could say, of mobile charge carriers near the junction between these materials.

Once left over on either side are stationary negatively charged atoms and stationary positively charged atoms. Because this region is now free of mobile charge carriers, it’s called the depletion region. It’s the two sides of this region that are indicated by the red and blue areas in our answer options. With our diode as is, no external voltage being applied, we say that it’s experiencing zero bias. We can say then that under the condition of zero bias, we do expect to have a depletion region in our diode. From this, we can change the size of that depletion region by applying external voltage.

Say we connect up the ends of our diode to a cell like this. Oriented this way, the cell will conventionally send positive charge moving clockwise and negative charge moving counterclockwise through this circuit. Considering the incoming positive charge on the left side of our circuit, we know that this will repel the mobile positive charges in the p-type side of our diode. As a result, those mobile positive charges will push into the depletion region. Likewise on the right, these incoming negative charges in our current will push on the mobile negative charges on the n-type side of our diode. And through this mutual repulsion, the mobile negative charges will push into the depletion region from the right.

The overall result of this is that the depletion region will shrink. And if the voltage applied by the cell is great enough, the depletion region will disappear entirely. This is what we see depicted in answer option (C). When an externally applied voltage shrinks a depletion region down, the diode is said to be in a condition of forward bias. Set up this way, it’s much easier for mobile charge to flow from one end of the diode to the other. To see what reverse biasing looks like, let’s imagine we have a switch in our circuit that we open so that charge stops flowing in the circuit. When this happens, our depletion region returns as before.

Now let’s say that we take our cell and we flip it around so that now the positive terminal faces to the right. If we close the switch so that charge again starts to flow in the circuit, we know that now positive charge will effectively flow counterclockwise and negative charge clockwise. On the right side of our circuit, our incoming positive charges will attract the mobile negative charges on the n-type side of the diode. Likewise on the left, the incoming negative charges will attract the mobile positive charges on the p-type side. On both sides of the diode, mobile charges will move away from the depletion region and that region will grow.

Note that the larger the depletion region gets, the less likely it is that mobile charges can move across this region. In other words, the diode becomes a worse and worse conductor of electricity. The amount the depletion region grows depends on the magnitude of voltage being applied. What we’ve seen is that reverse biasing a diode creates a larger depletion region. Reviewing our answer options (A), (B), and (C), we can say that (C) shows us forward biasing, (B) shows us the depletion region under zero bias, while answer option (A) shows us the depletion region under reverse bias.

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