Video Transcript
The diagram shows a p-n junction in a series circuit. The positively charged side of the junction’s depletion region is shown in red, and the negatively charged side is shown in blue. When the battery terminal connections are reversed, does the resistance of the circuit greatly decrease, greatly increase, or stay approximately the same?
In our diagram, we see two circuits that are identical to one another except that the polarity of the battery is opposite in the two cases. Each of these circuits has a p-n junction. And if we look closely at the depletion region of these junctions, indicated by red and blue, we see that they’re not the same. The depletion region in the circuit with the original battery polarity is much larger than that region in the circuit with a reversed battery polarity. As we’ll see, this is a sign that the resistance of the circuit has changed because we flipped the battery around.
To better understand the differences between these two depletion regions we see, let’s clear some space on screen and consider an up close view of a p-n junction that is not connected to an external circuit. In this instance, the left side of our junction is the n-side. This name comes from the fact that the mobile charge carriers on this side are free electrons which are negative, while the p-side has positively charged holes, or vacancies, as its mobile charge carriers.
Even though we’ve drawn a number of negative charges on our n-side and positive charges on our p-side, each one of these sides of our p-n junction overall is electrically neutral. That’s because the mobile charge carriers in each case pair with stationary ions, which effectively cancel out those charges. Here though, for clarity, we’re only showing the charges that can move. Because opposite electrical charges attract one another, the free electrons and holes near the boundary between these two sides of our p-n junction will tend to combine.
When this happens, the depletion region that’s created rather than having no electrical charge actually does have a net charge on either side of the region. This is due to stationary positive ions on the n-side of the depletion region and stationary negative ions on the p-side of that region. The positive red side of the depletion region and the negative blue side are indicated in our original diagram.
Say that we take this p-n junction and we connect it up to the circuit with the original battery polarity. That would look something like this, where we can see the positive terminal of the battery faces to the left and the negative terminal faces to the right. Conventionally, positive charge flows out of the positive terminal of the battery. In this circuit, that charge will travel in a clockwise direction, while the negative charge that comes from the negative terminal of the battery will travel counterclockwise.
Let’s now think about what will happen when these positive and negative charges approach opposite ends of our p-n junction. On the left, the n-side, many of our negative mobile charge carriers will be drawn to the left toward the positive incoming charges. The effect of that will be to expand the depletion region on this side of the junction. The mobile negative charges, we could say, are pulled away from the center of the junction, but then something very similar happens on the other side of the junction where the mobile positive charges are drawn towards the incoming negative charge. This also has the effect of expanding the depletion region.
We see that the depletion region has gotten larger. The wider this region is, the less likely it is that a mobile charge carrier from either side can make it across the region. Therefore, the bigger the depletion region, the smaller the current through this p-n junction. For our circuit with the original battery polarity then, we would expect a very small, perhaps even negligently small, current in that circuit. Let’s now imagine that we have a switch in our circuit that we’re able to open up. Doing this stops the flow of charge in the circuit, and our depletion region returns to its original size.
With our switch still open, let’s say that we now reverse the polarity of our battery. The positive terminal is now on the right and the negative on the left. If we then close our switch so the circuit is now complete, now positive charge will start to flow in a counterclockwise direction, while negative charge goes the other way, clockwise. Notice that now we have a positive charge in our current approaching mobile positive charges on the p-side of our junction and likewise a negative charge in our current approaching mobile negative charges on the inside.
Because like electrical charges repel one another, the positive charge coming in from the right in our circuit will push these mobile positive charges toward the depletion region. And likewise, the negative charge coming in on the opposite side of our p-n junction will push the mobile negative charges also towards the center of the junction. The effect of this is to shrink the depletion region. In fact, if the voltage supplied by the battery is high enough, the depletion region might disappear entirely. Therefore, there is now a relatively smaller region, if there is such a region at all, in our p-n junction where mobile charge carriers cannot go. This means there’s now much less resistance to the flow of mobile charge carriers across the junction. With many such charges moving, the current in the circuit will be relatively large.
By setting up the battery polarity this way, relative to our p-n junction, we’ve significantly increased the current in the circuit. We’ve done this by decreasing very strongly the resistance of our p-n junction. This tells us then how to answer our question of the effect on the overall circuit resistance of reversing our battery’s polarity. By doing this, we greatly decreased the resistance in the circuit. And therefore, the current in the circuit goes from a very small value to a large value. This is the effect of reversing the battery’s polarity.
