Video: Finding the Resistance from an 𝐼-𝑉 Graph of a Diode

The graph shows the 𝐼-𝑉 characteristics of a diode. At which of the points marked on the graph is the resistance of the diode highest? At which of the points marked on the graph is the resistance of the diode lowest?

09:52

Video Transcript

The graph shows the 𝐼-𝑉 characteristics of a diode. At which of the points marked on the graph is the resistance of the diode highest? At which of the points marked on the graph is the resistance of the diode lowest?

Okay, so, what we can see in this question is that we’ve been given a graph which shows the 𝐼-𝑉, or current voltage, characteristics of a diode. So, on this graph, we can see the potential difference, or voltage, across the diode on the horizontal axis and the current through the diode on the vertical axis. We can also see that there are four different points marked on this curve, point 𝑃, point 𝑄, point 𝑆, and point 𝑇.

Now based on the two questions that we’ve been asked over here, we need to choose from points 𝑃, 𝑄, 𝑆, and 𝑇, the points at which the resistance of the diode is highest and the point at which the resistance of the diode is lowest.

Now to calculate the resistance of a component in an electric circuit, we need to recall that the resistance is defined as the potential difference across that component divided by the current through that component. Now at this point, we’ll see this equation and immediately think that this is just a rearranged version of Ohm’s law, which is 𝑉 is equal to 𝐼𝑅. In other words, the potential difference across a component is equal to the current through that component multiplied by the resistance of that component.

However, this is a slight misconception. Because actually what Ohm’s law is telling us is that the potential difference across a component for an ohmic conductor is directly proportional to the current through the conductor. And actually, the constant of proportionality is the resistance of that conductor. Therefore, it is true that for an ohmic conductor 𝑉 is equal to 𝐼𝑅. However, it’s important to remember that this 𝑅 is constant for an ohmic conductor.

Whereas in this case, we’ve got a diode which is a non-ohmic conductor. Therefore, this value that we’re trying to calculate, the resistance of the diode at points 𝑃, 𝑄, 𝑆, and 𝑇, is not a constant value. It’s actually a function of the potential difference or equivalently a function of the current through the diode. And hence, this equation is not actually Ohm’s law because the resistance value in this equation is not a constant. So, that’s one misconception cleared up.

Now the other misconception is that the resistance value of the diode, in this case, is directly related to the slope of this curve. For example, let’s say we wanted to calculate the resistance of the diode at this point over here. And in order to do so, we calculated the slope of the 𝐼-𝑉 characteristics curve at that point. In other words, we drew a tangent to the curve at that point, and then worked out what the change in the vertical axis of this value was for that tangent, as well as the change in the horizontal axis’ value. Basically, the slope of the curve is equal to the change in current on that tangent divided by the change in potential difference. And so, we might think that this looks slightly similar to this equation here. And we might take the reciprocal of both sides of this equation.

In other words, we might say that one divided by the slope of the curve is equal to Δ𝑉 divide by Δ𝐼. We’ve just flipped this fraction here. And then we might think that Δ𝑉 divide by Δ𝐼 looks something like 𝑉 divide by 𝐼. And so, one divided by the slope must be the resistance of the diode at this point here. However, that is not true. In fact, Δ𝑉 divided by Δ𝐼 in general is not equal to 𝑉 divided by 𝐼. The only time that the slope of the 𝐼-𝑉 characteristics is equal to the resistance is when we have an ohmic conductor.

In other words, Δ𝑉 divide by Δ𝐼 is only equal to 𝑉 over 𝐼 when 𝑅, the resistance of the conductor, is constant. And as we’ve seen already, this is not the case.

So, if we want to calculate the actual resistance of the diode at points 𝑃, 𝑄, 𝑆, and 𝑇, what we’re going to have to do is to take the value of 𝑉 at that particular point and divide it by the value of 𝐼 at that particular point. In other words, we have no reason to find the slope of the curve.

So, now that we’ve discussed these two misconceptions, let’s actually go about calculating the resistance value at points 𝑃, 𝑄, 𝑆, and 𝑇. But how are we going to go back doing that when our axes are not labelled. We don’t know what each one of these large grids represents in terms of potential difference. And similarly, we don’t know what each one of these large grids represents, in terms of current.

However, this actually doesn’t matter. What we can do is to firstly label the origin of the curve, which is at this point over here, And then we can say that each large box on the horizontal axis represents a potential difference of 𝐴 volts. So, if the first large box represents 𝐴 volts. Then the second large box will represent two 𝐴 volts. The third one will be three 𝐴 volts. And the fourth will be four 𝐴 volts, and so on and so forth.

And we can do a similar thing for the vertical axis. We can say that each large box represents a current of 𝐡 amps. So, the first one is 𝐡 amps. The second one is two 𝐡 amps. The third one is three 𝐡 amps. The fourth one is four 𝐡 amps, and so on.

Using this information, we’ll be able to calculate the resistance of the diode at each one of the points 𝑃, 𝑄, 𝑆, and 𝑇, in terms of these arbitrary values for the potential difference 𝐴 and for the current 𝐡. So, let’s start by calculating the resistance of the diode at point 𝑃.

Let’s say that this resistance, which we’ll call 𝑅 subscript 𝑃, is equal to the potential difference at point 𝑃 divided by the current at point 𝑃. Now to find the potential difference at point 𝑃, we have to draw a line downwards to the horizontal axis. So, we start at point 𝑃, and then we draw a dotted line down to the horizontal axis. And we see that it intersects the horizontal axis at this point here. In other words, the potential difference at point 𝑃 is somewhere between two 𝐴 and three 𝐴.

Specifically, we can see that there are four smaller grey divisions in between the two large ones. And so, the distance between this point here and this point here has been divided into five. Therefore, each small grey division represents 0.2 lots of 𝐴. In other words, if this point represents a potential difference of two 𝐴, then this point is 2.2𝐴. This is 2.4𝐴. This is 2.6𝐴. And this is 2.8𝐴. And so, we can say that the potential difference at point 𝑃 is 2.8 𝐴.

And then we need to divide this by the current at point 𝑃. To do this, we simply draw a dotted line across to the vertical axis this time. And we can see that this dotted line intersects the vertical axis at the point representing a current of five 𝐡 amps. Therefore, we take our 2.8𝐴 and divide it by five 𝐡. So, when we simplify this fraction, we can write it in terms of 𝐴 divided by 𝐡, where it’s worth noting, by the way, that 𝐴 is this arbitrary potential difference value here, and not to the unit amps.

So, to write this fraction in terms of 𝐴 divided by 𝐡, we just need to find out what 2.8 divided by five is. 2.8 divided by five ends up being 0.56. And so, we can see that the resistance of the diode at point 𝑃 is 0.56 lots of 𝐴 divided by 𝐡, whatever 𝐴 and whatever 𝐡 is. So, let’s copy this resistance value up here. And let’s now move on to finding the resistance of the diode at point 𝑄.

So, at point 𝑄 we can say that the resistance, which we’ll call 𝑅 subscript 𝑄, is equal to, once again, the potential difference, this time at point 𝑄, divided by the current at point 𝑄. We can see that the vertical dotted line that we draw downwards towards the potential difference axis crosses somewhere in between 2.2𝐴 and 2.4𝐴. So, let’s say that the potential difference at point 𝑄 is 2.3𝐴. And, once again, this 𝐴 is not amps. It’s just this arbitrary potential difference value here.

And then we need to divide this value by the current at point 𝑄. So, the horizontal dotted line that we draw intersects the vertical axis at a value of 1.2𝐡, 1.4𝐡, 1.6𝐡, 1.8𝐡. And hence, we can say that the resistance of the diode at point 𝑄 is equal to 2.3𝐴 divided by 1.8𝐡. And once again, we’ll have this resistance in terms of 𝐴 divided by 𝐡. In fact, to three significant figures, this ends up being 1.28𝐴 divided by 𝐡.

And so, when we write that value down up here, we can see now that we’ve got the resistance of diode at point 𝑃, in terms of 𝐴 divided by 𝐡. And we’ve got the resistance at point 𝑄, in terms of 𝐴 divided by 𝐡. And then, we can see that the resistance at point 𝑃 is only 0.56 lots of 𝐴 divided by 𝐡, whatever 𝐴 divided by 𝐡 is. Whereas the resistance at point 𝑄 is 1.28 lots of 𝐴 divided by 𝐡. Hence, even though we don’t know the values of 𝐴 and 𝐡, we know that the resistance at point 𝑃 is smaller than the resistance at point 𝑄. And that is the reason why we didn’t need to know the values of 𝐴 and 𝐡 individually.

So, using this method, once again, let’s work out the resistance of the diode at point 𝑆. We can see that at point 𝑆 the potential difference is about 1.2𝐴. And the current is about 0.2𝐡. Hence, the resistance of the diode at point 𝑆 becomes 1.2𝐴 divided by 0.2𝐡, which, in other words, is six lots of 𝐴 divided by 𝐡. And so, we can write that down over here. At which point, we can work out the resistance at point 𝑇. So, that’s this point over here.

Now we can see that the potential difference value is actually negative three 𝐴 at point 𝑇. So, we can say that the resistance at point 𝑇 is negative three 𝐴 divided by the current value. But this current value actually happens to 𝐡 zero amps. And so we’ve got zero lots of 𝐡 in the denominator. But then zero lots of anything is simply zero. So, we’ve actually got zero in the denominator.

Now, if the value in the denominator is really really small, it’s close to zero, then the value of the fraction is extremely large. And if the value in the denominator actually is zero, then we say that the fraction is approaching infinity. In other words, then we can see that the resistance of the diode at point 𝑇 is going to infinity. It’s as large as the resistance can be.

And at this point, we won’t even worry about the negative sign here. Because we could say that the resistance at point 𝑇 is approaching negative infinity. Which simply means a negative resistance in the opposite direction, in terms of current flow. But the point is that the resistance at point 𝑇 is extremely large. It’s as large as it can be. And so, at this point, we’ve got enough information to find the answers to these questions.

So, the first question says, at which of the points marked on the graph is the resistance of the diode highest? What we’ve just seen that at point 𝑇, the resistance of the diode is as large as it can possibly be. And so, we say that the point on the graph at which the resistance of the diode is highest is point 𝑇. Similarly, we need to find at which of the points marked on the graph is the resistance of the diode lowest.

Well, comparing the other values for resistance, the resistance at point 𝑃 is 0.56 lots of 𝐴 divided by 𝐡. Whereas the resistance at point 𝑄 is larger, at 1.28 lots of 𝐴 divided by 𝐡. And the resistance at point 𝑆 is larger still, at six lots of 𝐴 divided by 𝐡. So, the smallest resistance we’ve got is at point 𝑃. And at that point, we’ve got the answer to both of our questions.

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