Lesson Video: Current Rectification | Nagwa Lesson Video: Current Rectification | Nagwa

# Lesson Video: Current Rectification Physics

In this video, we will learn how to identify the output of a current-rectification circuit based on the design of the circuit.

15:27

### Video Transcript

In this video, we will be learning about a process known as current rectification and how it’s achieved in practice.

Now, let’s start by imagining a simple circuit consisting of an AC source and a resistor. Now, the AC source will produce a time-varying sinusoidal potential difference. In other words, if we would just take a voltmeter in parallel with the resistor so as to measure the potential difference across the resistor, as time progressed, the voltmeter would show a sinusoidally varying potential difference across the resistor. And this is why the source is known as an AC source or alternating current source because it outputs a sinusoidally varying potential difference. And so, whenever its potential difference output is positive, there is a resultant current flow in one direction in the circuit; let’s say in this direction. And then as soon as the potential difference output becomes negative, the current starts flowing in the opposite direction. And so, what we’ve got is a back-and-forth flow of charge in this circuit, depending on what point in the cycle the output potential difference is.

Now, alternating current sources are great. They can be used in conjunction with transformers to step up or down their voltage. And this fact allows us a great deal of flexibility. However, sometimes, we have components in circuits that need direct current. For example, some of our laptops or lightbulbs will run on direct current. But as we can recall, mains electricity is AC. In that situation, what we need to do is to take the AC from the mains which looks a little bit like this and somehow turn it into direct current if we want to power these appliances. By the way, it is worth recalling the alternating current is defined as an electric current that periodically at regular intervals reverses its direction. However, a direct current is a current that flows in one direction only.

So coming back to our circuit diagram here, we’ve got an alternating potential difference that results in an alternating current within this circuit. And this alternating potential difference is similar to what is generated in the mains. So if we’ve got an appliance that runs on direct current, then we somehow need to convert this alternating potential difference into a direct source of potential difference. Now, there are two ways to do this.

The first way is to take all the negative parts of this potential difference curve and to get rid of them entirely. In other words, if we still have a potential difference source that acts kind of like a sinusoid for the first half of the cycle and then for the half of the cycle where it should be negative it actually becomes zero and then the cycle repeats again so the positive part stays the same and the negative part actually becomes zero, then we technically have a direct potential difference source. It’s no longer an alternating potential difference source because at some points, the potential difference is zero and at other points, the potential difference is positive. But at no point is the potential difference negative. And therefore, if a positive potential difference generates a current in the circuit in this direction for example, then at no point will it generate a current in this direction because there is no negative potential difference at any point.

Now, the process that converts this kind of potential difference across a component in a circuit to this kind of potential difference is known as rectification. And more specifically, this kind of rectification is known as half-wave rectification because originally the voltage source produces a sinusoidal voltage. And after rectification, we can see that half of the sinusoidal voltage is reproduced whereas during the other half of the cycle, the voltage is zero. And so, only half of the initial sinusoid is reproduced in this case. And hence, it’s called half-wave rectification. And the point is that if this is the potential difference across a component in the circuit, then a graph showing the current for that component will have a similar shape.

And so, if we now plug the current for that component against time, then the shape of that curve would be the same as the shape of this curve. Because assuming the component across which we’re measuring the current behaves like a resistor, we can use Ohm’s law to remind us that the voltage across a component is directly proportional to the current through that component. And hence, the shape of the current curve is going to be the same as the shape of the voltage curve. And because the current for this component is always either positive or zero, positive or zero, positive or zero, and so on and so forth, the current for that component is a direct current because there is no reversing in direction of that current.

So as we’ve seen, we can use half-wave rectification to take a sinusoidally oscillating voltage and turn it into a direct voltage source. That’s resulting in a direct current. But how exactly do we go about doing this? Well, what we can do is to take our circuit and stick a diode in series with the current source and the resistor. Now, a diode is an electrical component that only allows current to pass through it in one direction and does not allow any current to pass through in the other direction. Therefore, if the diode was not present in the circuit, then the current through the circuit would look something like this, whereas when we put the diode in the circuit, it does not allow any current in the negative direction.

In other words, current can’t flow this way. But even though the potential different source is emitting a sinusoidal voltage, the diode prevents any current in the negative direction. And so, if we measure the potential difference across the resistor in this circuit, then it would look something like this because the current through the resistor looks something like this, a rectified alternating current. And so, putting a diode in series with the component of the circuit results in half-wave rectification. However, half-wave rectification is not the only way to go from AC to DC. There is something known as full-wave rectification which takes a sinusoidal potential difference source and reverses the polarity of all the negative parts of the cycle.

In other words, a cycle that should have been doing this is now instead doing this. And we can see that this is called full-wave rectification because this time even the initially negative parts of the cycle are present in the rectified voltage just as positive parts of the cycle. We no longer have the potential difference through half of the cycle being zero. So how do we go about achieving full-wave rectification? Well, as we can guess, diodes are involved once again. And in fact, there’s a way to achieve full-wave rectification with two diodes in the circuit. But this is a little bit more complicated. The method that we will be looking at uses a bridge circuit. And this bridge circuit will consist of four diodes.

Now, the circuit in question looks something like this. And in the circuit, we can see one, two, three, and four diodes connected in what’s known as a bridge circuit. Now, this circuit is going to help us achieve full-wave rectification. And here’s how. Let’s imagine that our AC source is putting out a sinusoidally varying potential difference as we were expected it to. Now, during the first half of a sinusoidal cycle, we can see that the potential difference is positive. And therefore, let’s go by the convention that a positive potential difference will induce a current in the clockwise direction in this circuit.

Well, in this case, we can see then that current is flowing in this direction in the circuit. And it does so until it reaches this branch point here. Now, when it reaches this branch point, it can easily travel in this direction. However, the current cannot travel in this direction because this diode here stops that from happening. Remember that the direction of the arrow in the diode circuit symbol shows the direction in which conventional current is allowed to flow through that diode. But anyway, so current can flow through this branch but not through this branch. And then once again, the current reaches a branching point. At which point, the current can easily flow in this direction, but it cannot flow in this direction because there’s another diode stopping it from going that way. And so, what we have is current flowing in this direction through our resistor and coming back round this way.

Now, it’s worth noting by the way that this low curvy part of the circuit shows that the orange wire and the blue wire are not connected to each other. The orange wire is passing over the blue wire. And this means that the current continues to flow in this direction along the circuit. And then upon reaching this branching point, the current is now about to flow both in this direction and in this direction. Looking at the current flow in this direction, we see that it reaches this branch point once again. And we’ve seen what happens at that branch point already.

Looking at the current flow in this direction through the diode, we see that it then reaches this branching point, at which point current flows clockwise once again, around this orange part of the circuit. And at this point, we’ve looked to the complete circuit. But the thing to keep in mind is that during the first half of the cycle, the current through the resistor is in this direction. Let’s keep this current direction in mind and think about what happens if we now look at the negative part of the cycle. That is the potential difference we’ll try and induce a current flow in the opposite direction in the circuit. Well, in that case, the current flowing through the circuit will be counterclockwise now. So now, what we’ve got is current flowing this way along the circuit up and along the orange wire at which point it reaches a branch point.

Now, at this branch point, current can go in this direction because the diode direction allows it but not in this direction. And so, current flows along this branch and reaches another branch point. At that branch point, current is allowed to flow in this direction, but not along this diode. And so, current flows along this blue wire through the resistor and down through the resistor as we’ve drawn it and comes back round this way. Then, the current continues to flow and reaches a branch point. At this branch point, current is allowed to split and flow along both branches because that is allowed by those diodes. Now, the current in this branch reaches a branch point here and then is allowed to flow through this diode and also in this direction, at which point we’ve once again considered the whole circuit.

But here’s the interesting part. During the positive part of the cycle, we saw that the current through the resistor was flowing in this direction. And as we’ve just seen, during the negative part of the cycle, current is once again flowing in this direction through the resistor. And so what we’ve successfully managed to do is to create a circuit in which current is always flowing in the same direction through the resistor. Now, the magnitude of that current will change with time depending on the magnitude of the potential difference generated by the AC source at any given point in time.

So for example, at this point in time, the magnitude of the potential difference generated is not quite maximum. And so, the current value is not going to be quite maximum. However, at this point in time, the potential difference value is maximum. And so, the current through the circuit is going to be maximum as well. Now, the end result of all of this is that if we were to place an ammeter in series with the resistor and have the ammeter measure the current through the resistor, what we’d see as time passes is that we’ve got a full-wave rectified current through the resistor. And using Ohm’s law once again, we can see that because the resistor will have a constant resistance value, the potential difference across that resistor will be directly proportional to the current through that resistor. And therefore, if we stick a voltmeter parallel to the resistor to measure the potential difference across it, what we would see is this: a full-wave rectified potential difference. And so that is another way in which we can convert AC into DC, an alternating current source resulting in a direct current through a resistor.

Now, there are still problems with both of these methods, half-wave rectification and full-wave rectification as well. For one, normally a DC voltage would be a smooth flat line. However, rectified potential differences are definitely not a smooth straight line. And there are ways to get around this using capacitors for example. However, we won’t be looking at that here. Instead, let’s get some practice in understanding current rectification by looking at an example question.

Diagram (a) shows a circuit that can be used to rectify an alternating current. If the input voltage is that shown in diagram (b), which of the following graphs shows the output voltage as measured by the voltmeter in the circuit diagram? [A] Diagram 1 [B] Diagram 2 [C] Diagram 3 [D] Diagram 4

Okay, so in this question, we’ve got diagram a, which shows the circuit that we’re considering, and diagram b, which is the input voltage that’s produced by the AC source in this circuit. And as well as this, we’ve been told that the circuit in diagram a can be used to rectify an alternating current. Now, based on this information, we need to work out which one of these four graphs here shows the output voltage as measured by the voltmeter in the circuit.

Okay, so to answer this question, let’s first consider what we have in the circuit diagram in diagram a. What we’ve got is an AC source, a diode, a resistor, and a voltmeter. Now, the AC source, the diode, and the resistor are in series. And the voltmeter is measuring the potential difference across the resistor. Now, the AC source is producing a sinusoidal potential difference across the circuit. And if we had a circuit where there was no diode, but instead we just had an AC source, the resistor, and the voltmeter, then the sinusoidally varying potential difference produced by the AC source would be the same as the potential difference across the resistor. In that situation, we would have a sinusoidally varying current through the resistor as well.

However, coming back to the circuit in diagram a, that’s not what we’re going to see because remember we have a diode. Now, a diode is a circuit component that only allows current flow through it in one direction. Now, we can choose to say that anytime the potential difference produced by the AC source is positive, that is going to generate a current flow in this direction that’s clockwise around the circuit. And then, we can see that that current flow is actually allowed to be sustained because current can flow in this direction through the diode. We can see which direction a diode allows a current through it by looking at the arrow in that circuit diagram.

And so, anytime the potential difference is positive produced by the AC source, there is a current in the clockwise direction in the circuit. Therefore, there is a current through the resistor in this direction in the circuit. And then for the resistor, we can recall that the potential difference across the resistor which is the value measured by the voltmeter is equal to the current through that resistor multiplied by the resistance of the resistor itself. So the point is that anytime the potential difference from the source is positive, a current is flowing clockwise through the circuit which is allowed to flow because of the diode. And so, there is a nonzero current for the resistor. And if there’s a nonzero current through the resistor, then the potential difference also is nonzero, where that potential difference is the voltage measured by the voltmeter. And this is because we’re looking at Ohm’s law just for the resistor and the resistance of the resistor is a constant.

Therefore, we can say that the voltmeter will measure the same potential difference across the resistor as is produced by the source whenever the source is actually producing a positive voltage. However, as soon as we look at the negative portion of the curve, then it’s a whole different story because when the voltage from the AC source becomes negative, that negative voltage tries to generate a current in the circuit in the opposite direction to before. In other words, this AC source is now trying to set up a current in the counterclockwise direction, which would be fine if it weren’t for the diode because the diode does not allow a current through it in the opposite direction. And hence, there is no current in this circuit because there aren’t any other branches for the current to flow through that don’t end up at the diode at some point.

And so, whenever the potential difference from the source is negative, there is no current in the circuit. So the current of the resistor is zero. And then coming back to Ohm’s law once again, if the current is zero, then the potential difference across the circuit, which is remember being measured by the voltmeter, is also going to be zero. So in other words, for the entirety of the time that the potential difference source is producing a negative voltage, the voltage measured by the voltmeter itself is going to be zero because there is zero potential difference across the resistor.

Then, we go back once again to the positive part of the cycle. And in that situation, once again, we will have a positive potential difference measured by the voltmeter. And that potential difference will follow the voltage produced by the AC source. And once again, we return to the negative part which is going to result in a zero voltage measured by the voltmeter. And so, all in all, what we expect to see is something like this when it comes to the potential difference measured by the voltmeter. And at this point, we can identify this as a half-wave rectified potential difference.

But more importantly, out of the four diagrams that we have to choose from, we can see that the first of these diagrams correctly matches what we’re expecting to see in terms of the voltage measured by the voltmeter in diagram a.

So now that we’ve looked at an example question, let’s summarize what we’ve talked about in this lesson.

We firstly recalled that diodes are circuit components allowing charge flow. That is, there can be a current through them in one direction, but not in the reverse direction. And this is the circuit diagram. Secondly, we saw that diodes can be used in rectifier circuits to convert AC to DC. And finally, we looked at two different types of rectification, maybe half-wave rectification, where sinusoidal voltage source is converted into having positive peaks but then a zero voltage, where the negative half of the cycle would otherwise be. And half-wave rectification can be done by placing a single diode in series with the AC source.

We also looked at full-wave rectification, where a sinusoidally varying AC source is modified so that the positive peaks stay exactly the same, but the negative peaks are mirrored in the horizontal axis. Or in other words, they take on the same value, but the opposite polarity to what they normally would have. One way to achieve full-wave rectification is using a bridge circuit. And that consists of using four diodes placed in a very specific orientation so that the potential difference across this part of the circuit, for example, this resistor here, is full-wave rectified. And so, that is an overview on current rectification.

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