Welcome to you all to this recording. My name is Nanubu FI. I will be presenting to you today industrial electronics and foam. The chapter that we will be dealing with is diode allocation. so we do that application together so here let's just assume we have on the primer side we apply for example your VAC we say this is equal to maybe 20 volts right VAC is 20 volts so this is the secondary side so you also can be given the tens ratio they can say this is ten this is two this is one, they give you something like that and they say to you calculate the secondary voltage, right so the secondary voltage is the voltage that you are going to measure across the secondary widings that is the secondary voltage the wave that is there so we are stepping down to it into a certain wave we call it 5 volt for example, or it could be 10 because you are using the step down transform.
Right? So the primary voltage can be measured across the primary windings. So the secondary voltage can be measured across the secondary windings.
So this is actually the voltage that will supply anything that you see, the diode and the secondary windings. Right? So before we have done calculations, we can now calculate the secondary windings if we are given information like this. But before that, I just want us to look at this diode as it operates.
We know that it can operate when it is connected in the forward bias mode. That's where it operates as a conductor. When it is connected in the reverse bias mode, it operates as an insulator. We see it does not operate when it is connected in the forward bias mode.
not conduct right so now let's look at this second this is a half wave rectifier right so you are converting ac quantity into dc quantity right so meaning that now this dc quantity you would have cancelled now only one direction as you know that ac current is a current that changes direction but dc current is one direction it flows in one direction it doesn't doesn't change direction so here we are talking about the AC that we are going to apply we are going to consider both directions right so if we are going to have let's say for example the input signal we have an input signal for example sine wave this is your input signal right we know this is the positive point we have the negative halfway so this is what you have so this is the input right this is your input so this is your input this is your AC this is what you want to want this is what this is what you want to convert right this is what you want to complete right what you're going to do is that you are going to step down you are going to step down this bearing if it's 12 volts you are going to step it down and then now you have a secondary a secondary voltage the voltage that is taken from the secondary wire is then supplied to whatever is on the tech on the secondary side right so So you will then see that it is always very very important for you to calculate the secondary voltage because your rectifier is connected on the secondary side. So we use the transformer to just step down the primary voltage. Step down the primary voltage and then we can now supply the secondary voltage to whatever it is that is connected.
on the second hand side in this case we are going to supply the voltage to our rectifier and the load so you can see The voltage that you apply here is AC. Now the secondary voltage before the diode is still AC, right? Because we are going to use that diode, right, as our rectifier to convert from AC to DC. So now you will then see that any voltage that will be measured after the diode, like for example, the voltage across the node will be Vdc. What you are going to measure here will be Vdc.
AC. But the voltage that is supplied by the secondary wire net is AC. It is the one that maximum is converted.
But remember, what you are going to convert here, I believe you know guys about iron is represented as maximum or peak. So if you are giving values like this here, you are then required to start by calculating the secondary voltage. Now, when you have the secondary voltage, right, if they do not specify, they do not say anything, they just give you a result, 20 and 10 is to 1, and then you apply that equation, the transformational formula, to calculate the secondary voltage, right?
You assume that what you have calculated is an iron is A, right? But what you are going to supply to your rectifier, which is the diode, it has to be the maximum value, right? The maximum value of the secondary voltage. The maximum value of the secondary voltage. That's what you are going to apply to the diode.
Right. So, let's assume AC. Let's assume this is AC input to... our rectifier.
Let's not talk much about the transformer because the transformer is only there to step down. Let's just say this is the step, the value that is now reduced from doing it because we are using the spectral transformer. This is the value that is now reduced from 20. We can give it a magnitude and say maybe this is now for example 10 volts. You got 10 volts.
I don't know if you are going to use this value to calculate. So let's assume our second-order voltage is 10 volts. Now we are going to take this 10 volts on the second-order side, these 10 volts, as what? As iron.
So this will be V. the and they did not specify they just gave you values 10 to 1 and 20 and we are using this 20 and 10 to 1 which are the tens ratio to calculate the secondary voltage you get 10 right? this 10 is also the parameters of A so you must make it A in my team right? so that's what we are going to compare so let's assume this is a wave that is suffering Let's assume this is the wave that is applied on the input of the diode. That is applied there on the input or that is applied to the diode. The secondary voltage.
I just want us to know how to draw the input wave and the output wave. So if this is AC, you want to convert AC into AC. Obviously, the output is not expected to have both directions.
So we are going to only have... remember we are converting from AC to DC so this is actually AC but on the other side we expect what? DC so what is the other thing that is going to do the job for us? It's our diode, the rectifier so now if we are going to say this is A and then this is B as you can see A and B right?
and we say now here we have a plus then we have a minus right and then we know the diode that is the positive terminal of the diode and that is the negative terminal of the same diode so that's what we have right so if we check during the positive half circle we are now trying to find out how do you draw the output if you are given the input as a c how do you the DC output. Now during the positive half circle, where A is positive and B is negative, this diode is connected in a forward bias mode during the positive half circle. half circle where A is positive B is negative the diode D is connected in forward bias mode then it conducts right so because it conducts to lift the positive half circle it will then generate a half wave on the positive right so this is what is going to happen during the positive cycle we are going to learn a half way right generated by by t right this is during the positive cycle Good Let's continue Now, do you know the negative half circle? Do you know the negative half circle?
We now check the terminals We make A to be negative and we make B to be positive Now this is the negative half circle the negative half circle now during the negative half circle let's check what is it that happens during the negative half circle during the negative half circle as you can see the diode is connected in a reverse bias mode right This is what we talked about, negative to positive, positive to negative. This is reverse biasing. The diode operates as an insulator. Then you have diode operating as a conductor.
So during the negative half circle, where A is negative and then B is positive, this diode operates as an insulator. Right? so it will then blow the flow of current in other words the diode during the negative answer where A is negative and B is positive the diode has got the current We will expect a negative wave to be generated, right?
But it does not conduct. So there is no negative wave that is generated for halfway. So the current is safe.
Therefore, we are going to move here with the reference line. So we change again, we say during the positive half circle, what happens? We know that this diode is now connected to the forward bias knot again. So forward bias knot, positive 2, positive. Negative 2, negative.
negative the diagram comes down again during the positive half circle therefore it will generate another half wave this is the half wave generated by T so you see so this is the input this is the output so for a half wave if you are given an input that looks like this you just have to click out the negative half waves the negative half wave but and the clip must be clipped off can you see? so it doesn't matter I can continue here and make more circles can you see that? so you just look at the input and then you remove the negative half waves that will be your output this is only applicable if you are taking with A a half wave let's now move on to a full wave now, if you use this two diodes. You can see you have D1 and D2, right?
If I can say this is A and this is B, right? I say this terminal is now positive, this is negative. This is the positive terminal of the diode, this is the negative terminal of the same diode, right?
We know this is the positive terminal of this diode and then this is the negative terminal of it, of D2. right so during the positive half circle before we get down to our problem let me just draw the input let's have the input let's have the same input you know so this is plus plus this is minus minus so these are the directions positive half and negative half so how can you draw the output A let's try it let's try it we want the positive half the DC output for a full wave remember what we said, we said if you are using a half wave and you are given this as the input you just have to remove the negative half waves Right? You leave only the positive halfway.
That is your output. Right? Let's now... Ah, look at this one here. Now, during the positive half-self, during the positive half-self, let's look at...
This is like T1 You have plus there, you have minus there Right? If this can be known as you have plus there, you have minus here You have connected the plus to the plus Right? Remember in this case, this way is are the ones that are going to supply the voltage to the diode so you see the positive terminal of this supply is connected to the positive terminal of the angle of the diode right and then the negative terminal is connected to the negative terminal of the supply. So, you can see D1 is now connected in forward bias mode. Therefore, it conducts.
But when we move on to D2, D2, let's look at D2 carefully. D2, you have negative that side, as you can see. You have a negative there.
You also have, you have A positive. Let's check our supply. So, you have. negative connected to positive can you see that and then you have negative then connected to positive so D2 during the positive cluster where A is positive with respect to B D2 is connected in the reverse bias mode, D1 is connected in the forward bias mode, therefore D1 conducts D2 does not conduct, it operates as an insulator.
So during the positive absolute the only diode that will conduct is D1 right, during the positive half-sone so we will then expect to have a positive half-wave generated on the positive right that will be the wave generated by D1 because now the current flows through D1 so we have a wave for D1 during the positive half circle where A is positive and B is negative so we can say this is B 1 right now during the negative half circle now let's change the terminus during the negative half circle let's make a to be negative and then b to be positive during the negative half circle half circle so let's have a look so we have that one it's negative and that one is positive so during the negative half circle you can see that now you have connected this one is negative you connect it on the positive terminal of the diode this one is negative it is now connected to the positive terminal of this supply so you can see that during the negative half circle the diode d1 Now, the diode D1 is connected in a reverse bias mode. It operates as an insulator. It does not conduct.
It blocks the flow of current. But D2, let's check during the negative flux. D2, let's check.
You have a plus there, connected to a plus. And you have a minus, connected to a minus. So, this diode D2 is connected in a forward bias mode. It will then come out.
it will allow current to pass through because there is no negative uniform bias but now you see this T2 it operates during the negative half circle it operates during the negative half circle but you expect now to have a negative half wave on the negative half circle sorry a negative half wave right generated by T2 because it operates during the negative half circle but what you want to do is what will happen is that instead of having this half wave on the negative this half wave will then be inverted for a full wave instead of having it here on the negative it goes to the positive so we have 1 d2 right so for d1 for d2 right instead of having it on the negative it flips to the positive or it is inverted can you see that? from negative to positive right so you continue during the positive fast circle where A is positive and then B is negative what happens? D1 conducts D2 operates as an instigator therefore D1 will conduct another half way so it's D1 you keep on changing so you will be doing the same fast circle D2 D2 conducts, D1 is operating as an insulator. Then, during the negative half circle, the D2 is going to conduct another half wave, but it will be on the positive, right?
So, this is D2. So, for a full wave, you can see that what happens is that if you are getting that, this is your input. For the input, you leave the negative half, sorry, the positive half.
positive half way right you just feed the negative into the positive can you see it you take this one up you take this one up so you will have only 4 positive half way so you just invent all half ways can you see it so this is the input this is the output 4 and following this is the input this is the output 4 you see