Practice English Speaking&Listening with: Module - 5 Lecture - 8 SCR Applications

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the earlier classes we have seen how the current flows in a purely resistive load in an AC-AC

converter. Now let us see what happens when instead of a purely resistive load we have

a complex load having resistance and inductance in series. As there is inductance, the current

through this load will not be immediately zero when the thyristor turns off. It will

continue to flow for sometime even after the thyristor stops conducting. The voltage and

current waveforms if we plot, then the current will start after firing angle alpha which

is the angle at which the thyristor is fired and it will even continue after the thyristor

turns OFF.

Let us find out what will be the current in such a load having inductance in series with

the resistance. We will consider the input voltage to the

converter as small v equal to root 2V sin omega t, where V is the RMS value. Here V

is the RMS value of the voltage. So, root 2 into V means peak value is that much; it

is Vm sin omega t, which is equal to root 2V sin omega t.

If we consider the current in the circuit, the current in the circuit when one turn thyristor

is firing say thyristor T1 fires in the positive half cycle at an angle of alpha, then what

will be the current? That can be found out using Thevenin's voltage law, sorry Kirchhoff's

voltage law in this circuit and the law will give us the equation which is L di dt, the

drop in the resistance plus the drop in the inductance that will be equal to input voltage.

Writing the Kirchhoff's voltage law, L di dt plus Ri equal to v, which is the input

voltage and that is a sinusoidal voltage, so root 2V sin omega t.

We have to solve this equation. It is a first order differential equation. We know that

solving this differential equation will lead to the current which is a summation of the

complementary function of the particular integral. We have two portions in the solution. The

complementary function will be given by Ce; C is a constant into e to the power minus

R by L into t and the particular integral iP is equal to root 2V by root over R square

plus omega square L square into sin omega t minus phi. This is a standard solution I

am using; instead of solving from the first principle, I am just writing down the equations

solution because it is first order differential equation. This phi is the phase angle which

is given by tan phi equal to omega L by R. We need to know the constant C. So, we have

to find out this constant C from the first principle. That is initial condition.

We will write the current equation i equal to C into e to the power minus R by L into

t plus root 2V by Z. Z is nothing but root over R square plus omega square L square.

This is the impendence of the load that is representing the Z and sin omega t minus phi

term which is the steady state component, basically. Now we will apply the initial condition

to find out the constant C. Applying initial condition means the current in the load will

be zero till the thyristor is triggered and thyristor is triggered at omega T equal to

alpha. We will use that condition to find out C. So, i will be equal to zero when omega

t is equal to alpha. Putting here the value of omega t is equal to alpha gives us value

of t equal to alpha by, if we replace here C into current, which is equal to zero and

zero equal to root 2V by Z sin omega t equal to alpha; so, we will write omega t as alpha

minus phi. The value of C can be found out.

Here C is equal to minus root 2V by Z sin alpha minus phi. Now we can write down the

value of i. So, i equal to C is root 2V by Z sin alpha minus phi into e to the power

minus R by L t. So, e to the power minus R by L t plus the other term which is the stead

state component. Now from this, we will get the equation for the this current i equal

to minus root 2V by Z sin alpha minus phi e to the power minus R by L t plus root 2V

by Z sin omega t minus phi. Finally we get an equation which gives the value of i and

we can see that the exponential term is there and because of this exponential term the current

in the inductance will fall very slowly.

It is not immediately falling to zero and depending upon the R and L, the time constant

of the inductive circuit the falling of the current will take place. If time constant

is high in comparison with the input cycle period, then there will be fall of this current

very instantaneously. But if they are not comparable, if the time constant is not so

high as compared to the period of the input voltage waveform then there will be significant

lagging of the current and the current in the inductor will continue for considerable

period. We have discussed about the AC to AC converter and we have seen how controlled

power can be supplied to a load using the AC to AC converter which employs a thyristor.

Another application of thyristor is in rectification. Earlier we have seen rectification in diode

circuits. The diodes were used for rectification and half wave rectifiers, full wave rectifiers

we have discussed earlier; but thyristors can also be used for rectification. Here as

it is dealing with high power, so sometimes it is easier and cheaper to use thyristors

in rectification than using other metals. We will now discuss about some basic rectifier

circuits using thyristor which will give the rectified output and using half wave rectifier

and full wave rectifiers also we can get the rectified output, which basically employ thyristor

in the circuit.

First of all let us take a simple half wave rectifier using resistive load.

Here instead of the diode which we are using earlier in half wave rectification, we will

have a thyristor and initially let us assume that the load is purely resistive. The input

is sinusoidal Vm sin omega t which is given as the input. Let us take the voltage across

the thyristor as VT and the resistance has a voltage across it as VL. We will try to

find out the waveforms VT and VL also in this circuit.

Input is sinusoidal VS is equal to Vm sin omega t and we are having a single thyristor

firing at an angle of alpha, say. When the thyristor fires at an angle of alpha, the

voltage across the thyristor becomes zero. In comparison with the voltage at input, we

can assume that the thyristor is having a voltage drop of zero because the voltage drop

is very small and it can be neglected. Once it is ON or in conduction, the thyristor will

have a voltage of zero across it. Up to the point of conduction the thyristor is not firing.

The voltage across the thyristor will be the equal to input voltage because we can see

in this circuit that the voltage across the thyristor when it is not conducting there

is no current flow. So, voltage which is at the input will be equal to VT. I am using

symbol VT and VL here. What will be VT? Initially it will be equal to the VS. As soon as it

starts conducting, it becomes zero. It will conduct till pi because after that the input

voltage becomes negative. If the input voltage is negative that means there cannot be any

conduction in the thyristor since the anode is now negative and cathode is positive.

This is reverse biased, so it will not conduct and that is why after this pi the thyristor

is OFF and that is why the voltage across the thyristor is same as the input voltage.

After this point again alpha angle is exceeded because alpha is the delay angle. After this

point again it will have to wait for alpha and till then the input voltage will be equal

to the voltage across the thyristor. Now what about the voltage across this load resistance?

Up to the point of alpha that is till the thyristor conducts, the voltage across this

resistance RL will be equal to zero. Since no current flows, the thyristor is OFF. The

voltage is zero because the voltage across the resistance equal to current into the resistance,

so it will be zero till this point. As soon as the thyristor conducts or fires then what

will happen? The voltage across this resistance will be equal to the voltage at the input

because this drop equal to zero we are assuming; thyristor conduction voltage is zero. So,

this voltage is equal to voltage across this resistance. It will follow this shape of the

input voltage and after the pi, phase angle that will not conduct; the thyristor will

not conduct. The current will be zero, so voltage will be also zero.

Again from this point till alpha angle is crossed the voltage will continue to be zero.

Since there is no current flow in the resistance, the thyristor is not firing. The voltage is

zero and once the thyristor fires again in the positive half cycle after this angle alpha

is crossed, then the voltage across the resistance will be again equal to input voltage. These

are the waveforms of the voltage across the thyristor and the voltage across the load.

We can find out the voltage values. Assuming zero voltage drop across the thyristor when

it is switched or fired we can find out the DC value of the voltage, DC value or average

value because we are having a rectifier. It is a half wave rectifier.

Using the expression for finding out the mean value or the DC value over the whole period

2 pi, we are getting the output voltage VL within this region only. This is the region

where we are having an output voltage in between from zero to 2 pi. From alpha to pi we are

having an output voltage. The integration in the expression from alpha to pi will be.

It is Vm sin theta because the output waveform is exactly following the input waveform Vm

sin theta.

This integration Vm sin theta d theta between alpha to pi has to be done but multiplied

by 1 by 2 pi because we are having the input voltage between zero to 2 pi if we consider

output voltage will be between this portion only and the rest is zero in the whole period.

In order to find out the DC value or mean value, we will use this expression and that

is equal to, Vm by 2 pi constant part can be taken out; integration sin theta d theta

will be give minus cos theta between the limits alpha and pi which is equal to Vm by 2 pi

minus cos theta. If we put the limit it will be, minus cos phi minus cos alpha; cos phi

is minus 1, minus 1 minus cos alpha so, 1 plus cos alpha. So, Vm by 2 pi 1 plus cos

alpha. This is the DC value of the voltage for this resistive load.

Similarly if we want to find out the current IDC, by dividing the VDC by RL we will easily

get the DC value of the current. That is what is done here. IDC equal to 1 by 2 pi integration

alpha to pi Vm sin theta d theta by RL; RL is the load resistance and it can be written

as Vm by 2 pi RL; integration sin theta is minus cos theta within the limits alpha to

pi. Putting the limits we get Vm by 2 pi RL 1 plus cos alpha. This is just a follow up

of the DC value of the voltage. Dividing this DC value of the voltage by RL, we get the

DC value of the current and if we want to find out the RMS value of the current, then

we have to do it according to the way RMS value is found out.

Square of the RMS value if we take, it will be 1 by 2 pi integration alpha to pi square

of this expression Vm square by RL square sin square theta d theta. In this integration,

replacing this sin square theta in terms of cos twice theta, we will get finally the value

of RMS square equal to Vm square by 4 pi RL square; putting the limit we get pi minus

alpha minus sin twice alpha by 2.

If we want to find out the average output power to the load, output power is nothing

is IRMS square into RL; average output power to the load that is obtained as IRMS square we have found

out. Replacing this expression here and multiplying by RL will cause the square in the denominator

of RL to vanish. What will be here is Vm square by 4 pi RL into pi minus alpha minus sin twice

alpha by 2.

This example was using a resistive load. If the load is an inductive one that means instead

of having a pure resistance we are having an inductance in series with the resistance

as the load. What will happen to the half wave rectifier? What will happen here is that

this inductance will not let the current to stop immediately to zero once the thyristor

turns OFF. But it will allow the current to continue for some more time because the current

cannot immediately become zero in an inductance. It will follow exponential curve.

If we plot these waveforms in this circuit again this VT and VL and VS, VS is the same sinusoidal waveform, then the output

voltage across this inductive load is equal to the voltage at input. But here after the

thyristor stops conducting what will be this voltage? After the thyristor stops conducting,

still there will be current flow for sometime. 2519 The thyristor stops conducting after

pi, from alpha to pi it will conduct. But after pi also, if we project this point, after

pi also, there will be a current flow due to which there will be a voltage available

across the load and that voltage is equal to the input voltage. If we look into this

waveform of the input voltage, from alpha to pi the thyristor is conducting. It is fired

at an angle of alpha, it will conduct till pi. After pi, the input voltage waveform becomes

negative that is why the thyristor will be turned OFF.

It should have current zero, but it will not have current zero. There will be current flowing

in the circuit. That is why there will be voltage available across the load which is

equal to the supply voltage. This is the supply voltage. It will follow the supply voltage

and then only it will become zero. Till when the current flow will continue that will be

decided by the load. The current waveform if we see from this point, there will be conduction

of current. Current will start flowing and then after this point till this point also

current will be flowing and it will become zero at this point. After this becoming zero,

it will continue to be zero again till alpha is crossed. Again the current will start to

flow and it will continue beyond zero value of the input voltage and after sometime only

the current will become zero.

We have seen that there is voltage available across the load even after the zero of the

input voltage is crossed to the negative side. This is a disadvantage or actually this should

not happen because we are having a controlled half wave rectifier. But here in the negative

half also there is voltage available across the load. In order to get rid of this negative

voltage in the output what we can do is that, we can use a diode which is known as flywheel

diode which will prevent the voltage across the load becoming negative. How it does that?

Let us consider the circuit.

Here in addition to the thyristor which we were having in the half wave rectifier circuit,

we are also having a diode. This is a normal or ordinary diode connected in the circuit.

Basically what does this diode do? Let us see and try to understand the behavior of

the circuit. We are having an inductive load as it is; a resistance and an inductance in

series is the load. The current through this circuit is iL say, voltage across this load

is VL. Now the thyristor fires at angle of alpha that is the delay angle. What will happen

in the positive half cycle? After the alpha angle is crossed the thyristor will fire and

it will cause the current flow through the load in this direction. The diode is reverse

biased. It will not conduct in the positive half cycle anyway and we will get an output

voltage across this resistance and inductance in series which is the load, which is VL that

is equal to the supply voltage.

This is the input voltage waveform VS, it is VT, this is VL. When we have non conduction

of the thyristor, input voltage and thyristor voltage are same because it is not conducting.

At this point it starts to conduct. It conducts or fires at this point. By using triggering

voltage we are firing the thyristor and then it turns ON. Thyristor voltage becomes zero

and that is the reason why in this region the thyristor is conducting. The voltage across

the resistance will be now equal to the supply voltage, so VL and VS are same.

What will happen when the supply voltage crosses into the negative half cycle? When it enters

the negative half cycle, then in the negative half cycle what will happen is that this anode

is negative. But what will happen to the current through the load? Because we see that when

the thyristor conducts, the load current is the thyristor current.

Thyristor has a current flow through it that is say iT. We are naming this current as iT;

iT is the current in the thyristor. When it is in the negative half let us consider this

negative half; in the negative half cycle this diode is now forward biased. The current

will be bypassed through this diode. The diode is conducting. The voltage across this diode

is zero. This also we are assuming zero because it is very small 0.7 volt; it is almost zero.

The voltage across the load is the voltage across this diode which is zero.

When it is in the negative half cycle, the voltage across this load is zero. Mind it

the current flows, but that current will be now bypassed by this diode in this circuit

consisting of load and the diode and the supply will not allow any current. We do not get

the supply current through the thyristor but we are having a current in the load due to

the diode current. That is the difference that the flywheel diode is making.

Earlier what we have seen is that in the inductive load, even when it was in the negative half

cycle, the supply was in the negative half cycle at that time we were getting a voltage

across the load which was negative. The supply voltage and load voltage were same and the

current was flowing. In the negative half cycle also there was current in the load but

here what we are getting is that there is current flowing in the load but that current

is bypassed by this diode. It will not flow through the thyristor; earlier it was flowing

through the thyristor. The current in the load was drawn from the supply. It was flowing

through the thyristor in the load even when the voltage was in the negative. It was going

into the negative half cycle because of the current in the inductor not becoming zero

immediately; it was following an exponential behavior. That was the difference.

But here we have seen in this case, use of flywheel diode has stopped the voltage across

the load becoming negative as was the earlier case. Now the voltage is zero but the current

flows in the load. We are getting the current in the load. But that current is not through

the thyristor, it is bypassed by this diode.

Here what is happening is that if we consider the load voltage or the voltage across this

load, when the thyristor was not conducting it was zero. When it starts conducting it

becomes equal to the supply voltage. But as soon as the supply voltage enters the negative

region, then the thyristor stops conducting. Why the thyristor stops conducting is because

this anode is positive and in the negative half cycle the anode is negative. So, it will

stop to conduct and the voltage which we will obtain across this thyristor will be the voltage

across this supply because if we apply the voltage law here, then the voltage across

this is equal to the voltage across this, because this diode drop is zero. We get the

voltage across the thyristor as the supply voltage; when again it starts conducting then

again it will be zero and the voltage across the load will be zero. When the thyristor

fires, then it will be equal to the supply voltage. As soon as the supply voltage becomes

negative this voltage across the load will become zero because it is the diode drop,

diode drop is zero. Again it will be equal to the supply voltage after the thyristor

fires.

Now what about the load current? Load current is flowing throughout the period but this

is contributed by two different parts. One is the diode current which flows when you

have the negative half cycle. In the negative half cycle, as we know the current is bypassed

through the diode and load current flows due to the current in the diode only and that

is the current here, the diode current and it is exponentially decaying because it is

flowing through the resistance and inductance in series. Because of this time constant of

this load, the current which is contributed by the diode in the negative half cycle will

fall exponentially and in the positive half cycle this is due to the thyristor current

which is drawn from the supply but in this portion the current is due to the bypassing

of the current by the diode. We will have a current which will rise and fall like this.

The rising means it is because thyristor is conducting. When conducting, it is flowing

through the load and thyristor only and that is why it will rise from zero onwards. Exactly

this value is not going to become zero because it will fall exponentially.

We have assumed that the time constant of the load is sufficiently high as compared

to the time period of the input voltage. Under that assumption it will not fall down to zero

and it will fall and again it will rise like this. This rectified output is half wave rectified

output and it is not present during the negative half cycle. That has been achieved by using

a flywheel diode which is bypassing the load current in the negative half cycle.

We consider now a full wave rectifier using a center tap transformer. We were discussing

earlier using diodes we were using a center tap transformer. Here also similarly we are

using that type of circuit using a center tap transformer and two thyristors.

When we have two thyristors what will happen is alternately the conduction will take place;

also we are having a flywheel diode. Here we are connecting the diode; as we have connected

in the half wave rectifier, here also there is a flywheel diode and the load is an inductive

load or a complex load R and L in series. The voltages across these thyristors are assumed

to be VT1 and VT2. We are applying sinusoidal voltage Vm sin omega t. In the positive half

cycle of the input signal, the thyristor which will conduct is the upper one because in the

positive half cycle of AB we are considering this secondary side of the transformer and

we will analyze the waveform.

Let us consider the voltage across AB. The point A when it is positive with respect to

B, it is a center tap transformer. It will be like plus and it will be minus, this point

is the ground or center point is the ground. In this situation, the upper one thyristor

that is T1 will be forward biased or anode will be positive and it will conduct at angle

of alpha, say. Both are identical thyristors, we are assuming. Conduction angle or delay

angle is alpha. When it conducts after an angle of alpha this current will flow through

the load and the voltage across this load, this voltage will be equal to this. This point

is ground.

We are getting the voltage across this load following the voltage at the input that means

at the secondary side.

If we consider the voltage VA here and VB here, VA and VB will be just out of phase

by 180 degree and so VA will be like this and VB will be like this. They are out of

phase by 180 degree. Now what will be the output voltage? That will depend upon the

current flowing through it. When the current flows through this load, the voltage across

this load will be equal to this VA when this T1 conducts. That is happening here. When

this T1 conducts after angle of alpha, the voltage across the load will follow this Vm.

In the negative half cycle when the VA enters negative half cycle, VB becomes positive.

But the conduction will not take place in the lower one thyristor T2 till another angle

alpha beyond pi. That is it will conduct at pi plus alpha only. Within this region that

is from pi to pi plus alpha VA enters negative half cycle. What will happen when VA enters

negative half cycle.

Now this becomes negative with respect to B. This is ground. What will happen is that

if we look into the diode D, then this n is negative; that means the n is negative with

respect to this p and this means it is forward biased. As soon as this VA enters the negative

half cycle, then the diode D will be forward biased. The load current which is flowing,

IL is the load current, will be contributed by the diode current only. The current will

not flow through this transistor but we will have a load current which is contributed because

of the diode. Since the diode is forward biased it will bypass that load current in this way,

in this fashion and what will be output voltage across this load? This voltage will be equal

to the diode voltage and which is equal to zero; this voltage equal to zero.

In the full wave rectifier here we have seen that during the portion when the VA goes into

negative half cycle before pi plus alpha angle, this is pi plus alpha, we have the voltage

across this load zero, but the current which is flowing in the load that is almost constant

because here the current which is flowing is the diode current. When the thyristor T1

was conducting, the current was due to this iT1; iT1 means the thyristor current which

was drawn from the supply because the current flow was from the point A through this load

to this point, center or ground and that current we are denoting by iT1. But when VA enters

into the negative half cycle the diode conducts. The thyristor T1 is now turned OFF since anode

is negative and the load current is due to the diode current. If the R and L, the time

constant of this circuit is properly chosen then we will have almost a constant current

like this although contributed by two different components. Each time we are not getting the

same contribution from the same component but here it is the diode current, here it

is the thyristor current but we will have almost a constant current.

In the other half cycle, that means when VB is in the positive half cycle and after pi

plus alpha then the thyristor T2 will fire. The thyristor T2 will conduct now because

it is positive, anode is positive. When this thyristor T2 is conducting, then the load

current will flow from positive like this and it will conduct and current will flow

through the load and it will enter the ground. Through the load, the current enters or flows

in the same direction in the both the cases. The current in the load is due to the thyristor

T2 that is iT2 and again when it will be entering the negative half cycle then your VA will enter the positive half cycle.

When VA is entering positive half cycle, then the diode current will flow.

If we consider now positive half cycle, and it is below this alpha. That means it is below

the firing angle. Then also the diode current will flow and so, it will continue.

What is important here is that it is full wave rectifier. We are not getting a negative

voltage across the load as we were getting without the flywheel diode. But here during

the positive as well as negative half cycle of each signal we are getting output waveform

across this load. That means we are getting the output voltage across load and in between

the conduction by the two thyristors the voltage is zero. When the diode is conducting, the

diode voltage is zero. This is typically the output voltage waveform for a full wave rectifier

using center tap transformer and flywheel diode. Here we have seen that the rectification

using this thyristor will have output waveform across the load with no negative portion.

Similarly there are also circuits for full wave rectifier using bridge.

If we consider a bridge rectifier bridge type rectifier using the thyristors, we will have

thyristors. In addition to those thyristors we will also use diodes - the circuit for

a bridge rectifier. We are using single phase bridge rectifier, we are till now considering

single phase only but it can be extended to three phase also. What will be the shape of

the bridge rectifier? It is having diode and thyristor in series like this. We will have

pairs of diode and thyristors; they will conduct in pairs. Here this is say D1, this is say

T1; this is D2, T2 and the supply is given between A and B which is a sinusoidal supply

VS. Apart from this diode thyristor pairs, we will have a single diode connected in this

way across this load R and L. This is a single phase bridge rectifier circuit. The voltage

across this load is VL say.

Here what will happen? This vS when it is in the positive half cycle we can see that

the diode D1 and the transistor T2 will conduct. In the positive half cycle, this diode D1

and T2 will conduct. These pair will conduct and in the negative half cycle, when this

is negative, in the negative half cycle it is like this. We are having the diode D2 and

T1 conduct. In pairs they will conduct and the current will flow through the load after

firing of the thyristor. Because only after firing of this thyristor T2 this loop will

be complete and the current flow through this load will occur.

When this thyristor say T2 is conducting in the positive half cycle, the voltage across

this load VL if we consider, if this is the input voltage vS, then if we draw the voltage

VL that is the voltage across this load, here say alpha is the angle and pi; this is again

pi plus alpha and this is 2 pi like this. So, the VL will follow the input voltage like

this and from here to here, the other pair will conduct and we will get the voltage like

this. We will get this output voltage, the rectified output voltage but the difference

between the earlier rectifier using center tap transformer and this bridge rectifier

is that we are having pair of diode and thyristor which will conduct one pair in one half cycle

and this is the diode which is there to prevent the output voltage to become negative.

Today we have seen the application of the thyristor in rectification. We have studied

about the half wave and full wave rectifier and we have also seen for an inductive load

how the flow of current in the load takes place for the presence of a diode and how

the voltage across the load which goes negative when the thyristor does not stop conduction

can be prevented by using a flywheel diode. Using a flywheel diode we have seen that the

voltage across the load is stopped from going down to negative values and we get proper

rectified output across the load. The thyristor in this way is used in many of the applications

which are useful in power control.

The Description of Module - 5 Lecture - 8 SCR Applications