Lecture – 4 Power Electronics


Hello, in my last class I told you some of
the very important applications of power electronics and some of the significant events in the
history of power electronics till 1957, the year in which silicon controlled rectifier
or thyristor was invented. As I go long, of course I will tell you the other significant
events and we started discussing about the power semiconductor devices. I told you that
power semiconductor devices are heart and soul of power electronic equipment. These
are used as switches that means when they are on, we need to operate them in saturation
and when they are off, we need to operate in: sorry, when they are off, they are in
cut off region. There are three types of switches. One is
an uncontrolled switch, uncontrolled because whether a switch is on or off it depends on
the circuit operating conditions. Say, diode for example
and there is a semi controlled switch. A silicon controlled rectifier or thyristor is a semi
controlled switch because it can be turned on by supplying some current to the gate or
to the control terminal. But then, having turned on the device, you cannot turn it off
using the gate. Now, the SCR which is conducting, it turns
off depending upon the circuit condition. If the current becomes zero in the circuit,
thyristor will turn off and third one is a controlled switch. It can either turn it on
or off using the control terminal. Say bipolar junction transistor, by supplying a base current,
you can turn it on. You make ib is equal to zero, base current zero, transistor turns
off. Now we will study the various devices in detail. First one is diode, a two terminal device.
All of us know, we have been studying may be since the class eleven, anode and cathode.
So if VAK, voltage across the diode VAK, should be positive then it is said to be in forward
bias mode and diode starts conducting. So for a silicon diode, a low power diode, it
could be, VAK could be of the order of 0.7 volts or so and this value increases as the
power rating increases. Say, for a power diode it could be of the order of 1.5 volts or so
and when the diode is on, all of us know that current in the circuit is limited by the load.
See the ideal characteristics are along the y axis. That is, it can carry any current
assuming which is less than the rated and it can block a negative voltage, it cannot
a block positive voltage. These are the characteristics of a non ideal diode. May be, a cut in voltage
somewhere at this point and this the current that is limited by the load. This is forward
biased and this is during reverse biased mode, VAK is negative. Now as long as the applied voltage is less
than VBD, see VBD is this point, break down, junction break down, diode will block that
voltage. In other words, the maxim reverse voltage that is applied across the diode should
be less than VBD, we need to ensure that. Say, the input voltage is 230 volts, so negative
peak is going to be 230 into root 2. So, the voltage rating of the diode should be higher
than 230 into root 2, if you are using in a single phase circuit assuming that load
is a passive one. But then, if the applied voltage is higher
than VBD and if you are not able to control the current that is flowing, then the device
will get damaged. Remember, if the applied voltage is higher than VBD and you are not
able to control the current that is flowing through the diode, it is going to be destructive.
It is going to be destructive. So, during on state, there is a finite voltage drop across
the diode. I will call it as Vf and Ia is the current that is flowing through the diode.
So, on state loss or the conduction loss is given by Vf into Ia that depending upon Vf
and Ia. The magnitude of Vf and Ia say, if it is higher than a certain value, you need
to go in for or you need to mount the diode on a heat sink. Invariably power diodes are
mounted on a heat sink. A small signal diode, it would not need to or generally is not mounted
on a heat sink. Now, all of us know the on state behavior
of the diode. Now, in the power electronics circuit, in the off state, the diode, the
way it turns off is a bit important. Why so, I will tell you. See, when the current in the circuit, of course
this is the current that is flowing in the circuit depending upon the input voltage or
so current starts following. At this point, current becomes zero but the diode continuous
to conduct. A reverse current starts flowing till, say trr, that is because the minority
carriers they require a certain time to recombine with the opposite charge and to get neutralized.
I will repeat, the minority carriers which are there they require a certain time to recombine
and that takes place in the time trr. trr is the reverse recovery time. It is a time
from this point, the first zero crossing to say 25% of the maximum value of the current
that is flowing in the reverse direction. So, the maximum value of current in the reverse
direction is irr. So, trr is the time from this point to say
0.25 of irr. I am assuming that this point is 0.25 of rr and this current is the leakage
current which is very small and this is the charge, reverse recovery charge.
Now, what is the reverse recovery charge? It is the area of this plot. I am assuming
that this is a linear one. I am just determining the area of the triangle to find out the reverse
recovery charge. The area of this shaded part is nothing but irr, is a peak into trr divided
by 2. I am assuming this to be a triangle one, area of the triangle. So, reverse recovery
time and reverse recovery charge, they are very important, two important parameters.
Why are they important? See, even the current has become zero it continuous
to conduct for or goes to offset only after trr.
So if the circuit in which the frequency of operation is very high, this trr will play
an important role or in other words trr will determine the upper frequency of operation
and this qrr or the reverse recovery current, it is flowing in the opposite direction, it
is flowing from cathode to anode. See this current is flowing from cathode to anode so
either it has to flow to the source or it may have to flow through other switches. Those
switches have their own current carrying capacity. So, in addition to the load current, that
device may have to carry this reverse recovery current. See, yesterday I showed you this power module.
I said, there are two control switches. The name of the switches, I will tell you sometime
later and in parallel there are two diodes. These are the two diodes, this module has.
The current rating of this, each diode is 75 amperes and it can block 1200 volts. The reverse recovery time of this diode is
of the order, say 0.2 micro second. That is what the specifications, the data sheet says.
It could be of the order of 0.2micro second and depending upon the junction temperature,
qrr the reverse recovery charge is of the order of 1 micro Coulombs. This is all given
in the data sheets. So, what about the other ratings? important ratings of the diode. What
are the other important specifications of the diode? They are; one is average forward current,
see here, it is required to assess the suitability with the power circuit, so depending upon
the input voltage, the load and the duty cycle. So average current can be determined, I am
saying duty cycle because diode may not conduct for the entire cycle. The second point, reverse
blocking voltage, the reverse voltage that is appearing across the diode should be less
than the break down voltage. Third, on state voltage drop, it is required to determine
the conduction loss and therefore the heat sink size or the cooling requirement. The
fourth one, trr reverse recovery time of the diode is another very important factor in
the diode used in high switching applications or it is required to assess the high frequency
switching capability. The fifth one is the surge current rating.
To explain this, consider this circuit. What I have done is there is an ac sinusoidal source,
diode. I am connecting a switch and a capacitor, the remaining part I have not drawn. Usually
that capacitor is completely discharged and I close the switch at omega t is equal to
pi by 2. At omega t is equal to pi by 2, input voltage set the positive peak. Capacitor voltage
is zero and it cannot change instantaneously. So what happens when I close the switch? There
is going to be a large inrush current. Why? Input is at the positive peak and the output
voltage is zero and that large inrush current has to flow through the diode and it has to
withstand that current and let me tell you that this surge current is much much higher
compared to the average forward current. May be sometime later, I will show you a data
sheet wherein as a average forward current, RMS current and surge current ratings are
given. Next one is I square t rating, current squared time rating. It is an indication of
or it is the short time surge energy that the diode can withstand. This is required
to choose a suitable fuse in the diode circuit. When this is important? So take this example,
I have a input AC source, a diode and a capacitor. Now I have not drawn the remaining part of
the circuit and I have a switch here. Something may be a main switch or so. Assume that I am closing the switch at omega
t is equal to pi by 2 or in the positive peak. Capacitor is totally discharged, in other
words there is a short circuit here. Capacitor voltage cannot change instantaneously. Voltage
at this point is instantaneous value is at the peak and we have closed the switch at
that time. What happens? A large surge current will flow because voltage across the capacitor
cannot change instantaneously. It was zero just prior to closing the switch. This is
at the peak, you have closed the switch. A large surge current will flow and diode should
be able to withstand this surge current which is higher than the average current rating.
So that is about the diodes, the so called uncontrolled switch. Now, what are the various types of diodes?
The first one is a rectifier diode or a slow diode. They are suitable for line frequency
application, so you want to rectify the input AC which is 50 hertz. So you can use the conventional
slow diodes or rectifier diodes. The reverse recovery time is generally is not specified,
they are basically slow diodes. So, the maximum rating is of order of 6kv, the diode can block
as high as 6kv and current rating is of the order of 4500 amperes are available. I will
repeat, diode which can carry say, 4500 amperes of current and can block 6kv, they are available.
These are conventional rectifier diodes or slow diodes. There is second type the one known as the
fast recovery diodes. Fast recovery diodes, these are generally used in high frequency
application. Sometime later in the course, we will find that fast recovery diode should
be used in the circuit. You cannot use the conventional rectifier diodes. So, the rating
is of the order of say 6kv and current of the order of 1.1 kilo ampere. They are available
in the market, 6kv it can block and current ratings, 1.1 kilo amperes and the reverse
recovery time could be of the order of say, 0.1micro second, could be. I am not saying
that 6kv 1.1 kilo diode, has a reverse recovery time of 1.1micro second, no. The reverse recovery
time of a fast recovery diode could be of the order of say, 0.1micro second. Just now,
example which I showed you is of the order of 0.2micro seconds. The third one is the short key diodes. What
are the features? They have a very low on state voltage drop
basically these are only majority carriers. Current only due to majority carriers but
then the voltage rating is low. Maximum voltage rating is of the order of 100 volts or so
and current is of the order of 300 amperes and the last one, the fourth one is what is
known as the silicon carbide diode. They have ultra low power loss. See, Schottky
diode has a very low on state voltage drop. That is what I told you. Therefore on state
power loss also is low but silicon carbide diode, it has an ultra low power loss, ultra
fast switching behavior, very fast diodes, highly reliable, no temperature influence
on the switching behavior, looks like we are etching almost everything in silicon carbide
diode. Ultra low power loss, ultra fast switching behavior, highly reliable and what is the
problem? Why are they not very popular? You find only silicon diodes not silicon carbide
diodes. The limitation is they are very expensive. The process is also very expensive, that summarizes. Just coming to the second type of switches,
semi controlled switch, a thyristor or a silicon controlled rectifier. The moment I said semi
controlled device, it has to be a three element device; anode, cathode and gate. Anode and
cathode, they form the power circuit terminal, they are connected in series in the power
circuit and the gate current is supplied with respect to or gate signal is applied with
respect to the cathode, say something like this. Our source, as I have shown you, a AC source,
a load and a thyristor. So, this is the symbol of a thyristor diode with a small gate line
here, anode, cathode and gate and can be controlled or can be turned on
using positive gate current IG or control signal is applied with respect to the cathode.
There are 4 layers say, P N P N and therefore there are three junctions J1, J2 J3. P N P
N; first P is anode, N is cathode and the gate
is connected to this P. This is the gate. N2 or this layer is very thin and it is highly
doped. This N layer is very thin and it is highly doped. I will call this as P2, this
layer is thicker than this or thicker than N and it is less highly doped. N1 is the thickest
of all, this N is a thickest of all and it is less doped. Highly doped, less highly doped,
less doped. Doping level is very low in this and a very thick layer and this P is same
as this P, slightly thicker and less highly doped. So, therefore J3, see N is very highly
doped and a very thin layer, this P is relatively highly doped and slightly thicker. So, junction
J3 has a very low breakdown voltage in either direction, something similar to base emitter
junction in a transistor. Base emitter junction in a transistor is lightly
low, very highly doped. So, it cannot block a voltage in either direction, in the sense
it can block a very low voltage. In the class, may be small signal diode, you need to apply
a 0.7volts to the base to turn it on and reverse voltage it can block is very low, very small.
It cannot block a very high voltage. So, therefore J3 cannot support high reverse voltage, cannot
support a very high reverse voltage. Now what are the various characteristics when
VAK is positive? In other words, the device is forward biased and the current is zero,
the device current or the current that is flowing from anode to cathode is very small,
the device is said to be in forward blocking mode. I will repeat, device should be forward
biased, VAK is positive, a very small current equal to the leakage current will flow through
the device. So, this mode is known as forward blocking mode. Now, what happens in forward
blocking mode? There are three junctions J1 J 2 J 3. I said VAK is positive, see look here, VAK
is positive. So P is connected to, as said, positive of the battery. N is negative, so
definitely J1 is forward biased and J3 is forward biased. Junction J minus forward biased,
J3 is forward biased and J2 is reverse biased. So, in the forward blocking mode, junction
J2 is reverse biased. So the entire voltage appears across J2. Now, what happens when the devices in reverse
biased mode? When the junction is, when the VAK is negative, something similar to the
diode, VAK is negative, at that time J1 and J3 are reverse biased. See, when VAK is positive,
J1 and J3 are forward bias and J2 is reverse biased and when VAK is negative, J1 and J3
are reverse biased and J2 is forward biased. In the forward blocking mode, since J1 and
J3 are forward biased, the entire voltage appears across J2 or J2 should block the voltage
in the forward blocking mode, remember. Now, when at the device in reverse bias mode,
J1 and J3 are reversed biased. But I told you that J3 cannot support a high reverse
voltage because N2 is very thin and it is highly doped. So, therefore when it is reverse
biased, entire voltage should be blocked by J1. Now I will show you the characteristics. Before
that maybe, I will try to show you the two transistor analogy. See, I told you there
are 4 layers, P1 N1 P2 N2. Now, what I will do is, I will cut this and this. Say, N1,
N1, P2, P2. Now, what is this? Say, here P1, N1 and P2. Now this forms a PNP transistor.
See here, P1, N1 and P2. This is P2, a PNP transistor and again here I have N1, P2, N2.
It forms a NPN transistor. See here, NPN transistor. Now how do I interconnect T1 and T2? Base
of PNP transistor, say, base of T1 is same as collector of T2, NPN. Base of T1 is same
as collector of T2 and collector of T1 is same as the base of T2 and this is the gate
terminal. So, there is a PNP transistor, PNP, NPN transistor, NPN transistor inter connect
them. Base of PNP is same as the collector of NPN, collector of PNP is same as base of
NPN. So, this is anode, this is cathode, this is gate and here are the SCR characteristics,
I will explain to you now. In the reverse bias mode, characteristics are almost similar
to that of a diode, almost similar. VBR, reverse break over voltage, so you should ensure that
applied voltage is less than VBR. Same thing, for diode I called as VBD and here I am calling
as VBR. In the forward bias mode, if the current is
very small that is equal to the leakage current, I told you the device is in forward blocking
mode, again it is stable that corresponds to
P to Q or P to Q1or P to Q2. Assume that you are not supplying any gate current, as of
now the gate is open and the device is forward biased. When the device is forward biased
J1 and J3 are also forward biased. Junction J1 and J3 are forward biased and J2 is reverse
biased. So, when it is in forward biased mode the entire voltage J2 blocks and in case the
applied voltage is higher than the break down voltage of the junction J2, device goes into
conduction mode. I will repeat, I assume that gate current
is 0 and if the applied voltage is higher than VBO, see here VBO with IG is equal to
0. VBO that is break over, forward break over, what happens? Device goes into conduction
mode. But then how does it go? In what is the path that does it takes? I told you that
just prior to device goes into conduction mode, the voltage is may be, approximately
equal to VBO. It is high the moment the device goes into conduction mode. The voltage will
drop to a very low value. It could be of the order of say 1.5 volts or so 1.5 to 1, 2 volts.
Till 0 to VBO, current that is flowing is very small. It is equal to the forward leakage
current. Once the device is in conduction mode, current
is limited by the load. Now, what path does it takes? It
was blocking a relatively higher voltage, the moment it goes to conduction mode voltage
drops to a very low value, the paths taken by this is this. I have shown this is to be
dotted one. Why did I show this to be a dotted? Why it is not firm like P to Q? Why it is
not firm? This is because this region is an unstable region. Why it is unstable? because
it is a negative resistance region. See, voltage is falling, current is increasing,
the slope of this line is negative resistance region. So, it is unstable and it goes to,
see, RS is the forward conduction mode, current, the operating point here depends only on the
load and now see, there is a third element what is known as a gate and therefore by supplying
the gate current it should be able to trigger the device. So, definitely with the finite
gate current, the voltage at which it goes to conduction mode should reduce. So see,
as the gate current increases, the voltage at which device goes into conduction mode
also drops. IG0, so voltage applied should be higher than
VBO forward break over, VBO is forward break over. So, with the finite IG1, device goes
into conduction mode at a relatively lower voltage and if I increase IG further, it goes
to conduction mode somewhere at this point, say PQ for 0 IG, P1Q1 for some IG1 which is
less than IG2 and for that the path is PQ2 and once the device current is higher than
the latching current, gate has no control. So I will repeat, by supplying a positive
gate current, device goes into a conduction mode and by supplying a gate current the voltage
at which device goes into conduction mode also reduces. But then, the gate signal should be present
till the current through the device is higher than the latching current. The device should
be able to latch, till then gate signal should be present and having gone into conduction
mode, the gate has no control. You can withdraw or you can make IG is equal to 0. In fact,
it is advantageous to make IG is equal to 0. Why? Gate has no control, if there is a
constant IG flowing, definitely there is going to be a dissipation in that junction. So,
in a way, it is an advantageous to make IG is equal to 0. Once the current through the
device is higher than the latching, because having gone into conduction mode if the current
is higher than the latching, gate has no control and you can withdraw the gate signal. But then to turn the device off, current through
the device should fall to a value which is less than the holding current. Somewhere here,
see here, IH is the holding current. So, to turn off the device, I will repeat, the current
that is flowing through the device should fall to a value which is less than the holding
current. How does it happen? We will see, whether it happens naturally or do we need
to do something different to reduce this current. We will study during appropriate time. So,
that is about the SCR characteristics. This is all the explanation that I have given. See, VBO is the forward break over voltage
and this voltage is determined by, VBO is determined by J2 because J1 and J2 are J1
and J3 are, they are forward biased. So, VBO is completely blocked by J2. Now here, I have just shown you the plot of
the variation of VBO with IG. So, here is a simple circuit, I am just closing the switch
till the device goes into conduction motor or till the current through the device, this
load current higher than the latching, then I can open the switch and I am varying in
this resistor to vary the current that is flowing through the gate circuit. So, as IG increases, the voltage at VBO with
very low value of IG is almost the same. This is the voltage, this voltage is determined
by junction J2 itself. Above a certain value, this VBO the voltage at which device goes
into conduction mode reduces. So, it goes on reducing, why? It is because IG, the gate
current reduces the depletion layer around J2. I will repeat, the gate current it reduces
the depletion layer around the junction J2. Now how do we analyze the process by which
the device goes into conduction mode using 2 transistor analogy? I just showed you PNPN
is equivalent to 2 transistors, there is a PNP transistor and there is an NPN transistor
and they are interconnected. Now, using this transistor analogy, can we find out or can
we or is it possible to understand the turn on process of the thyristor? Answer is yes,
I will explain to you now. See, for any transistor collector current
IC is given by alpha into IE plus ICBO. Collector current IC, IE is the emitter current, alpha
is the common base current gain, it is approximately equal to IC divided by IE and what is ICBO?
It is the leakage current of the collector base junction. So, this is a common equation
IC is equal to alpha into IE plus ICBO. Now, for this transistor T1 transistor, what
is IE emitter current? Emitter current is nothing but the anode current IA. So, therefore
IC1 is equal to alpha into IA because IE is equal to IA, plus ICBO1. I am calling ICBO1
is the leakage current of collector to base junction of transistor T1. Now, how about
for T2? Now, for T2, IE, T2 is a NPN transistor whereas T1 is a PNP transistor. IE is equal
to IK, the cathode current IK. So, therefore IC2 is equal to alpha two, the gain of common
base current, gain of T2 into IK plus ICBO2. Now, we all know that IE, emitter current
is collector current plus IB, base current. Now, what is IE1? IE1 is equal to IA itself,
anode current. Emitter current of T1 is anode current and what is IB1, base current of transistor
T1? Base current of transistor T1 is same as collector current of T2, IC2. I showed
you in the beginning, base of T1 and collector of T2, they are same, they are tied together.
So, I will add up those 2 equations. Which are the 2 equations? These are the two equations,
I will add them up. What do I get? IC1 plus IC2 is equal to alpha one into IA plus alpha
two into IK plus ICBO1 plus ICBO2. So, here is the equation I have written, as
simple as IC2. Now, substitute here, IC here is IC1 itself for transistor T1, IB is IB
for T1 is IC2. So, IC1 plus IC2 is equal to IA. So, these are the 2 current IC1 plus IC2
should be equal to IA, a PNP transistor that is equal to alpha one into IA plus whatever
that equation we are adding to. What about for transistor T2? Listen to me
carefully, here what is IE? That is IK itself. It is IB2, base current plus IC2.This current,
IB2 plus IC2. Now, if there is finite IG, what is base current? Base current is IG plus
IC1 plus IC2 is equal to I emitter, IE2 that is equal to IK. See, I will repeat, emitter
current of T2 is this current, base current plus collector current. Now what is base current?
You apply KCL at this point for finite IG, IB2 is equal to IC1 plus IG. So, therefore
cathode current is IC1 plus IG plus IC2. Now we all ready found that IC1 plus IC2 is
equal to IA for this transistor. IC1 plus IC2 is equal to IA. So, therefore IK is equal
to IA plus IG. Now, you substitute this value in this equation, IA is equal to alpha one
into IA1 plus CBO1 plus this equation and what do I get? I get this equation, IA is
equal to alpha two into IG plus ICBO1 plus ICBO2 divided by 1 minus alpha one plus alpha
two. So, I have an expression here for anode current in terms of gate current, the leakage
current of collector to base junction and common base current. Now using this expression,
how to understand the turn on process of the thyristor? We will do it in our next class. Thank you.

22 thoughts on “Lecture – 4 Power Electronics

  1. Btw it is also semi-controlled since you set the angle where it is starting to conduct by a trigger pulse, versus the diode which is non-controlled.

  2. Sir, u r telling that if we apply voltage greater than Vbo the scr comes to on state.is it posiable to turn on without gate triggering?

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