1  

Power Electronics

Prof. B. G. Fernandes

Department of Electrical Engineering

Indian Institute of Technology, Bombay

Lecture - 4

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.

2  

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3  

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4  

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6  

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.

7  

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8  

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9  

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10  

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11  

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

12  

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ard break ovJ2 are J1 and

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ow does it hfferent to reharacteristic

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13  

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w do we anar analogy? r and there ican we find

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any transistoemitter curr

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0:48:38 min

or collector crent, alpha ishat is ICBO?n IC is equal

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o, this

ut the CBO1. I , how ual to

base

14  

(Refer Sl

Now, weIE1? IE1 isbase currshowed ySo, I wilwill add plus ICBO

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lide Time: 5

e all know ths equal to IA

rent of transiyou in the bell add up thothem up. W

O1 plus ICBO2.

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hat IE, emitteA itself, anodistor T1? Baeginning, baose 2 equati

What do I get?

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er current is de current. Ese current ofse of T1 and ons. Which? IC1 plus IC

collector cuEmitter curref transistor Tcollector ofare the 2 eq

C2 is equal to

urrent plus IB

ent of T1 is aT1 is same asf T2, they arequations? Tho alpha one i

B, base curreanode currens collector cue same, they hese are the into IA plus

ent. Now, whnt and what iurrent of T2,

are tied togetwo equatio

alpha two in

hat is is IB1, , IC2. I ether. ons, I nto IK

15  

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.