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So, we
have been discussing the Bipolar Junction Transistor.
In the previous lecture we started explaining the common emitter current voltage characteristics.
Let us look at the slide which we were trying to explain.
So, we want to explain the family of curves. In fact we started with a single curve, we
took one of these curves and then we qualitatively explained how we get the various points on
this curve. So for example this was the curve we obtained
for one value of IB. We obtained this curve by plotting the excess carrier distributions
or different voltage and current conditions. So for VEC = 0 then for IC = 0 and then this
point corresponds to VCB = 0 which is the collector base voltage = 0.
And then we went beyond this point and then we also arrived at an explanation as to why
the current goes on increasing in this region instead of saturating.
What we said is that the beta of the device goes on changing as you change your VEC, this
is because of base width modulation. In fact this was the formula for beta as a
function of beta0 where beta0 is the beta when collector base voltage is 0.
The ΔXB is the base width modulation the change in the depletion layer of the collector
junction within the base which is this quantity and this is the ΔXB.
This ΔXB is also ΔWB because as this depletion region shifts the base width also gets modulated.
So 1 - ΔXB/WB0 where WB0 is the base width for collector base voltage = 0.
The reciprocal of this quantity multiplied by beta0 is the beta for any collector base
voltage. And then we made a calculation for the device
in the solved example and we showed that beta for collector base voltage of - 9 volts is
approximately = 1.4 beta0 This means that from VCB = 0 two VCB = 9 volts
and the change in beta is 0 .4 times beta0. So this is almost forty percent of the value
of beta and this is what we explained. If we try to put this information on this
curve here then it would mean that when you go to the point corresponding to VCB = - 9
volts then this change in current will correspond to forty percent change in the beta.
So this is your ΔIC which is equal to Δ beta × IB. We are neglecting the contribution
because of IC0. So the equation is IC = beta times IB + beta
+ 1 times IC0 and then you take increments you neglect this part because this is small.
So when you take increments you get this equation so this is your ΔIC.
Now obviously this diagram is not to scale because this Δ beta is forty percent of beta0.
In fact this value is beta0 × IB because this is VCB = 0 so beta corresponding to this
point is beta0 so it is beta0 × IB. Whereas this is beta times IB and Δ beta
× IB is this increment ΔIC. This explains why the current does not saturate
and it goes on increasing. Now, in our slide here we do not find that much of increase
in current. You take any one of these lines. Supposing you take the line corresponding
to 36 microamps IB the increment in the collector current from
this point to this point is much less than what we have estimated here in the solved
example. Here the change is forty percent whereas there
the change appears to be only about some ten or fifteen percent that is this increment
is some ten or fifteen percent of this in our slide.
This obviously means that the practical transistor for which those current voltage characteristics
are obtained has a higher value of doping in the base.
When the doping in the base is higher then this ΔXB will be lower. The modulation of
the base width will be lower and the collector base depletion region will be smaller.
Therefore beta will not change that much with collector base voltage.
This is how the base width modulation and the slope of these characteristics here is
related to the doping in the base. Now proceeding further we find that this curve
is going to breakdown at some point. Now what is the breakdown point? At what voltage
will this breakdown occur? This is the next point we want to explain.
Now, to understand why the breakdown occurs let us look at the picture inside the device.
This is your transistor and this is your collector to base depletion layer so this is p + +,
n + and p. Now let us show the currents which are flowing
according to the flow diagram. You have the hole current being injected and in the modern
transistor in order to arrive at a simple picture we neglect that recombination in the
base so this currents goes right through like this so this is the hole current.
Now there is a part of the hole current recombining here with the electron current which is injected
and there is an IC0 here which is the generated current which will either be shown within
the depletion layer or here outside. In fact in both these places there will be generation.
There will be generation within the depletion layer and there will be generation in the
neutral region. Right now we are not bothered about exactly
where the generation is so we will just show the generation within the depletion layer
for example. This is IC0 so these electrons move from collector
to base whereas here the electrons are moving from base to emitter.
This is reverse biased whereas this junction is forward biased and the base supplies the electrons. This
is the picture below breakdown. We know from the PN junction theory that,
as you go on increasing the reverse bias across the junction impact ionization within the
depletion layer sets in and because of the impact ionization the reverse current that
is flowing through the depletion layer starts getting multiplied.
We will show this here as follows: When this current moves through the depletion
layer it gets multiplied so this we show as follows.
Multiplied means electron holes pairs are generated.
The holes which are generated will move up and join these holes whereas electrons which
are generated will cross as something like this.
This is the additional current because of impact ionization.
Similarly there will be a multiplication of this hole current also which is injected from
the base to the collector. Here you have holes which are generated, we
join up with these holes and the electrons which are generated will join here.
So let us show all these joining the same point.
These are the extra electrons the electron hole pair which are generated because of multiplication.
So, it is the multiplication due to impact ionization as shown here.
It is this and this. Now we need to see the result of this multiplication. We can write
a simple equation to explain the effect of this multiplication.
But before you write the equation let us understand qualitatively why the breakdown occurs.
Breakdown means a sudden change in current, it is a rapid change in current.
Here if you see carefully what is happening is the following.
It is because of the multiplication in the depletion layer you have extra electrons coming
in where all the electrons join together along with the base current and IC0 and all these
electrons are getting injected here. Now since this electron current that is injected
from base to emitter is increasing because of impact ionization in the collector the
hole current which is this current injected from emitter to base also increases because
the ratio of the hole current to electron current across a forward bias junction depends
only on the ratio of the doping levels on both sides and also on other parameters such
as the diffusion coefficient lifetime and so on.
Since this parameters are not changing here on this side and other side of the junction
the ratio of the hole current and electron current the ratio of these two currents should
remain a constant. Therefore if the electron current increases
the hole current here that is injected also increases.
Therefore the current that is moving in here is increasing and this increasing current
is further getting multiplied giving rise to more electron hole pairs and these electrons
again come back to the emitter and they increase hole current in turn. Therefore here you find
a positive feedback. So the increasing current giving rise to more
current which further gives rise to more current and so on.
So there is a feedback from emitter to collector and collector to emitter.
This cycle sets in and as this cycle sets in the current starts increasing very rapidly.
So this positive feedback is present because of the transistor action and because of the
presence of a reverse bias junction in close proximity to the forward bias junction.
That is the reason why the impact ionization is causing a rapid change in current in the
transistor and this is the reason for breakdown.
When there is no multiplication or impact ionization then we can write this in the form
of an equation as follows: The equation is IC = α times IE + IC0 that
is this white line and this white line. This white line is α times IE and this is I.
And this quantity is IC0 and this quantity is α times IE.
So, there is no impact ionization, IC is α IE + IC0 we know this equation because we
have discussed this in the context of basic transistor action.
Now, in the presence of multiplication what is happening? This α IE current is getting
multiplied so you have this additional current shown here by this yellow electron hole pair
generation. Similarly IC0 also is getting multiplied because of impact ionization.
So, in the presence of impact ionization we can simply write the equation as IC equal to this quantity multiplied
by M. So it is IC = M × α IE + IC0 where M is
the multiplication factor due to impact ionization, so this is the difference between the situation
in the presence of impact ionization and in the absence of impact ionization. The M = 1
when there is no impact ionization. Now please note that we must start with this
equation and we cannot start with this equation in the presence of impact ionization.
This equation is alright when the impact ionization is absent but in the presence of impact ionization
you cannot have IC as beta times IB + this other current.
This is because here the situation has changed. You have additional electron current coming
in because of the generation within the depletion layer because of this impact ionization.
Therefore that equation gets modified so we should not start with that equation but you
must start with this equation. Now when we write this equation you can easily
transform this equation to show when the collector current will start increasing rapidly.
For this purpose we must write IE in terms of IC and IB because IB is being maintained
constant. When you are tracing this hole curve your IB is constant as shown here.
That is why you must express this IE in terms of IB and IC. When we do that let us see what
happens. So IC = M and IE α IE + IC0 where this IE
is IC + IB so we get IC = M times α times IC + IB + M times IC0
Now you shift this IC on this side and you will get 1 - M times α × IC = M times α
× IB + M times IC0. In other words IC = M × α IB + IC0/1 - Mα.
So this is your equation for IC in the presence of multiplication. This equation shows the
so called positive feedback action. In fact one can easily correlate the form
of this equation to the form of a feedback amplifier.
Now we will not do it in this course but when you do the feedback amplifier in the circuits
course you can easily see that this is the loop gain M × α is a loop gain.
M is a forward gain and M × α is the loop gain.
Now, coming back to breakdown here we see that IC tends to infinity when Mα tends to
1 in this denominator. or this is same as saying M tends to 1/α
So whenever M tends to 1/α multiplication = 1/α. Please note that α is very close
to 1 but less than 1. So 1/α is a quantity which is slightly greater
than 1. So even if there is a slight amount of multiplication
what it shows is that the current IC will tend to infinity.
Now this is what is happening here. At this point your multiplication is becoming equal
to 1/α and that point the breakdown occurs. This is how you can trace a complete IC versus
VEC curve for one value of IB. Now, to complete the picture we must trace
the curve for another value of IB and for the condition IB = 0.
Now what is the curve for the condition IB = 0? That can be seen very easily from here.
Note that IC = beta times IB + beta + one time IC0 is valid in the active region
So from here that is VCB = 0 to breakdown you have the active region and this region
is
called the saturation region. We will see shortly why the word saturation is used here.
But right now let us return to the active region and this is the equation:
If IB = 0 your equation is IC is simply = beta + one times IC0 so IC is not equal to 0 when
IB = 0 there is a small amount of IC flowing. So that will be the condition here when IB
= 0 obviously IC will be 0 for 0 VEC. And here you will have something like this
going and of course this will break down. Now the breakdown voltage here is not the
same as the breakdown voltage here for higher value of IB.
Now we will remove the other things, this is for IB = 0. The breakdown voltage for IB
= 0 is more than breakdown voltage for any other value of IB. We will discuss this issue
separately. Now we just want to plot this curve for IB
= 0. So this value is beta + one time IC0 the value
of the current here, this current is beta + one time IC0. There is a small slope here;
it is not exactly plot because beta is changing with VEC.
Now let us plot for any other value of IB as supposing we want to plot for a lower value
of IB than this. Now obviously here what is going to happen
is, this point is going to shift up for a smaller value of IB and this point here is
also going to shift to the left because for a smaller value of IB the collector base voltage
= 0 condition occurs the emitter base voltage will be smaller.
For small value of IB the emitter base voltage that you have to maintain or to give the particular
value of IB is smaller. Since this is the emitter base voltage, for
small value of IB the emitter base voltage is smaller so your point will shift left so
we start here at a point which is higher because this IB is smaller and then here we end up
with a point that is lower because your IB is lower beta times IB is smaller and also
this VEB is lower for a lower value of IB. Now when we join we will find that this curve
will almost go through the same point. We will not discuss this issue in much detail
because it is not so important We have seen that these voltages are very
small. In fact this is the curve for a lower value of IB up to the point where the saturation
region ends. Now in the active region will the slope of
the curve here be more than this or less
than this?
Now it can be easily seen that this slope should be less as compared to this slope because
we have seen that the ΔIC = Δ beta × IB so ΔIC/ΔVEC is the slope here.
We can write this in this form and Δ beta/ΔVEC will not be very different for these two cases
because when we estimated the Δ beta if you recall we wrote down the fact that beta = beta0
× 1/1 - ΔXB/WB0. Now in the entire calculation the IB did not
enter into picture. We found that Δ beta is dependent only on the change in the collector
base voltage. And since emitter base is forward biased the change in the emitter base voltage
is rather small because the change in the emitter collector voltage is almost the same
as the change in the collector base voltage. So this quantity does not change much and
this is almost constant therefore this ratio depends on IB, it is proportional to IB.
Therefore the slope will be smaller for a smaller value of I which is very important
to note. Therefore you will find that for higher and higher values of IB your slope
will go on increasing.
As we said earlier the breakdown voltage goes on decreasing as your IB increases from IB
= 0. Of course it decreases and then after sometime
it becomes constant and then its starts increasing again.
So this change in the breakdown voltage with IB is happening because of the change in the
beta as a function of IC. So, as the collector current changes the beta
does not remain constant but it changes. So far we have been assuming an ideal Bipolar
Junction Transistor where the beta was independent of collector current. Please look at an equation
for your beta it contains various parameters none of which depend on the collector current
So in practice however this is a non ideality that for a practical transistor beta depends
on the collector current and that is the reason why the α also depends on the collector current.
And if you use this equation for breakdown that is M 10 × 1/α then if α is a function
of IC M also is a function of IC so breakdown will
occur at different voltages for different values of collector current.
To summarize our discussion here we can go to the slide. Now we have explained how your
IB increases and when the IB increases your collector current goes on increasing, for
IB = 0 your current is beta + one time IC0 collector current and then the collector current
almost increases linearly with base current. That is how you have the increase in currents.
And here you find that this slope of the characteristics goes on increasing.
So when you start with the lowest curve you have a certain slope whereas you find a slope
of this curve for - 24 microampere IB which is more than the slope of the curve for IB
= 0 and finally a slope of 60 microampere curve IB = 60 microamps is more than this
slope, so continuously the slope of the curve is increasing.
Now let us separate the various regions of operation here.
This region to the left of this dotted line is the saturation region and between this
dotted line and this dotted line here is the active region and beyond this dotted line
to the right of this dotted line is the breakdown portion.
Now, we want to explain why the breakdown voltage is decreasing here as your collector
current is increasing. Before we do that let us shortly discuss why
this region is called the saturation region. The reason for this is very simple.
In this region here the collector base voltage is such that the collector base junction is
forward biased. Now, when the collector base junction is forward
biased you have a large number of excess carriers injected into the collector where the collector
has got saturated with excess carriers. It is this large number of excess carriers
for which we use the word saturation. So, the collector is saturated with excess
carriers and that is why it is termed as saturation. Therefore as the collector base junction is
forward biased there is excess carrier concentration in the collector that creates the saturation
region. In other words the collector is saturated with excess carrier creating the saturation
region.
Now let us move to the explanation of the variation of the breakdown voltage with the
collector current. For this purpose let us look at this slide, beta versus IC characteristics.
This slide shows how the beta changes with the collector current. So you find that for
small collector current the beta is low and as you increase the collector current the
beta starts increasing it reaches a peak at some value and then it falls again.
So this curve has been shown for a particular value of VCE. You can have similar curves
for other values of VCE. Why does this happen? Why does the beta change with IC?
So beta changing with IC also means α changing with IC. So you can either explain variation
of beta with IC or we can explain variation of α with IC.
Let us explain why α varies with IC. Look at your device again, the internal picture.
Now we will show the emitter base and collector base depletion regions clearly.
Let us show the currents. Now you will assume that the collector base
voltage is 0 for simplicity to explain the variation α with IC, we do not need the collector
base voltage so this is the picture. These are the holes and these are the electrons,
we have neglected the recombination in the base because that is the situation in modern
transistor. So α = gamma × base transport factor, injection
efficiency × base transport factor is approximately gamma itself for modern BJT particularly for
modern small signal BJT. So concentrate on injection efficiency.
What is injection efficiency? Injection efficiency = IEP/IE so this is IE, this is IEP and this
quantity is IEN. For very low collector currents what happens
I, the recombination in the depletion layer which we have neglected is significant as
compared to this current. So for low currents the picture is this.
This current is normally denoted as Igr generation recombination current.
When talking about PN junctions we have explained this which is the deviation of the characteristics
of a PN junction from ideal diode characteristics.
So real diode characteristics differ from the ideal diode characteristics and in that
you must take into account the generation recombination within the depletion layer.
Now that becomes important when the collector current is low.
So when you write this equation for gamma this is equal to IEP/IE = IEP/IEP + IEN +
Igr So IE = this quantity + this quantity + this
quantity + Igr. So far we have neglected this Igr whereas
if you want to find out the gamma at low currents then Igr should be taken into account and
that is why the gamma is low. And since gamma is low alpha is also low.
So α is small α low at low IC due to Igr. At very high values of IC again alpha is low
and that is because gamma is low. Let us see why?
In this high IC condition your gamma is again given by IEP/IEP + IEN because Igr is not
so important. So this is low IC. In low IC Igr is important
but for high IC Igr is not important. However, when you take the ratio you can write
this as 1/1 + IEN/IEP. So when you take the ratio IEN/IEP this ratio
IEN/IEP is not the same at very high currents and moderate currents.
This is because of the fact that since base is lightly doped compared to emitter at high
currents there is high injection level in the base. When the injection level in the
base is high the amount of electron current that is injected from the base to the emitter
is more than what you calculate assuming low injection level in the base.
This is directly followed from the Boltzmann relation.
When you go to the PN junction theory from there you can easily show that when one region
of the junction is in high injection level the amount of current that particular region
injects into the low injection level region is higher.
So the amount of IEN injected is higher as compared to moderate current conditions.
So IEN/IEP ratio rises for higher values of IC. So gamma falls for high IC due to increase
in IEN/IEP and therefore α also falls, hence α falls for high IC.
Now we can show this on a graph as follows. If you plot alpha the alpha is a function
of IC where IC is plotted on a log scale. This is
because you must show a wide range of IC and only then for very high IC and very low IC
you can show the behavior which looks as something like this.
This is for alpha. So in both these regions α is low because
of low injection efficiency. Here it is because of generation recombination in the depletion
layer here it is because of high injection level in the base.
So generation recombination in emitter depletion layer and beyond this here it is high injection level in the base.
Now when alpha changes like this and if you plot the change in beta it will look something
like this. So beta appears to be changing everywhere
on this IC axis although alpha appears to be fairly constant over a wide range.
This difference in shapes of alpha versus IC and beta versus IC has to do in the fact
that beta is very sensitive to change in alpha. So even for a small change in alpha there
is a large change in beta. So the equation is as follows: beta = α/1 - α and then
try to express d beta/dIC as a function of dα/dIC just by differentiating and you can
very easily show that even for a small change in alpha there is a significant change in
beta. That is why even though in this range alpha
appears to be almost constant but the beta changes.
That is the reason for the variation of beta with IC or variation of alpha with IC.
Now once we know this variation we can easily explain why the breakdown voltages are decreasing
when your collector current is raising to start with.
If you look at the breakdown region in this region the breakdown occurs when multiplication
factors tends to 1/α. Now alpha is low for low values of IC or low
values of IB so multiplication factor required for a breakdown is higher, 1/α is higher
so M is higher. As your collector current goes on increasing your alpha goes on increasing
therefore multiplication factor required for breakdown goes on falling therefore the breakdown
voltage falls. This means that this region here corresponds
to this region here that we are talking about. We have really not shown the IC VC characteristics
for this region, that is a very high IC region.
So in the slide here this collector current range that is shown is the range where the
alpha is increasing with IC or beta is increasing with IC so that explains why your breakdown
voltage falls. With this we have completed the explanation of the IC versus VC characteristics.
Now the breakdown voltage corresponding to IB = 0 this is called VECO or VCEO.
Look at this slide again, this voltage is the VCEO.
Why is it called VCEO is because it means that it is the collector to emitter breakdown
with the base open. Please note that IB = 0 for that case and
0IB means the base is open that is the transistor is connected like this.
The base is open and this is the transistor connection.
This is the bias when IB = 0. So about all this to show the condition for
IB = 0, now IB is 0. So that is why this breakdown voltage here
is VECO or VCEO depends on how you show the axis.
It is the collector to emitter breakdown with base open.
There are two types of breakdown. Normally we come across a breakdown voltage called
VCEO and a breakdown voltage called VCBO. Now this VCBO here is higher then VCEO.
What is VCBO? This is collector to base breakdown with emitter open.
So, breakdown voltage for this condition is much higher than breakdown voltage for this
condition, this is for VCEO. For VCBO, when you are talking of breakdown
you normally use the B here so BV stands for breakdown.
For BVCBO the condition is as follows: You have this transistor and the emitter is
open so you are measuring the breakdown voltage between collector and base.
So this is VCE and this is VCD. This breakdown voltage VCBO is much higher than this breakdown
voltage VCEO. What is the reason? The reason is very simple
this breakdown voltage is the breakdown voltage of the collector base junction.
So it is as though there is no transistor action. Breakdown voltage of this device is
the same as the breakdown voltage of this device because the emitter is open which means
the transistor breakdown is the same as the PN junction breakdown.
So here breakdown voltage occurs when multiplication in this junction reaches infinity at breakdown
whereas here breakdown occurs when multiplication reaches 1/α.
Clearly you will need a higher voltage to make the multiplication factor infinity because
of in impact ionization than this case when you only have to reach 1/α which is very
close to 1. To reach a multiplication factor which is just more than 1 you need a much
lower voltage. This is the reason why breakdown voltage between
collector and base with emitter open is much higher than breakdown voltage between collector
and emitter with base open. In fact there is a relation between the two
which we can derive very easily as follows. For this purpose we use an empirical relation
for the multiplication factor due to impact ionization.
The expression is M = 1/1 - voltage across the junction by breakdown voltage across the
junction whole power M where M normally lies between 4 and 6, this is an empirical relation
for multiplication factor. You find that when the voltage becomes equal
to breakdown voltage this multiplication factor tends to infinity.
Now using this basic relation for a PN junction let us see how we can express the relation
between VCBO and VCEO. Now, you go here VCEO occurs when M tends
to 1/α that is this quantity is 1/α. So I write 1/α at breakdown when the base
is open. M = 1/α = so this quantity
is 1 - voltage across the collector base junction by the breakdown voltage of the collector base
junction. So it is M = 1/α = 1 - (V/VBR)m. Now we can write breakdown voltage of the
collector base junction as VCBO. Now we need to find out for what value of
voltage this quantity will become 1/α. This voltage here is the collector base voltage
of this device in this particular configuration. Please note that VCBO is the breakdown voltage
of the collector base junction when it is isolated which is same as the collector base
breakdown when the emitter is open. That is why VBR is being replaced by VCBO.
But this V is nothing but the voltage across the collector base junction in this particular
circuit configuration when the base is open. So, if you add this collector base voltage
to the very small value of forward bias across the emitter base voltage please note that
this is the polarity when you apply a voltage like this such as the reverse bias and forward
bias which is very small. So really the voltage here is approximately
equal to this total voltage. That is why we can replace this by the VCEO because at breakdown
the voltage is VCEO.
So this is the relation between VCEO and VCBO. When you solve this you will get VCEO/VCBO
= (1 - α)1/m. Now 1 - α is nothing but 1/beta + 1 and
therefore you have VCEO = VCBO/1 + beta whole power 1/m.
This is the relation between the collector to emitter breakdown voltage of a transistor
when the base is open and the collector base breakdown voltage of the same transistor.
What is the factor here? if you have beta which is about 99 so 1 + beta is 100 and if
M is for example 4 then 1/4th root of 100 the √100 is 10 and then √10 so you will
have the VCEO as VCBO/√10. So VCEO can be as low as one third of the
VBCBO as shown here on the slide. Note here that the VCO corresponds to the
condition IB = 0. But as your collector current in the device
increases your breakdown voltage falls because of change in alpha with IC.
And in fact here a slight raise is also shown for very high collector currents which is
again because of fall in alpha. This region was not shown in our family of
current voltage characteristics we showed. With this we have completed the output characteristics
of the device in the emitter configuration. We shall continue with some more aspects of
the DC characteristics in the next class following which we will discuss the small signal characteristics.