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This is the 28th lecture of this course and the 3rd lecture on bipolar junction transistors.
Now, in the previous lecture we basically explained what transistor action is.
We said that transistor action is the transfer of current from one junction which is forward
bias to a nearby junction. Now this transfer is complete if this nearby
junction which is called the collector junction is very close to the emitter junction as defined
by this particular relation WB is much less than LB.
WB is the width of the base region here that is the distance between the two depletion
edges. And secondly another condition for the transfer
to be complete is that the doping NE that is the doping in the emitter should be much
greater than the doping in the base, this is the relation.
So, if these two conditions are satisfied then you will have almost complete transfer
of the current from the emitter junction to the collector junction or IC will be very
close to IE. IC is equal to alpha times IE where alpha
tends to 1. Now what is the condition for this transfer
on the bias across the collector junction? What we said is that is this bias is 0 that
is the collector base is sorted or if the collector base junction is reverse bias then
in both these cases the transfer is almost complete. But if the collector junction is
forward bias, that is, if it is like this the p region positive with respect to the
n region then the transfer is seriously affected and the transistor action therefore is seriously
affected. So we must ensure that if you want complete
transfer of the current from the emitter to collector this junction should either be zero
bias or reverse bias. For the zero bias case the relation is IC
is equal to alpha times IE so this is when VBC = 0.
VBC is the voltage between base and collector. And IC is equal to alpha times IE plus a small
current if VBC is greater than 0. That is, the base is positive with respect
to collector or collector is reverse bias. So in both these cases you have the complete
transfer. This is the only small difference in the reverse bias and zero bias case, this
current IC here. Now in this lecture we will see how the transistor
action can be used for the purpose of amplifying small signals.
So the topic in this lecture is small signal amplification.
Now let us look at the biasing arrangement for this purpose.
The biasing arrangement is as follows. Small signal means the emitter base junction
is forward biased and this forward bias is incremented by a small value ΔVEB.
Now as a result of this forward bias change there will be a change in the collector current
ΔIC. Of course there will be changes in emitter
and base currents as well. So, for example, the base current would become IB + ΔIB and
the emitter current would become IE + ΔIE. So they are the changes in response to the
change in the controlling voltage that is the emitter base voltage.
So, the first step of amplification is to relate this change in collector current to
the emitter base voltage. Let us see what is this change ΔIC in the
response to ΔVEB. We can then build up other changes also that
is ΔIB and ΔIE. And then we will see how this change in collector
current can help us to achieve the amplification. Now to begin with we will assume that the
collector base junction is zero bias. So here this is shorted so what we will do
now is, if you start with the equation IC = alpha times IE then we know that ΔIC = alpha
times ΔIE. Now this is the first step.
Now what is ΔIE in response to ΔVEB is what we need to see.
It can be easily shown that the relation between ΔIE and ΔVEB is nothing but the diode relation
that is the relation between the current and the voltage across a diode and this can be
shown easily as follows. Let us draw the minority carrier distributions
in the emitter or base and collector. We are assuming that the base width is very
small as compared to the diffusion length LB. So this distribution is almost a straight
line as we have pointed out in the previous lecture and this distribution is exponential.
So this is emitter this is base and this is collector this is ΔpE and this is ΔnE this
is p-type region, this is n-type region and this is again p-type region.
So this is the excess minority carrier distribution when the emitter base junction is forward
bias and the collector junction is zero bias. We have drawn this in the previous lecture.
Using this distribution let us explain the relation between the emitter current and the
emitter base voltage. Now, ΔpE according with the law of the junction
can be written as ΔpE = pn0 e(VEB/Vt - 1) where the pn0 is the equilibrium concentration
of minority carriers in the base region. We can similarly write ΔnE = np0(e(VEB/Vt
- 1) where this is equilibrium concentration of minority carriers in the emitter region.
Now, we have been using the symbol n suffix p0 also for the electron concentration in
the collector under equilibrium. That is minority carrier concentration in
the collector because collector is also p-type like the emitter
Now we may therefore think that there can be some confusion.
Depending on the context it will be clear whether np0 corresponds to emitter or the
collector. For example here we are talking about ΔnE
that is the excess concentration of electrons in the emitter at the depletion edge. Therefore
obviously np0 should correspond to emitter. So depending on the context it will become
clear whether np0 is related to the emitter or to the collector.
Now, proceeding further how can we write the emitter current in terms of these concentrations?
We again follow the PN junction theory and we can write IE = IEP + IEn so IE is the current
across this particular emitter junction. IEP is due to these holes which are injected from
emitter into base. And IEn is due to these electrons which have been injected into the
emitter from the base region. So we can write IEP which is q × the diffusion
coefficient of the carriers in the base × ΔpE/WB, ΔpE/WB is nothing but DP/Dx in the base that
is the slope of this line so WB is this width. Now here we have not shown the depletion edges
or rather the depletion layer in the emitter and the collector to avoid complication in
the diagram. It is understood that this concentration corresponds
to the concentration at the depletion edge. And similarly these other concentration also
corresponds to concentration at the depletion edge.
So q × DB × ΔpE/WB is the current density because of diffusion where DB is the diffusion
coefficient of holes. Let us explain this nomenclature. We are always
going to consider the diffusion coefficient of minority carriers when we talk about the
currents in a PN junction. So we have only one suffix there which shows
a region in which the minority carriers are being considered.
So, DB would imply the diffusion coefficient of minority carriers in the base.
Now we should multiply this by the area of the emitter to get the current.
So this is IEP in terms of ΔPE that is this current. Similarly we can write IEn as area
of the emitter × q × diffusion coefficient in the emitter into ΔnE/LE where LE is the
diffusion length of electrons in the emitter, this an exponential decay so you take the
diffusion length of the electrons there. Now, we can combine these relations for ΔpE
and ΔnE. And then we can write the expression for IE
as IE = AE × q[DB × pn0/WB + DE × np0/LE] (eVEB/Vt - 1) because this term will be common
for both these terms. Now one can easily recognize that this particular
form is nothing but the reverse saturation current of the emitter junction and we will
therefore represent this as I suffix E0. So this is nothing but the diode law IE as
a function of VEB the exponential and this is a reverse saturation current.
So, as compared to the PN junction theory we have discussed earlier the only difference
is that here for this term instead of the diffusion length we are having the width of
the particular region. And this is because this is like a short region and therefore
the diffusion length in this region is being replaced by WB that is the width of that region
because as we have said in our transistor WB is much less than LB for the device to
act like a good transistor or to have the efficient transistor action. That is why LB
is being replace by WB here and this is the difference.
So now we can write this formula IE = IE0 (eVEB/Vt - 1).
And coming back to our relations here we need to obtain ΔIC which is given by alpha times
ΔIE in response to ΔVEB. So now we can get ΔIE in response to ΔVB using this formula.
So we can write ΔIE = IE0 (eVEB)/Vt/Vt × ΔVEB where in
if VEB/Vt is more than about three times Vt which will practically be the case (eVEB/Vt)
is much greater than 1 and therefore IE0 e(VEB/Vt) is nothing but IE itself.
So we can write this as approximately equal to IE × ΔVEB/Vt.
The increment in the emitter current is proportional to the increment in the emitter base voltage
and the proportionality constant is IE/Vt. So substituting this relation in this formula
here we can write ΔIC is equal to alpha times IE/Vt × ΔVEB where in alpha times is IE nothing but IC.
So this can be further simplified to IC/Vt × ΔVEB.
So ΔIC is nothing but IC/Vt × ΔVEB. So increment in the collector current is proportional
to the increment in the emitter base voltage. This is a consequence of the exponential dependence
of the emitter current on the emitter base voltage. And since most of the emitter current
is transfer to the collector the collector current also depend on the exponentially on
emitter base voltage. Therefore when you differentiate or when you take the increments you end up
getting a linear relation between the increment in the collector current and the increment
in the emitter base voltage. This term IC/Vt has dimensions of 1/resistance
or conductance and therefore we can represent this using a symbol gm and we can write this
as gm ΔVEB This gm is called the transconductance.
So that is the relation between the increment in the collector current and the increment
in the emitter base voltage. Now, let us see how this increment in the
collector current can be used for purposes of amplification.
Suppose we pass this current to a resistor R and we try to find out the voltage change
across this resistor as compared to the change in the emitter base voltage.
Let us try to relate the change in the emitter base voltage to the change in voltage across
this resistor R because of the increment in collector ΔIC.
Now we shall call the increment in the voltage across the resistance as ΔV0 that is we shall
assume that the voltage across the resistance is the output voltage.
The VEB emitter base voltage is the input voltage and voltage across this resistor is
the output voltage. So we will denote this as ΔV0 and we write
ΔV0 = R × ΔIC. Now, expressing ΔIC in terms of ΔVB we can
write the relation ΔV0 by ΔVEB as R × gm where ΔIC is nothing but gm. So ΔV0/ΔVEB
is simply R(gm) where gm is IC/Vt. Now IC corresponds to the voltage VEB.
So when you make an increment in the emitter base voltage what you find is that there is
an increment in the voltage across the resistor which depends on gm and R.
Now, if R × gm is greater than 1 then we find that we have amplification because ΔV0
is more than ΔVEB. The change in the output voltage is more than the change in the input
voltage. Let us put some typical values and see how
much can ΔV0/ΔVEB be in practice. Let us assume these values which are typical
IC = 1 mA supposing we set up 1 mA of current in the transistor.
Let us take room temperature so Vt is .026 volts and let us assume a resistance of 1
kilo ohm. R × gm is given by 1000 ohms × 1 mA upon
0.026 volts. So ohms into ampere by volts this cancels
giving a dimensionless quantity 10 to the power - 3 and this cancels so 1/.026 that
is 1000/26. So R times gm = 1000/26 so this is close to
1000/25 that is about 40 maybe less than 40 and that maybe around 38 or something.
We want just an approximate figure so it is about 40.
So you find that the change in the voltage across a resistance is forty times the change
in the voltage across the emitter base junction. This is what is meant by amplification.
This is what small signal amplification is. So, ΔVEB is the input small signal and ΔV0
is the output small signal. And the ratio between these two small signal
voltages is 40 so you are getting voltage amplification here.
We emphasize that this amplification is for small signal. So we are not taking the ratio
between V0 and VEB, this V0 is the voltage across the resistance are because of this
current IC + ΔIC and VB is the DC voltage so we are not taking the ratios of the DC
current the total current and total voltage. We are taking the ratios of the increment
in the voltage across resistor and the increment in the emitter base voltage. This is therefore
an incremental picture. This is what is meant by small signal amplification.
Now there is one catch here and that is, if you pass this current IC through the resistor
it develops a voltage drop which is like this. Now ΔV0only correspond to IC so strictly
speaking this is V0 + ΔV0. Since we are only concerned about the ΔV0
in response to ΔIC we showed as ΔV0. So this is the voltage drop because of IC
+ ΔIC. Now obviously this voltage drop is going to come here, it appears like this across
the junction. The moment you put a resistance here in this
collector to base lead actually your collector base junction has got forward bias by this
much amount. And if there is a forward bias across the
collector base junction then your transistor action is destroyed. We cannot write IC = alpha
IE where alpha IE is very close to 1. Now, that being the case, actually this formulae
we have derive are not valid for this particular circuit unless we do something to bring this
voltage back to zero bias. If you can bring this voltage across the collector
base junction, that is, back to zero bias then all that we have discussed is valid.
Now this is what is important for using the transistor as a small signal amplifier
One simple way of doing that is you include a battery whose polarity is opposite to that
of the battery here. So that would mean we must include a battery
which is positive on the base side and negative on the collector side.
Now you can see that when you go like this then this voltage will compensate this voltage
and if you choose this battery to be exactly equal to V0 + ΔV0 then this voltage will
return to 0. Let us call this voltage Vcc that is c stands
for collector. So this is actually a collector voltage between collector and base.
Now if Vcc is exactly equal to V0 + ΔV0 then the collector base voltage is zero and then
this formula is valid. Now can you always maintain the Vcc = V0 +
ΔV0? Obviously this is not possible because ΔVEB
means you are changing the emitter base voltage and this is going to change the ΔIC and ΔV0.
So, as you go on changing your emitter base voltage your ΔV0will change so you have to
keep changing your Vcc in conformity with the emitter base voltage.
So if VEB for example is a sinusoid your Vcc will also have to be a sinusoid of appropriate
amplitude because it must compensate this voltage. Now it is obviously not possible,
you cannot have a battery whose voltage is going on changing with time, this is practically
not possible so what you do? Now this is where we make use of another result
that we obtained in the previous lecture. That is, even if the collector base junction
is reverse bias the transistor action is not seriously affected and we can still write
IC is equal to alpha times IE except that there is going to be a small current ICO.
But when we take increments ΔIC in response to ΔIE because of ΔVEB there will not be
any significant change in ICO. This current is a reverse current of the collector base
junction which is reverse bias. And once a reverse bias is more than three
times Vt any change in the character base reverse bias does not change this ICO. And
therefore we can still write ΔIC is equal to alpha times ΔIE when we take increments
because ΔICO is 0 and if that is the case this formula will still hold even if this
is reverse biased. Now, what this means is that we can choose
a Vcc whose value is such that it compensates this voltage whenever it is maximum.
So this Vcc = V0 + ΔV0 the maximum value of this. And when this voltage reduces because
of change in the emitter base voltage then the junction will get reverse bias to a certain
extent. For example, let us take Vcc = V0 + ΔV0.
Supposing this is what we have chosen and when your ΔVEB increment this maximum then
this is the voltage and at that point this is zero bias.
Now suppose this voltage becomes VEB and therefore this current becomes IC and therefore this
becomes V0 then you can see that a bias equal to ΔV0 will appear here but it would be reverse
bias because this voltage is more than this voltage, the bias with this polarity.
And under that condition this equation will be valid but for incremental purposes same
old equation we considered will be valid and therefore our entire analysis will hold.
Therefore the moral of the hole story is that you will have to include a power supply in
this lead whose magnitude is equal to the maximum voltage drop across the resistor then
your emitter base voltage is changing. Now this is the circuit that will behave and
it will give you amplification. Is there any effect of reverse bias across a collector
on the amplification? We said that as far as this equation is concerned the collector
current is equal to alpha times IE + IC0. The IC0 is the only addition as far as DC
is concerned. Now, as far as AC is concerned, that is when you make increments in the currents
is the amplification not affected by the presence of a change in reverse bias across the collector
base junction? So let us examine this issue in little detail.
The effects of reverse bias across the collector base junction. There are two main effects we will show.
The first is the following. Now because of the presence of ICO the difficulty is that
if the temperature goes on changing which is what can happen in practice then this IC
will be changing with temperature even if you keep your IC constant by maintaining VEB
constant. So supposing you consider the situation when
there are no increments so you have set up a VEB here and as a result you have a current
set up in the transistor. Now for some reason the temperature starts
changing, this happens in practice. What are the reasons because of which temperature can
change? One reason is that the transistor is dissipating power and this power is dissipated
as heat. After all you have voltages and currents in the transistor and obviously the voltage
into current that much power is dissipated in the device.
And that power is dissipated as heat therefore the temperature of the transistor can raise.
Similarly ambient temperature can rise. The temperature of the room in which you are setting
up this device as an amplifier can rise. So because of such reasons temperature can
change and in such a case this ICO will change rapidly with temperature. We know that I CO
is the reverse saturation current of a PN junction and as we have explained it doubles
approximately for around every 8 to 10°C rise in temperature for silicon diodes.
So for silicon transistors also similarly ICO will change rapidly with temperature.
Now as a result of this the bias point of the transistor under DC conditions can shift
which is also called as the quiescent point. So it is the shift in quiescent point with
temperature.
I want to emphasize what is meant by the quiescent point.
Quiescent point means the collector current and the collector base voltage when there is no signal when there
is no disturbance. The word quiescent means calm.
This means the condition when there is no ΔVEB.
When there is no signal you have a certain collector current here and then you have a
certain collector base voltage. Both these IC and VCB is referred to as the quiescent
point. so what is happening is that when your temperature
increases the ICO changes rapidly as a result of which IC changes because of which V0 changes
because of which the Vcc - V0 which is the voltage drop here goes on changing.
Now what is the effect of that? The effect is that your swing capacity of the amplifier
is reduced. That is the maximum signal amplitude that can be amplified is reduced.
What is the maximum signal that you can amplify? This depends on the difference between Vcc
and V0 when there is no signal. If the IC is more this voltage drop or V0
is more the difference between this Vcc and V0 is less. Therefore you can only increase
V0 until it becomes equal to Vcc. If V0 becomes more than Vcc then this will
be forward bias and the transistor action will be destroyed.
So the difference between Vcc and V0 that is the difference between the voltage drop
across the resistor and the collector power supply represents the maximum voltage that
you can get at the output as result of the signal amplification.
So signal voltage that you can get at the output depends on the difference between Vcc
and voltage drop across R which depends on IC.
If IC is more and that difference is less then the maximum swing of the amplifier, the
maximum input voltage that it can amplify or the maximum output signal voltage you can
get is reduced. So shift in quiescent point temperature reduces
the swing of the amplifier. This is one effect of the reverse bias across
the collector base junction. This is happening because this IC0 cannot
be control by any means. This is dependent on temperature and this cannot be controlled.
And even if I want to maintain IE constant I cannot control this IC0.
So this lack of control of this current that is coming about because of reverse bias is
the cause of the shifting quiescent point with temperature.
The next important thing that happens is the reduction in amplification because of what
is called base width modulation. Let us understand what is this base width
modulation. The base width modulation is the change in
WB with change in VEB or IC because of change in the depletion width here.
The voltage drop across this junction depends on the difference in the voltage of the power
supply and the voltage drop across R. When you introduce a signal VEB changes, IC
changes, this voltage changes and this voltage here is changing.
So when the amplification is taking place and signal amplification is taking place collector
base voltage is changing with time. Therefore the depletion width is also changing
with time. Now because of a change in depletion width
the WB is also changing. So this variation in WB in the presence of
signal is what is called base width modulation. Let us see how the base width modulation can
reduce the amplification. When the collector current is at the maximum the difference in
these voltages is the minimum here. Therefore the depletion width across the collector junction
is small. When the depletion width is small the base
width is large. Let us show this effect by exploding this particular portion of the diagram.
so we will exaggerate the collector base depletion width for one. This is the collector junction,
this is the emitter junction, this is the collector base depletion width.
This is emitter p + +, n + and p. So this is the collector depletion width for
one reverse bias across the collector base junction.
This is another collector depletion width for another reverse bias across the collector
base junction a higher reverse bias. So, when the collector current is more VCB
is less so let us say this dotted line corresponds to the current IC then the solid line depletion
region will correspond to IC + ΔIC because when the collector current is more this voltage
is more and the difference in this voltage which is this voltage is less so the depletion
width is less. So IC + ΔIC is solid line, this and IC is
the dotted line. Therefore this is the difference in the width
of the base region. So we have not shown the emitter depletion
layer here but you can show that. The change in this is small so we are not showing the
change here.
So this difference from here to here is the base width so this is the ΔWB.
Now this is the base width modulation effect. As a result of this base width modulation
your amplification will reduce. How do we show that? We know that the change
in the base width is going to affect the alpha of the transistor.
So, in this particular equation ΔIC = alpha ΔIE is not exactly correct when the base
width modulation is present because this assumes that alpha is constant when you make the change.
So the correct equation in the presence of base width modulation is ΔIC = alpha ΔIE
+ Δ alpha IE. Now, depending on this Δ alpha your ΔIC
in presence of base width modulation can be more or less.
For example, if this Δ alpha is negative then this term will subtract from this and
therefore in the presence of base width modulation your increment in the collector current will
be less because of increment in the emitter base voltage.
We will exactly show what happens, Δ alpha is negative when you increase the collector
current the alpha decreases and Δ alpha is negative.
How do we show that? For this purpose we must derive an equation for alpha.
This equation can be derived as follows. The equation for alpha is gamma into b where gamma
is injection efficiency and b is base transport factor where gamma is 1/1 + IEn/IEP and B
is 1 - Ir/IEP. The base transport factor b is nothing but
ICP the hole current reaching the collector divided by the hole current injected from
the emitter. And that hole current reaching the collector
is nothing but the current injected from the base minus the recombination taking place
in the base.
So, from there we get this relation. Now you can write equations for these terms using
the equations for minority carrier concentration.
Now IEn/IEP will depend on ΔnE and ΔpE. We can write this as ΔnE × q × DE/LE.
I shift this LE here, IEP is q ΔpE × DB/WB. Of course you also have the emitter areas
coming both in numerator and denominator. Now this q cancels and AE cancels and ΔnE/ΔpE
we can write in terms of the doping levels. So ΔnE/ΔpE will be in the reverse ratio
of the doping in the emitter and base. That is, Δp is inversely proportional to doping
here. So doping in the base will come in the numerator
and doping in the emitter will come in the denominator.
So IEN/IEP is, we can remove this and write it the other way.
So, by transferring that information here this is equal to 1/1 + NB DE WB/NE DB LE.
Similarly we can write Ir/IEP. Ir is the recombination current which is related
to this area so we can write Ir/IEP here. Now this area is half of ΔPE × WB this is
the area under the triangle this difference is WB the base width.
Now this area is multiplied by the charge q multiplied by the area of the emitter AE.
So AE × WB is the volume and the recombination current will depend on this stored charge
divided by the lifetime in this region. Let us call the lifetime in the base region as
tau B so charge by lifetime this is from the law of the junction, it is a recombination
in this region. So divided by IEP we can rewrite that equation
which we wrote for IEP as Q ΔPE DB × WB because the slope of this line is delta P
by WB so this is delta P by WB.
Of course the area of the emitter also comes there. Now we can see that we can cancel this
q, we can cancel this ΔPE, can cancel this AE and then this is equal to WB2/2 DB × tau
B which is nothing but WB2/2LB2 and we got WB2 because of this WB × WB.
So Ir/IEP is WB2/2LB2. So we can write this here.
This is the expression for alpha and it shows how alpha depends on base width WB.
You can clearly see that if WB is much less than LB this quantity is negligible and therefore
it becomes close to 1. Similarly if WB is much less than wb and also
if the base doping is much less than emitter doping this quantity is very small and again
this tends to 1 and that is how alpha tends to 1 for NB much less than NE and WB much
less than WB. Now what is important to see from here is
that, as your WB increases your alpha this term will increase and also this hole term
will become less and therefore your alpha is going to reduce. So as WB increases alpha
reduces. Therefore because of increase in WB the increment
in alpha will be negative. So when you increase ΔIC what you have seen
is that the collector base voltage reduces which means the depletion width reduces and
ΔWB therefore is increasing. So, base width modulation is increased in
the base width when your collector current is increasing.
And increase in the base width causes alpha to fall. Therefore delta alpha in response
to ΔIC will be negative. So we can write this as alpha ΔIE - modulus
of Δ alpha × IE. Now this is a positive quantity subtracting
from this quantity and therefore it clearly shows ΔIC is negative. Therefore the increment
in the collector current is reduced, it is not negative it is still positive because
this is more than this but the increment in the collector current is reduced. Therefore
your amplification is affected it is reduced because amplification depends on ΔIC.
So this explains how base width modulation reduces amplification.
We can write this statement as WB increases alpha falls therefore the base width modulation
we will abbreviate as BWM base width modulation. Therefore BWM reduces amplification.
These are the two effects of the reverse bias. So base width modulation affects amplification
to some extent and also it results in change in the quiescent point width temperature which
affects the swing. But for these two effects your gain of the
transistor is quite large as we have shown that the voltage gain is large and therefore
the device works very well as amplification. Now please note that it is a small signal
amplifier and we are only amplifying small signals.
So you are super imposing a small signal over a DC voltage and you are amplifying that small
signal or disturbance. Now, if there is amplification then it means
there is a power gain. So how can there be gain in power? Your output signal power is
more than input signal power. How can this happen because conservation of
energy should be there. It is to be understood here that there is
an increase in the AC power. If you see the input power of the small signal
here and compare that with the small signal output power the output small signal power
is more than input small signal power. But this output extra small signal power is coming
from the power supply. So what the amplifier is doing is it is converting
DC power into AC power that is why it is called an active device. Transistor is called an
active device because it converts DC power into AC power.
The diagram is something like this. You have a DC power as input you also have a AC power
as input. Now you have AC power output.
Now this small signal AC power Po and if this is Pi Po is greater than Pi. But there is
this DC power that is coming in. So you are supplying energy from the power supply.
And one should not think that you are getting something out of nothing.
You are getting AC power amplification but definitely you are supplying DC power and
only in the presence of DC power the amplification can take place.
So unless you have DC conditions maintained you have VEB, IC and Vcc you cannot have this
small signal amplification. So with this we complete the important application
of the transistor action namely small signal amplification.
We have explained how you can get the voltage gain. But you can also show how you can get
a current gain, and we will see this in the next class.