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Today we shall just wind up our discussions
on models of diodes and bipolar transistors.
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That is we have seen some models of diodes
and bipolar transistors and the model consists
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of certain parameters in terms of which the
currents are expressed in terms of the terminal
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voltages. So we have to see how to extract
these parameters that is we must know the
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values of these parameters in order to able
to use these models properly. For example
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the basic idea is that you may use these models
to simulate a circuit, to find out how the
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circuit behaves and there these models are
used and the parameter values must be known
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properly in order to get the proper response
of the circuit to understand how the circuit
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behaves. So how to extract these parameters,
so what you have to do is basically you have
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to take a sample device, do some small experiments
and extract these parameters. So we shall
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just talk about these things because that
is important in order to able to use the models
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properly.
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First let us take the diode. In the diode
you have the two terminals the anode and the
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cathode and the current is expressed as this.
So here you have the parameter IS, so if you
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know the value of IS you would know how the
diode current varies with the applied voltage,
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vD is actually the anode to cathode forward
voltage. Now so how do you do it? So basically
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to do that, normally if you plot the current
versus voltage in a linear scale, the characteristics
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is going to be something like this. So the
diode current is initially very small as this
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is an exponential relationship. So this current
is you see is very small but as we increase
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the value of vD it becomes appreciable. Now
in order to extract this, the way to do it,
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you see that if vD is much larger than vt,
vt is only about 26 milli volts as we said
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at room temperature. So if it is say a few
hundred milli volts vD, this term becomes
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much larger than one so we can neglect this
one and so we have an exponential relationship.
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So if we now plot, if one is neglected you
have ln of ID is equal to this, now if you
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plot ln of ID you should get a straight line,
as we said that you should do the experiment
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sufficiently higher high values of vD compared
to vt and then this line if you extend it,
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the value at this point should give you ln
of IS.
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Now I must tell you that in the actual diode
model there is one small correction factor
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which is introduced here which is called the
Etta which is called the ideality factor and
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this value has in the range of 1:2 because
when we develop the model, we did not take
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into account certain non-idealities like recombination
generation in the depletion region certain
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leakage currents. So these give rise to a
factor which is Etta here so if you have Etta
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here this becomes Etta, you have Etta here.
So the slope of this line is one by Etta vE
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and vt you know is a constant depending on
the temperature. So one can easily evaluate
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Etta so this is how one can extract the parameters
for the diode.
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So once you know the value of IS for the particular
diode you can use the model. In fact all the
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circuit simulation packages will have some
default parameter that is if you don’t specify
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the value of IS it will take some value of
IS and do the simulations but finally what
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you get may not be what you expect, when you
do an actual fabrication because that IS value,
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the default parameter value may not match
with the exact values, actual values. So you
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must in order to get the exact values one
must actually do this experiment, find out
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IS and use that value in the circuit simulation.
So this is called extraction of parameters
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so if you have a model there must be a mechanism
to extract these parameters, so that you can
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use them in the circuit simulation. Of course
in addition you also have the diode model
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we have already discussed, we also have the
capacitances
between the anode and the cathode you have….
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Basically draw this model once again, you
have a current force. This
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is the
diode voltage and then you have the two capacitances,
one is the diffusion capacitance and one is
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the junction capacitance. (Student: sir in
Etta how do you fix a number, when you say
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that Etta fixes the non-idealities, how do
you fix particular number to that?) Now when
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you find out the slope of this curve, now
this has a particular slope. Now from the
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slope one can find the value of Etta say suppose
if Etta is equal to one then the slope will
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be 1 by vt. If the slope is say, if Etta is
two it is going to be one by twice vt so this
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is done experimentally, it is very difficult
actually to model the exact value of Etta
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from the model because this is due to some
non-idealities and one has to really know
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what are all these non-idealities in the device
properly to get the exact value of Etta from
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a model. So it is usually extracted once by
doing the experiment. So this value of Etta
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usually lies between one and two.
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So these are the two capacitances in the device
and we have already seen that the diffusion
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capacitance is equal to the transit time del.
Once you substitute this value here you will
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get almost equal to, we have already seen
that and so this diffusion capacitance is
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proportional to the diode current and the
proportionality factor is toud by vt. So this
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is the transit time, it is the diode transit
time. So one has to know this diode transit
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time in order to know the diffusion capacitance
and this diode transit time is usually extracted
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by the transient experiment which we have
already discussed. When you have a transient,
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the time required for the diode to recover
that is once you have the diode in the forward
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biased region and when you reverse bias that
the storage delay time is dependent on this
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diode transit time. So it can be extracted
from that, you also have the junction capacitance
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which is given by 1 - vD.
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So this is the diode voltage, cjo is the junction
capacitance for zero bias when vD is equal
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to v0, cj is equal to cjo. I have just written
this in terms of m when we discussed we have
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put half, that is for a particular case that
is when the p n junction is an abrupt junction
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that is the doping changes from p to n abruptly
but it can have graded junctions when this
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value changes for linearly graded it becomes
one third. Suppose this is half so again we
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have to extract the values of cjo and vBI,
vBI is actually the built in potential of
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the p n junction. So once we know these values,
we will know the value of the junction capacitance
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versus the voltage that is what we want to
know and also m.
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Now for this suppose m is equal to half that
is for an abrupt case then if you plot one
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by cj square and this side if you plot the
reverse voltage that is reverse voltage is
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actually the negative of vD. This vD is always
taken as a forward voltage that is the convention.
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If you reverse bias the diode and you increase
the reverse voltage what happens, this is
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going to be a straight line
and if you extend this straight line, the
point of intersection here that is equal to
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cjo that is equal to zero bias junction capacitance
and if you extend this point further so that
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it meets this axis what happens, this is going
to give you vBI because when vD is equal to
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vBI, cj goes to is infinitely large so 1 by
cj is zero.
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So you can extract both vBI and cjo from this
mechanism. The problem is you actually do
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not know if you are just given a diode whether
it is going to behave in a, m is going to
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be half or one third. So maybe you do this
experiment, you may find that this is not
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a straight line then you have a problem in
your hands, it may be one third. So what you
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have to do in that case? You have to try out
different things if it is one third then you
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have to take one by cj cube which will be
a straight line and then you have to do this
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experiment and you get the values. So this
is how you extract the parameter, so once
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you know these parameters these diode parameters
for example in this case you have IS and here
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you have the transit time cjo, vBI, m, you
have properly defined the different model
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parameters. Once you know the different model
parameters you can use it properly and get
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the proper device, you can simulate the device
in a proper fashion.
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You can use this now to simulate a bigger
circuit. Otherwise you have to depend on the
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default parameters in any simulator which
may not be the exact values for your device
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and so you may not get the proper results
for the circuit. So extraction of parameters
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is very important to know how to extract the
parameters. Similarly we can now look at the
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transistor model, we have already seen that
the transistor model, the simple Ebers Moll
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model of a transistor consists of three current
sources.
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So this is the collector terminal, this is
the base terminal and this is the emitter
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terminal. This current source value is given
by IS exponential, this current source is
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given by IS by betaR exponential vBC by vt
-1 and this current source is given by IS
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by betaF exponential. So you have these three
current sources and from that knowing the
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two terminal voltages, if you know the terminal
voltages vBE and vBC you can evaluate the
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collector current which is the difference
of this current source and this one, emitter
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current sum of these two and base current
which is the sum of these two currents. So
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again you have to know certain parameters.
What are these parameters? IS betaF and betaR.
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So how do you extract these parameters? You
do a similar experiment as you have done in
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the case of a diode to extract IS. You see
that what you have to do is you make for the
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transistor because vBC is equal to zero. The
base collector voltage is made equal to zero
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and then you plot IC and IB versus vBE. Now
what is IC going to be equal to? This current
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source is zero, vBC is zero so this is zero,
so there also exponential vBE becomes IS,
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IC is equal to I¬S exponential vBE by vt
-1. So again now just like in a diode if vBE
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is slightly greater than vt, few times 4 or
5 times greater than vt then you can neglect
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the one and if you plot on a log scale so
basically you have to plot on a log scale.
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This is going to be a straight line, IC is
going to be a straight line. What about IB?
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IB is also going to be a straight line.
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It is IC by beta exponential vBE by vt -1.
So you get two straight line so this is, if
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this is ln IC this is say ln IB this when
you extend this is going to be ln of IS and
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so this one is going to be ln of IS by betaF.
So you have two straight lines and the ratio
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of these two currents is going to be equal
to betaF. So you see when vBC is equal to
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zero the base current and the collector current
are related or the ratio is betaF. So the
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collector current is betaF times the base
current so that is the forward beta of the
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transistor. So this is how you can find out
IS and betaF. How do you find out betaR? You
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have to do exactly a similar experiment but
this time you have to make vBE is equal to
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zero and plot IE and IB versus vBE and again
you will get two straight lines and the ratio
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is going to be betaR. So you can evaluate
again IS from that also, as well as betaR.
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So that is how you can extract these parameters.
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So an interesting thing is that you see for
yourself here that when vBC is zero the collector
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current is related to the base current betaF
times the base current. Also if vBC is negative
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then also you can neglect those terms exponential
vBC by vt terms with respect to exponential
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vBE by vt terms and so when vBC is becoming
negative it actually means that vCE is increased.
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So it doesn’t really have any effect on
the IC by IBE ratio, it still remains more
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or less equal to betaF. On the other hand
if vBC is made positive which means that vCE
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is reducing then collector current is going
to decrease. At the same time base current
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goes on increasing because the current is
sum of these two. So keeping vBE constant
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if you increase vBC, make it positive you
are going to reduce vCE with the result that
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the beta of the transistor, I mean you should
not call it the beta of the transistor the
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collector current to the base current ratio
reduces. So you can say that the effective
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forward gain of the transistor has reduced.
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Some other small effects which actually not
taken into account for that some base width
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modulation also takes place which is going
to if vBC changes but that is not really taken
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into account in this. So one has to actually
make some corrections to this model, I think
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what you are talking of is the early effect.
So this is the basic model which is used to
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understand many effects but if you really
want to do exact analysis some other corrections
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taken into account other effects like the
early effect has to be done because what she
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is referring to is if you reverse bias the
base collector junction the base collector
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depletion width is going to change which changes
the effective base width.
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So all this betaf is dependent on the actual
base width so that is going to change so that
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is not a constant actually but dependent on
vBC also to some extent. So this model is
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just the basic model and so we may have to
do some corrections for other effects. Now
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in this model of course the device also have
the capacitances which is just like a p n
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junction diode. So between each of these junctions
you have the junction capacitance as well
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as the diffusion capacitance. So you will
have if I draw it here, the two capacitances
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here and two capacitances. So this is cjc
I can call it, this is cDC, cJE, cDE. So you
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have
the two capacitances again, these are the
junction capacitances and they have a similar
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relation as in the case of a p n junction
diode and they can be extracted in a similar
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fashion by reverse biasing the particular
junction and evaluating the capacitances for
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the diffusion capacitances, they are again
related to the transit times. So the device
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actually in a bipolar junction transistor
you have two transit times which are defined,
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one is called the forward transit time and
the other is called the reverse transit time.
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What is the transit time? The transit time
as we had defined it is the ratio of the charge
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stored to the current. In this case, the forward
transit time is given by so when we define
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all the forward parameters are defined for
vBC equal to zero. So when you define touF
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which is the forward transit time it is for
vBC equal to zero, so you don’t have the
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reverse the charges due to the base collector
junction. So that is given by Q by the collector
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current, the stored charge so the forward
transit time is the stored charge by the collector
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current whereas the reverse transit time,
so this is when vBC is equal to zero. For
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the reverse transit time it is going to be
the stored charge versus IE with vBE is equal
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to zero and again the diffusion capacitance
for this case cDE will be given in terms of
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the forward transit time touF times just as
in the case of a diode del ICc del vBE which
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is almost equal to IC by vt.
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So this is of course when vBC is equal to
zero and for this case cDC will be equal to
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touR. So this is the reverse transit time,
so you have the two transit times, each junction
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as associated with two capacitances one is
the diffusion capacitance which is due to
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the stored charge in the bulk region and the
junction capacitance which is due to the charge
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stored in the depletion region. So this gives
the complete model of the transistor. So these
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again can be extracted so the touF and touR
are again can be extracted from the transient
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analysis. We shall take up the bjt transient
analysis later on and I just wanted to tell
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you about how the model parameters are extracted.
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So once when you discuss a model, it is very
important to know because the model is defined
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in terms of certain parameters and it is important
to know how to extract these parameters so
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that you can use that model in a proper fashion
and you are not dependent on the default parameters
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provided in any simulator because they may
be wide of the mark because they may be just
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some representative values and not the exact
values for the given transistor. So going
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back to the bipolar transistor as such, we
have seen that we look at the current voltage
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relationship where the output characteristics
of the bipolar transistor, we have seen that
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if we plot IC versus vCE, now when vBC is
equal to zero for example we have seen that
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IC and IB follow a definite relation. That
is IC is beta times IB and if you go on increasing
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the reverse bias on the base collector region
that is effectively increasing vCE then the
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relationship doesn’t change much, it remains
is the same. So for different values of IB
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you will get IC’s they are separated almost,
if you change IB by given increments the IC
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is like this so these are for constant IB
but when you go to lower values of vCE which
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means that basically you are tending to make
the base collector junction go to forward
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bias.
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What happens to IC? It reduces, so for the
same IB effectively the ratio between IB and
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IC is going to be reduced. So
you have the characteristics something like
this, so this is the output characteristics
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of the bipolar transistor IC versus vCE and
of course we shall be using this characteristics
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quite a number of times to analyze given circuit.
So now from this we move on to an actual circuit,
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up to now we have been discussing the devices
we have been looking at the device. So for
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the first time we will now draw a circuit
and see how this behaves. The circuit which
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we shall take up is the simplest circuit one
can think of but it is very important in the
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case of especially for digital circuits. So
you have a voltage here vCC, this is the collector
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resistance you have a base resistance and
you have an input voltage. Suppose you have
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a input voltage here and let this be the output
voltage
so this is the given circuit.
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In order to understand the circuit or to find
out so this let us come to this output characteristics.
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So this gives us the collector current for
different values of base current versus vCE.
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Now if you are increasing the base current,
the collector current increases. So for a
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given base current this is how the collector
current can vary depending on vCE.
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Now if you want to find out the exact operating
point of the device for a given base current,
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so suppose we have given a base current what
is the exact point in which it operates, it
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can be operating in any point here. So you
have to draw another line which is known as
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the load line, the equation for the load line
is so this is vCE plus IC rC is
equal to vCC. This equation must also be obeyed,
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this is vCE plus IC rC is equal to vCC, so
this equation must also be obeyed. So now
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let us draw this line here on this output
characteristics. So this is obviously an equation
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of a straight line, let it be like this. So
this point when IC is equal to zero vCE equal
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to vCC and this point is when vCE is equal
to zero from that equation IC is equal to
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vCC by rC. So both these equations must be
obeyed.
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Now suppose IB is very small, it’s almost
zero what is the output voltage equal to?
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So IB is very small means you have a characteristic
curve like this at this point and this is
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the point of intersection. So it is operating
here so vCE is going to be this and actually
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the vCE is itself the v output, the output
voltage is going to be almost equal to vCC.
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Now if you go on increasing the base current
the point of intersection if IB changes to
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IB1, the point of intersection is here so
vCE is going to be equal to this value here
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and the remaining part you see that sum of
these two is equal to vCC. So the remaining
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part is the drop across the collector resistance.
If IB is increased to IB2 so this is vCE and
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the remaining part is the drop across the
collector resistance.
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Basically the collector current is increasing
so IC rC is increasing, you go on increasing
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to IB3. So this is the operating point. So
what is happening is vCE is reducing that
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is important and more and more voltage is
dropped across the collector resistance. Then
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you increase it to IB4, this is equal to vCE
the output voltage the remaining part is dropped
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across vCC. Now increase the base current
to IB5, what happens? The point of intersection
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remains almost the same.
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So the collector current saturates. Initially
as you are increasing IB, IC was increasing
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but once you reach this point here the collector
current no longer increases with increase
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in base current. So you can say that the collector
current saturates so that is why in a bipolar
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transistor, it is this portion of characteristics
which is called the saturation region. It
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is in terms of a circuit it’s not, that
is once you reach this portion of the characteristics
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with increase in base current collector current
is going to saturate. It is not going to increase
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any further, so that is why this portion of
the characteristics is called the saturation
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region.
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So once you reach here that’s it, the collector
current becomes fixed. The collector emitter
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voltage across the transistor also becomes
fixed. This is the saturation region and in
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this region where the collector current increases
with increase in base current is called the
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active region evaporation and where the base
current is almost zero is called the cut off
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portion. So you have the three regions of
operation, the cut off region where the collector
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current is almost zero then you have the active
region where the collector current increases
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with increase in base current and then you
have the saturation region where the collector
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current is fixed, it does not increase with
increase in base current. So in a digital
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circuit we actually operate the device either
in the cut off or in the saturation region.
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So it is either here or there whereas in an
analog circuit usually the transistor is biased
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in the active region, so that is the difference.
So in this active region the transistor behaves
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as an amplifier whereas here it is not amplifying
but it is fixed at two extremes of operation.
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Now if you come back to this circuit, suppose
we will make some assumptions before we go
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ahead and analyze the circuit. Suppose we
say that it is an exponential relation with
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the base current and the base emitter voltage.
So let us say that for cut off vBE is less
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than 0.6 volts. In the active region so you
are increasing the base current, so the base
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emitter voltage should increase. So let us
say it lies between 0.6 and 0.7 volts and
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say in the saturation region let us say vBE
when it reaches saturation it is equal to
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0.7 volts let us say this is just an assumption.
Just for the sake of doing some analysis.
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So when vin is less than 0.6 volts or just
up to 0.6 volts what is the base current?
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Base current is zero. So what is the collector
current? Zero, so the output voltage is equal
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to vCC. So let me draw the input output characteristics
here. So output voltage is equal to vCC when
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vin is zero. Now when you increase vin up
to say 0.6 volts it is the same, it is going
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to be output voltage is going to be equal
to vCC. Now once you exceed 0.6 volts, what
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happens? This base current keeps on increasing
so as the base current keeps on increasing,
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collector current keeps on increasing, vC
is falling so the output voltage keeps reducing.
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So this characteristic is going to go down
like this. Up to what value? Up to here, so
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this value is called vCE sat that is the vCE
at saturation so it is going to go down up
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to this value.
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So this vCE sat in normal cases is around
0.2 volts so we can write it straight away
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around 0.2 volts. So this goes down to 0.2
volts and after that what happens? If you
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go on increasing the input voltage output
voltage does not change. So this is the characteristics
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of an inverter because when input voltage
is low, output voltage is high. When input
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voltage is high, output voltage is low. So
this is the characteristics of an inverter
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and this is the basic circuit element in any
bipolar digital circuit. So this transistor
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has an inverter, now you see that using this
circuit we have got an inverter characteristics.
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Now this circuit here is also in this circuit
the transistor is also behaving as a switch.
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It is also called the transistor as the switch,
the reason is what happens when you have a
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switch? You have say a power supply, a resistance
and a switch. When the switch is on, the switch
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is closed then the current flowing through
the switch is equal to vCC by rC and the voltage
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drop across the switch is almost zero.
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Here also you see when the transistor goes
to saturation, the drop here is around 0.2
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volts which is very much smaller compared
to vCC and you can neglect it. So as if it
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is a closed switch, the drop across this is
zero the current flowing through it is going
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to be almost equal to vCC by rC but when the
switch is open what happens, you have the
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power supply all the voltage will be dropped
across the switch and current is almost is
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equal to zero. So that happens when the transistor
is in the cut off region. When the transistor
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is cut off this switch is open and what is
happening is, so the switch is open, current
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is zero and the entire voltage power supply
voltage is dropped across this switch.
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So basically you see that this is behaving
as a switch so either when it is in the saturation
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region it behaves as an on switch and when
it is in the cut off region it behaves as
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an off switch. So basically you see in initial
days the digital circuit began with switches
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only, you had electromechanical switches like
relays which were used to do digital functions.
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Now the advantage is you have a bipolar transistor
which is doing the same thing. What is the
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difference between a bipolar transistor and
say for example an electromagnetic switch?
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The advantage is an electromagnetic switch
on and off just may be a few times in a second
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whereas here you can switch it on and off
in millions of time per second with the applied
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voltage is changing. The switch can be turned
on and off millions of times of second. So
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that is the advantage, one derives from a
bipolar transistor. So what we shall do later
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in the course is we shall see how I mean basically
you have the different logic families. This
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is you will find that this circuit is coming
over and over again, you have different peripheral
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circuits which gives rise to different logic
families but this is almost an important component
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of almost all bipolar logic families, the
transistor has the switch. So next class we
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shall actually see the switching performance
of these transistors.
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