Module - 1 Lecture - 3 Diode Equivalent Circuits

In the last class we have seen the PN junction diode and its characteristic curve. That is
with the forward biasing how the current rises almost exponentially and in the reverse voltage
when it is applied current is almost zero except for the reverse saturation current
due to minority carriers. In order to be applied in a circuit we must get the equivalent circuit
for the PN junction diode. That is we have to model the diode electrically. We will be
now considering the circuit model for a PN diode. The diode equivalent circuit is a combination
of elements properly chosen to best represent the characteristics of a device in a particular
operating region.
We have this PN junction which is formed from P type and N type materials on silicon or
germanium semiconductor and this is the diode symbol that we have seen in the last class.
We have to now form an equivalent circuit by a combination of electrical elements. These
components must be properly chosen in order to represent the characteristic which is shown
by the diode and that should be also in the operating region where we are going to apply
the diode.
How to form the equivalent circuit model for a diode? If we consider the characteristic
of a diode the forward region we will be seeing this curve which is a non-linear curve. Up
to the threshold voltage 0.7 volt due to the barrier potential, current is zero and as
soon as the voltage applied across the diode is beyond or greater than 0.7 volt for silicon,
we are taking an example of the silicon, the current rises sharply and this is the characteristic
curve and in the reverse bias this is almost zero.
That is why it will be coinciding with the zero line. We do not consider the reverse
saturation current. Because it is in the micron range, very small it will not be significant.
If we study this characteristic curve of the PN diode we see that this is non-linear and
it is not easy to model a non-linear characteristic. Here we have to do an approximation. That
is we will be considering this whole characteristic as a Piecewise Linear characteristic. Piecewise
Linear characteristic assumption has to be made in order to correctly model the diode.
In this case you see that up to 0.7 volt current is zero and then it is rising. We can assume
that up to this voltage 0.7 it is zero and then this portion where it is a non-linear
characteristic. But when the current rises after sometime you see that it is rising almost
linearly. That is why we can approximate this region as a linear characteristic and the
overall characteristic can be basically divided into 2 regions. One is up to point 0.7 volt
when the current is zero and then a linear characteristic which will be determined by
the slope. The average resistance of this diode is the reciprocal of the slope. The
slope is actually the y by x and we will be finding out the average resistance by taking
its reciprocal because it is voltage by current.
This average resistance of the diode, if we know the two operating point then we can find
this out. We will approximate this as the Piecewise Linear characteristic where this
average resistance of the diode is to be found out. If you know one point which is say 0.8
volt and 10 milliampere in the operating region that is the point can be known from the data
sheet which says about the voltage and current levels for a particular operating point. If
we know one point then it is enough to find out this average resistance because r average
is del VD by del ID that is the rate of change of voltage by rate of change of current between
two operating points.
At this operating point if we know the voltage say 0.8 volt and current is 10 milliampere
which can be known easily if we see the data sheet of particular diode that we are using,
the other point can be taken as this point where the voltage is point 0.7 volt and current
is zero. This is a threshold point after which the current rises. As this region is linear
that we are approximating by Piecewise Linear approximation we can find from these two points
the average resistance. That can be found out if we divide the change of voltage by
the change of current between these two points.
One point is this one, one point is this one here the voltage is 0.8 volt and this threshold
voltage up to which the current is zero is 0.7 volt. The difference is 0.8-0.7 volt divided
by at this point the the current is 10 milliampere and at this point the current is zero. This
difference between the currents at the respective points is 10 milliampere minus zero milliampere
which is nothing but 10 milliampere. We get this ratio between 0.1 volt and the 10 milliampere
that will give us 10 ohm. The average resistance for this particular case we can find out as
10 ohms. This is the average resistance or the order of this resistance of a diode is
around this ohm. That is it will be in the order of ohm. Here we have got 10 ohm. The
information of this average resistance and this cut in voltage or threshold voltage which
is around 0.7 volt for silicon and 0.3 volt for germanium, these two things are enough
information to model the diode. That is the cut in voltage and this average resistance.
Piecewise-linear model of the diode forward characteristic is giving us the equivalent
circuit representation which is shown here in figure b. This figure ‘a’ is the Piecewise
Linear model. This VDO means this is the threshold voltage or cut in voltage because of barrier
potential. Up to this the current is zero and then this is a piecewise linear approximation
that we have done where the slope of this line is 1 by rD. This is the average resistance
of the diode. This straight line is having a slope of 1 by rD that is enough for finding
of the equivalent model. If we draw the equivalent circuit for this Piecewise-Linear model of
the diode we show it by two things one is this cut in voltage and one is this average
resistance and the direction of conduction in the diode is shown by this ideal diode
symbol. This ideal diode symbol is to show the current conduction part and as this barrier
potential or this threshold voltage is opposing this conduction of current in this direction
this opposition is shown by this battery which is VDO. That is this barrier potential or
threshold voltage or cut in voltage the same name denotes this cut in voltage only. Different
names are used for this voltage and this is the average resistance. This whole part will
be the diode representation and in an electrical circuit we denote it by this model.
This is VDO you know that is 0.7 volt for silicon and this average resistance we have
to find out. Given two operating points we can find this out and this ideal diode is
just to symbolize this direction of conduction of this diode. This model we have found will
be representing the characteristic of the diode in the piecewise-linear approximation.
Another approximate model which will be coming from this model again we can further approximatize
this. You can further simplify this model which will be giving rise to Constant Voltage
Drop Model.
This model is Constant Voltage Drop Model because of the fact that in most of the circuit
applications where you generally apply a diode you will find that the resistance of the diode
is quite small as compared to the other elements of the network. We can ignore this resistance
of the diode without violating much and this assumption will be used for further simplification
of this Piecewise-Linear model which gives rise to the constant voltage drop model. That
means we do not have this average resistance of the diode into consideration. We do not
take that into consideration and we ignore this. That means rD is ignored and it is made
zero. In the characteristic if we now see there will be this 0.7 volt for silicon which
is threshold or cut in voltage and the characteristic curve will be rising vertically because this
resistance is zero. It will be rising to infinitely high current and this representation we get
after simplifying the piecewise-linear model with the assumption that rD is zero which
can be ignored or neglected.
Our new characteristic which will be boiling down from this earlier Piecewise- Linear model
is this one. To represent this model we have only to represent it with this cut in voltage
which is 0.7 volt for silicon and this ideal diode will be there to denote the flow of
the current while the diode is conducting. Here this rD will be absent. This is the simplified
or Constant Voltage Drop Model and this model is very much used in practical circuits and
here this resistance is not there. So the model becomes very simpler and in cases where
the calculation of current is not so much important and only the application of diode
is important that is for example in rectified circuits where we are concerned about the
effect of using the diode not the exact calculation of the currents as such then we can use this
model and this is mostly used in those cases and that means we are getting 3 equivalent
circuits. That is one is the ideal assumption where the characteristic will become like
this. That means when the applied voltage is reverse biasing it will be totally zero
current and when it just increases beyond zero, that means we applied forward bias the
current will rise immediately to infinite value and there will be no voltage drop across
the diode.
This is the Ideal Equivalent Circuit for the diode and other model which we have just now
found is Piecewise-Linear model. In this case this characteristic will be obtained up to
cut in voltage VT that is a threshold voltage 0.7 volt for silicon, the current will be
totally zero and then it will be rising and the rise of this current will be determined
by the slope that is the average resistance which is the reciprocal of the slope. This
average resistance will determine the rise of this current. The current will rise as
per the slope given by this average resistance. This is the Piecewise-Linear model.
This is first model and further simplification from this piecewise-linear model will be the
Constant Voltage Drop Model. It is approximate equivalent circuit from the Piecewise-Linear
model. In this Constant Voltage Drop Model we have the cut in voltage 0.7 volt and the
ideal diode will be denoting the conduction part of the diode.
In the Diode Circuit Analysis when you will be applying a diode in an electrical circuit
having other components as well the diode will be in the ON state if the current established
by the applied sources is such that its direction matches to that of the arrow in the diode
symbol and if the voltage applied across the diode is greater than or equal to 0.7 volt
for silicon and 0.3 volt for germanium then the diode will be ON.
In a circuit when we apply the diode we have to see how much voltage is being applied across
the diode and in what direction. Whenever it is beyond 0.7 volt for silicon and 0.3
volt for germanium then the diode will conduct and the direction of current will be given
by the direction of the diode current when it is in the ON state that we can find out.
If we now consider a diode circuit, very simple diode circuit let us take having an applied
voltage VDD that is a battery and the circuit is having a diode and a resistance R. This
resistance is an external resistance.
This series circuit if we consider, the approach which is used for analyzing diode circuit
graphically is called Load Line Analysis. That means we will have to draw a Load Line.
If we consider only resistance as the component which is connected in the diode circuit then
the load will be the only resistance and for simplicity we will be considering a resistive
circuit. This is a dc circuit. Only component which will matter is resistance and if we
consider the series circuit, we will be analyzing this circuit graphically, drawing a load line.
How the load line can be drawn? What is the way of drawing this load line that we will
discuss first.
In the series circuit we apply the Kirchoff’s voltage law which is the law applied to a
series circuit and all of us know what Kirchhoff’s voltage law is. The algebraic summation of
the voltage drop in this loop is equal to zero. That law if we apply in the circuit
we will get VDD minus ID into R. ID is the current in this circuit, diode current and
minus VD the drop across this diode is equal to zero. It is coming from the Kirchoff’s
voltage law applied here in this circuit. Simplifying this we get VDD equal to IDR plus
VD. In this equation if we now set VD is equal to zero, we get the maximum diode current
that can flow in the circuit given by ID equal to VDD by R. That will be one extreme point
that denotes the maximum diode current that can flow in the circuit and if we make ID
equal to zero in equation one, this equation if we make ID equal to zero then we get the
maximum voltage across this diode. That is VD equal to VDD.
These two points will be defining a straight line because in order to draw a straight line
we need only two points and this straight line is called the load line and if we now
draw the characteristic curve and superimpose on it the load line then there will be a point
of intersection between these two graphs and that intersecting point will be the point
of operation of the system. That means we can find the point of intersection and that
point will denote the point of operation for the circuit. Let us take an example and see.
Before that this is the diode characteristic.
This plot will be the diode characteristic, non-linear characteristic. We draw the load
line in the same graph. These two points, one is VDD, one extreme point and the other
is VDD by R, the other extreme point for this load line. Connect these two points and we
get the straight line. This straight line and the characteristic curve are intersecting
at this point which is the Q point. It is called the Q or quiescent point or it is silent
point. It is also called the steel point meaning that only a dc voltage is applied. If a small
signal voltage is applied along with the dc then the Q point will shift up or down but
here will be the only the shift condition and that will be called the silent point or
Q point.
This is the operating point for your circuit. From this Q point, if I project on the x-
and y-axis the voltage and current axis, I get these two points VDQ and IDQ and these
two points will determine exactly the Q point. What is the Q point? That point, is determined
by this voltage VDQ and the current IDQ. I know now the operating point having this voltage
across this diode and the current through this circuit of the diode is IDQ. If we change
this resistance in the circuit then slope of this load line will change. We will be
getting a different load line giving a different Q point. Consider a circuit, a series diode
circuit given in figure 1.
Here the applied voltage VDD is 10 volt, resistance applied externally in the circuit is 1 kilo
ohm and this is the diode which is a silicon diode; it is mentioned here. This diode which
is used in this circuit is having this characteristic given by the figure 1(b). This is the characteristic
curve of the diode. You can see that after 0.5 volt, the current starts to rise and at
0.8 volt it has 10 milliampere current flowing through it. This characteristic curve is given
to you. For this circuit employing this diode characteristic you have to determine the Q
point as well as the voltage across this resistance VR. These two are to be found out. For that
first of all you have to find out the two extreme points of the load line. We have already
seen the extreme current IDmax will be given by VDD by R when you make this voltage drop
across the diode zero and for this particular example it is 10 volt by 1 kilo ohm.
So it is 10 milliampere and VDmax, this voltage can be found out making this ID zero that
is given by the applied voltage VDD so this is 10 volt. These two points are known. We
can now draw the load line. Let us do that. This load line is drawn. One point is 10 volt
VDD max, maximum VD and this point is IDmax that is VDD by R which is 10 milliampere.
Connecting these two points I get a straight line. This is the load line which is drawn
along with this characteristic curve. As you can see here the point of intersection between
this characteristic curve, this load line is Q. If I now project vertically from this
Q point, I get the point of intersection with the VD axis at around 0.8 volt. This whole
plot has to be drawn to scale only then the exact point of intersection can be found out.
Otherwise there will be an error. If I project it on the other axis, current axis here it
intersects that at around 9.25 milliampere. This Q point is defined by 0.8 volt and 9.25
milliampere. This is the operating point. That means this circuit current at that particular
operating point is 9.25 milliampere and the voltage across this diode is 0.8 volt.
What about the voltage VR that has to be found out? The voltage VR as you can see in this
figure, this voltage drop is the current flowing through this resistance multiplied by this
resistance. Current flowing through it is IDQ at that particular operating point and
multiplied by resistance 1 kilo ohm we get this. VR that is equal to 9.25 milliampere
into resistance 1 kilo ohm which is 9.25 volt. This is one way of finding out VR. Another
way also we can find out this VR. This VR is nothing but this voltage minus this drop.
There is another way. Both the ways would actually match. If we correctly measure this
graph, this would be same. VR is equal to VDD minus VDQ. 10 volt minus 8 volt is 9.2
volts. There is a difference. But ideally if we measure the graph correctly both will
be same. This error comes because there is an inaccuracy in reading the graph. When you
measure it to scale it should be same. Both the results should be same ideally. That should
be same in both the cases.
In this whole analysis we have seen that it is very important to draw the characteristic
curve and measure everything accurately. That is tedious or in some cases you do not really
need so much of accurate calculations. We can also use an approximate equivalent model
for a diode. If we consider again this characteristic curve which is non-linear, the exact characteristic
curve is this one. But this can be approximated by the constant voltage model. Constant voltage
model we have seen that 0.7 volt is the cut in voltage. If I use that model instead of
this whole non-linear characteristic curve which is very difficult to accurately measure
and it is not needed also. We draw this approximate equivalent characteristic that is 0.7 volt
and current is rising infinitely at this point. After 0.7 volt is overcome it is rising infinitely.
That is the approximate equivalent model. Then let us see where this point of intersection
is lying?
Here we can see that Q point is defined by 0.7 volt and this current is also found to
be 9.25 milliampere. This is the upper Q point we have got using Approximate Equivalent Model.
Using Approximate Equivalent Model whatever Q point we have got that is not very much
different from using exact equivalent circuit because here earlier we got this voltage as
0.8 volt but here we are getting 0.7 volt but the current is same. In this case we can
see very clearly that using approximate equivalent model will also serve our purpose and where
this exact calculations of current is not required or exact voltage calculation is also
not required we can use this Approximate Equivalent Model and carry on our analysis. You will find that
in most applications this Approximate Equivalent Model or Constant Voltage Model is used.
What will happen if you use ideal model? In ideal diode model current will be raising
at zero point itself and the point of intersection will be at 10 volt. This is quite different
than the results of approximate equivalent circuit because if I consider this line and
find out this intersecting point, the intersecting point is at 10 milliampere. 10 milliampere
intersecting point in a current point we get and voltage is zero whereas in this case we
have seen 0.7 volt and 9.25 voltage. This is very much different from what we will get
in ideal model. So ideal model is practically not very much useable and that also we will
be using only when we have to know the current conduction part. In that case we can use an
ideal model. Effectively we will be using mostly the Approximate Equivalent Model. But
what we are considering is about the dc signal, dc constant voltage. But what will happen
if we have a small ac signal superimposed on the dc voltage.
Then our whole model will have some extra terms not only the resistance because whenever
this change occurs in the voltage then capacitance effects will come. There are two capacitance
effects which will be present in a diode. Those are known as the junction capacitors
and the diffusion capacitors. Junction capacitance basically is related to the space charge region
and the capacitance will be developed in the space charge region and the voltage across
this space charge region or depletion region changes. Let us consider again the PN junction.
Go back to PN junction and what will be the depletion layer width?
We already know in an open circuited PN junction this will be the depletion layer width given
by this expression which we have earlier discussed. If we have forward biased this PN junction,
these majority charge carriers they will be repelled. These holes will be repelled by
this positive terminal and electrons will be repelled by this negative terminal. They
will come towards the junction and in the process they will be neutralizing these ions
as a result of which the depletion layer width will reduce.
That means there will be less uncovering of the fixed charges. The depletion layer width
reduces. The depletion layer width will be given by in the earlier expression of open
circuit instead of Vbi we will be having Vbi minus VF because the potential barrier is
now reduced. The potential hill is reduced by this voltage. The depletion layer width
is reduced and if we consider the reverse biased case the depletion layer width will
be more since more and more charge carriers will be attracted by this positive and negative
terminal of the reverse bias. Electrons will be attracted away from the junction holes
will be attracted by this negative terminal away from the junction. Uncovered charges
of the ions will be more and more. The charges will be more here, uncovered charges and the
depletion layer width also will be affected because the voltage across this will not be
Vbi but Vbi plus VR. This potential hill is now increased.
If you visualize this width there are charges on the right positive there are charges on
the left negative which are fixed charges and the number of fixed charges are changed
by this forward and reverse voltage because forward voltage will be reducing the number
of charges accumulated under depletion layer and the reverse bias will cause more charge.
That means it is like charging and discharging taking place across a capacitor. If we consider
as a parallel plate capacitor, we can visualize that there is a parallel plate capacitor,
the width between these two plates are W depletion and we have this in between material silicon
and so it is like a parallel plate capacitor, C which is given by Es A by Wdep.
Es is the permittivity of silicon, A is the cross sectional area of the junction which
is perpendicular to this plane of the paper and this Wdep is the depletion layer width.
I already know Wdep. I am taking this case specifically for reverse biased PN junction
because this junction capacitance will be dominant for reverse bias PN junction not
for the forward bias PN junction. After replacement the whole expression comes like this and now
in the numerator if we consider this whole term I can replace it by Cj0 means the capacitance
when applied potentially zero, open circuit. The whole term will be Cj0 by root over 1
plus VR by Vbi.
This expression is just found out after simplification of this and replacing the numerator part by
Cj0. We get junction capacitors given by this. Here VR is the reverse voltage. This is the
open circuit built in potential. This is called junction capacitance or transition capacitance
also. This is very prominent in the case of reverse biasing PN diode and there is another
diffusion capacitance which is prominent for forward bias PN junction and this diffusion
capacitance Cd is the rate of change of injected charge with voltage and if we consider the
small signal model of a diode apart from the resistance, average resistance, there will
be two other factors. One is junction capacitance, one is the diffusion capacitance.
The overall circuit representation for small signal model will be this resistance rd, small
rd to denote that it is small signal resistance and Cj Cd these two are the capacitances,
junction capacitance and diffusion capacitance. We have this small signal model. When you
have a small signal along with the dc voltage that I am applying then we have to consider
this model. But for the time being, we will concentrate more on the dc voltage applied
only. Let us do an example.
You have a circuit given by this. There are two resistances, two diodes. You have to find
out this current flowing through this diode D1, current flowing through this diode D2
as well as current flowing through this resistance IR1 and current flowing through this resistance
R2. The voltage applied in the circuit is 20 volt and it is given that the diodes are
made of silicon. This information is important for finding out the cut in voltage and you
are told that you have to assume constant voltage drop model of the diode. Constant
voltage drop model will have this cut in voltage. As it is silicon it will be having 0.7 volt
cut in voltage and there will be no resistance considered in the diode. Since the constant
voltage drop model does not have the resistance in its model that means it is ignored. Let
us solve this example.
We again draw the circuit. Here this voltage is given as 20 volt and this one is diode
D1 and this is D2. R1 is given as 3.3 k ohm. R2 is given as 5.6 kilo ohm. What will be
the currents ID1, ID2, IR1 and IR2? This is the problem. You have to use constant voltage
drop model. First thing is to find out whether the diodes will be conducting or not. As you
can see this 20 volt is applied which will be forward biasing the diodes and so this
diode will be conducting. When the diode conducts 0.7 drop will be there but 19.3 volt will
be available across this. This diode will also be conducting. Both these diodes are
conducting. That means we have now the current flowing in this diode also, in this diode
also. We have to find out what will be current? The drop across this D1 which is VD1 the voltage
drop across D1 diode and it is a constant voltage drop model so 0.7 volt. Similarly
VD2 will be also 0.7 volt. These two diodes are made of silicon. So they have same drop
across them.
If we consider the voltage drop across this resistance R1 the voltage drop between these
two points is 0.7 volt. If we know this volt divided by this resistance 3.3 k, I can find
the current IR1. IR1 is equal to VD2 by R1 and VD2 is 0.7 volt divided by 3.3 k ohm.
This will give you 0.212 milliampere. One current we have found out that is IR1. In
this circuit this part this Kirchhoff’s Voltage Law is applied. If you apply Kirchoff’s
voltage law in this loop that is 20 volt minus VD1 that is the drop across this diode minus
VD2 minus VR2 where VR2 is this drop.
Use KVL in this loop. Then using KVL we get 20 minus VD1 minus VD2 minus VR2 that is equal
to zero; everybody knows what Kirchoff’s voltage law is. Algebraic sum of the voltage
around the loop is zero. We are just simply applying that. You substitute VD1, VD2 that
you know. 20 minus 0.7 minus 0.7 minus VR2 is equal to zero. We can find out what is
VR2? 20-0.7-0.7; that is equal to 20-1.4 and so 18.6 ohms is available across this resistance
R2. If I know this voltage I can find this current IR2. What is this current? VR2 divided
by R2. IR2 is equal to VR2 divided by R2. Just substitute these values. 18.6 volt divided
by R2 is given as 5.6 k. That will be 3.32 milliampere. This is IR2.
What is left to find out? Let’s see. ID1 is to be found out. ID2 also we have to find
out. We have found out IR1 and IR2. This current ID1 is same current IR2 because you have seen
here the current which is leaving this positive terminal of this battery must enter this positive
and negative terminal of this battery. The current which is coming out from this terminal
is ID1. The current which is entering this terminal is IR2. This must be equal. So your
ID1 is nothing but IR2. IR2 you have already found out and that is equal to 3.32 milliampere.
What is left is ID2. This ID2 has to be found out. Come to this point and you apply Kirchoff’s
current law to this point.
Kirchoff’s current law says that the sum of the incoming and outgoing currents at a
node is equal to zero. What are the incoming currents and outgoing currents at this node?
Incoming current is ID2 that we have to find out and another incoming current is IR1. This
is IR1 and outgoing current is IR2. So this algebraic sum must be equal to zero. We follow
this rule of entering current and leaving current are having opposite signs this leaving
current is say a positive IR2. Entering current IR1 and ID2 are negative. Then we get IR2
minus IR1 minus ID2 equal to zero. What is ID2? That is the question. We have to find
out ID2. That is nothing but IR2 minus IR1. IR2 we already know is 3.32 milliampere minus
IR1 is 0.212 milliampere. These two we have found out. This subtraction will lead to 3.108
milliampere. So all the currents are now known. In this example we have used constant voltage
model. Let us do another example. Try to solve this problem.
In this circuit shown in figure 2 determine the current ID, the current which flows through
this diode if the voltage V1 is given as 10 volt, V2 is given as 15 volt and assume the
diode drop to be 0.7 volts. That means here also you have used constant voltage drop model
because you have to neglect forward resistance of the diode. In this circuit we have to find
out ID. We have to first of all examine what is the voltage being applied across the two
points of the diode. This is P this is N. What voltage will be applied across these
two points? Let us try to solve this problem. Again we draw. This is the diode and the resistances
are like this. This is ground, this is ground; this is V2, this is V1. The resistances are
10 k, 10 k. All are 10 kilo ohms and we have to find out this diode. This V1 is equal to
10 volt. V2 is equal to 15 volt. What voltage is being applied across this diode? VD is
the voltage across this diode. If I consider this voltage VD what will the voltage?
At this side you have these voltages 15 volt and this side it is 10 volts. The potential
across these two points is 15 minus 10 which is 5 volts. VD is 5 volts and as it is greater
than 0.7 volts the diode will conduct. This diode is conducting and that has been ascertained
first. We have to find out this current ID. To find out this current we will be taking
help of nodal analyses. Let us consider this node and let us name this node as ‘a’,
this node as ‘b’. The voltage at this node with respect to ground is Va, the voltage
at this node with respect to the ground is Vb and the Kirchoff’s current law will be
applied at these two nodes. Let us consider the node ‘b’ and find out according to
Kirchoff’s current law. To the node ‘b’ apply KCL, Kirchoff’s current law. Kirchoff’s
current law is sum of the incoming and outgoing current is equal to zero. The outgoing current
is what; Vb minus V2 by 10k. Assuming that this outgoing current is in this direction
Vb minus V2 by 10k plus Vb minus zero; this is ground divided by 10k. Another outgoing
current is ID. Summation of all this is equal to zero. All are in the same direction which
is outgoing. Applying Kirchoff’s law at node ‘b’ Vb minus V2 by 10k; I am removing
k. All are in kilo ohms order. So I am omitting this kilo plus Vb by 10 plus ID equal to zero.
This is equation number 1.
Similarly if I apply the Kirchoff’s current law at node ‘a’ what will I get? The sum
of incoming and outgoing current is equal to zero. Incoming current is ID and let us
take the outgoing current as Va by 10k plus Va minus V1. This is the drop across these
two points divided by 10k will give me this current.
In this second case at node a if i apply KCL, I get Va by 10 one outgoing plus Va minus
V1 by 10. If I consider the diode current ID as it is incoming it will minus ID that
is equal to zero. This is equation number 2. We have to solve number 1 and number 2.
I write equation number 1 again. What was the equation? Vb minus V2 by 10 plus Vb by
10 plus ID equal to zero. I can sum them up and eliminate the term ID in solving of this
equation. This is equation number 1. Summing up 1 and 2, I get Vb minus V2 by 10 plus Vb
by 10 plus Va by 10. This and this will cancel plus Va minus V1 by 10 is equal to zero. This
V2 and V1 we know. What is Va; Vb minus this drop. This drop is 0.7. So Vb minus 0.7, the
diode drop that is equal to Va the difference in the potential. Vb minus this drop is equal
to this potential. We can now replace this Va in terms of Vb. Vb minus 0.7. Constant
voltage drop model we are using and it is given that V1 is equal to 10 volt, V2 is equal
to 15 volt. If I substitute in this equation now I can solve it. From equation 3 I will
be having Vb minus V2 is equal to 15.
Vb minus 15 by 10 plus Vb by 10; keep Vb as Vb. I will substitute Va; plus Va is Vb minus
0.7. Here I substitute. I want to keep only one unknown plus Va is Vb minus 0.7. Vb minus
0.7 minus 10; V1 is 10 divided by 10 equal to zero. Here only one unknown is there which
is Vb. We can very easily solve this. It will be further simplified Vb by 10 minus 1.5 plus
Vb by 10 plus Vb by 10 minus 0.07 plus Vb by 10; that is 10.7 by 10 is 1.07; that is
equal to zero. I get 1, 2, 3, 4Vb by 10 and the other side if we do this calculation I
will be getting 2.64. That gives me Vb equal to 2.64 into 10 is 26.4 divided by 4 that
is giving me 6.6 volt. Now I get Vb equal to 6.6 volt.
If I know Vb, I can find out what is ID? What is this ID? Here I can substitute Vb which I have found out.
So what will be ID? Transfer all this left side to the right. It will be V2 minus Vb
by 10 minus Vb by 10. ID is equal to V2 minus Vb by 10 minus Vb by 10. Replace these values
V2 minus Vb by 10. V2 we have already given in the circuit as 15 volt. 15 minus Vb we
have found out to be 6.6 volt; divided by 10 minus again Vb by 10 is 6.6 by 10. That
is 10 in the denominator and in the numerator 15 minus 6.6 minus 6.6. Do this calculation.
It will give you the numerator as 15 minus 13.2 by 10. So you will be getting 1.8 volt
in the numerator. In the denominator remember the 10 is 10 kilo ohms actually. I have omitted
the kilo in the earlier stages; voltage by kilo will give the unit in milliamps. What
will be the result? 0.18 milliamps. ID, direct current which I have got is 0.18 milliampere.
In this example we have taken help of only the primary laws or principles like Kirchoff’s
voltage law, Kirchoff’s current law. Kirchoff’s voltage only we have been using. Whenever
you analyze the circuit all those basic primary laws, fundamental laws of electrical engineering
like Kirchoff’s current law, voltage law, nodal analysis all these will be applied and
the diode should be replaced by its model to fit into the circuit. It is not a physical
diode. We will consider the approximate equivalent model of the circuit that is the electrical
circuit which will be representing the diode.
Today we have learnt about how modeling is done on the physical diode, PN junction diode.
Approximate equivalent model of the diode has been derived which is most commonly constant
voltage model that is being used in electrical analysis of this circuits having diode. When
you have to only consider the diode in application only that is we are not very much concerned
how accurately we find out the current and voltages that is if it is not a prime concern
we will use approximate model and even if approximate then the constant voltage model
is the ideal model of the diode which is very basic and it will be only used for those cases
of circuits when the applied voltage is much, much more or the other components of the circuit
is much higher in values than the resistance of the diode then we will be ignoring the
diode resistance. We will only consider the ideal model of the diode without considering
the cutting voltage as well as the resistance. We will be using it only as a switch like
it is either conducting or non-conducting. In that respect we will use ideal diode but
generally we use constant voltage model. We will use this model, constant voltage model
as well as the ideal model as well also the Piecewise linear model of the diode. What
is the average voltage, average resistance if it is mentioned then we will use the Piecewise
linear model if the compulsion is there that we have to use the average resistance also.
Otherwise we will be sticking to the constant voltage drop model of the diode and next we
will be going into the detailed discussion on application of diode in the circuits like
rectifier circuits, etc.