Uploaded by nptelhrd on 07.10.2008

Transcript:

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.

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.