Understanding Metal Oxide Semiconductor Capacitance and Voltage Characteristics

[Music] [Music] oh sh

on me [Music] in the last few lectures we have been

discussing about the metal oxide semiconductor Junction or

capacitor we shall continue this discussion in this lecture let me remind you that our goal

of these discussions on this topic is to explain the CV characteristics shown on this

slide we are coming closer to this explanation as a first step we said that we will explain the charge conditions in

the MOs capacitor as a function of the DC applied voltage and from there we will derive the CV

characteristics in the last class what what we explained was the following in a MOS

capacitor when your voltage is greater than zero you have depletion and inversion

conditions and we discussed these conditions in detail and said the following when you increase the voltage

from zero first the depletion region starts expanding the charge in the

semiconductor required to sustain this high voltage uh sustain this applied voltage

comes from the depletion region okay because of ionized impurity atoms now this proceeds

until you reach the threshold Beyond threshold what happens is this depletion region stops expanding and instead you

have mobile electrons being attracted to the surface in large numbers and Beyond threshold it is this mobile electron

concentration or inversion layer charge concentration that goes on increasing and the depletion layer

concentration saturates now this was shown on a graph like

this so here the depletion charge is saturating and the inversion charge Qi is rapidly expanding beyond the

threshold point we also said that if you look at the

voltages scox and SS that is the potential drop across oxide and silicon

as two components of the applied voltage then when you increase the voltage from zero

sco keeps monotonically increasing now C also increases in the beginning but beyond threshold voltage

when the depletion layer almost saturates the Cs also saturates so this is what we said about CS the surface

potential Beyond threshold it saturates to a value 5

t Okay the saturation is approximate of course whereas the sco keeps increasing continuously because the Silicon charge

as a whole which is some of the inversion charge and the depletion charge that keeps increasing though the

depletion charge saturates Beyond threshold now we also derived an expression for VT and as well as this

ft now this expression was VT = to < tk2 Q na Epsilon s 5T by C which is oxide capacitance plus 5 this

is the scox and this is CS at inversion we must remember two numerical values for any calculation related to

Moss capacitors one is the oxide capacitance for .1 Micron oxide thickness which is about 35.5 nanofarad

per cm square and the f t is approximately 7 volts it would be useful to remember

these two values now in this lecture we would like to derive expressions

for QD as a function of voltage and Qi as a function of voltage that is our first exercise in this lecture now

before we take that up let me explain one doubt that students often get that is how is it

that the cus almost saturates when your inversion charge is increasing that is the doubt is when you increase the

voltage Beyond threshold this electron concentration is going on increasing so NS is going on

increasing now if NS is going on increasing how can CS be

constant you know that we had said that Cs and ns are related by the bman relation SS

is equal to VT long NS by n that

is if you look at this diagram SS is given by the ratio of this surface

concentration to n KN and logarithm of this ratio multiplied by the threshold voltage so if n s

increases s should also increase but we are saying C is almost becoming constant how is this possible well what we are

saying is the following that c depends logarithmically on NS so even if NS increases the increase in C will be very

very small because NS is logarithmic function of uh the C is logarithmic function of NS on the other hand if you

look at the inversion charge Qi it increases almost linearly with NS the function is not

exactly linear but approximately we can say if that function of the function SS of NS is logarithmic this is almost

linear Qi is linear function of this NS so when NS increases Qi increases directly with NS

whereas C increases logarithmically with NS that is why even though the inversion charge increases rapidly the surface po

itial increases very slowly and therefore we are treating it as a constant now let us return to our

discussion related to derivation of equations for QD and Qi now the starting point for this is the same

as that we used for deriving the special voltage expression so we can

write V is equal to sco plus SS sco is given

by minus the Silicon charge by the oxide capacitance this is from the

field picture okay please note that Qs by c not with a minus sign because when the applied voltage is positive the

charge in the Silicon is negative charge in the Silicon is of opposite polarity to the voltage applied to the gate that

is why you have minus so this is generally true now depletion as well as inversion

so in um in the expression for Qs you will have QD as well as

Qi look at this graph we said the Q silicon silicon charge is obtained by summing up Qi and QD

so let us understand this with the help of a field versus distance

diagram now so we draw the E versus X okay in the oxide and semiconductor so at

threshold suppose this is the field distribution so this is four in our

nomenclature now Beyond threshold what happens is that you have a step in the electric

field here because you have the inversion charge which is almost like a

sheet and then your oxide electric field will be three times this electric

field so this in inversion this step is

Qi by Epsilon s with a modulus or you can put a negative sign because the charge is

negative this is by gs's law so the inversion charge contributes to a step in the electric field so we can draw the

charge picture here underneath this is the depletion charge under depletion approximation and this is the

inversion charge we are not showing the charge in the metal is always positive whenever

this charge is negative so corresponding to this sheet charge you have this

step this on the other hand this distance is minus QD by Epsilon

s so this total thing is Qs by Epsilon s with a negative sign

silicon charge by Epsilon S as you go on increasing your voltage since depletion layer doesn't expand

Beyond threshold so we are talking of V greater than or equal Thresh should because the depletion WID doesn't expand

this triangle will almost remain the same whereas this step will appear here and this will go on

increasing as your V increases Beyond VT but now what we are interested in is what is the oxide and the Silicon

drop the oxide drop is given by this shaded area so what you find is this

electric field is nothing but this electric field into Epsilon s by Epsilon Ox okay because you are shifting from

Silicon to silicon dioxide so this value is minus Qs into minus Qs by Epsilon s into Epsilon

s by Epsilon Ox which is nothing but Qs by Epsilon ox and this area therefore

is Qs by Epsilon Ox into T because T is this width now this is how we have derived

the thr voltage expression also okay so that is the same thing what we want to show here is that even when

inversion charge is present okay the relation for the oxide potential drop is the same so this is nothing but minus Qs

by C so if you show this potential on potential variation with X diagram

this is how it will be so in depletion suppose this is your potential at

threshold let us draw this to scale for the particular calculation we did we found that this is about 7 potential

drop whereas the remaining one is about 3.2 so which means is almost 4.5 *

this so this is approximately to scale this is your SS and this is

scox this is four that is at threshold this sius will be

5T that is this area is

5T so that is a link between this diagram and this diagram and this diagram okay so from charge you get the

electric field by gs's law and by integrating the electric field or taking the areas in the electric field you get

the potential versus X this scos is this area and that is minus Qs

by C now in inversion what will happen is this part will almost remain the same only your oxide potential drop is going

to change okay this is what will happen now coming back to the equation

here Therefore your conditions are sco is equal to minus Qs by C where Qs is QD +

Qi in depletion Qs is approximately equal to QD whereas in

inversion Qs is equal to the value of QD at inversion

plus Qi now starting from here these two equations

and substituting this into this and then this back into this we can get the charge versus voltage relation first let

us take the depletion region QV relationship so we can write V is equal to minus QD by C plus SS

now we want QD as a function of v so to do that we must Express CS also as a function of

QD now if you recall when we wrote down the thr voltage expression okay we wrote down the

depression charge at threshold as follows QD is equal to minus of square < TK 2 Q Epsilon S NA a into

5T this was at inversion please look into the

derivation okay so basically this derivation is related to

this so we must relate this area QD and this electric field here and then this electric field is

related to the area under this which is the potential CS okay so following the same

approach if you try to derive the QD as a function of C all that we need

to do is replace 5 T by SS and here this is in general valid until

threshold okay which is in the depletion region so once you know this you know that SS

is given by QD s

by 2 Q Epsilon S NA by squaring both

sides so we can substitute this here and you will get an equation between V and QD so this is a quadratic equation so

your relation is V is equal to - QD by C plus plus QD ² by 2 Q Epsilon S

NA this is the equation for depletion region okay now how do you know

that this equation is valid well it depends on what is your value of Cs or V so depletion region means V less than or

equal to VT but greater than or equal to zero or SS is less than or equal to 5T

greater than or equal to zero let us write this here that is a

depletion now similarly let us take the inversion condition what is the charge voltage

relation under inversion now incident this uh quadratic equation represents this particular

curve okay please note how do we get this shape from

here the first term is a linear term okay and then the second term is a quadratic term basically subtracting

from the first term if you take the brackets so in Brackets if you write you can write this as

as follows so now this whole quantity will be on the negative axis

okay so a linear term increasing but then you have a quadratic term subtracting from the linear term so

that is what is seen here so you have a linear term a linear increase but then since the quadratic

term is there and subtracting the magnitude goes on falling and that is how you get this kind of shape

okay now let us return to the inversion charge so in inversion we can write V is equal

to QD inversion plus Qi with a negative sign upon C plus

5 T because inversion is characterized by V greater than or equal to VT or

SS equal to 5T so we must replace S by 5T which we can write as minus

Qi by C plus minus QD inverse inversion by C +

5T now this sum is nothing but threshold voltage VT

therefore you write this as V minus VT is equal to - Qi by C or

transforming this equation you get Qi is equal to minus C into V minus VT so you see Qi is linearly

increasing with voltage but please note this formula is only valid for V greater than

VT so one must not say that if V less than VT Qi will be positive there is no positive Qi for

V less than VT this formula itself is valid only for V greater than and VT okay in other words now you can see that

this particular shape is a straight line starting from VT okay that is a shape here that is the

shape for Qi so what we derive is that the inversion charge increases linearly with

the difference between the applied voltage and the threshold voltage now with this we have completed the QV

characteristics of an ideal MOS capacitor so let us summarize what we have obtained so

far so this is the Q versus V curve the various segments the equations are as follows for this accumulation

segment that is this segment here you can write

QA is equal to minus C into V that is this segment then for the depletion segment

that is between Z and VT you can write V is equal to minus QD by C

+ QD s by 2 Q Epsilon S NA now for this particular depletion charge segment Beyond threshold Beyond

threshold this value is minus QD inverse and what is this so QD is equal

to QD inverse which is nothing but minus of 2 Q Epsilon S NA a into 5T where 5

T is equal to 2 * VT long Na by n i then you have the inversion segment

that is this inversion charge segment and for this segment your equation

is Qi is equal to Min - c into vus VT the threshold voltage expression VT is given

by minus QD inverse by C which is oxide potential drop that

is squ < TK 2 Q Epsilon S NA a into 5 T upon C plus the Silicon drop that is 5T so these are the result for an ideal

MOS capacitor okay now we must see what happens

when the capacitor is not ideal or it is real now as we explained in a real capacitor the extra things that we must

take into account are firstly this metal or polysilicon gate and

semiconductor between them there is a contact potential which we have not taken into

account and also in addition you have a number of charges in oxide as well as at the

interface so the charges we listed where qit that is interface trapped charge q that is oxide trap

charge qf that is a charge at the interface fixed charge at the interface and qm that is a mobile charge

in oxide now we will not discuss the origin of all these

charges that is beyond the scope of our present work so we will concentrate on the

charge qf alone effect of qf now it turns out that

at silicon silicon dioxide interface you have a positive charge which is said to be fixed because the

value of this charge doesn't change with voltage it is always present whenever you create a silicon silicon dioxide

interface so right here at the interface you have this charge okay usually this charge is

shown in a box because because this is a fixed charge it cannot move nor can it vary with

voltage now very briefly the origin of this qf is as follows the chemical composition of the

oxide layer or silicon dioxide layer is S2 okay so this is Si O2 whereas this is Si now you cannot

have an Abrupt transition from SI to sio2 okay in practice so what is going to happen is that if you look at this

interface in detail so let us expand this interface so on this side you have

silicon and on this side you have silicon dioxide so there will be a region in which the composition will

gradually change from SI to S2 so here it is sio2 and this region it is sio X where X varies from

0 to 2 so here it is zero X is zero and here it is two so there is this

layer transition layer now because you do not have two oxide oxygen atoms for one

silicon okay so all the Silicon atoms are not satisfied by bonding and therefore there

is a charge that is created a positive charge that is created okay and this positive charge is the charge that is

called qf so in our non- ideal Mass capacitor which is the next topic that we will

discuss we will try to take into account the effect of qf

and what is called the contact potential or as we show that contact potential is nothing but work function difference

denoted 5ms this is called work function difference

now the question is how do we take both of these into account let us start with the effect of

qf what effect will qf have now you can see that this positive charge qf will tend to find negative

charge here okay in the nearest silicon

substrate so what does this mean this means that if you have a p type substrate here

then even when you don't apply any voltage because of this fixed charge you'll already have a depletion layer

so you have charge in Silicon even when there is no applied voltage recall that in ideal Moss

capacitor for no applied voltage there was no charge in Silicon whereas here you have a negative charge now if the qf

is very large you can even have inversion charge here in addition to depletion

charge so if you want to create the condition of zero charge in

Silicon what you'll have to do is you'll have to apply a voltage to the gate so that these

charges instead of terminating on this side with an electric field like this they

terminate on the gate with an electric field directed in the opposite way so you need electric field like this so

that this positive charges we will not look for negative charges in Silicon in other words you have to put a negative

charge equal in magnitude to the positive charge in the gate so which means you must apply a

negative voltage to the gate to create zero charge in

Silicon now zero charge in Silicon we said corresponds to flatband conditions in semiconductor if

the semiconductor is uniformly doped because then the conduction and balance bands are constant with

distance when there is no charge so therefore what we can say is that to create flatband conditions in

Silicon you will have to apply a voltage in a non-ideal moss capacitor and if you want to take into account only the

effect of qf then you will have to apply a flatband voltage which is negative and what is

the value of this flatband voltage that also we can determine since the charge is

qf okay the voltage to be applied from the field and charge picture is vfb is equal to

minus qf upon c not this is the simple Capa parallel plate capacitor relation that

is the charge is equal to voltage into capacitance okay so one can draw the charge and uh field picture here it

would be like this so you have fixed charge positive on the semiconductor and equal negative

charge on the gate so this is a charge condition both are sheet charges and now you have electric

field directed from this positive charge to negative charge so this is your E versus X and

the value of the electric field is nothing but qf upon Epsilon

Ox you recall that when we wrote the electric field in oxide because of a certain silicon charge Qs we wrote the

oxide field as minus Qs by Epsilon Ox so similarly here it is minus qf by Epsilon ox

qf is this charge area under this so now area under this is the applied

voltage so following this approach you get flatband voltage equal to minus qf by C so the effect of a fixed charge is

to shift the flatband voltage of an ideal Moss capacitor which is

zero to a negative value okay so the flatband voltage is shifted from zero to a negative value in a non- idial MOS

capacitor because of qf now how do you take into account the effect of

FMS to take into account effect of FMS we will have to go to the energy band diagram now this is an important point

that we arrive at here that whenever you want to discuss a

junction between dis similar materials okay then you will have to use the energy band diagram you cannot avoid the

energy band diagram if you recall when we discussed the PN Junction we said that when you have a PN

Junction made in Silicon that is in the same material one can do a great deal of analysis of this Junction and derive the

current voltage characteristics without the aid of energy band diagram although the energy

band diagram approach gives you some additional insights but with without energy band diagram one can go quite far

in understanding a p Junction which is a homo Junction okay so we can generally say to

understand a homo Junction we can avoid the use of energy bind diagram to a great extent but while understanding a

hetero Junction or a junction between dissimilar materials which is what a moss Junction is because material here

is different from the material here and of course the material here is also different okay so in such cases we need

to use the energy bind diagram to derive the effects of the contact potential or work function difference between the one

material and another material so next topic that we need to discuss in the context of non- adial moss capacitors is

the energy band diagram now let us look at the energy band diagram when all the components of the

MOs capacitor are isolated that means you have the metal then you have the

oxide and you have the semiconductor and suppose these are all isolated now uh without any loss of

generality we'll assume that the metal is also sufficiently thick this just to enable us to draw the

energy band diagram that is all now how do we draw the energy band diagram showing the bands in all the

three we must choose a reference we know that the reference is normally the vacuum

level so we first draw e the various steps in the drawing of energy band diagram were given in the

context of P injunctions We are following the same steps now in the metal

you cannot have conduction or bance bands you only have the form level so this is

EFM and this difference is normally referred to as the work function the difference between the vacuum level and

the from Level so this is Q * 5m where 5m is the work function in volts is energy

energy band in the oxide the oxide is an insulator it has a very wide band Gap so energy band diagram is something like

this this is the conduction and Valance band in fact you can show it even larger okay please uh remember that

these three are isolated right now when we draw the band diagram so there is no connection between in them so we are

drawing the structure individually in each of these three so this is

EC and this is EV now we come to the Semiconductor in the semiconductor you

will first locate the EC and then EV the band gap of semiconductor is

obviously much less than that of the oxide this EC is located at a distance given by Q * the electron

affinity then this EV is located below EC this is energy gap now since we are considering P type

silicon your firy level will be below the intrinsic level which is in the middle

here so this is intrinsic level so your for level is somewhere here so this is EF in the semiconductor s stands for

Semiconductor now here we have shown a metal whose work function is more than the electron affinity in the

semiconductor it is not necessary that this should always be the casee you can have the from Level Above This level

here or below even balance band level it depends on the metals in analogy to the work function

in metal you can define a work function in semiconductor

also so that would be this difference Q * 5 S although the the work function in

semiconductor doesn't have um that much real meaning as it has in a metal because you can remove an

electron from the form level out of a metal okay so when you remove a electron out of a metal it is most likely to be

at the form level whereas when you remove an electron in a semiconductor it is unlikely to be at

the form level because fir level is in the Forbidden Gap and you don't have any allowed states in the Forbidden gap for

but still it may help us in writing the equations that is why we are introducing

this work function in semiconductor so fir level in the metal is above the firm level in the

semiconductor as we have drawn here now from this starting point how do we proceed to the situation

when you put all this three components together and and join them by a wire so that means now we draw the band diagram

when it is like this so we are joining them together please note that just if you join them

together like this the band diagram will be no different from this diagram okay that is

because since there is an insulator here it prevents any interaction between the metal and the semiconductor so no

exchange of charges can happen so even when these three components are struck to each other they are still as isolated

as this case when they are separate so when we talk of the energy band diagram of a moss capacitor under

equilibrium it is necessary to note that the metal here is joined to the semiconductor here by a

wire only then the mass is said to be in equilibrium that is all these three form a system because in a system the various

parts should be able to interact with each other otherwise it is not a system at all

so in this case what is the energy band diagram we can start from here and we will note that since fromy level is

higher here than in this semiconductor the electrons will tend to be transferred in this direction so

electrons are transferred from metal to semiconductor now please note that electrons are being transferred from

metal to semiconductor because of from level in metal is higher than that in semiconductor one must not think that

because metals are in higher concentration uh the electrons are in higher concentration in metal than in

semiconductor therefore they are being transferred okay so many times students have this

means conception that because electrons concentration is higher in metal then in semiconductor always you

will have transfer of electrons from metal to semiconductor this is not correct for example here in this case if

the formy level was below this formy level here somewhere here like

this then even if this is a p type semiconductor which has a very small number of electrons when you join the

metal and semiconductor by a wire the electrons would get transferred from semiconductor to

metal okay because the from Level here is higher than here so it is a fir level difference that

decides because fir level represent the average energy of electrons and electrons always move from regions of

higher energy to regions of lower energy just as water goes from a point which is at a greater

height to a point which is at a lower height okay so even if you have a container like

this which is small and you have a big container

here at a lower height than this then when you join these two by a tube the water will flow from this to

this even though the container is small because it is at a greater height than here so it is a height that decides the

transfer of water similarly here it is the fromy level that decides the transfer of electrons direction of

transfer of electrons now how are the electrons transferred well even though the

electrons have been shown to be transferred like this it doesn't mean that they're going through the oxide

because the oxide would not allow any electrons so basically this transfer is taking place Through the Wire here so

this is how the electrons are transferred now after the transfer takes place and equilibrium is

established you know that both these form levels should be aligned because in equilibrium the form level should be a

constant therefore the band diagram would look something like this you should draw a constant from

your level now please note that we are not drawing a form level in

oxide because oxide is an insulator really speaking there is no form level okay having drawn the form level

now you identify a space chart region you are following the same sequence of steps which we followed for p

injunction now we ignore the width of the space charge region in the metal because that is really very

thin now here you see that if electrons are transferred here they would Rec combine with holes on the P

side and in effect what you are having is therefore depletion of holes so you are creating a depletion

layer therefore the extent of this region would be much more than a sheet charge because

you are having a depletion layer so whole concentration is going to decrease as compared to bulk now you

draw the energy band diagram in the bulk in the bulk it is bulk means away from the space charge

region so this is what is the energy B diagram in semiconductor away from the bulk and

now you must complete the picture the whole concentration is reducing as you come to the surface so it is something

like this okay and you can complete the picture for vacuum level this is e not

this is EC this is EV and this is EF so something like this is is your e not

now what about the vacuum level in the metal vacuum level in the metal will be at the same distance from the form level

as it was here now one way of understanding how this bending

occurs in energy band diagram is to build up this diagram as follows so here you find that the electrons will

move into the semiconductor and therefore the fromy level here will come down okay so we assume that the from

level is coming down here and it is trying to getting it is aligning with this

metal actually this is coming down and that is going up but for Simplicity we'll take this as the reference and

then we'll say this is coming down so along with this the E not also will come down so you have e not here somewhere

here so that is how the diagram looks note that we have not drawn the diagram in

the oxide you can draw it in the oxide also okay but that adds to the complexity and

for a first course we need not bother about this diagram in the oxide in fact whatever effects we are going to discuss

those effect depend on the energy band diagram in the oxide they depend only on the diagram in the metal and

semiconductor so this is the picture under equilibrium so you see what has happened is because of the contact

potential even when the applied voltage is zero you have a charge in the Semiconductor in this case it is a

negative charge so this is the effect of the contact potential that even when the applied voltage is zero you have a

charge in the semiconductor because of charge trans between metal and semiconductor originating from

difference in the from a levels in the two regions so that is the effect of non-

adality so which means if you want to create a zero charge condition in the semiconductor now you will have to apply

a voltage to the gate in other words the flatband voltage will not be zero that is the effect of the work function

difference now how much voltage should we apply so that the charge conditions are

zero when the charge when you have zero charge here the energy band diagram would be the same as this that is a flat

band condition okay

so to achieve this flat band condition you have to remove the negative charges which are there in P

type so from here we start and we have to remove the charges which are there in P type so to remove the charges

in negative charges in P type semiconductor you must apply a negative voltage to the metal so which means you

must lift this from level up because increasing energies here these are electronic

energies so when a from Level moves up it means that you are applying a negative

voltage so in order to get flatband condition what you will have to do is you'll have to introduce

a power supply here with a negative polarity because in this particular case you have negative charge

in the semiconductor so once you do that provided the power supplies of a proper magnitude we'll shortly see what the

magnitude is all these negative charges will be repelled so that is

interesting that you have zero charge condition in the metal for a given applied voltage so this is the flat bind

voltage now the difference in the fromy levels here so this is the condition now for V

equal to vfb so the band diagram for V equal to vfb is the same as the band diagram when

the three regions are isolated or not joined by a wire then this difference between the

form levels indicates the amount of Applied voltage because under equilibrium the form levels are Al

this is the flat blind voltage or the voltage you have to apply to the

gate now evidently this voltage is equal to the difference in the work function of the semiconductor and the metal okay

so we can therefore say that the effect of the work function depence FMS is that the flat bind voltage equal

to 5ms should be applied to create flatband conditions so we write vfb effect of 5ms is to create a

flatband voltage equal to 5ms 5ms is nothing but 5 m minus 5S okay so please note I repeat from here

that this voltage is equal to is the difference of these two okay but in magnitude it is negative so this voltage

is obtained as this minus this now the we have already seen that the effect

of qf was to create a flatband voltage equal to minus qf by

C so this was assuming that there was no FMS

so effect of qf when FMS is zero is this and effect of FMS when qf is zero so we considered no qf in this

case is this so whenever you want to understand the effect of two parameters you must understand them one at a time

that is what we are doing here okay so from these two we can easily get the effect of FMS and qf as

follows in presence of 5ms and qf you can simply write the FL bband

voltage is equal to the superposition of these two that is FS minus qf by C so we are

simply summing up these two to get the effect when both FMS and qf are present okay we can summarize all this

discussion on the effect of qf and FMS in a non- adal capacitor as follows this is the picture we had for the ideal

capacitor so zero charge condition in the Silicon at 0 volts and then Threshold at a voltage

given by this formula now the zero charge condition occurs at flatband voltage in other

words if you simply change this V to V minus vfb then this whole picture is valid same thing you do here also the

the charge was Zero for voltage equal to zero but now the charge is zero when voltage is equal

to vfb in other words if you change this to V minus vfb this whole diagram becomes valid and it becomes applicable

to a real capacitor so we can therefore say that the effect

of non idealities of qf and FMS is to Simply change the zero charge condition to Flat bind voltage and the threshold

voltage to the the following formula so the threshold voltage is now given by VT is equal to the threshold

corresponding to Ideal capacitor plus vfp right because V minus vfb is

this okay so therefore V is vfb plus this so VT of a real capacitor is VT of an ideal capacitor plus vfp we shall

continue the discussion in the next class [Music]

and