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Today we will discuss about the MOSFET circuit under DC. When we apply a DC voltage between
the gate and source then what will happen is the currents in the different branches
can be found out. That DC operation we will take up today. In the last classes we have
discussed about the different types of MOSFET. We have discussed about the enhancement type,
depletion type and in the enhancement and depletion type MOSFETs there are two types
again that is n-channel and p-channel. In our discussion today we will take examples
of both these MOSFET’s and try to solve examples dealing with the DC operation. For
example let us take one circuit. Here we are having a circuit. By the symbol of the MOSFET
we know what type of a MOSFET it is.
This MOSFET is n-channel depletion type MOSFET. The voltage at the drain is denoted as VDS
because S is the source which is grounded. So VDS is also represented by VD because source
is grounded. A 20 volt supply is there and the resistance 1.5 kilo ohm is connected between
the voltage and the drain terminal and here the gate terminal is shorted to the source.
We are asked to find out the VDS. What will be the voltage between the drain and source
that is to be found out? Given that IDSS is equal to 10 milliampere and we know IDSS,
the drain current or drain to source current when the voltage between gate and source is
zero. This is a value which is specified in the datasheet for a particular depletion type
of MOSFET. For this example we are having IDSS value as 10 milliampere. Also specified
is that the MOSFET device has a threshold voltage -4 volt. We have to find out VDS.
In order to find out the drain to source voltage we have to solve this circuit.
In this circuit which is again drawn we are denoting the drain current by ID that is the
drain to source current and one thing to be noted is that the gate terminal is connected
to source as a short circuit and the source is grounded. We now apply the Kirchoff’s
voltage law to this circuit because we know that this circuit can be solved using the
typical laws of electrical engineering like Kirchoff’s voltage law, Kirchoff’s current
law etc. So we will apply Kirchoff’s voltage law in the loop. From the end of the 20 volt
supply voltage if we travel downwards to ground it will complete one loop starting from here
to ground. Traveling along this path if we apply the Kirchoff’s voltage law what will
it be? It will be equal to V is 20 volt minus ID into 1.5 kilo ohm –minus VDS is equal
to zero.
You have to look into this circuit. Here we are having VGS equal to zero because the circuit
itself shows that the voltage between the gate to source is zero. VGS is zero and that
is why the drain to source current is the specified IDSS. For this particular circuit
as we are having the gate to source voltage zero, the current between the drain and the
source is the specified IDSS which is given as 10 milliampere. We can now find out VDS,
the drain to source voltage or simply we can write it as VD. VD or VDS, this is the same
thing because the source is grounded. The voltage between drain and source is nothing
but the voltage between the drain and the ground. It is very simple to find out this
voltage VDS or VD which is nothing but 20 minus this drop and this drop is IDSS into
1.5 k. That if we calculate replacing the value of the current which is 10 milliampere,
for this particular MOSFET we are given that IDSS is equal to 10 milliampere, this calculation
yields a voltage of 5 volt; so 20-15 is 5 volt. That means the voltage at the drain
end with respect to the source or ground is 5 volt. This circuit or example is employing
the depletion type MOSFET having n-channel that is we are using an n-channel depletion
type MOSFET which is very clear from this symbol of the MOSFET. Let us try another example
which is given in the circuit shown in the figure. Here you have to design the circuit
as shown in figure 2 so that the transistor operates at ID drain current is equal to 0.4
milliampere and VD is equal to 1 volt.
The MOSFET has a threshold voltage VT is equal to 2 volt and it is in saturation. Given the
value of k is equal to 0.4 into 10 to the power -3 ampere per volt square. Here this
point is important that the MOSFET is in saturation. As it is in saturation we have to consider
that the voltage VD must be greater than VGS-VT and the threshold voltage is given as 2 volt.
So we will find out VGS corresponding to that condition of saturation.
Writing down the relation of the ID current with respect to VGS and VT which is this non-linear
expression we will find out VGS and this quadratic equation gives two solutions. But out of these
two solutions, only one solution will be feasible which will satisfy the conditions for saturation
that VD must be greater than VGS-VT. We will now solve this quadratic equation and find
out that value of VGS. This and this cancel, 0.4 and 0.4; we get an equation 1 equal to
VGS minus 2 whole square. Simplifying this we get VGS square minus 4 VGS plus 4 equal
to 1. That gives us VGS square minus 4 VGS plus 3 equal to zero. We need to solve this
equation for finding out the gate to source voltage for this circuit to be under saturation.
For the ON condition we have to find out what is the VGS? This equation is a quadratic equation
as we see. It will have two values of VGS. If we solve this equation it will be VGS equal
to 4 plus minus under root 16 minus 12 divided by 2. What will it be? It is 4 plus minus
2 by 2. Basically we get 6 by 2 and 2 by 2. That means we have two values 3 volt and 1
volt. But if we have VGS is equal to 1 volt, then our condition that VGS minus V threshold
must be less than VDS, that condition will not be satisfied because VGS-VT will be then
how much? 1 minus VT, we have been given as 2 volt; so 1-2 means it will be -1 but we
have VDS is equal plus 1 volt, so this is ruled out.
If VGS is equal to 3 volt this condition is met; that is VDS is greater than equal to
VGS-VT that will be met when this is 3 volt. VGS is equal to 3 volt is the critical condition
which will make the circuit to be under saturation. We will take that value of VGS which is equal
to 3 volt. With this VGS equal to 3 volt we will now proceed to find out the resistances.
In this circuit we now solve in the output circuit VDD minus ID RD minus VD is equal
to zero. VD means with respect to ground. VD has been given to us as 1 volt. Putting
down the values of this VDD and ID we can find out what is RD because others are known
except for only RD.
VDD is equal to 5 volt. So 5 minus ID given to us is 0.4 milliampere, keeping that as
0.4 only then the answer will be in kilo ohm, minus VD is equal to 1 volt; the voltage at
the drain is 1 volt. This gives us the solution of RD equal to 10 kilo ohm. If we solve this
equation it will be 10 kilo ohm. We know what is RD? What will be RS? In order to find out
RS we will come through the bottom loop. This is zero minus VGS. This is VGS between this drain
and gate. This is drain, this is gate and this is source. With respect to source terminal
this voltage is VGS. This is ground point. We will go through this loop zero minus VGS
minus ID into RS minus VSS is equal to zero. Applying KVL in the bottom loop we get this
equation.
VGS has been found to be equal to 3 volt and ID is also known. VSS is -5 volt. We will
now solve this equation. Only unknown is RS. RS is equal to 5 kilo ohm. For this circuit
we have found out the two resistances RD and RS. In this way the operation of this MOSFET
device when only DC voltages are there is illustrated in these two examples.
Why do we use a MOSFET device or for what applications we use a MOSFET device? We use
MOSFET for amplification. Earlier when we were discussing BJT, bipolar junction transistor
we have studied about the application of the BJT as amplifier. We have seen how amplification
of the weak signal takes place. Similarly a MOSFET device also is used as an amplifier.
Now we will study about the use of MOSFET as an amplifier. Take for example a simple
circuit to illustrate how a MOSFET device is used for amplification. 1856
Let us now take this circuit having an n-channel enhancement type of MOSFET device. We are
restricting our use to enhancement type MOSFET in this particular discussion but we have
depletion type MOSFET also which is used as amplifier. But as we are going to discuss
about the basic amplification and other related matters, we will take up the example of enhancement
type MOSFET and we will discuss other aspects related to amplification. Instead of discussing
different, different type of MOSFET devices let us concentrate on the application of enhancement
type and see how the amplification and other related things can be understood. It is not
necessary that only enhancement type of MOSFET device be used for amplification but we are
taking this example. The amplification and biasing, etc can be also explained for depletion
type of MOSFET but we will discuss one particular type of MOSFET and focus on it. We are taking
an n-channel enhancement type of MOSFET. What is the amplification? We already know from
our earlier studies that amplification means we are trying to increase or enhance the magnitude
of a weak signal without distorting its shape. If we have a small or weak signal we will
use an amplifier so that at the output we get an amplified form of the input signal.
The shape of the input signal should not vary. We need or we desire faithful amplification
but its magnitude should be enhanced. That is the basic principle for amplification.
As we were discussing in BJT, the concept of amplification is similar. For this MOSFET
device also we will proceed in a similar way. Here we are having this MOSFET which is an
n-channel enhancement type MOSFET. In our study we are taking that as an example. We
are having DC sources. We are also having a small signal so that we finally get an amplified
form of this weak signal. Our aim is to amplify the small signal. This is the gate terminal,
this is source and this is the drain. This amplifier is amplifying a signal between gate
to source which is small vgs. Apart from this signal there are DC sources which are needed
for biasing. We will discuss about the biasing schemes later but in general this figure typically
shows one arrangement for amplifying a small signal. The gate to source voltage is if it
is varying we know that drain to source current will vary. We have earlier studied about the
transfer characteristic or and VI characteristics and we have seen there are three different
regions of operation. One is the cut off region one is the triode amplifier region and the
other is the saturation region. For saturation region the condition to be satisfied is that
the drain to source voltage must be greater than VGS minus V threshold.
In this circuit let us show that drain current is iD. One important thing to note is that
here a combination of both small as well as a capital letter is used just to signify the
instantaneous value of voltage and current we use both the small as well as capital symbol
because we have two situations here; one is the DC and one is the AC. The DC condition
is for biasing and the AC condition is for the signal. If we do not have the signal,
as we have earlier discussed in the examples which we did just now, if we have only DC
quantities or DC voltages and currents if we have only DC situation then we denote that
voltage and current by capital letter say VD, ID, etc and if we purely have AC signal
we denote this signal by small v small d, small i small d, like this. Now we are having
both the DC and AC conditions. What we will have? We will have a superimposition of AC
on DC. In the absence of the AC signal we will have a constant DC voltage say VD or
if we consider the gate to source voltage we have VGS which is a DC signal DC voltage.
If we now apply this AC signal small v small gs, we will have on this DC voltage superimposed
AC voltage like this, small v small gs. At any instant you consider the total voltage
between the gate and source. Suppose at this instance what will be the voltage? It will
be the total voltage which is having the DC part and an AC part. To know that or to express
that instantaneous value of this voltage we use the symbol small v capital GS. Similarly
the current component also if we consider, if we have only the gate to source DC value
of voltage which is capital VGS then corresponding drain current is say capital I capital D,
and when we have the signal applied small v small gs the corresponding current in the
drain is small i small d. Then with both of them being applied VGS plus small vgs as is
shown in this figure we have corresponding drain current as capital I capital D, plus
this small i small d and this quantity is denoted by small i capital D, iD. That is
shown here in this figure.
This small i capital D actually means combination of DC current as well as the AC current, DC
ID and AC iD. When we want to do an amplification that means if we have a small signal here
at the drain end we will have to get amplified signal. The voltage at the drain end with
respect to ground is small v capital D. By that also we are meaning instantaneous voltage
which is a combination of both DC and AC. This circuit is a very simplified version
to actually understand the amplification. In practice you may not get such a simple
circuit but for understanding purposes we are taking it in a simple manner like this.
But in actual while we discuss about biasing, etc we will have a complicated or a different
version than this particular small circuit but for the time being we will consider a
smaller circuit or a simpler circuit for first of all understanding the basics behind the
amplification. At this point we are having a combination of both DC and AC. As it is
not having any capacitor, etc to block the DC components we are having both the DC and
AC being present here.
One important thing to understand or to remember before going to discuss the amplifier in details
is that to operate as an amplifier the MOSFET device must be biased in the saturation region.
This is an important thing to note and always remember that we will operate the MOSFET in
saturation when you want to apply it as an amplifier. It is just like when we were considering
the BJT as an amplifier we operated the transistor in the linear region that is in the active
region. The first and foremost condition which has to be met when a transistor BJT is operated
as an amplifier is the BJT must be operated in active region. Similarly the MOSFET must
be operated or it must be in saturation condition to work as an amplifier. That condition has
to be always satisfied that VDS is greater than VGS-VT so that the MOSFET is under saturation.
Here what we are doing is we are varying the gate to source voltage VGS. This is the controlling
quantity. In BJT we were controlling the collector current by means of the base current. Here
we will control the gate to source voltage and control the drain current iD. This is
the idea which will be applied in amplification.
That is we have to control VGS to control iD. In order to study the amplifier in details
we will consider the DC and AC conditions separately. In BJT also we considered the
DC biasing condition and then when the signal was applied we combined both of them. We want
to analyse the DC condition and then we will go to the small signal model of the MOSFET
device so that we can analyse it for being used as an amplifier to know the parameters.
But before that let us now see what will be the condition when you do not have the signal
and only have the DC condition.
In this particular example first of all we do not have the signal. Without the signal
only purely DC condition is prevailing. Then the circuit will have the DC voltages VGS
and VDD and the small vgs is absent. In the absence of small vgs we are having this circuit.
The current which will flow in the drain to source is capital I capital D now. Without
signal if we find out the drain current capital I capital D we know that this is an enhancement
type of MOSFET and the relationship from the current and voltage is given by ID equal to
k into VGS minus VT whole square. This expression we will have to always keep in mind. What
is this voltage? You note that this is DC voltage. We are having the DC condition only.
This capital V capital D signifies or denotes the DC voltage only. What is VD? We can apply
Kirchoff’s voltage law directly; VDD minus capital I capital D into RD equal to VD.
This is as simple as that because we are not having the signal. We are having only DC condition
so we are analysing the DC circuit and as the amplifier has to be in saturation at any
point during operation it must be satisfying the condition for saturation. So VD must be
greater than the VGS-VT. We are applying a signal also, that has to be kept in mind.
When you apply the signal your VD whatever we will be getting at the drain end that voltage
should be such that at no point the condition for saturation is violated. That is because
we are having a signal swing because of the swing in the input AC voltage. This is vgs.
If we have a DC VGS here because of the signal we will have the drain current small i small
d superimposed on the capital I capital D which is due to VGS only. For only application
of capital V capital G capital S that means the DC VGS, suppose we have a VDS as this
one that is VDS. Capital V capital DS means it is that DC voltage due to this VGS only.
Because of the application of this signal we have a small i small d. Due to this small
i small d we will have a corresponding small v small d. So this will come on it. At no
point of time the VDS or VD must be less than this quantity that has to be kept in mind.
Because we are dealing with amplification you know that the output voltage will be greater.
It will have high magnitude. The biasing should be proper so that under no condition this
is violated.
As we have seen in BJT at no point it should go to cut off or saturation. It should be
always in the active region so that the input is forward biased. The emitter base junction
is forward biased and collector base junction is reverse biased that has to be maintained
always. Similarly in MOSFET this condition has to be satisfied for enhancement type of
MOSFET which we are dealing with so that it is always in the saturation region because
the MOSFET amplifier operates in saturation region. Without the signal we have discussed
the DC conditions. If this signal is applied now what will happen is that we will have
current at the drain which is denoted by small i capital D because small i capital D means
it is DC plus AC.
When we have the signal along with the DC signal VGS equal to capital V capital G capital
S plus small vgs is the total instantaneous gate to source voltage. Due to this gate to
source voltage we have the drain current which is denoted by small i capital D meaning that
we are having both DC and AC. That is given by the current law. What is the current law?
That is ID equal to k VGS minus VT whole square. This is if we have the DC condition. We are
not having DC condition because we are having both the DC and AC. If you want to find the
instantaneous value of the drain current then the total voltage between gate to source we
have to take care of and that voltage is capital VGS plus small vgs. So instead of this VGS
we will have to put capital VGS plus small vgs minus V threshold that is specific for
a particular MOSFET device which is given in the data sheet itself. Now our relation
for the current is this one. Instead of only VGS we are having this voltage.
This Kn is the specification constant for the n-channel enhancement type of MOSFET and
this is basically a constant which is given or which can be computed given the parameters
like length of the channel, width of the channel, etc. In this particular discussion we are
using it as a constant. Instead of going into the computation of Kn let us keep in mind
that we are using a value for the constant which we are assuming as known. Basically
the way to find out the value for K is another topic. We can find this out provided we know
about certain other parameters like length, width of the channel mu n etc. We are not
going into details but let us assume that we already know or it is given to us in the
datasheet.
If we now expand this current expression in the right side it is A+B whole square type
of quantity. If we want to break up, A square plus 2AB plus B square kind of expression
we will get. So that is equal to Kn into let us keep VGS-VT together; this whole thing
under the bracket is VGS plus small vgs minus VT but let us take the quantity VGS–VT together
and small vgs separately. That will be now expanded as a square. So it is A square plus
twice A into B plus B square. If we closely look into this whole expression, first expression
Kn into VGS-VT whole square is nothing but the DC biasing current capital I capital D.
Because if we look into the quantities, this is the DC quantity of the gate to source voltage
and this is the threshold voltage. This whole quantity is nothing but the DC drain current.
Last term in this expression Kn vgs square, is a square term. Small vgs is the signal.
This square means actually it is introducing a non linearity. Because it is a square term
it will be a nonlinear term. So it will be causing distortion. In the whole current expression
if we have a nonlinear term like this one, the last one will cause a distortion in the
current. We have to reduce the effect of this distortion of the nonlinear term and that
we can do if we choose the term carefully. That means in comparison with this term suppose
we have to neglect this Kn VGS square. So then what we will do?
We will make the VGS term or this small signal vgs so small that the square of that small
term which is again further smaller and it should be very, very small as compared to
the other term which is this one. We are now focusing on these two terms because we did
not want the nonlinear distortion which is introduced by this square of the vgs term.
We want to get rid of this. This should be very, very small so that we can neglect it
in comparison with this one. This plus this is almost equal to this so that this term
should be neglected. When can we neglect it? If it is so small in comparison with this
other term, then only we can neglect. We must make or we must choose this vgs sufficiently
small; sufficiently it should be small so that if it is very small its square will be further smaller; it will be smaller term.
Knvgs square is small in comparison with other term twice Kn into VGS minus VT into vgs or
if we simplify little further it boils down to vgs is less than equal to twice VGS-VT
because this and this cancel and this square term and this term cancel. Finally we get
a condition that vgs is very, very less as compared to twice into VGS-VT.
Then if this small signal condition is satisfied then we can neglect the term Knvgs square.
This is an important condition which we will keep in mind that if the signal which we want
to amplify is very, very less as compared to the right hand side expression two times
capital VGS-VT, then the distortion effect can be neglected or it is not causing distortion
to the drain current. Then small i capital D or the instantaneous value of the drain
current is equal to the original expression if we now again write down and ignore the
Knvgs square term then we get this is the expression for the drain current. What is
this current, the drain to source current whose instantaneous value is small i capital
D and that is equal to first term is capital I capital D that is the DC drain current and
the second term is the AC term, small i small d where this small i and small d is representing
this whole term twice Kn into VGS- VT into small vgs.
We do one step further and find out the ratio between small id and small vgs. If we find
this then that is equal to twice Kn VGS minus VT and that signifies an important parameter
which is known as the MOSFET transconductance which is denoted by small g small m, gm.
The MOSFET transconductance small g small m is given by small i small d by small vgs
which is equal to twice Kn into capital VGS-VT where this Kn is the fabrication constant
for the MOSFET device and this is the DC gate to source voltage and the VT is the threshold
voltage which is known. By transconductance we know a very important parameter for the
MOSFET under operation or the MOSFET device which we are using. That transconductance
is equal to this right hand term which actually characterises the MOSFET. This will be very
important when we go to discuss about the amplifier small signal model.
The characteristic between drain current and the gate to source voltage is nothing but
the transfer characteristic. We are familiar with the transfer characteristic which is
a nonlinear curve which can be drawn if we know certain points or by joining the points
we can draw a nonlinear curve. The starting point will be the threshold voltage when the
current is zero and some other points, 3 points or 4 points will be necessary to draw this
transfer characteristic and this is the operating point Q. The operating point will be fixed
by the biasing circuit and that we are going to discuss very soon. If we have the operating
point denoted by Q, then this is having the coordinates VGS and ID; both are capital to
denote the DC condition. For a VGS the corresponding ID is shown here and the slope of this transfer
characteristic at the operating point which is id by small vgs that is actually the transconductance.
From the transfer characteristic, finding the slope at the operating point we can know
the transconductance.
In today’s class we have started discussing about the MOSFET as an amplifier. Basically
the MOSFET device is used for amplification of weak signal by controlling the gate to
source voltage and the MOSFET device must be under saturation to be used as an amplifier.
That is one very important condition and we will have the normal biasing arrangements
for fixing the operating point and then we will apply a small weak signal between the
gate and source to amplify it at the output end or the drain end. Basically we will get
a voltage at the drain which will be in an amplified form but we do not want distortion
of the input signal which is applied at the gate to source terminal. We will later on
discuss about details of amplifier circuits.