Handout 4

MOSFET

Sheikh Sharif Iqbal

(Ref: Text book and

KFUPM Online course of EE-203)

(Remember to solve all the related examples,

exercises problems as given in the Syllabus)
Chapter 4 – MOS Field-Effect Transistors (MOSFETs)
Text book: “Microelectronic Circuits by Sedra and Smith
- Metal-Oxide semiconductor Field-Effect Transistors (MOSFETs):
- MOSFET has been extremely popular since the late 1970s. Like transistors,
the current flow between two terminals (Drain to source) in MOSFET are
controlled by the third terminal (gate)
- Why MOS Transistors? - Most digital ICs use
• Takes smaller silicone area on the IC MOS technology.
• Simple to manufacture - Also recently more
• No need for biasing resistors. and more analog
• Used in VLSI (very-large-scale integration) circuits are
implemented in MOS
- Comparison between MOSFET & BJT??
technology for lower
• Can be made smaller /higher integration scale
cost integration with
• Easier to fabricate /lower manufacturing cost
digital circuits in the
• Simpler circuitry for digital logic and memory
same chip (IC)
• Inferior analog circuit performance (lower gain)
4.1:Device Structure of MOSFET: The name of MOS is apparent from figures

L= 0.15 to 10 µm,
W= 0.3 to 500µm,
MOS layer= 0.02 to 0.1 µm.

• Four Terminals are Gate, Drain, Source & Body

Figures from text book

• Unlike BJT, MOSFET is normally constructed as a symmetrical device (DS)

• Minimum achievable value of L in a particular MOS technology is often
referred as the feature size. Intel Pentium-4 uses 0.13 µm technology.
• Lately poly-silicon with high conductivity is used instead of metal to form gates
BASIC OPERATIONAL THEORY OF NMOS: N-channel MOSFET considered
• The current controlled mechanism (for drain current) is based on electric field
established by the voltage ‘VGS’ applied to control terminal (gate).
• Current (iD) is conducted by only one type of carrier “electrons (for NMOS or N-
channel MOSFET) or holes (for PMOS)”. So FET is also called unipolar transistors
Figures from text book

Dep. Reg. not shown

Physical Operation with No vGS: With no bias voltage is applied to gate, two back-to-
back diodes between drain & source prevent the flow of iD as vDS is applied. (RDS ≈ 1012 Ω)
Creating a Channel for iD flow: If ‘S’ & ’D’ are GNDed and a ‘+vGS’ is applied to ‘G’
holes are repelled from the channel region, leaving behind a carrier-depletion region.
Further increasing VGS attracts minority carrier (e-1’s) from P-substrate into the channel
region. When sufficient amount of e-1’s accumulate near the surface of the substrate
under the gate, an N region (N-channel) is created-called as the inversion layer.
Applying a Small vDS or if vDS ≈ (0.1 or 0.2 V) causes a current iD to flow through
the induced N-channel from D to S. The magnitude of iD depends on the density of
electrons in the channel, which in turn depends on vGS. For vGS = Vt (threshold
voltage), the channel is just induced and the conducted current is still negligibly
small. As, vGS > Vt, depth of the channel increases, iD will be proportional to (vGS –
Vt), known as effective voltage. Increasing vGS above Vt enhances the channel,
hence it is called enhancement type MOSFET. Note that iG = 0, due to M.O. layer

-Now since the vDS drops across the channel length, this voltages decreases from vDS
to 0 volt, as we travel along the channel from drain to source. Thus the voltage between
the gate and the points along the channel becomes: vGS-0 at source end and vGS-vDS at
the drain end. This shows that the channel don't have even depth, as the depth depends
on voltage. Now increasing vDS beyond vGS value causes channel to pinchoff

- THUS, (vGS – vDS ) > Vt, or vDS < (vGS - Vt) or vGD > Vt produce continues
channel depth at drain end and results the MOSFET to operate in Triode region.
Otherwise the MOSFET operates in Saturation region with pincoff channel and iD ∞ vDS
Channel length Modulation: If vDS is further increased from pinched-off channel
(vDSsat), the channel length is reduced (by moving from drain end). This phenomena
is known as “channel length modulation” & its affect on iD is incorporated by “λ”
Note: Most of the problems here will assume λ=0
iD-vDS curve for MOSFET for Figures from text book
small vDS the device operates
as vDS controlled resistor
iD-vGS curve for enhancement-type
Book figure 4.4
NMOS transistor in saturation
Book figure 4.12
Physical Operation of Enhancement NMOS: For increasing vDS .
Thus, vDS appears as a voltage drop across the channel. Voltage across the oxide
decreases from vGS at ‘S’ to (vGS - Vt) at ‘D’. The channel depth will be tapered and
become more tapered as vDS is further increased. So for continuous channel in triode, V <(V )
DS GS –Vt

Eventually, when (vGS - vDS)= Vt, the channel will be pinched off (see figures 4.5 & 4.7)
Increasing vDS beyond this value has no effect as iD saturates. Thus, MOSFET is now
operating in the saturation region. Thus, vDSsat= vGS - Vt
MOSFET transconductance
k'n=µnCox is constant depend on
the fabrication process.
µn = channel e-1 mobility
Cox= cap. of unit area of channel
vGD>Vt

Note here
Figure from text book
Book Figure 4.6
4.2: for Enhancement type NMOS 4

t
i

Î Switch
V and VGD >Vt
(V ) Î VDS <VGS -Vt

iD

VDS ≥ VGS -Vt

ÎAmplifier
4
Figures from text book
Example: Use triode expression of iD, Exercise 1:For Enhancement type NMOS with Vt
given in eq 4.5(a), to calculate rDS = 1V and k'n(W/L)= 0.5 mA/V2 , find iD and
whether the circuit below is operating as
a switch or an amplifier.
(a) if VGS = 4v and VDS = 2v
(b) if VGS = 4v and VDS = 6v

if other parameter are given - Remember for symbol and similar circuit
We can solve this rDS for PMOS, read book pg 256 and 258

CMOS: Cross section of a

complementary MOS integrated
circuit. Note that the PMOS
transistor is formed in a separate n-
type region, known as an n well.
Another arrangement is also
possible in which an n-type body is
used and the n device is formed in
a p well.
Figures from text book
Physical Operation of Enhancement PMOS: P-channel MOSFET.
To operate in Triode region:

(a) (b)

(a) Simplified PMOS circuit symbol with connected To operate in Saturation region:
source & body. (b) PMOS circuit. Note that vGS and or vGD>Vt
vDS are negative and iD flows out of drain
Since in PMOS, , So is used
to induces a channel. Thus, Neglecting λ,

Thus to recap PMOS operation, the gate voltage has to be made lower than that of the source by
at least |Vt|. To operate in Triode region, the drain voltage has to exceed the gate voltage by at
least |Vt|, other wise the PMOS operates in Saturation region.

The figure is given in next page: Saturation until VD + |Vt| > VG

See book pg 268 for solution
4.2.5 & 4.2.6: some Practical Considerations of Enhancement MOS

PMOS circuit in previous Exercise 4.6

Body effect can cause considerable degradation in

circuit performance (as shown in chapter 6 of book)
4.11: Depletion Type NMOS or n-channel MOSFET’s:
The depletion type MOSFET has similar structure to
that of enhancement type but with a physically
implanted channel (instead of an induced channel). Thus D D
an n-channel depletion-type MOSFET always has an
G
n-type silicone region connecting the source and drain G

(both +n) at the top of the type substrate. Thus, for any S S
vDS applied between the drain and source, iD flows even if vGS = 0. Thus, the channel
depth and hence its conductivity is controlled by vGS. Applying a ‘+ vGS’ enhances the
channel by attracting more e-1’s. Applying ‘– vGS’ is said to deplete/reduce the channel.

& Vt =- 4 V here

Figures from text book

4.3: MOSFET circuits at DC:
4.20,

as vGD<Vt

4.20,
4.3: MOSFET circuits at DC:
Fig.1

Î Assume Saturated Î

as {vGS=(vG-vS)}>Vt

Exercise-2: Solve the above problem in Fig.1, after replacing NMOS with PMOS (P-
channel MOSFET) with Vt(PMOS)= -1V. Hint: see example 4.5 (NMOS) & 4.6 (PMOS) solutions
CMOS DC circuits:
(see pg 269)

Find

or VDS< (vGS – Vt )

Figures from text book

Biasing
Using
Constant
Current
source
Summary of DC biasing a MOS amplifier in discrete circuits:

See text book

4.6: Small signal models for MOSFET amplifier:
(a) neglecting the dependence of iD on vDS in saturation region of operation (channel-
length modulation effect);
(b) including the effect of channel-length modulation modeled by output resistor (ro)
(c) T-model with output resistance, ro = |VA|/ID = 1/(λ.ID)

Remember that for PMOS:

Vt=“-” and
Figures from text book
Calculating small signal parameter for MOSFET Amplifier:

Remember for PMOS, the calculation of

gm , ro and K'n is calculated using |(Vgs-Vt)|,
|VA| or |λ| and replacing µn with µp ,
respectively. See book page 297
Here for NMOS Î K'n= µnCox
MOSFET As An Amplifier – Small-Signal Analysis:
4.10 4.38

As IG=0, VRG=V10M=0, Thus VG=VD. Since VS=0; VGS=VDS

as VDS>(VGS-Vt)

DC

So, VD = (15 – iD*10K) = 4.4 v ÎVDSÎVGS

• Remember channel length modulation is neglected in this solution.
• Solve exercise 4.24, 4.28 and hand-in next class.
4.7: Common Source (CS) Amplifier: Single stage MOS Analysis
DC

Figures from text book

4.7.4: Common Source (CS) Amplifier with source resistance (Rs):
DC and Rout=RD

As ig=0,

vgs ∝ 1/Rs

Rs introduce ‘-’ feedback,

that ↑ the BW but ↓ the
gain by (1+gmRs) w.r.t CS

r0 is neglected

Figures from text book

4.7: Common Gate (CG) Amplifier: acts as Unity gain current amplifier
DC

Figures from text book

; ;

;
4.7: Common Drain (CD) Amplifier: acts as voltage amplifier Figures from text book

DC

; ;
4.4.4: Operation as a linear Amplifier: see page 279 of book for more explanation
Load line is drawn between the two extreme biasing As vi varies, vGS also
points; (1) when iD=0, v0=VDD, (2) since slope = 1/RD; varies and the Q-
iD=VDD/RD (when max iD is flows) point moves along the
load line. Thus.
wrong Q-point will
cause distortion in id

Figures from text book

Triangular vi is
superimposed on a
DC bias voltage

=10V

1.8k

FIG