The MOS Transistor

We consider an n channel transistor. When a positive potential is applied to the gate with respect to the source, a depletion region is formed under the gate, resulting from holes being pushed away from the silicon-silicon dioxide interface. The depletion region consists of fixed ions that have a negative charge using one dimensional analysis , the charge density of depletion region is given by

=e(-NA)(1)

The electric field resulting from this charge is E(x)=/ dx== (-eNA/si) dx

= (-eNA/si)x+c(2)

The integration constant c is determined by evaluating E(x) at the edges of the depletion region (x=0 at

the Si-SiO2 interface; X=Xd at the boundary of the depletion region in the bulk)

E(0)=E0=(-eNA/si).0+c=cE(Xd)= 0 =(-eNA/si).Xd+c=c

E(x)=eNA. (Xd-X)/si ...................... (3)

Applying the relationship between potential and electric field yields

F Xd

d = -E(x)dx = -eNA (Xd-x)/si dx s 0

Integrating both sides with appropriate limit of integration gives

d = -eNA (Xd-x)/si dx= -eNAXd

2/2si = F - S

or S F = eNAXd2/2si ...................... (4)

Here F is the equilibrium electrostatic potential (Fermi potential) in the semiconductor, S is the surface potential of the semiconductor and Xd is thickness of depletion region.For p -type semiconductor, F is given by

F = +VT ln(ni/NA)= - VT ln(NA/ni) where VT = KT/q

and for n type semiconductor F is given by F = VT ln(ND/ni)

Equation(4) can be solved for Xd assuming | S - F|>=0 to get

Xd=[2si| S F| / eNA]1/2 ...................... (5)The immobile charge due to acceptor ion that have been stripped of their mobile holes is given by

Q = -eNAXd

Q = -eNA[2si| S F|/eNA]1/2

= -(2eNAsi| S F|)1/2 ...................... (6)

When the gate voltage reaches a value called the threshold voltage, designated as VTh,the substrate underneath the gate becomes inverted, i.e. it changes from p type to an n type semi conductor. Consequently an n type channel exists between the source and the drain that allows carriers to flow.In order to achieve this inversion,the surface potential must increase from its original negative value(S= F) to zero (S=0) and then to a positive value(S= - F) . The value of the gate-source voltage necessary to cause this change in surface potential is defined as the threshold voltage,VT.This condition is known as strong inversion.

With substrate at ground potential the charge stored in the depletion region between the channel under the gate and the substrate is given by equation(6) where S has to be replaced by F to account for the fact that VGS=VT.This charge Qb0 is written as

Qb0 = (2eNA Si| 2F |)1/2 ...................... (7)If a reverse bias voltage is applied across the pn junction, equation(7) becomes

Qb = (2eNA Si| 2F + VSB|)1/2 ...................... (8)

An expression for the threshold voltage can be developed by breaking it down into several components.First the term MS(the metal to- silicon work function) must be included to represent the difference in the work function beteen the gate material and bulk silicon in the channel region.The term MS is given by

MS = F(substrate) - F(gate)

Second, a gate voltage of [ 2 F Qb/Cox] is required to change the surface potential and offset . The depletion region charge Qb.

Lastly these is always an undesired positive charge Qss present in the interface between the oxide and the bulk silicon .This charge is due to impurities and imperfection at the interface and must be compensated by a gate voltage of Qss/Cox. Thus the threshold voltage for the mos transistor can be expressed as

VT = MS +( 2F Qb/Cox) + (- QSS/Cox )= MS 2F -Qb0/Cox QSS/Cox - (Qb - Qb0 )/Cox

The threshold voltage can be rewritten as

VT = VTO + ((|-2F + VSB|)1/2 - (|-2F|)1/2) ...................... (9)

where VTO = MS 2 F - Qb0/Cox - QSS/Cox ...................... (10)and the body factor,body-effect coefficient, or bulk threshold parameter is defined as

= (2eSiNA)1/2 / Cox ...................... (11)

The signs for the various quantities in the Threshold Voltage function:

Parameter N-channel P-channel(p-type substrate) (n type substrate)

MS- -Metal- -n+ poly+ +p+ poly

F - +Qb0,Qb - +

QSS + +

VSB + -

+ -

Problem:

Find the threshold voltage and body factor for an n-channel transistor with an n+ poly silicon gate if tox

= 200,NA= 3*10-16

cm-3

,gate doping ,ND = 4*10-19

cm-3

and if the no. of positively charged ions at

the oxide silicon interface open area is 1010 cm-2.