vcs2 > fej

Figure 11-43 I Family of In versus Vos curves for an n-channel depletion-mode MOSFET.

In the next section we will derive the ideal current-voltage relation for the d-

channel MOSFET. In the nonsaturation region, we will obtain

Id - [2iVGs - VtWds - Vs

and, in the saturation region, we will have

(11.41)

The operation of a p-channel device is the same as that of the n-channel device, except the charge carrier is the hole and the conventional current direction and voltage polarities are reversed.

11.3.3 Current-Voltage Relationship-Mathematical Derivation

In the previous section, we qualitatively discussed the current-voltage characteristics. In this section, we will derive the mathematical relation between the drain current, the gate-to-source voltage, and the drain-to-source voltage. Figure 11.44 shows the geometry of the device that we will use in this derivation.

In this analysis, we will make the following assumptions: j

*

1. The current in the channel is due to drift rather than diffusion.

2. There is no current through the gate oxide.

3. A gradual channel approximation is used in which 3Ey/dy ;:§> ЭЕд /Эл:. i This approximation means that Ej is essentially a constant.

4. Any fixed oxide charge is an equivalent charge density at the oxide-semiconductor interface.

5. The carrier mobility in the channel is constant.

Channel S inversion

Induced D depletion region

Figure 11.44 t Geometry of a MOSFET for Id versus Yds derivation.

We start the analysis with Ohms law, which can be written as

(11-42)

where a is the channel conducrivity and Ед is the electric field along the channel created by the drain-to-source voltage. The channel conductivity is given by - epnniy) where д„ is the electron mobility and n{y) is the electron concentration in the inversion layer.

The total channel current is found by integraring over the cross-secrional area in the y- and -directions. Then

We may write that

Jx dy dz

(11.43)

en(y) dy

(11.44)

where is the inversion layer charge per unit area and is a negative quantity for this case.

Equation (11.43) then becomes

(11.45)

where Wis the channel width, the result of integraring over z.

Two concepts we will use in the current-voltage derivation are charge neutrality and Gausss law. Figure 11.45 shows the charge densities through the device for Yas > Vt- The charges are all given in terms of charge per unit area. Using the

Gate

p-type semiconductor

Figure 11.45 I Charge distribution in the n-channel enhancement mode MOSFET

forVc5 > Ут-

Tixed oxide charge, Q,

Inversion layer charge Q]

Figure 11.46 I Geometry for applying Gausss law.

concept of charge neutrality, we can write

O; + + G;, + QoCmax) = 0 (11..

The inversion layer charge and induced space charge will be negarive for this channel device.

Gausss law can be written as

E,dS=QT

(11.47)1

where the integral is over a closed surface. Qj is the total charge enclosed by the surface, and E is the outward directed normal component of the electric field crossing the surface 5. Gausses law will be applied to the surface defined in Figure 11.46. Since the surface must be enclosed, we must take into account the two end surfaces in the л'-у plane. However, there is no г component of the electric field so these two end surfaces do not contribute to the integral of Equation (11.47). :

Now consider the surfaces labeled 1 and 2 in Figure 11.46. From the gradual channel approximation, we will assume that is essentially a constant along the channel length. This assumption means that into surface 2 is the same as E out of surface 1. Since the integral in Equarion (11.47) involves the outward component of the E-field, the contributions of surfaces I and 2 cancel each other. Surface 3 is in the neutral p-region, so the electric field is zero at this surface.

Surface 4 is the only surface that contributes to Equation (11.47). Taking into account the direction of the electric field in the oxide, Equation (11.47) becomes

wtiere €ox is the permittivity of the oxide. The total charge enclosed is

Qt - (GL + Qn + G5D(ma) (11-49)

Combining Equations (11,48) and (11.49), we have

-oxEox = + a: + Gndax) (1L50)

We now need an expression for Eox- Figure 11.47a shows the oxide and channel. We will assume that the source is at ground potential. The voltage is the potential in the channel at a point x along the channel length. The potential difference across the oxide at x is a function of Vcs У.х^ iind the metal-semiconductor work function difference.

The energy-band diagram through the MOS structure at point x is shown in Figure 11.47b, The Fermi level in the p-type semiconductor is Efp and the Fermi level in the metal is £f . We have

Erp-Ef e{Vcs-V.) (11.51)

Considering the potential barriers, we can write

Vv - (0; + Vox) -

- + ф/р

(11.52)

which can also be written as

Vcs -Vx = Vox + 2фfp + Фms

where ф^ is the metal-semiconductor work function difference, and ф^ the inversion condition.

The electric field in the oxide is

(11.53) 2ф/р for

Eqx -

(11.54)

Л

4>x

Figure 11.47 I (a) Potentials at a point x along the channel, (b) Energy-band diagram through the MOS structure at the point x.

Combining Equations (11.50), (11.53), and (11.54), we find that

CfiY E,

ox *-OX

= G; + e; + Qi-fiCmax) (11.55)

The inversion charge density, Gh- from Equation (11.55) can be substituted ii Equation (11.45) and we obtain

[iVoS - Vv) - Vj]

where E = -dVy/dx and Vj is the threshold voltage defined by Equation (11.27). We can now integrate Equation (11.56) over the length of the channel. We have

fvad Jv. (0)

ox , \iVGS-yT)-VAdV,-

(11.57)1

We are assuming a constant mobility д„. For the n-channel device, the drain current enters the drain terminal and is a constant along the entire channel length. Letfing I о = -I.xy Equation (11.57) becomes

Id =

2(Vas - Vt)Vds - Vfjs

(11.58)

Equation (11.58) is valid for Vcs Vt and for 0 < Vps 5 Vp5(sat).

Figure 11.48 shows plots of Equation (11.58) as a function of Vos for several values of Vcs- We can find the value of Vos at the peak current value from Ыо/bVos = (- Then, using Equation (11.58), the peak current occurs when

Vos = Vr.s - V

(11.59)

This value of Vos is just /(sat), the point at which saturation occurs. For Vos > Vo5(sat), the ideal drain current is a constant and is equal to

/о (sat) =

2(Vas - VV)VoHsat) - Viat)

(11.60)

vgsi

Figure 11.48 I Plots of Id versus V from Equation (11.58).

Using Equation (U.59) for V£)s(sat), Equation (11.60) becomes

/(sat) = --{Vcs - VtY

(11.61)

Equation (11.58) is the ideal current-voltage relationship of the n-channel MOSFET in the nonsaturation region for 0 < Vos /(sat), and Equation (11.61) is the ideal current-voltage relationship of the n-channel MOSFET in the saturation region for Vos > VD5(sat). These J-V expressions were explicitly derived for an n-channel enhancement mode device. However, these same equations apply to an n-channel depletion mode MOSFET in which the threshold voltage Vj is a negative quantity.

Objective

To design the width of a MOSFET such that a specified current is induced for a given applied bias.

Consider an ideal n-channel MOSFET with parameters L - 1.25 /m, = 650cm-A-s, Cox = 6.9 x 10- F/cm, and Vt = 0.65 V. Design the channel width W such that /(sat) = 4 mA for Vos = 5 V.

Solution

We have, from Equation (11.61),

T / \ 1/71 C,x XT \2

/(sat) = -~-[Vcs - VtY

DESIGN EXAMPLE 1L8

Then

, H4650)(6.9 X 10-**)

4 x 10- =----- (5 - 0.65)- = 3.39W

2(1.25 X 10-)

W = 11.8 дт

Comment

The current capability of a MOSFET is directly proportional to the channel width tV. The current handling capability can be increased by increasing W.

TEST YOUR UNDERSTANDING

E1L13 The parameters of an n-channel MOSFET are = 650 cm-A/-s, / х = 200 A, W/L = 50, and W = 0.40 V. If the transistor is biased in the saturation region, find the drain current for Vcs = 1.2, and 3 V. (УШ 6! pu TO 1 = I suy)

Ell. 14 The n-channel MOSFET in El 1.13 is to be redesigned by changing the W/L ratio such that I/j = 100/iA when the transistor is bia.sed in Ihe saturation region with Vas = 1Л5 V. (9L6 0 = 7/Ж suy)

Very small

Figure 11.49 I (a) > versus Vcs (fer small Vs) cnhanccmem mode MOSFET. (b) Ideal УТ versus Vcs in saturaiion region for enhancement mode (curve A) and depletion mode (curve B) n-channcl MOSFETs,

We can use the f-V relations to experimentally determine the mobility ai threshold voltage parameters. From Equation (11.58), we can write, for very si values of Vps,

(Vgs-Vt)Vds

(11.6:

Figure 11.49a shows a plot of Equation (11.62a) as a function of Vos for constant Vol A straight line is fitted through the points. The deviation from the straight line at lo values of Vgs S due to subthreshold conduction and the deviation at higher values Vc5 is due to mobility being a function of gate voltage. Both of these effects will be со sidered in the next chapter. The extrapolation of the straight line to zero current givd the threshold voltage and the slope is proportional lo the inversion earner mobility. If wc take the square root of Equation (11.61), we obtain

TMsat)

Wm C

{Vgs-Vt)

(11.6:

Figure 11.49b is a plot of Equation (11.62b). In the ideal case, we can obtain the sai information from both curves. However, as we will see in the next chapter, the thresh-J old voltage may be a function of Yds in short-channel devices. Since Equation (11.62b)i applies to devices biased in the saturation region, the V/ parameter in this equation may differ from the extrapolated value determined in Figure 11.49a. In general, the nonsat-1 uration current-voltage characteristics will produce the more reliable data.

Objective

To determine the inversion carrier mobility from experimental results.

Consider an n-channcl MOSFET with W = \5 fim, L = 2fim, and C = 6.9 x 1C~ F/cm*. Assume that the drain current in the nonsaturation region for Vs = 0.10 V is /5 = 35 M A at Vcs = 1 -5 V and ) = 75 м A at Vs = 2.5 V.

Solution

From Equation (11.62a), we can write

Id2 ~ hn - --(Уая2 - VGs\)yns

so that

75 X 10--35 X iO

9 X io j(2.5- (.5)(aia)

which yields

We can then determine

= 11Ъ cm/V-s

Vr = 0.625 V

Comment

The mobility of carriers in the inversion layer is less than that in the bulk semiconductor due to the surface scattering effect. Wc will discuss this effect in the next chapter.

The cuirent-voltage relarion ship of a p-channel device can be obtained by the same type of analysis. Figure 11.50 shows a p-channel enhancement mode MOSFET. The voltage polariries and current direcrion are the reverse of those in the n-channel device. We may note the change in the subscript notarion for this device. For the current direcrion shown in the figure, the I-V relarions for the p-channel MOSFET are

2iysG + yr)ysD-Vlo

(11.63)

Figure 11.50 1 Cross section and bias configuration for a p-channel enhancement-mode MOSFET.

for 0 < VsD < V5D(sat), and

/z)(sat) = (V5c + V,)

for VsD VsD(sat), where

VsDCsat) = VsG + Vj

(11.65)

Note the change in the sign in front of Vj and note that the mobihty is now the mo-bihty of the holes in the hole inversion layer charge. Keep in mind that Vr is negi tive for a p-channel enhancement mode MOSFET and positive for a depletion mc p-channel device.

TEST YOUR UNDERSTANDING

Ell.15 The parameters of a p-channel MOSFET are ju = 310 cm/V-s, / х = 220 A,

W/L = 60, and Vr = -0.40 V. If the transistor is biased in the saturation region, find the drain current for Vsc = С E5, and 2 V. {YVLZ puis ц-\ 95*0 = о/ suy)

Ell Л 6 The p-channel MOSFET in El 1.15 is to be redesigned by changing the (W/L) ratio such that Id = 200 /xA when the transistor is biased in the saturation regie with VsG = 1.25 V. (17T I = 7/M suy)

One assumption we made in the derivation of the current-voltage relationsl was that the charge neutrality condition given by Equation (11.46) was valid over the entire length of the channel. We implicitly assumed that б'J(max) was constant along the length of the channel. The space charge width, however, varies between source and drain due to the drain-to-source voltage; it is widest at the drain when Vds > 0. a change in the space charge density along the channel length must be balanced by a corresponding change in the inversion layer charge. An increase in the space charge width means that the inversion layer charge is reduced, implying that the drain current and drain-to-source saturation voltage are less than the ideal values. The actual saturation drain current may be as much as 20 percent less than the pre-j dieted value due to this bulk charge effect.

11.3.4 Transconductance

The MOSFET transconductance is defined as the change in drain current with respect to the corresponding change in gate voltage, or

The transconductance is sometimes referred to as the transistor gain.

If we consider an n-channel MOSFET operating in the nonsaturation region, then, using Equation (11.58), we have

din Wp C

gml - - -- у os

eVas L

(11.67)

The transconductance increases linearly with Vos but is independent of Vc,s in the

nonsaturation region.

The I-V characteristics of an n-channel MOSFET in the saturation region were given by Equation (11 -61). The transconductance in this region of operation is given by

8/;,(sat) W/i,C

dVc<

iVas-Vi)

(11.68)

In the saturation region, the transconductance is a linear function of V; and is independent of Vos-

The transconductance is a function of the geometry of the device as well as of carrier mobility and threshold voltage. The transconductance increases as the width of the device increases, and it also increases as the channel length and oxide thickness decrease. In the design of MOSFET circuits, the size of the transistor, in particular the channel width W. is an important engineering design parameter.

11.3.5 Substrate Bias Effects

In all of our analyses so far, the substrate, or body, has been connected to the source and held at ground potential. In MOSFET circuits, the source and body may not be at the same potential. Figure 1 K51 a shows an n-channel MOSFET and the associated double-subscripted voltage variables. The source-to-substrate pn junction must always be zero or reverse biased, so Vsh must always be greater than or equal to zero.

If Vsh 0, threshold is defined as the condition when = 2фfp as we discussed previously and as shown in Figure 11.51b. When Vsn > the surface will still try to invert when - 2( ,. However, these electrons are at a higher potential

p substrate

Body(BI V,

еф, = e{2фJ, + V)

Figure 11.511 (a) Applied voltages on an n-channel MOSFET. (b) Energy-band diagram at inversion point when Vsn = 0. (c) Energy-band diagram al inversion point when Vb > 0 is applied.