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2.1 CMOS Transistors</H1>
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Figure 2.3 illustrates how electrons and holes abandon their dopant atoms leaving a <SPAN CLASS="Definition">
depletion region</SPAN>
around a transistor’s source and drain. The region between source and drain is normally nonconducting. To make an <SPAN CLASS="EmphasisPrefix">
n</SPAN>
-channel transistor conducting, we must apply a positive voltage <SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
GS</SUB>
(the gate voltage with respect to the source) that is greater than the <SPAN CLASS="EmphasisPrefix">
n</SPAN>
-channel transistor <SPAN CLASS="Definition">
threshold voltage</SPAN>
, <SPAN CLASS="EquationNumber">
V</SPAN>
<SUB CLASS="Subscript">
t</SUB>
<SUB CLASS="SubscriptVariable">
n</SUB>
(a typical value is 0.5 V and, as far as we are presently concerned, is a constant). This establishes a thin (<SPAN CLASS="Symbol">
ª</SPAN>
50 Å) conducting channel of electrons under the gate. MOS transistors can carry a very small current (the <SPAN CLASS="Definition">
subthreshold current</SPAN>
—a few microamperes or less) with <SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
GS</SUB>
< <SPAN CLASS="EquationNumber">
V</SPAN>
<SUB CLASS="Subscript">
t</SUB>
<SUB CLASS="SubscriptVariable">
n</SUB>
, but we shall ignore this. A transistor can be conducting (<SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
GS</SUB>
> <SPAN CLASS="EquationNumber">
V</SPAN>
<SUB CLASS="Subscript">
t</SUB>
<SUB CLASS="SubscriptVariable">
n</SUB>
) without any current flowing. To make current flow in an <SPAN CLASS="EmphasisPrefix">
n</SPAN>
-channel transistor we must also apply a positive voltage, <SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
DS</SUB>
, to the drain with respect to the source. Figure 2.3 shows these connections and the connection to the fourth terminal of an MOS transistor—the <SPAN CLASS="Definition">
bulk</SPAN>
(<SPAN CLASS="Definition">
well</SPAN>
, <SPAN CLASS="Definition">
tub</SPAN>
, or <SPAN CLASS="Definition">
substrate</SPAN>
) terminal. For an <SPAN CLASS="EmphasisPrefix">
n</SPAN>
-channel transistor we must connect the bulk to the most negative potential, GND or VSS, to reverse bias the bulk-to-drain and bulk-to-source <SPAN CLASS="EmphasisPrefix">
pn</SPAN>
-diodes. The arrow in the four-terminal <SPAN CLASS="EmphasisPrefix">
n</SPAN>
-channel transistor symbol in Figure 2.3 reflects the polarity of these <SPAN CLASS="EmphasisPrefix">
pn</SPAN>
-diodes. </P>
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<IMG SRC="CH02-3.gif" ALIGN="BASELINE">
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<P CLASS="TableFigureTitle">
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FIGURE 2.3 An <SPAN CLASS="EmphasisPrefix">
n</SPAN>
-channel MOS transistor. The gate-oxide thickness, <SPAN CLASS="EquationNumber">
T</SPAN>
<SUB CLASS="Subscript">
OX</SUB>
, is approximately 100 angstroms (0.01 <SPAN CLASS="Symbol">
m</SPAN>
m). A typical transistor length, <SPAN CLASS="EquationNumber">
L</SPAN>
<SPAN CLASS="EquationNumber">
=</SPAN>
<SPAN CLASS="EquationNumber">
2</SPAN>
<SPAN CLASS="Symbol">
l</SPAN>
. The bulk may be either the substrate or a well. The diodes represent <SPAN CLASS="EmphasisPrefix">
pn</SPAN>
-junctions that must be reverse-biased.</P>
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<P CLASS="Body">
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The current flowing in the transistor is </P>
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<P CLASS="TableEqnCenter">
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current (amperes) = charge (coulombs) per unit time (second).</P>
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<P CLASS="TableEqnNumber">
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(2.1)</P>
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</TABLE>
<P CLASS="Body">
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We can express the current in terms of the total charge in the channel, <SPAN CLASS="EquationVariables">
Q</SPAN>
(imagine taking a picture and counting the number of electrons in the channel at that instant). If <SPAN CLASS="EquationVariables">
t</SPAN>
<SUB CLASS="SubscriptVariable">
f</SUB>
<SPAN CLASS="EquationVariables">
</SPAN>
(for <SPAN CLASS="Definition">
time of flight</SPAN>
—sometimes called the <SPAN CLASS="Definition">
transit time</SPAN>
) is the time that it takes an electron to cross between source and drain, the drain-to-source current, <SPAN CLASS="EquationVariables">
I</SPAN>
<SUB CLASS="SubscriptVariable">
DSn</SUB>
, is </P>
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<P CLASS="TableEqnCenter">
<A NAME="pgfId=366814">
</A>
<SPAN CLASS="EquationVariables">
I</SPAN>
<SUB CLASS="SubscriptVariable">
DSn</SUB>
= <SPAN CLASS="EquationVariables">
Q</SPAN>
/<SPAN CLASS="EquationVariables">
t</SPAN>
<SUB CLASS="SubscriptVariable">
f</SUB>
.</P>
</TD>
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<P CLASS="TableEqnNumber">
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</A>
(2.2)</P>
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</TABLE>
<P CLASS="Body">
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We need to find <SPAN CLASS="EquationVariables">
Q </SPAN>
and <SPAN CLASS="EquationVariables">
t</SPAN>
<SUB CLASS="SubscriptVariable">
f</SUB>
. The velocity of the electrons <SPAN CLASS="EquationNumber">
v</SPAN>
(a vector) is given by the equation that forms the basis of Ohm’s law: </P>
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<P CLASS="TableEqnCenter">
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</A>
<SPAN CLASS="Vector">
v</SPAN>
= –<SPAN CLASS="Symbol">
m</SPAN>
<SUB CLASS="SubscriptVariable">
n</SUB>
<SPAN CLASS="Vector">
E</SPAN>
,</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnNumber">
<A NAME="pgfId=366872">
</A>
(2.3)</P>
</TD>
</TR>
</TABLE>
<P CLASS="BodyAfterHead">
<A NAME="pgfId=203253">
</A>
where <SPAN CLASS="Symbol">
m</SPAN>
<SUB CLASS="SubscriptVariable">
n</SUB>
is the <SPAN CLASS="Definition">
electron mobility</SPAN>
(<SPAN CLASS="Symbol">
m</SPAN>
<SUB CLASS="SubscriptVariable">
p</SUB>
is the <SPAN CLASS="Definition">
hole mobility</SPAN>
) and <SPAN CLASS="Bold">
E</SPAN>
is the electric field (with units Vm<SUP CLASS="Superscript">
–1</SUP>
). </P>
<P CLASS="Body">
<A NAME="pgfId=301928">
</A>
Typical <SPAN CLASS="Definition">
carrier mobility</SPAN>
values are <SPAN CLASS="Symbol">
m</SPAN>
<SUB CLASS="SubscriptVariable">
n</SUB>
= 500–1000 cm<SUP CLASS="Superscript">
2</SUP>
V<SUP CLASS="Superscript">
–1</SUP>
s<SUP CLASS="Superscript">
–1</SUP>
and <SPAN CLASS="Symbol">
m</SPAN>
<SUB CLASS="SubscriptVariable">
p</SUB>
= 100–400 cm<SUP CLASS="Superscript">
2</SUP>
V<SUP CLASS="Superscript">
–1</SUP>
s<SUP CLASS="Superscript">
–1</SUP>
. Equation 2.3 is a vector equation, but we shall ignore the vertical electric field and concentrate on the horizontal electric field, <SPAN CLASS="EquationVariables">
E</SPAN>
<SUB CLASS="SubscriptVariable">
x</SUB>
, that moves the electrons between source and drain. The horizontal component of the electric field is <SPAN CLASS="EquationVariables">
E</SPAN>
<SUB CLASS="SubscriptVariable">
x</SUB>
= –<SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
DS</SUB>
/<SPAN CLASS="EquationNumber">
L, </SPAN>
directed from the drain to the source, where <SPAN CLASS="EquationNumber">
L</SPAN>
is the channel length (see Figure 2.3). The electrons travel a distance <SPAN CLASS="EquationNumber">
L</SPAN>
with horizontal velocity <SPAN CLASS="EquationVariables">
v</SPAN>
<SUB CLASS="SubscriptVariable">
x</SUB>
= –<SPAN CLASS="Symbol">
m</SPAN>
<SUB CLASS="SubscriptVariable">
n</SUB>
<SPAN CLASS="EquationVariables">
E</SPAN>
<SUB CLASS="SubscriptVariable">
x</SUB>
, so that </P>
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<P CLASS="TableEqnRight">
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</A>
</P>
</TD>
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<P CLASS="TableEqnCenter">
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</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=366905">
</A>
L</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=366907">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=366952">
</A>
L<SUP CLASS="Superscript">
2</SUP>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqn">
<A NAME="pgfId=366909">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqn">
<A NAME="pgfId=366911">
</A>
</P>
</TD>
</TR>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnRight">
<A NAME="pgfId=366913">
</A>
<SPAN CLASS="EquationVariables">
t</SPAN>
<SUB CLASS="SubscriptVariable">
f</SUB>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=366940">
</A>
=</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=366915">
</A>
<SPAN CLASS="EquationVariables">
–––</SPAN>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=366917">
</A>
=</P>
</TD>
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<P CLASS="TableEqnCenter">
<A NAME="pgfId=366954">
</A>
–––––––</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqn">
<A NAME="pgfId=366919">
</A>
.</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnNumber">
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</A>
(2.4)</P>
</TD>
</TR>
<TR>
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<P CLASS="TableEqnRight">
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</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
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</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=366925">
</A>
<SPAN CLASS="EquationVariables">
v</SPAN>
<SUB CLASS="SubscriptVariable">
x</SUB>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=366927">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=366956">
</A>
<SPAN CLASS="Symbol">
m</SPAN>
<SUB CLASS="SubscriptVariable">
n</SUB>
<SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
DS</SUB>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqn">
<A NAME="pgfId=366929">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqn">
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</A>
</P>
</TD>
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</TABLE>
<P CLASS="Body">
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Next we find the channel charge, <SPAN CLASS="EquationVariables">
Q</SPAN>
. The channel and the gate form the plates of a capacitor, separated by an insulator—the gate oxide. We know that the charge on a linear capacitor, C, is <SPAN CLASS="EquationVariables">
Q</SPAN>
= <SPAN CLASS="EquationNumber">
C</SPAN>
<SPAN CLASS="EquationVariables">
V</SPAN>
. Our lower plate, the channel, is not a linear conductor. Charge only appears on the lower plate when the voltage between the gate and the channel, <SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
GC</SUB>
, exceeds the <SPAN CLASS="EmphasisPrefix">
n</SPAN>
-channel threshold voltage. For our nonlinear capacitor we need to modify the equation for a linear capacitor to the following: </P>
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<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=366976">
</A>
<SPAN CLASS="EquationVariables">
Q</SPAN>
= <SPAN CLASS="EquationVariables">
C</SPAN>
(<SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
GC</SUB>
– V<SUB CLASS="Subscript">
t</SUB>
<SUB CLASS="SubscriptVariable">
n</SUB>
) .</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnNumber">
<A NAME="pgfId=366978">
</A>
(2.5)</P>
</TD>
</TR>
</TABLE>
<P CLASS="Body">
<A NAME="pgfId=202772">
</A>
The lower plate of our capacitor is resistive and conducting current, so that the potential in the channel, <SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
GC</SUB>
, varies. In fact, <SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
GC</SUB>
= <SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
GS</SUB>
at the source and <SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
GC</SUB>
= <SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
GS</SUB>
– <SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
DS</SUB>
at the drain. What we really should do is find an expression for the channel charge as a function of channel voltage and sum (integrate) the charge all the way across the channel, from <SPAN CLASS="EquationVariables">
x</SPAN>
= 0 (at the source) to <SPAN CLASS="EquationVariables">
x</SPAN>
= <SPAN CLASS="EquationNumber">
L</SPAN>
(at the drain). Instead we shall assume that the channel voltage, <SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
GC</SUB>
(<SPAN CLASS="EquationVariables">
x</SPAN>
), is a linear function of distance from the source and take the average value of the charge, which is thus </P>
<TABLE>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=376082">
</A>
<SPAN CLASS="EquationVariables">
Q</SPAN>
= <SPAN CLASS="EquationVariables">
C </SPAN>
[ (<SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
GS</SUB>
– V<SUB CLASS="Subscript">
t</SUB>
<SUB CLASS="SubscriptVariable">
n</SUB>
) – 0.5 <SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
DS</SUB>
] .</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
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