📄 ch02.1.htm
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2 (6/60) (850 <SPAN CLASS="Symbol">
¥</SPAN>
10<SUP CLASS="Superscript">
–6</SUP>
)</P>
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––––––––––––––––––</P>
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(–3.0 – (–0.85) )<SUP CLASS="Superscript">
2</SUP>
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3.68 <SPAN CLASS="Symbol">
¥</SPAN>
10<SUP CLASS="Superscript">
–5</SUP>
AV<SUP CLASS="Superscript">
–2</SUP>
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The next section explains the signs in Eq. 2.14. </P>
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<SPAN CLASS="Bold">
(a)</SPAN>
</P>
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<IMG SRC="CH02-4.gif">
</DIV>
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(b)</SPAN>
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FIGURE 2.4 MOS <SPAN CLASS="EmphasisPrefix">
n</SPAN>
-channel transistor characteristics for a generic 0.5 <SPAN CLASS="Symbol">
m</SPAN>
m process (G5). (a) A short-channel transistor, with W = 6 <SPAN CLASS="Symbol">
m</SPAN>
m and L = 0.6 <SPAN CLASS="Symbol">
m</SPAN>
m (drawn) and a long-channel transistor (W = 60 <SPAN CLASS="Symbol">
m</SPAN>
m, L = 6 <SPAN CLASS="Symbol">
m</SPAN>
m) (b) The 6/0.6 characteristics represented as a surface. (c) A long-channel transistor obeys a square-law characteristic between <SPAN CLASS="EquationVariables">
I</SPAN>
<SUB CLASS="Subscript">
DS</SUB>
and <SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="Subscript">
GS</SUB>
in the saturation region (<SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="Subscript">
DS</SUB>
= 3 V). A short-channel transistor shows a more linear characteristic due to velocity saturation. Normally, all of the transistors used on an ASIC have short channels.</P>
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<SPAN CLASS="Bold">
(c)</SPAN>
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<DIV>
<H3 CLASS="Heading2">
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2.1.1 P-Channel Transistors</H3>
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The source and drain of CMOS transistors look identical; we have to know which way the current is flowing to distinguish them. The source of an <SPAN CLASS="EmphasisPrefix">
n</SPAN>
-channel transistor is lower in potential than the drain and vice versa for a <SPAN CLASS="EmphasisPrefix">
p</SPAN>
-channel transistor. In an <SPAN CLASS="EmphasisPrefix">
n</SPAN>
-channel transistor the threshold voltage, <SPAN CLASS="EquationNumber">
V</SPAN>
<SUB CLASS="Subscript">
t</SUB>
<SUB CLASS="SubscriptVariable">
n</SUB>
, is normally positive, and the terminal voltages <SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
DS</SUB>
and <SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
GS </SUB>
are also usually positive. In a <SPAN CLASS="EmphasisPrefix">
p</SPAN>
-channel transistor <SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="Subscript">
t</SUB>
<SUB CLASS="SubscriptVariable">
p</SUB>
is normally negative and we have a choice: We can write everything in terms of the magnitudes of the voltages and currents or we can use negative signs in a consistent fashion. </P>
<P CLASS="Body">
<A NAME="pgfId=205327">
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Here are the equations for a <SPAN CLASS="EmphasisPrefix">
p</SPAN>
-channel transistor using negative signs: </P>
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<SPAN CLASS="EquationVariables">
I</SPAN>
<SUB CLASS="SubscriptVariable">
DSp</SUB>
</P>
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=</P>
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<A NAME="pgfId=378409">
</A>
–k<SUP CLASS="Superscript">
'</SUP>
<SUB CLASS="SubscriptVariable">
p</SUB>
<SPAN CLASS="EquationNumber">
(W/L)</SPAN>
[ (<SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
GS</SUB>
– V<SUB CLASS="Subscript">
t</SUB>
<SUB CLASS="SubscriptVariable">
p</SUB>
) – 0.5 <SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
DS</SUB>
]<SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
DS</SUB>
; <SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
DS </SUB>
> <SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
GS</SUB>
– V<SUB CLASS="Subscript">
t</SUB>
<SUB CLASS="SubscriptVariable">
p</SUB>
</P>
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(2.15)</P>
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<SPAN CLASS="EquationVariables">
I</SPAN>
<SUB CLASS="SubscriptVariable">
DSp</SUB>
<SUB CLASS="Subscript">
(sat)</SUB>
</P>
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</A>
=</P>
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–<SPAN CLASS="Symbol">
b</SPAN>
<SUB CLASS="SubscriptVariable">
p</SUB>
/2 (<SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
GS</SUB>
– V<SUB CLASS="Subscript">
t</SUB>
<SUB CLASS="SubscriptVariable">
p</SUB>
)<SUP CLASS="Superscript">
2</SUP>
; <SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
DS</SUB>
< <SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
GS</SUB>
– V<SUB CLASS="Subscript">
t</SUB>
<SUB CLASS="SubscriptVariable">
p</SUB>
.</P>
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In these two equations <SPAN CLASS="EquationNumber">
V</SPAN>
<SUB CLASS="Subscript">
t</SUB>
<SUB CLASS="SubscriptVariable">
p</SUB>
is negative, and the terminal voltages <SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
DS </SUB>
and <SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
GS </SUB>
are also normally negative (and –3 V < –2 V, for example). The current <SPAN CLASS="EquationVariables">
I</SPAN>
<SUB CLASS="SubscriptVariable">
DSp</SUB>
is then negative, corresponding to conventional current flowing from source to drain of a <SPAN CLASS="EmphasisPrefix">
p</SPAN>
-channel transistor (and hence the negative sign for <SPAN CLASS="EquationVariables">
I</SPAN>
<SUB CLASS="SubscriptVariable">
DSp</SUB>
<SUB CLASS="Subscript">
(sat)</SUB>
in Eq. 2.14). </P>
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<H3 CLASS="Heading2">
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</A>
2.1.2 Velocity Saturation</H3>
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</A>
For a deep submicron transistor, Eq. 2.12 may overestimate the drain–source current by a factor of 2 or more. There are three reasons for this error. First, the threshold voltage is not constant. Second, the actual length of the channel (the electrical or effective length, often written as <SPAN CLASS="EquationNumber">
L</SPAN>
<SUB CLASS="Subscript">
eff</SUB>
) is less than the drawn (mask) length. The third reason is that Eq. 2.3 is not valid for high electric fields. The electrons cannot move any faster than about <SPAN CLASS="EquationNumber">
v</SPAN>
<SUB CLASS="Subscript">
max</SUB>
<SUB CLASS="SubscriptVariable">
n</SUB>
= 10<SUP CLASS="Superscript">
5</SUP>
ms<SUP CLASS="Superscript">
–1</SUP>
when the electric field is above 10<SUP CLASS="Superscript">
6</SUP>
Vm<SUP CLASS="Superscript">
–1</SUP>
(reached when 1 V is dropped across 1 <SPAN CLASS="Symbol">
m</SPAN>
m); the electrons become <SPAN CLASS="Definition">
velocity saturated</SPAN>
. In this case <SPAN CLASS="EquationVariables">
t</SPAN>
<SUB CLASS="SubscriptVariable">
f</SUB>
= <SPAN CLASS="EquationNumber">
L</SPAN>
<SUB CLASS="Subscript">
eff</SUB>
/<SPAN CLASS="EquationNumber">
v</SPAN>
<SUB CLASS="Subscript">
max</SUB>
<SUB CLASS="SubscriptVariable">
n</SUB>
, the drain–source saturation current is independent of the transistor length, and Eq. 2.12 becomes </P>
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</A>
<SPAN CLASS="EquationVariables">
I</SPAN>
<SUB CLASS="SubscriptVariable">
DSn</SUB>
<SUB CLASS="Subscript">
(sat)</SUB>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=378527">
</A>
=</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnLeft">
<A NAME="pgfId=378529">
</A>
Wv<SUB CLASS="Subscript">
max</SUB>
<SUB CLASS="SubscriptVariable">
n</SUB>
C<SUB CLASS="Subscript">
ox</SUB>
(<SPAN CLASS="EquationVariables">
V</SPAN>
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