ch07.2.htm
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</P>
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<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=47438">
</A>
=</P>
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<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnLeft">
<A NAME="pgfId=47440">
</A>
R<SUB CLASS="Subscript">
P2</SUB>
(C<SUB CLASS="Subscript">
P2</SUB>
+ C<SUB CLASS="Subscript">
2</SUB>
+ 3C<SUB CLASS="Subscript">
P1</SUB>
) + (R<SUB CLASS="Subscript">
P2</SUB>
+ R<SUB CLASS="Subscript">
P1</SUB>
)(3C<SUB CLASS="Subscript">
P1</SUB>
+ C<SUB CLASS="Subscript">
3</SUB>
+ C<SUB CLASS="Subscript">
P2</SUB>
) </P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnNumber">
<A NAME="pgfId=47442">
</A>
(7.9)</P>
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<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnRight">
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</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=47446">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnLeft">
<A NAME="pgfId=47448">
</A>
+ (2R<SUB CLASS="Subscript">
P2</SUB>
+ R<SUB CLASS="Subscript">
P1</SUB>
)(C<SUB CLASS="Subscript">
P2</SUB>
+ C<SUB CLASS="Subscript">
4</SUB>
) .</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqn">
<A NAME="pgfId=47450">
</A>
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<P CLASS="Body">
<A NAME="pgfId=13322">
</A>
If <SPAN CLASS="EquationNumber">
RP1</SPAN>
= <SPAN CLASS="EquationNumber">
RP2</SPAN>
, and <SPAN CLASS="EquationNumber">
CP1</SPAN>
= <SPAN CLASS="EquationNumber">
CP2</SPAN>
, then </P>
<TABLE>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnRight">
<A NAME="pgfId=47534">
</A>
<SPAN CLASS="Symbol">
t</SPAN>
<SUB CLASS="SubscriptVariable">
D</SUB>
<SUB CLASS="Subscript">
</SUB>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=47536">
</A>
=</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnLeft">
<A NAME="pgfId=47538">
</A>
(15 + 21)R<SUB CLASS="Subscript">
P</SUB>
C<SUB CLASS="Subscript">
P</SUB>
+ (1.5 + 1 + 4.5)R<SUB CLASS="Subscript">
P</SUB>
C<SUB CLASS="Subscript">
LX</SUB>
.</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnNumber">
<A NAME="pgfId=47540">
</A>
(7.10)</P>
</TD>
</TR>
</TABLE>
<P CLASS="Body">
<A NAME="pgfId=13339">
</A>
We need to know the pass-transistor resistance <SPAN CLASS="EquationNumber">
RP </SPAN>
. For example, suppose <SPAN CLASS="EquationNumber">
RP</SPAN>
= 1k<SPAN CLASS="Symbol">
W</SPAN>
. If <IMG SRC="CH07-7.gif" ALIGN="BASELINE">
k<SUP CLASS="Superscript">
'</SUP>
<SUB CLASS="Subscript">
n</SUB>
= 50 <SPAN CLASS="Symbol">
m</SPAN>
AV<SUP CLASS="Superscript">
–2</SUP>
, then (with <SPAN CLASS="EquationNumber">
Vt</SPAN>
<SPAN CLASS="EquationVariables">
n</SPAN>
= 0.65 V and <SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="Subscript">
DD</SUB>
= 3.3 V) </P>
<TABLE>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnRight">
<A NAME="pgfId=47575">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=47577">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=47579">
</A>
1</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=47581">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=47583">
</A>
1</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=47673">
</A>
</P>
</TD>
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<A NAME="pgfId=47585">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqn">
<A NAME="pgfId=47587">
</A>
</P>
</TD>
</TR>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnRight">
<A NAME="pgfId=47589">
</A>
W/L</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=47591">
</A>
=</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=47593">
</A>
––––––––––</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=47595">
</A>
<SUB CLASS="SubscriptVariable">
=</SUB>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=47597">
</A>
––––––––––––––––––––––––––</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=47675">
</A>
<SUB CLASS="SubscriptVariable">
=</SUB>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnLeft">
<A NAME="pgfId=47599">
</A>
7.5 .</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnNumber">
<A NAME="pgfId=47601">
</A>
(7.11)</P>
</TD>
</TR>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnRight">
<A NAME="pgfId=47603">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=47605">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=47607">
</A>
k<SUP CLASS="Superscript">
'</SUP>
<SUB CLASS="Subscript">
n</SUB>
R<SUB CLASS="Subscript">
p</SUB>
(<SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
DD</SUB>
– V<SUB CLASS="Subscript">
t</SUB>
<SUB CLASS="SubscriptVariable">
n</SUB>
)</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=47609">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=47611">
</A>
(50 <SPAN CLASS="Symbol">
¥</SPAN>
10<SUP CLASS="Superscript">
–6</SUP>
)(1 <SPAN CLASS="Symbol">
¥</SPAN>
10<SUP CLASS="Superscript">
3</SUP>
)(3.3 – 0.65)</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=47677">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnLeft">
<A NAME="pgfId=47613">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqn">
<A NAME="pgfId=47615">
</A>
</P>
</TD>
</TR>
</TABLE>
<P CLASS="Body">
<A NAME="pgfId=13345">
</A>
If L = 1 <SPAN CLASS="Symbol">
m</SPAN>
m, both source and drain areas are 7.5 <SPAN CLASS="Symbol">
m</SPAN>
m long and approximately 3 <SPAN CLASS="Symbol">
m</SPAN>
m wide (determined by diffusion overlap of contact, contact width, and contact-to-gate spacing, rules 6.1a + 6.2a + 6.4a = 5.5 <SPAN CLASS="Symbol">
l</SPAN>
in <A HREF="/Humuhumu/from Antibes/Prof.htm#23193" CLASS="XRef">
Table 2.7</A>
). Both drain and source areas are thus 23 <SPAN CLASS="Symbol">
m</SPAN>
m<SUP CLASS="Superscript">
2</SUP>
and the sidewall perimeters are 14 <SPAN CLASS="Symbol">
m</SPAN>
m (excluding the sidewall facing the channel). If we have a diffusion capacitance of 140 aF<SPAN CLASS="Symbol">
m</SPAN>
m<SUP CLASS="Superscript">
–2</SUP>
(area) and 500 aF<SPAN CLASS="Symbol">
m</SPAN>
m<SUP CLASS="Superscript">
–1</SUP>
(perimeter), typical values for a 1.0 <SPAN CLASS="Symbol">
m</SPAN>
m process, the parasitic source and drain capacitance is </P>
<TABLE>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnRight">
<A NAME="pgfId=47681">
</A>
C<SUB CLASS="Subscript">
P</SUB>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=47683">
</A>
=</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnLeft">
<A NAME="pgfId=47685">
</A>
(140 <SPAN CLASS="Symbol">
¥</SPAN>
10<SUP CLASS="Superscript">
–18</SUP>
)(23) + (500 <SPAN CLASS="Symbol">
¥</SPAN>
10<SUP CLASS="Superscript">
–18</SUP>
)(14)</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnNumber">
<A NAME="pgfId=47687">
</A>
(7.12)</P>
</TD>
</TR>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnRight">
<A NAME="pgfId=47716">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=47718">
</A>
=</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnLeft">
<A NAME="pgfId=47720">
</A>
1.022 <SPAN CLASS="Symbol">
¥</SPAN>
10<SUP CLASS="Superscript">
–14</SUP>
F .</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqn">
<A NAME="pgfId=47722">
</A>
</P>
</TD>
</TR>
</TABLE>
<P CLASS="Body">
<A NAME="pgfId=13334">
</A>
If we assume <SPAN CLASS="EquationNumber">
CP</SPAN>
= 0.01 pF and <SPAN CLASS="EquationNumber">
CLX</SPAN>
= 0.075 pF (<A HREF="CH07.2.htm#38131" CLASS="XRef">
Table 7.3</A>
), </P>
<TABLE>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnRight">
<A NAME="pgfId=47732">
</A>
<SPAN CLASS="Symbol">
t</SPAN>
<SUB CLASS="Subscript">
D</SUB>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=47734">
</A>
=</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnLeft">
<A NAME="pgfId=47736">
</A>
(36)(1)(0.01) + (7)(1)(0.075)</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnNumber">
<A NAME="pgfId=47738">
</A>
(7.13)</P>
</TD>
</TR>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnRight">
<A NAME="pgfId=47740">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=47742">
</A>
=</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnLeft">
<A NAME="pgfId=47744">
</A>
0.885 ns .</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqn">
<A NAME="pgfId=47746">
</A>
</P>
</TD>
</TR>
</TABLE>
<P CLASS="BodyAfterHead">
<A NAME="pgfId=31535">
</A>
A delay of approximately 1 ns agrees with the typical values from the XACT delay calculator and is about the fastest connection we can make between two CLBs.</P>
<TABLE>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableFigure">
<A NAME="pgfId=31563">
</A>
</P>
<DIV>
<IMG SRC="CH07-8.gif">
</DIV>
</TD>
</TR>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableFigureTitle">
<A NAME="pgfId=31566">
</A>
FIGURE 7.7 <A NAME="26676">
</A>
The Xilinx EPLD UIM (Universal Interconnection Module). (a) A simplified block diagram of the UIM. The UIM bus width, <SPAN CLASS="EquationVariables">
n</SPAN>
, varies from 68 (XC7236) to 198 (XC73108). (b) The UIM is actually a large programmable AND array. (c) The parasitic capacitance of the EPROM cell.</P>
</TD>
</TR>
</TABLE>
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