📄 ch03.1.htm
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<P CLASS="TableFigure">
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(b)</SPAN>
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<A NAME="pgfId=184136">
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FIGURE 3.2 <A NAME="34786">
</A>
CMOS inverter characteristics. (a) This static inverter transfer curve is traced as the inverter switches slowly enough to be in equilibrium at all times (<SPAN CLASS="EquationVariables">
I</SPAN>
<SUB CLASS="SubscriptVariable">
DSn</SUB>
<SPAN CLASS="Symbol">
</SPAN>
=<SPAN CLASS="Symbol">
</SPAN>
–<SPAN CLASS="EquationVariables">
I</SPAN>
<SUB CLASS="SubscriptVariable">
DSp</SUB>
). (b) This surface corresponds to the current flowing in the <SPAN CLASS="EmphasisPrefix">
n</SPAN>
-channel transistor (falling delay) and <SPAN CLASS="EmphasisPrefix">
p</SPAN>
-channel transistor (rising delay) for any trajectory. (c) The current that flows through both transistors as the inverter switches along the equilibrium path.</P>
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<A NAME="pgfId=180270">
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<SPAN CLASS="Bold">
(c)</SPAN>
</P>
<DIV>
<IMG SRC="CH03-4.gif">
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<P CLASS="Body">
<A NAME="pgfId=180567">
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The input waveform, <SPAN CLASS="BodyComputer">
v(in1)</SPAN>
, and the output load (which determines the transistor currents) dictate the path we take on the surface of Figure <A HREF="#34786" CLASS="XRef">
3.2</A>
(b) as the inverter switches. We can thus see that the currents through the transistors (and thus the pull-up and pull-down resistance values) will vary in a nonlinear way during switching. Deriving theoretical values for the pull-up and pull-down resistance values is difficult—instead we work the problem backward by picking the trip points, simulating the propagation delays, and then calculating resistance values that fit the model.</P>
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<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="Table">
<A NAME="pgfId=180098">
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<SPAN CLASS="Bold">
(a)</SPAN>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="Table">
<A NAME="pgfId=210966">
</A>
</P>
<DIV>
<IMG SRC="CH03-5.gif">
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<TD ROWSPAN="2" COLSPAN="1">
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<A NAME="pgfId=180108">
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<P CLASS="Table">
<A NAME="pgfId=184502">
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<SPAN CLASS="Bold">
(b)</SPAN>
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<IMG SRC="CH03-6.gif">
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<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="Table">
<A NAME="pgfId=180113">
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<SPAN CLASS="Bold">
(c)</SPAN>
</P>
<DIV>
<IMG SRC="CH03-7.gif">
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<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="Table">
<A NAME="pgfId=184458">
</A>
<SPAN CLASS="Bold">
(d)</SPAN>
</P>
<DIV>
<IMG SRC="CH03-8.gif">
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</TD>
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<TD ROWSPAN="1" COLSPAN="2">
<P CLASS="TableFigureTitle">
<A NAME="pgfId=180121">
</A>
FIGURE 3.3 <A NAME="19386">
</A>
Delay. (a) LogicWorks schematic for inverters driving 1, 2, 4, and 8 standard loads (1 standard load<SPAN CLASS="Symbol">
</SPAN>
=<SPAN CLASS="Symbol">
</SPAN>
0.034<SPAN CLASS="Symbol">
</SPAN>
pF in this case). (b) Transient response (falling delay only) from PSpice. The postprocessor Probe was used to mark each waveform as it crosses its trip point (0.5 for the input, 0.35 for the outputs). For example <SPAN CLASS="BodyComputer">
v(out1_4)</SPAN>
(4 standard loads) crosses 1.0467<SPAN CLASS="Symbol">
</SPAN>
V (<SPAN CLASS="Symbol">
ª</SPAN>
0.35<SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
DD</SUB>
) at <SPAN CLASS="EquationVariables">
t</SPAN>
<SPAN CLASS="Symbol">
</SPAN>
=<SPAN CLASS="Symbol">
</SPAN>
169.93<SPAN CLASS="Symbol">
</SPAN>
ps. (c) Falling and rising delays as a function of load. The slopes in pspF<SUP CLASS="Superscript">
–1</SUP>
corresponds to the pull-up resistance (1281<SPAN CLASS="Symbol">
W</SPAN>
) and pull-down resistance (817<SPAN CLASS="Symbol">
W</SPAN>
). (d) Comparison of the delay model (valid for <SPAN CLASS="EquationVariables">
t</SPAN>
<SPAN CLASS="Symbol">
</SPAN>
><SPAN CLASS="Symbol">
</SPAN>
20<SPAN CLASS="Symbol">
</SPAN>
ps) and simulation (4 standard loads). Both are equal at the 0.35 trip point.</P>
</TD>
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</TABLE>
<P CLASS="Body">
<A NAME="pgfId=180541">
</A>
Figure <A HREF="#19386" CLASS="XRef">
3.3</A>
shows a simulation experiment (using the G5 process SPICE parameters from Table 2.1). From the results in Figure <A HREF="#19386" CLASS="XRef">
3.3</A>
(c) we can see that <SPAN CLASS="EquationVariables">
R</SPAN>
<SUB CLASS="SubscriptVariable">
pd</SUB>
<SPAN CLASS="Symbol">
</SPAN>
=<SPAN CLASS="Symbol">
</SPAN>
817<SPAN CLASS="Symbol">
W</SPAN>
and<SPAN CLASS="EquationVariables">
R</SPAN>
<SUB CLASS="SubscriptVariable">
pu</SUB>
<SPAN CLASS="Symbol">
</SPAN>
=<SPAN CLASS="Symbol">
</SPAN>
1281<SPAN CLASS="Symbol">
W</SPAN>
for this inverter (with shape factors of 6/0.6 for the <SPAN CLASS="EmphasisPrefix">
n</SPAN>
-channel transistor and 12/0.6 for the <SPAN CLASS="EmphasisPrefix">
p</SPAN>
-channel) using 0.5 (input) and 0.35/0.65 (output) trip points. Changing the trip points would give different resistance values. </P>
<P CLASS="Body">
<A NAME="pgfId=299676">
</A>
We can check that 817<SPAN CLASS="Symbol">
W</SPAN>
is a reasonable value for the pull-down resistance. In the saturation region <SPAN CLASS="EquationVariables">
I</SPAN>
<SUB CLASS="SubscriptVariable">
DS</SUB>
<SUB CLASS="Subscript">
(sat)</SUB>
is (to first order) independent of <SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
DS</SUB>
. For an <SPAN CLASS="EmphasisPrefix">
n</SPAN>
-channel transistor from our generic 0.5<SPAN CLASS="Symbol">
m</SPAN>
m process (G5 from Section 2.1) with shape factor W/L<SPAN CLASS="Symbol">
</SPAN>
=<SPAN CLASS="Symbol">
</SPAN>
6/0.6, <SPAN CLASS="EquationVariables">
I</SPAN>
<SUB CLASS="SubscriptVariable">
DSn</SUB>
<SUB CLASS="Subscript">
(sat)</SUB>
<SPAN CLASS="Symbol">
</SPAN>
=<SPAN CLASS="Symbol">
</SPAN>
2.5<SPAN CLASS="Symbol">
</SPAN>
mA (at <SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
GS</SUB>
<SPAN CLASS="Symbol">
</SPAN>
=<SPAN CLASS="Symbol">
</SPAN>
3V and <SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
DS</SUB>
<SPAN CLASS="Symbol">
</SPAN>
=<SPAN CLASS="Symbol">
</SPAN>
3V). The pull-down resistance, <SPAN CLASS="EquationVariables">
R</SPAN>
<SUB CLASS="Subscript">
1</SUB>
, that would give the same drain–source current is </P>
<TABLE>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=299682">
</A>
<SPAN CLASS="EquationVariables">
R</SPAN>
<SUB CLASS="Subscript">
1 </SUB>
= 3.0 V / (2.5 <SPAN CLASS="Symbol">
¥</SPAN>
10<SUP CLASS="Superscript">
–3</SUP>
A) = 1200 <SPAN CLASS="Symbol">
W</SPAN>
.</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnNumber">
<A NAME="pgfId=299684">
</A>
(3.3)</P>
</TD>
</TR>
</TABLE>
<P CLASS="BodyAfterHead">
<A NAME="pgfId=299690">
</A>
This value is greater than, but not too different from, our measured pull-down resistance of 817<SPAN CLASS="Symbol">
W </SPAN>
. We might expect this result since Figure 3.2b shows that the pull-down resistance reaches its maximum value at <SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
GS</SUB>
<SPAN CLASS="Symbol">
</SPAN>
=<SPAN CLASS="Symbol">
</SPAN>
3V, <SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
DS</SUB>
<SPAN CLASS="Symbol">
</SPAN>
=<SPAN CLASS="Symbol">
</SPAN>
3V. We could adjust the ratio of the logic so that the rising and falling delays were equal; then <SPAN CLASS="EquationVariables">
R</SPAN>
<SPAN CLASS="Symbol">
</SPAN>
=<SPAN CLASS="Symbol">
</SPAN>
<SPAN CLASS="EquationVariables">
R</SPAN>
<SUB CLASS="SubscriptVariable">
pd</SUB>
<SPAN CLASS="EquationVariables">
</SPAN>
<SPAN CLASS="Symbol">
</SPAN>
=<SPAN CLASS="Symbol">
</SPAN>
<SPAN CLASS="EquationVariables">
R</SPAN>
<SUB CLASS="SubscriptVariable">
pu </SUB>
is the <SPAN CLASS="Definition">
pull resistance</SPAN>
<A NAME="marker=299694">
</A>
.</P>
<P CLASS="Body">
<A NAME="pgfId=184560">
</A>
Next, we check our model against the simulation results. The model predicts </P>
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<P CLASS="TableEqnRight">
<A NAME="pgfId=221579">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqn">
<A NAME="pgfId=233176">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=232908">
</A>
–<SPAN CLASS="EquationVariables">
t'</SPAN>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqn">
<A NAME="pgfId=232910">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqn">
<A NAME="pgfId=221581">
</A>
</P>
</TD>
</TR>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnRight">
<A NAME="pgfId=221583">
</A>
<SPAN CLASS="BodyComputer">
v(out1)</SPAN>
<SPAN CLASS="Symbol">
ª</SPAN>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqn">
<A NAME="pgfId=233178">
</A>
<SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
DD</SUB>
exp</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=232912">
</A>
–––––––––––––</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqn">
<A NAME="pgfId=232914">
</A>
for <SPAN CLASS="EquationVariables">
t</SPAN>
' > 0 .</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnNumber">
<A NAME="pgfId=221585">
</A>
(3.4)</P>
</TD>
</TR>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnRight">
<A NAME="pgfId=221587">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqn">
<A NAME="pgfId=233180">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=232916">
</A>
<SPAN CLASS="EquationVariables">
R</SPAN>
<SUB CLASS="SubscriptVariable">
pd </SUB>
(<SPAN CLASS="EquationVariables">
C</SPAN>
<SUB CLASS="SubscriptVariable">
out</SUB>
+ <SPAN CLASS="EquationVariables">
C</SPAN>
<SUB CLASS="SubscriptVariable">
p</SUB>
)</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqn">
<A NAME="pgfId=232918">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqn">
<A NAME="pgfId=221589">
</A>
</P>
</TD>
</TR>
</TABLE>
<P CLASS="BodyAfterHead">
<A NAME="pgfId=184411">
</A>
(<SPAN CLASS="EquationVariables">
t'</SPAN>
is measured from the point at which the input crosses the 0.5 trip point, <SPAN CLASS="EquationVariables">
t'</SPAN>
<SPAN CLASS="Symbol">
</SPAN>
=<SPAN CLASS="Symbol">
</SPAN>
0 at <SPAN CLASS="EquationVariables">
t</SPAN>
<SPAN CLASS="Symbol">
</SPAN>
=<SPAN CLASS="Symbol">
</SPAN>
20<SPAN CLASS="Symbol">
</SPAN>
ps). With <SPAN CLASS="EquationVariables">
C</SPAN>
<SUB CLASS="SubscriptVariable">
p</SUB>
<SPAN CLASS="Symbol">
</SPAN>
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