ch03.a.htm
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<SPAN CLASS="EquationVariables">
B</SPAN>
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=</P>
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<SPAN CLASS="BigMath">
∏</SPAN>
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<SPAN CLASS="EquationVariables">
b</SPAN>
<SUB CLASS="SubscriptVariable">
i </SUB>
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(3.47)</P>
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<SPAN CLASS="EquationVariables">
i </SPAN>
<SPAN CLASS="Symbol">
∈</SPAN>
path</P>
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<P CLASS="Exercise">
<A NAME="pgfId=55459">
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The <SPAN CLASS="Definition">
branching effort</SPAN>
<A NAME="marker=55467">
</A>
is the ratio of the on-path plus off-path capacitance to the on-path capacitance. The <SPAN CLASS="Definition">
path effort</SPAN>
<A NAME="marker=55563">
</A>
<SPAN CLASS="EquationVariables">
F</SPAN>
becomes the product of the path electrical effort, path branching effort, and path logical effort: </P>
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<SPAN CLASS="EquationVariables">
F</SPAN>
= <SPAN CLASS="EquationVariables">
GBH</SPAN>
.</P>
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(3.48)</P>
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<P CLASS="Exercise">
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Show that the path delay <SPAN CLASS="EquationVariables">
D</SPAN>
is </P>
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<SPAN CLASS="EquationVariables">
D</SPAN>
</P>
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<A NAME="pgfId=328054">
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=</P>
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<SPAN CLASS="BigMath">
∑</SPAN>
</P>
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<A NAME="pgfId=328058">
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<SPAN CLASS="EquationVariables">
g</SPAN>
<SUB CLASS="SubscriptVariable">
i</SUB>
<SPAN CLASS="EquationVariables">
b</SPAN>
<SUB CLASS="SubscriptVariable">
i</SUB>
<SPAN CLASS="EquationVariables">
h</SPAN>
<SUB CLASS="SubscriptVariable">
i</SUB>
</P>
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+</P>
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<A NAME="pgfId=328062">
</A>
<SPAN CLASS="BigMath">
∑</SPAN>
</P>
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<P CLASS="TableEqnLeft">
<A NAME="pgfId=328064">
</A>
<SPAN CLASS="EquationVariables">
p</SPAN>
<SUB CLASS="SubscriptVariable">
i</SUB>
.</P>
</TD>
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<P CLASS="TableEqnNumber">
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(3.49)</P>
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<SPAN CLASS="EquationVariables">
i </SPAN>
<SPAN CLASS="Symbol">
∈</SPAN>
path</P>
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<SPAN CLASS="EquationVariables">
i </SPAN>
<SPAN CLASS="Symbol">
∈</SPAN>
path</P>
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<P CLASS="Exercise">
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(***) Show that the optimum path delay is then </P>
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<A NAME="pgfId=328153">
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<SPAN CLASS="EquationVariables">
D^ </SPAN>
</P>
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<A NAME="pgfId=328155">
</A>
=</P>
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<SPAN CLASS="EquationVariables">
NF</SPAN>
<SUP CLASS="Superscript">
1/</SUP>
<SUP CLASS="SuperscriptVariable">
N </SUP>
</P>
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=</P>
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<SPAN CLASS="EquationVariables">
N</SPAN>
(<SPAN CLASS="EquationVariables">
GBH</SPAN>
)<SUP CLASS="Superscript">
1/</SUP>
<SUP CLASS="SuperscriptVariable">
N</SUP>
+ <SPAN CLASS="EquationVariables">
P</SPAN>
.</P>
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(3.50)</P>
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<P CLASS="ExerciseHead">
<A NAME="pgfId=142584">
</A>
3.16 <A NAME="18393">
</A>
(*Circuits from layout, 120 min.) Figure <A HREF="#24869" CLASS="XRef">
3.26</A>
shows a D flip-flop with clear from a 1.0<SPAN CLASS="Symbol">
m</SPAN>
m standard-cell library. Figure <A HREF="#18538" CLASS="XRef">
3.27</A>
shows two layout views of this D flip-flop. Construct the circuit diagram for this flip-flop, labeling the nodes and transistors as shown. Include the transistor sizes—use estimates for transistors with 45° gates—you only need W/L values, you can assume the gate lengths are all L<SPAN CLASS="Symbol">
</SPAN>
=<SPAN CLASS="Symbol">
</SPAN>
2<SPAN CLASS="Symbol">
l</SPAN>
, equal to the minimum feature size. Label the inputs and outputs to the cell and identify their functions. </P>
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<P CLASS="TableFigure">
<A NAME="pgfId=136312">
</A>
<IMG SRC="CH03-44.gif" ALIGN="BASELINE">
</P>
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</TR>
<TR>
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<P CLASS="TableFigureTitle">
<A NAME="pgfId=136315">
</A>
FIGURE 3.26 <A NAME="24869">
</A>
A D flip-flop from a 1.0<SPAN CLASS="Symbol">
m</SPAN>
m standard-cell library (Problem <A HREF="#18393" CLASS="XRef">
3.16</A>
).</P>
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</TR>
</TABLE>
<P CLASS="ExerciseHead">
<A NAME="pgfId=56420">
</A>
3.17 <A NAME="16235">
</A>
(Flip-flop circuits, 30 min.) Draw the circuit schematic for a positive-edge–triggered D flip-flop with active-high set and reset (base your schematic on Figure 2.18a, a negative-edge–triggered D flip-flop). Describe the problem when both SET and RESET are high.</P>
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<P CLASS="TableFigure">
<A NAME="pgfId=213724">
</A>
<IMG SRC="CH03-45.gif" ALIGN="BASELINE">
</P>
</TD>
</TR>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableFigure">
<A NAME="pgfId=213762">
</A>
</P>
<DIV>
<IMG SRC="CH03-46.gif">
</DIV>
</TD>
</TR>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableFigureTitle">
<A NAME="pgfId=213789">
</A>
FIGURE 3.27 <A NAME="18538">
</A>
(Top) A standard cell showing the diffusion (<SPAN CLASS="EmphasisPrefix">
n</SPAN>
-diffusion and <SPAN CLASS="EmphasisPrefix">
p</SPAN>
-diffusion), poly, and contact layers (the <SPAN CLASS="EmphasisPrefix">
n</SPAN>
-well and <SPAN CLASS="EmphasisPrefix">
p</SPAN>
-well are not shown). (Bottom) Shows the m1, contact, m2, and via layers. Problem <A HREF="#18393" CLASS="XRef">
3.16</A>
traces this circuit for this cell.</P>
</TD>
</TR>
</TABLE>
<P CLASS="Exercise">
<A NAME="pgfId=72352">
</A>
If we want an active-high set or reset we can: (1) use an inverter on the set or reset signal or (2) we can substitute NOR cells. Since NOR cells are slower than NAND cells, which we do depends on whether we want to optimize for speed or area. </P>
<P CLASS="Exercise">
<A NAME="pgfId=159526">
</A>
Thus, the largest flip-flop would be one with both Q and QN outputs, active high set and reset—requiring four TX gates, three inverters (four of the seven we normally need are replaced with NAND cells), four NAND cells, and two inverters to invert the set and reset, making a total of 34 transistors, or 8.5 gates.</P>
<P CLASS="ExerciseHead">
<A NAME="pgfId=106045">
</A>
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