ch03.a.htm

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3.18&nbsp;(Set and reset, 10 min.) Show how to add a <SPAN CLASS="Definition">
synchronous set</SPAN>
<A NAME="marker=106063">
 </A>
 or a <SPAN CLASS="Definition">
synchronous reset</SPAN>
<A NAME="marker=106064">
 </A>
 to the flip-flop of Figure&nbsp;2.18(a) using a two-input MUX.</P>
<P CLASS="ExerciseHead">
<A NAME="pgfId=123413">
 </A>
3.19&nbsp;(Clocked inverters, 45 min.) Using PSpice compare the delay of an inverter with transmission gate with that of a clocked inverter using the G5 process SPICE parameters from Table&nbsp;2.1.</P>
<P CLASS="ExerciseHead">
<A NAME="pgfId=185168">
 </A>
3.20&nbsp;(S-R, T, J-K flip-flops, 30 min.)&nbsp;The <SPAN CLASS="Definition">
characteristic equation</SPAN>
<A NAME="marker=105911">
 </A>
 for a D flip-flop is Q<SUB CLASS="Subscript">
t+1</SUB>
<SPAN CLASS="Symbol">
 </SPAN>
=<SPAN CLASS="Symbol">
 </SPAN>
D. The characteristic equation for a J-K flip-flop is Q<SUB CLASS="Subscript">
t+1</SUB>
<SPAN CLASS="Symbol">
 </SPAN>
=<SPAN CLASS="Symbol">
 </SPAN>
J(Q<SUB CLASS="Subscript">
t</SUB>
)'<SPAN CLASS="Symbol">
 </SPAN>
+<SPAN CLASS="Symbol">
 </SPAN>
K'Q<SUB CLASS="Subscript">
t</SUB>
. </P>
<UL>
<LI CLASS="ExercisePartFirst">
<A NAME="pgfId=123726">
 </A>
a.&nbsp;Show how you can build a J-K flip-flop using a D flip-flop. </LI>
<LI CLASS="ExercisePart">
<A NAME="pgfId=123730">
 </A>
b.&nbsp;The characteristic equation for a T flip-flop (toggle flip-flop) is Q<SUB CLASS="Subscript">
t+1</SUB>
<SPAN CLASS="Symbol">
 </SPAN>
=<SPAN CLASS="Symbol">
 </SPAN>
(Q<SUB CLASS="Subscript">
t</SUB>
)'<SPAN CLASS="Symbol">
 </SPAN>
. Show how to build a T flip-flop using a D flip-flop. </LI>
<LI CLASS="ExercisePart">
<A NAME="pgfId=123731">
 </A>
c.&nbsp;The characteristic equation does not show the timing behavior of a sequential element&#8212;the characteristic equation for a D latch is the same as that for a D flip-flop. The characteristic equation for an S-R latch and an S-R flip-flop is Q<SUB CLASS="Subscript">
t+1</SUB>
<SPAN CLASS="Symbol">
 </SPAN>
=<SPAN CLASS="Symbol">
 </SPAN>
S<SPAN CLASS="Symbol">
 </SPAN>
+<SPAN CLASS="Symbol">
 </SPAN>
R'Q<SUB CLASS="Subscript">
t</SUB>
. An S-R flip-flop is sometimes called a pulse-triggered flip-flop. Find out the behavior of an S-R latch and an S-R flip-flop and describe the differences between these elements and a D latch and a D flip-flop. </LI>
<LI CLASS="ExercisePart">
<A NAME="pgfId=123732">
 </A>
d.&nbsp;Explain why it is probably not a good idea to use an S-R flip-flop in an ASIC design.</LI>
</UL>
<P CLASS="ExerciseHead">
<A NAME="pgfId=58665">
 </A>
3.21&nbsp;(**Optimum logic, 60 min.) Suppose we have a fixed logic path of length <SPAN CLASS="EquationVariables">
n</SPAN>
<SUB CLASS="Subscript">
1</SUB>
. We want to know how many (if any) buffer stages we should add at the output of this path to optimize the total path delay given the output load capacitance. </P>
<UL>
<LI CLASS="ExercisePartFirst">
<A NAME="pgfId=123733">
 </A>
a.&nbsp;If the total number of stages is <SPAN CLASS="EquationVariables">
N </SPAN>
(logic path of length <SPAN CLASS="EquationVariables">
n</SPAN>
<SUB CLASS="Subscript">
1</SUB>
 plus <SPAN CLASS="EquationVariables">
N</SPAN>
<SPAN CLASS="Symbol">
 </SPAN>
&#8211;<SPAN CLASS="Symbol">
 </SPAN>
<SPAN CLASS="EquationVariables">
n</SPAN>
<SUB CLASS="Subscript">
1</SUB>
 inverters), show that the total path delay is  </LI>
<TABLE>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqn">
<A NAME="pgfId=328620">
 </A>
&nbsp;</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=328622">
 </A>
&nbsp;</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnLeft">
<A NAME="pgfId=328624">
 </A>
&nbsp;</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=328626">
 </A>
&nbsp;</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=328628">
 </A>
<SPAN CLASS="EquationVariables">
n</SPAN>
<SUB CLASS="Subscript">
1</SUB>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=328630">
 </A>
&nbsp;</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqn">
<A NAME="pgfId=328632">
 </A>
&nbsp;</P>
</TD>
</TR>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqn">
<A NAME="pgfId=328634">
 </A>
<SPAN CLASS="EquationVariables">
D^</SPAN>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=328636">
 </A>
=</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnLeft">
<A NAME="pgfId=328638">
 </A>
<SPAN CLASS="EquationVariables">
NF</SPAN>
<SUP CLASS="Superscript">
1/</SUP>
<SUP CLASS="SuperscriptVariable">
N</SUP>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=328640">
 </A>
+</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=328642">
 </A>
<SPAN CLASS="BigMath">
&#8721;</SPAN>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnLeft">
<A NAME="pgfId=328644">
 </A>
(<SPAN CLASS="EquationVariables">
p</SPAN>
<SUB CLASS="SubscriptVariable">
i</SUB>
 + <SPAN CLASS="EquationVariables">
q</SPAN>
<SUB CLASS="SubscriptVariable">
i</SUB>
) + (<SPAN CLASS="EquationVariables">
N</SPAN>
 &#8211; <SPAN CLASS="EquationVariables">
n</SPAN>
<SUB CLASS="Subscript">
1</SUB>
)(<SPAN CLASS="EquationVariables">
p</SPAN>
<SUB CLASS="SubscriptVariable">
inv</SUB>
 + <SPAN CLASS="EquationVariables">
q</SPAN>
<SUB CLASS="SubscriptVariable">
inv</SUB>
) .</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnNumber">
<A NAME="pgfId=328646">
 </A>
(3.51)</P>
</TD>
</TR>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqn">
<A NAME="pgfId=328648">
 </A>
&nbsp;</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=328650">
 </A>
&nbsp;</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnLeft">
<A NAME="pgfId=328652">
 </A>
&nbsp;</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=328654">
 </A>
&nbsp;</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=328656">
 </A>
<SPAN CLASS="EquationVariables">
i </SPAN>
<SPAN CLASS="Symbol">
= 1</SPAN>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=328658">
 </A>
&nbsp;</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqn">
<A NAME="pgfId=328660">
 </A>
&nbsp;</P>
</TD>
</TR>
</TABLE>
</UL>
<P CLASS="Exercise">
<A NAME="pgfId=58673">
 </A>
The optimum number of stages is given by the solution to the following equation:  </P>
<TABLE>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqn">
<A NAME="pgfId=328903">
 </A>
<SPAN CLASS="EquationVariables">
</SPAN>
&nbsp;</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=328905">
 </A>
&nbsp;</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnLeft">
<A NAME="pgfId=328907">
 </A>
&nbsp;</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=328909">
 </A>
&nbsp;</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnLeft">
<A NAME="pgfId=328911">
 </A>
&nbsp;</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqn">
<A NAME="pgfId=328913">
 </A>
&nbsp;</P>
</TD>
</TR>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnRight">
<A NAME="pgfId=328915">
 </A>
<SPAN CLASS="Symbol">
&#8706;</SPAN>
<SPAN CLASS="EquationVariables">
D^/&#8706;N</SPAN>
 </P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=328917">
 </A>
=</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnLeft">
<A NAME="pgfId=328919">
 </A>
<SPAN CLASS="Symbol">
&#8706;</SPAN>
<SPAN CLASS="EquationVariables">
/&#8706;N</SPAN>
 (<SPAN CLASS="EquationVariables">
 NF</SPAN>
<SUP CLASS="Superscript">
1/</SUP>
<SUP CLASS="SuperscriptVariable">
N</SUP>
 + (<SPAN CLASS="EquationVariables">
N</SPAN>
 &#8211; <SPAN CLASS="EquationVariables">
n</SPAN>
<SUB CLASS="Subscript">
1</SUB>
)(<SPAN CLASS="EquationVariables">
p</SPAN>
<SUB CLASS="SubscriptVariable">
inv</SUB>
 + <SPAN CLASS="EquationVariables">
q</SPAN>
<SUB CLASS="SubscriptVariable">
inv</SUB>
 ) )</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=328921">
 </A>
=</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnLeft">
<A NAME="pgfId=328923">
 </A>
0 .</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnNumber">
<A NAME="pgfId=328925">
 </A>
(3.52)</P>
</TD>
</TR>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqn">
<A NAME="pgfId=328927">
 </A>
&nbsp;</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=328929">
 </A>
&nbsp;</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnLeft">
<A NAME="pgfId=328931">
 </A>
&nbsp;</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=328933">
 </A>
&nbsp;</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnLeft">
<A NAME="pgfId=328935">
 </A>
&nbsp;</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqn">
<A NAME="pgfId=328937">
 </A>
&nbsp;</P>
</TD>
</TR>
</TABLE>
<UL>
<LI CLASS="ExercisePart">
<A NAME="pgfId=123734">
 </A>
b.&nbsp;Show that the solutions to this equation can be written in terms of <SPAN CLASS="EquationVariables">
F</SPAN>
<SUP CLASS="Superscript">
1/</SUP>
<SUP CLASS="SuperscriptVariable">
N^</SUP>
 (the optimum stage effort) where <SPAN CLASS="EquationVariables">
N^</SPAN>
 is the optimum number of stages:  </LI>
<TABLE>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqn">
<A NAME="pgfId=223046">
 </A>
<SPAN CLASS="EquationVariables">
F</SPAN>
<SUP CLASS="Superscript">
1/</SUP>
<SUP CLASS="SuperscriptVariable">
N^</SUP>
(1 &#8211; ln <SPAN CLASS="EquationVariables">
F</SPAN>
<SUP CLASS="Superscript">
1/</SUP>
<SUP CLASS="SuperscriptVariable">
N^</SUP>
 ) + (<SPAN CLASS="EquationVariables">
p</SPAN>
<SUB CLASS="SubscriptVariable">
inv</SUB>
 + <SPAN CLASS="EquationVariables">
q</SPAN>
<SUB CLASS="SubscriptVariable">
inv</SUB>
 ) = 0 .</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnNumber">
<A NAME="pgfId=223048">
 </A>
(3.53)</P>
</TD>
</TR>
</TABLE>
</UL>
<P CLASS="ExerciseHead">
<A NAME="pgfId=123554">
 </A>
3.22&nbsp;<A NAME="16983">
 </A>
(XOR and XNOR cells, 60 min.) Table&nbsp;<A HREF="#37834" CLASS="XRef">
3.4</A>
 shows the implementations of two- and three-input XOR cells in an ASIC standard-cell library (D1 are the 1X drive cells, and D2 are the 2X drive versions). Can you explain the choices for the two-input XOR cell and complete the table for the three-input XOR cell?</P>
<TABLE>
<TR>
<TD ROWSPAN="1" COLSPAN="3">
<P CLASS="TableTitle">
<A NAME="pgfId=212693">
 </A>
TABLE&nbsp;3.4&nbsp;<A NAME="37834">
 </A>
Implementations of XOR cells (Problem <A HREF="#16983" CLASS="XRef">
3.22</A>
).</P>
</TD>
</TR>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableFirst">
<A NAME="pgfId=212699">
 </A>
<SPAN CLASS="TableHeads">
Cell</SPAN>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableFirst">
<A NAME="pgfId=212704">
 </A>