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=<SPAN CLASS="Symbol">

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4 standard loads<SPAN CLASS="Symbol">

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=<SPAN CLASS="Symbol">

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4<SPAN CLASS="Symbol">

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pF<SPAN CLASS="Symbol">

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=<SPAN CLASS="Symbol">

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0.136<SPAN CLASS="Symbol">

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pF, </P>

<TABLE>

<TR>

<TD ROWSPAN="1" COLSPAN="1">

<P CLASS="TableEqn">

<A NAME="pgfId=295569">

 </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>

) = (38 + 817 (0.136)) ps = 149.112 ps .</P>

</TD>

<TD ROWSPAN="1" COLSPAN="1">

<P CLASS="TableEqnNumber">

<A NAME="pgfId=295571">

 </A>

(3.5)</P>

</TD>

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</TABLE>

<P CLASS="Body">

<A NAME="pgfId=221610">

 </A>

To make a comparison with the simulation we need to use ln<SPAN CLASS="Symbol">

 </SPAN>

(1/0.35)<SPAN CLASS="Symbol">

 </SPAN>

=<SPAN CLASS="Symbol">

 </SPAN>

1.04 and not approximately 1 as we have assumed, so that (with all times in ps)  </P>

<TABLE>

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<P CLASS="TableEqnRight">

<A NAME="pgfId=264835">

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&nbsp;</P>

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&nbsp;</P>

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<TD ROWSPAN="1" COLSPAN="1">

<P CLASS="TableEqnCenter">

<A NAME="pgfId=264839">

 </A>

&#8211;<SPAN CLASS="EquationVariables">

t'</SPAN>

</P>

</TD>

<TD ROWSPAN="1" COLSPAN="1">

<P CLASS="TableEqn">

<A NAME="pgfId=264841">

 </A>

&nbsp;</P>

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<P CLASS="TableEqn">

<A NAME="pgfId=264843">

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&nbsp;</P>

</TD>

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<TR>

<TD ROWSPAN="1" COLSPAN="1">

<P CLASS="TableEqnRight">

<A NAME="pgfId=264845">

 </A>

<SPAN CLASS="BodyComputer">

v(out1)</SPAN>

</P>

</TD>

<TD ROWSPAN="1" COLSPAN="1">

<P CLASS="TableEqnCenter">

<A NAME="pgfId=420232">

 </A>

<SPAN CLASS="Symbol">

&#170;</SPAN>

</P>

</TD>

<TD ROWSPAN="1" COLSPAN="1">

<P CLASS="TableEqnCenter">

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3.0&nbsp;exp</P>

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<P CLASS="TableEqnCenter">

<A NAME="pgfId=264849">

 </A>

&#8211;&#8211;&#8211;&#8211;&#8211;&#8211;&#8211;&#8211;&#8211;&#8211;&#8211;</P>

</TD>

<TD ROWSPAN="1" COLSPAN="1">

<P CLASS="TableEqn">

<A NAME="pgfId=264851">

 </A>

V</P>

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<P CLASS="TableLeft">

<A NAME="pgfId=264853">

 </A>

&nbsp;</P>

</TD>

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<TR>

<TD ROWSPAN="1" COLSPAN="1">

<P CLASS="TableEqnRight">

<A NAME="pgfId=264855">

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&nbsp;</P>

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<P CLASS="TableEqnCenter">

<A NAME="pgfId=420234">

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&nbsp;</P>

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&nbsp;</P>

</TD>

<TD ROWSPAN="1" COLSPAN="1">

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<A NAME="pgfId=264859">

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149.112/1.04</P>

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<P CLASS="TableEqn">

<A NAME="pgfId=264861">

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&nbsp;</P>

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<A NAME="pgfId=264863">

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&nbsp;</P>

</TD>

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&nbsp;</P>

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&nbsp;</P>

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&nbsp;</P>

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<A NAME="pgfId=420238">

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&nbsp;</P>

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&nbsp;</P>

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<P CLASS="TableEqnCenter">

<A NAME="pgfId=265091">

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&#8211;(<SPAN CLASS="EquationVariables">

t</SPAN>

 &#8211; 20)</P>

</TD>

<TD ROWSPAN="1" COLSPAN="1">

<P CLASS="TableEqn">

<A NAME="pgfId=265093">

 </A>

&nbsp;</P>

</TD>

<TD ROWSPAN="1" COLSPAN="1">

<P CLASS="TableEqn">

<A NAME="pgfId=265067">

 </A>

&nbsp;</P>

</TD>

</TR>

<TR>

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<P CLASS="TableEqnRight">

<A NAME="pgfId=265069">

 </A>

&nbsp;</P>

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<A NAME="pgfId=420240">

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=</P>

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<P CLASS="TableEqnCenter">

<A NAME="pgfId=265095">

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 3.0 exp</P>

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<A NAME="pgfId=265097">

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&#8211;&#8211;&#8211;&#8211;&#8211;&#8211;&#8211;&#8211;&#8211;</P>

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<TD ROWSPAN="1" COLSPAN="1">

<P CLASS="TableEqn">

<A NAME="pgfId=265099">

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&nbsp;&nbsp;&nbsp;for t &gt; 20 ps .</P>

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<TD ROWSPAN="1" COLSPAN="1">

<P CLASS="TableEqnNumber">

<A NAME="pgfId=265077">

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<A NAME="17393">

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(3.6)</P>

</TD>

</TR>

<TR>

<TD ROWSPAN="1" COLSPAN="1">

<P CLASS="TableEqnRight">

<A NAME="pgfId=265079">

 </A>

&nbsp;</P>

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<P CLASS="TableEqnCenter">

<A NAME="pgfId=420242">

 </A>

&nbsp;</P>

</TD>

<TD ROWSPAN="1" COLSPAN="1">

<P CLASS="TableEqnCenter">

<A NAME="pgfId=265101">

 </A>

&nbsp;</P>

</TD>

<TD ROWSPAN="1" COLSPAN="1">

<P CLASS="TableEqnCenter">

<A NAME="pgfId=265103">

 </A>

143.4</P>

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<P CLASS="TableEqn">

<A NAME="pgfId=265105">

 </A>

&nbsp;</P>

</TD>

<TD ROWSPAN="1" COLSPAN="1">

<P CLASS="TableEqn">

<A NAME="pgfId=265087">

 </A>

&nbsp;</P>

</TD>

</TR>

</TABLE>

<P CLASS="Body">

<A NAME="pgfId=184561">

 </A>

Equation <A HREF="#17393" CLASS="XRef">

3.6</A>

 is plotted in Figure&nbsp;<A HREF="#19386" CLASS="XRef">

3.3</A>

(d). For <SPAN CLASS="BodyComputer">

v(out1)</SPAN>

<SPAN CLASS="Symbol">

 </SPAN>

=<SPAN CLASS="Symbol">

 </SPAN>

1.05<SPAN CLASS="Symbol">

 </SPAN>

V (equal to the 0.35 output trip point), Eq.&nbsp;<A HREF="#17393" CLASS="XRef">

3.6</A>

 predicts <SPAN CLASS="EquationVariables">

t</SPAN>

<SPAN CLASS="Symbol">

 </SPAN>

=<SPAN CLASS="Symbol">

 </SPAN>

20<SPAN CLASS="Symbol">

 </SPAN>

+<SPAN CLASS="Symbol">

 </SPAN>

149.112<SPAN CLASS="Symbol">

 &#170; </SPAN>

169<SPAN CLASS="Symbol">

 </SPAN>

ps and agrees with Figure&nbsp;<A HREF="#19386" CLASS="XRef">

3.3</A>

(b)&#8212;it should because we derived the model from these results!</P>

<P CLASS="Body">

<A NAME="pgfId=290708">

 </A>

Now we find <SPAN CLASS="EquationVariables">

C</SPAN>

<SUB CLASS="SubscriptVariable">

p</SUB>

. From Figure&nbsp;<A HREF="#19386" CLASS="XRef">

3.3</A>

(c) and Eq.&nbsp;<A HREF="#10713" CLASS="XRef">

3.2</A>

 </P>

<TABLE>

<TR>

<TD ROWSPAN="1" COLSPAN="1">

<P CLASS="TableEqn">

<A NAME="pgfId=221812">

 </A>

<SPAN CLASS="EquationVariables">

t</SPAN>

<SUB CLASS="SubscriptVariable">

PDr</SUB>

 = (52 + 1281 <SPAN CLASS="EquationVariables">

C</SPAN>

<SUB CLASS="SubscriptVariable">

out </SUB>

) ps&nbsp;&nbsp;thus&nbsp;&nbsp;<SPAN CLASS="EquationVariables">

C</SPAN>

<SUB CLASS="SubscriptVariable">

pr</SUB>

 = 52/1281 = 0.041 pF&nbsp;&nbsp;(rising) ,</P>

</TD>

<TD ROWSPAN="1" COLSPAN="1">

<P CLASS="TableEqn">

<A NAME="pgfId=221814">

 </A>

&nbsp;</P>

</TD>

</TR>

<TR>

<TD ROWSPAN="1" COLSPAN="1">

<P CLASS="TableEqn">

<A NAME="pgfId=221816">

 </A>

&nbsp;</P>

</TD>

<TD ROWSPAN="1" COLSPAN="1">

<P CLASS="TableEqn">

<A NAME="pgfId=221818">

 </A>

&nbsp;</P>

</TD>

</TR>

<TR>

<TD ROWSPAN="1" COLSPAN="1">

<P CLASS="TableEqn">

<A NAME="pgfId=224448">

 </A>

<SPAN CLASS="EquationVariables">

t</SPAN>

<SUB CLASS="SubscriptVariable">

PDf</SUB>

 = (38 + 817 <SPAN CLASS="EquationVariables">

C</SPAN>

<SUB CLASS="SubscriptVariable">

out </SUB>

) ps&nbsp;&nbsp;thus&nbsp;&nbsp;<SPAN CLASS="EquationVariables">

C</SPAN>

<SUB CLASS="SubscriptVariable">

pf</SUB>

 = 38/817 = 0.047 pF&nbsp;&nbsp;(falling) .</P>

</TD>

<TD ROWSPAN="1" COLSPAN="1">

<P CLASS="TableEqnNumber">

<A NAME="pgfId=221822">

 </A>

<A NAME="33563">

 </A>

(3.7)</P>

</TD>

</TR>

</TABLE>

<P CLASS="BodyAfterHead">

<A NAME="pgfId=187479">

 </A>

These intrinsic parasitic capacitance values depend on the choice of output trip points, even though <SPAN CLASS="EquationVariables">

C</SPAN>

<SUB CLASS="SubscriptVariable">

pf </SUB>

<SPAN CLASS="Symbol">

 </SPAN>

<SPAN CLASS="EquationVariables">

R</SPAN>

<SUB CLASS="SubscriptVariable">

pdf</SUB>

 and <SPAN CLASS="EquationVariables">

C</SPAN>

<SUB CLASS="SubscriptVariable">

pr</SUB>

<SPAN CLASS="Symbol">

 </SPAN>

<SPAN CLASS="EquationVariables">

R</SPAN>

<SUB CLASS="SubscriptVariable">

pdr</SUB>

 are constant for a given input trip point and waveform, because the pull-up and pull-down resistances depend on the choice of output trip points. We take a closer look at parasitic capacitance next.</P>

<HR><P>[&nbsp;<A HREF="CH03.htm">Chapter&nbsp;start</A>&nbsp;]&nbsp;[&nbsp;<A HREF="CH03.htm">Previous&nbsp;page</A>&nbsp;]&nbsp;[&nbsp;<A HREF="CH03.2.htm">Next&nbsp;page</A>&nbsp;]</P></BODY>



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