📄 ch15.5.htm
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t</SUB>
<SUB CLASS="SubscriptVariable">
n</SUB>
) -0.5 <SPAN CLASS="EquationVariables">
V</SPAN>
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] <SPAN CLASS="EquationVariables">
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12 <SPAN CLASS="Symbol">
¥</SPAN>
10<SUP CLASS="Superscript">
–3</SUP>
</P>
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=</P>
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–––––––––––––––––––––––––</P>
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[(3.3 – 0.65) – (0.5) (0.5)] (0.5)</P>
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=</P>
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<A NAME="pgfId=202290">
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0.01 AV<SUP CLASS="Superscript">
–1</SUP>
</P>
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<P CLASS="Body">
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If the output buffer is switching at 100 MHz and the input rise time to the buffer is 2 ns, we can calculate the power dissipation due to short-circuit current as </P>
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<A NAME="pgfId=202539">
</A>
<SPAN CLASS="EquationVariables">
P</SPAN>
<SUB CLASS="Subscript">
2</SUB>
</P>
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<TD ROWSPAN="1" COLSPAN="1">
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<A NAME="pgfId=202541">
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=</P>
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<A NAME="pgfId=202543">
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(1/12) <SPAN CLASS="Symbol">
b </SPAN>
<SPAN CLASS="EquationVariables">
f t</SPAN>
<SUB CLASS="SubscriptVariable">
rf</SUB>
<SPAN CLASS="EquationVariables">
(V</SPAN>
<SUB CLASS="SubscriptVariable">
DD</SUB>
– 2 V<SUB CLASS="Subscript">
t</SUB>
<SUB CLASS="SubscriptVariable">
n</SUB>
)<SUP CLASS="Superscript">
3</SUP>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnNumber">
<A NAME="pgfId=202545">
</A>
(15.7)</P>
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<TR>
<TD ROWSPAN="1" COLSPAN="1">
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<A NAME="pgfId=202547">
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<A NAME="pgfId=202549">
</A>
=</P>
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<A NAME="pgfId=202551">
</A>
(0.01) (100 <SPAN CLASS="Symbol">
¥</SPAN>
106) (2 <SPAN CLASS="Symbol">
¥</SPAN>
10<SUP CLASS="Superscript">
–9</SUP>
) (3.3 – (2)(0.65))<SUP CLASS="Superscript">
3</SUP>
</P>
</TD>
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=</P>
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0.00133W or about 1 mW .</P>
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<P CLASS="Body">
<A NAME="pgfId=161289">
</A>
If the output load is 10 pF, the dissipation due to switching current is </P>
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<SPAN CLASS="EquationVariables">
P</SPAN>
<SUB CLASS="Subscript">
1</SUB>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=202616">
</A>
=</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnLeft">
<A NAME="pgfId=202618">
</A>
<SPAN CLASS="EquationVariables">
fCV</SPAN>
<SUP CLASS="Superscript">
2</SUP>
<SUB CLASS="SubscriptVariable">
DD</SUB>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqn">
<A NAME="pgfId=202620">
</A>
</P>
</TD>
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<TR>
<TD ROWSPAN="1" COLSPAN="1">
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</A>
</P>
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<A NAME="pgfId=202624">
</A>
=</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnLeft">
<A NAME="pgfId=202626">
</A>
(100 <SPAN CLASS="Symbol">
¥</SPAN>
10<SUP CLASS="Superscript">
6</SUP>
) (10 <SPAN CLASS="Symbol">
¥</SPAN>
10<SUP CLASS="Superscript">
–12</SUP>
)(3.3)<SUP CLASS="Superscript">
2</SUP>
</P>
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<P CLASS="TableEqn">
<A NAME="pgfId=202628">
</A>
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<A NAME="pgfId=202632">
</A>
=</P>
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0.01089 W or about 10 mW .</P>
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<P CLASS="TableEqn">
<A NAME="pgfId=202636">
</A>
</P>
</TD>
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<P CLASS="BodyAfterHead">
<A NAME="pgfId=161293">
</A>
As a general rule, if we adjust the transistor sizes so that the rise times and fall times through a chain of logic are approximately equal (as they should be), the short-circuit current is typically less than 20 percent of the switching current.</P>
<P CLASS="Body">
<A NAME="pgfId=161294">
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For the example output buffer, we can make a rough estimate of the output-node switching time by assuming the buffer output drive current is constant at 12 mA. This current will cause the voltage on the output load capacitance to change between 3.3 V and 0 V at a constant slew rate d<SPAN CLASS="EquationVariables">
V</SPAN>
/d<SPAN CLASS="EquationVariables">
t</SPAN>
for a time </P>
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<SPAN CLASS="EquationVariables">
C</SPAN>
<SPAN CLASS="Symbol">
D</SPAN>
<SPAN CLASS="EquationVariables">
V</SPAN>
</P>
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(10 <SPAN CLASS="Symbol">
¥</SPAN>
10<SUP CLASS="Superscript">
–12</SUP>
) (3.3)</P>
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</A>
</P>
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<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnRight">
<A NAME="pgfId=202725">
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<SPAN CLASS="Symbol">
D</SPAN>
<SPAN CLASS="EquationVariables">
t</SPAN>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=202727">
</A>
=</P>
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</A>
–––––</P>
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</A>
=</P>
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</A>
––––––––––––––––</P>
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(15.8)</P>
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<SPAN CLASS="EquationVariables">
I</SPAN>
</P>
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</A>
</P>
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</A>
(12 <SPAN CLASS="Symbol">
¥</SPAN>
10<SUP CLASS="Superscript">
–3</SUP>
)</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=202749">
</A>
</P>
</TD>
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</A>
</P>
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<P CLASS="BodyAfterHead">
<A NAME="pgfId=161299">
</A>
This is close to the input rise time of 2 ns. So our estimate of the short-circuit current being less than 20 percent of the switching current assuming equal input rise time and output rise time is valid in this case.</P>
</DIV>
<DIV>
<H2 CLASS="Heading2">
<A NAME="pgfId=161301">
</A>
15.5.3 <A NAME="10200">
</A>
Subthreshold and Leakage Current</H2>
<P CLASS="BodyAfterHead">
<A NAME="pgfId=192871">
</A>
Despite the claim made in Section 2.1, a CMOS transistor is never completely <SPAN CLASS="Emphasis">
off</SPAN>
. For example, a typical specification for a 0.5 <SPAN CLASS="Symbol">
m</SPAN>
m process for the <A NAME="marker=192872">
</A>
subthreshold current (per micron of gate width for <SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
GS</SUB>
= 0 V) is less than 5 pA<SPAN CLASS="Symbol">
m</SPAN>
m<SUP CLASS="Superscript">
–1</SUP>
, but not zero. With 10 million transistors on a large chip and with each transistor 10 <SPAN CLASS="Symbol">
m</SPAN>
m wide, we will have a total subthreshold current of 0.1 mA; high, but reasonable. The problem is that the subthreshold current does not scale with process technology. </P>
<P CLASS="Body">
<A NAME="pgfId=185316">
</A>
When the gate-to-source voltage, <SPAN CLASS="EquationVariables">
V</SPAN>
<SUB CLASS="SubscriptVariable">
GS</SUB>
, of an MOS transistor is less than the threshold voltage, <SPAN CLASS="EquationNumber">
V</SPAN>
<SUB CLASS="Subscript">
t</SUB>
<SPAN CLASS="EquationNumber">
,</SPAN>
the transistor conducts a very small subthreshold current in the <A NAME="marker=161308">
</A>
subthreshold region </P>
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