📄 ch15.5.htm
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</A>
<A HREF="CH15.5.htm#37555" CLASS="XRef">
Table 15.6</A>
shows specified maximum leakage currents of junction parasitic diodes as well as the leakage currents of the <A NAME="marker=161340">
</A>
field transistors (the parasitic MOS transistors formed when poly crosses over the thick oxide, or field oxide) in a typical 0.5 <SPAN CLASS="Symbol">
m</SPAN>
m process. </P>
<TABLE>
<TR>
<TD ROWSPAN="1" COLSPAN="4">
<P CLASS="TableTitle">
<A NAME="pgfId=182781">
</A>
TABLE 15.6 <A NAME="37555">
</A>
Diffusion leakage currents (at 25 °C) for a typical 0.5 <SPAN CLASS="Symbol">
m</SPAN>
m (<SPAN CLASS="Symbol">
l</SPAN>
= 0.25 <SPAN CLASS="Symbol">
m</SPAN>
m) CMOS process.</P>
</TD>
</TR>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableLeft">
<A NAME="pgfId=161352">
</A>
<SPAN CLASS="TableHeads">
Junction</SPAN>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableLeft">
<A NAME="pgfId=161354">
</A>
<SPAN CLASS="TableHeads">
Diode type</SPAN>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableFirst">
<A NAME="pgfId=161356">
</A>
<SPAN CLASS="TableHeads">
Leakage (max.)</SPAN>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableFirst">
<A NAME="pgfId=161358">
</A>
<SPAN CLASS="TableHeads">
Unit</SPAN>
</P>
</TD>
</TR>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableLeft">
<A NAME="pgfId=161360">
</A>
<SPAN CLASS="Emphasis">
n</SPAN>
-diffusion/<SPAN CLASS="Emphasis">
p</SPAN>
-substrate</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableLeft">
<A NAME="pgfId=161362">
</A>
area</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="Table">
<A NAME="pgfId=161364">
</A>
0.6</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="Table">
<A NAME="pgfId=161366">
</A>
fA <SPAN CLASS="Symbol">
m</SPAN>
m<SUP CLASS="Superscript">
–2</SUP>
V<SUP CLASS="Superscript">
–1</SUP>
</P>
</TD>
</TR>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableLeft">
<A NAME="pgfId=161368">
</A>
<SPAN CLASS="Emphasis">
n</SPAN>
-diffusion/<SPAN CLASS="Emphasis">
p</SPAN>
-substrate</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableLeft">
<A NAME="pgfId=161370">
</A>
perimeter</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="Table">
<A NAME="pgfId=161372">
</A>
2.0</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="Table">
<A NAME="pgfId=161374">
</A>
fA <SPAN CLASS="Symbol">
m</SPAN>
m<SUP CLASS="Superscript">
–1</SUP>
V<SUP CLASS="Superscript">
–1</SUP>
</P>
</TD>
</TR>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableLeft">
<A NAME="pgfId=161376">
</A>
<SPAN CLASS="Emphasis">
p</SPAN>
-diffusion/<SPAN CLASS="Emphasis">
n</SPAN>
-well</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableLeft">
<A NAME="pgfId=161378">
</A>
area</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="Table">
<A NAME="pgfId=161380">
</A>
0.6</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="Table">
<A NAME="pgfId=161382">
</A>
fA <SPAN CLASS="Symbol">
m</SPAN>
m<SUP CLASS="Superscript">
–2</SUP>
V<SUP CLASS="Superscript">
–1</SUP>
</P>
</TD>
</TR>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableLeft">
<A NAME="pgfId=161384">
</A>
<SPAN CLASS="Emphasis">
p</SPAN>
-diff/<SPAN CLASS="Emphasis">
n</SPAN>
-well</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableLeft">
<A NAME="pgfId=161386">
</A>
perimeter</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="Table">
<A NAME="pgfId=161388">
</A>
3.0</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="Table">
<A NAME="pgfId=161390">
</A>
fA <SPAN CLASS="Symbol">
m</SPAN>
m<SUP CLASS="Superscript">
–1</SUP>
V<SUP CLASS="Superscript">
–1</SUP>
</P>
</TD>
</TR>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableLeft">
<A NAME="pgfId=161392">
</A>
<SPAN CLASS="Emphasis">
n</SPAN>
-well /<SPAN CLASS="Emphasis">
p</SPAN>
-substrate</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableLeft">
<A NAME="pgfId=161394">
</A>
area</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="Table">
<A NAME="pgfId=161396">
</A>
1.0</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="Table">
<A NAME="pgfId=161398">
</A>
fA <SPAN CLASS="Symbol">
m</SPAN>
m<SUP CLASS="Superscript">
–2</SUP>
V<SUP CLASS="Superscript">
–1</SUP>
</P>
</TD>
</TR>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableLeft">
<A NAME="pgfId=161400">
</A>
Field NMOS transistor</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableLeft">
<A NAME="pgfId=161402">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="Table">
<A NAME="pgfId=161404">
</A>
100</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="Table">
<A NAME="pgfId=161406">
</A>
fA <SPAN CLASS="Symbol">
m</SPAN>
m<SUP CLASS="Superscript">
–1</SUP>
</P>
</TD>
</TR>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableLeft">
<A NAME="pgfId=161408">
</A>
Field PMOS transistor</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableLeft">
<A NAME="pgfId=161410">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="Table">
<A NAME="pgfId=161412">
</A>
30</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="Table">
<A NAME="pgfId=161414">
</A>
fA <SPAN CLASS="Symbol">
m</SPAN>
m<SUP CLASS="Superscript">
–1</SUP>
</P>
</TD>
</TR>
</TABLE>
<P CLASS="Body">
<A NAME="pgfId=161415">
</A>
For example, if we have an <SPAN CLASS="EmphasisPrefix">
n</SPAN>
-diffusion region at a potential of 3.3 V that is 10<SPAN CLASS="Symbol">
m</SPAN>
m by 4<SPAN CLASS="Symbol">
m</SPAN>
m in size, the parasitic leakage current due to the area diode would be </P>
<TABLE>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnRight">
<A NAME="pgfId=203650">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=203652">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnLeft">
<A NAME="pgfId=203654">
</A>
40 <SPAN CLASS="Symbol">
m</SPAN>
m<SUP CLASS="Superscript">
2</SUP>
<SPAN CLASS="Symbol">
¥</SPAN>
3.3 V <SPAN CLASS="Symbol">
¥</SPAN>
0.6 fA <SPAN CLASS="Symbol">
m</SPAN>
m<SUP CLASS="Superscript">
–2</SUP>
V<SUP CLASS="Superscript">
–1</SUP>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=203656">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqn">
<A NAME="pgfId=203658">
</A>
</P>
</TD>
</TR>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnRight">
<A NAME="pgfId=203666">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=203668">
</A>
=</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnLeft">
<A NAME="pgfId=203670">
</A>
(40) (3.3) (0.6 <SPAN CLASS="Symbol">
¥</SPAN>
10<SUP CLASS="Superscript">
–15</SUP>
)</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=203672">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqn">
<A NAME="pgfId=203674">
</A>
</P>
</TD>
</TR>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnRight">
<A NAME="pgfId=203676">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=203678">
</A>
=</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnLeft">
<A NAME="pgfId=203680">
</A>
7.92 <SPAN CLASS="Symbol">
¥</SPAN>
10<SUP CLASS="Superscript">
–14</SUP>
A ,</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=203682">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqn">
<A NAME="pgfId=203684">
</A>
</P>
</TD>
</TR>
</TABLE>
<P CLASS="BodyAfterHead">
<A NAME="pgfId=161420">
</A>
or approximately 80 fA. </P>
<P CLASS="Body">
<A NAME="pgfId=192041">
</A>
The perimeter of this drain region is 28 <SPAN CLASS="Symbol">
m</SPAN>
m, so that the leakage current due to the perimeter diode is </P>
<TABLE>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnRight">
<A NAME="pgfId=203688">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=203690">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnLeft">
<A NAME="pgfId=203692">
</A>
28 <SPAN CLASS="Symbol">
m</SPAN>
m <SPAN CLASS="Symbol">
¥</SPAN>
3.3 V <SPAN CLASS="Symbol">
¥</SPAN>
2.0 fA <SPAN CLASS="Symbol">
m</SPAN>
m<SUP CLASS="Superscript">
–1</SUP>
V<SUP CLASS="Superscript">
–1</SUP>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=203694">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqn">
<A NAME="pgfId=203696">
</A>
</P>
</TD>
</TR>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnRight">
<A NAME="pgfId=203698">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=203700">
</A>
=</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnLeft">
<A NAME="pgfId=203702">
</A>
(28) (3.3) (2.0 <SPAN CLASS="Symbol">
¥</SPAN>
10<SUP CLASS="Superscript">
–15</SUP>
)</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=203704">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqn">
<A NAME="pgfId=203706">
</A>
</P>
</TD>
</TR>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnRight">
<A NAME="pgfId=203708">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=203710">
</A>
=</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnLeft">
<A NAME="pgfId=203712">
</A>
2.848 <SPAN CLASS="Symbol">
¥</SPAN>
10<SUP CLASS="Superscript">
–13</SUP>
A ,</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=203714">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqn">
<A NAME="pgfId=203716">
</A>
</P>
</TD>
</TR>
</TABLE>
<P CLASS="BodyAfterHead">
<A NAME="pgfId=161426">
</A>
or approximately 0.2 pA, over twice as large as the area-diode leakage current.</P>
<P CLASS="Body">
<A NAME="pgfId=161427">
</A>
As a very rough estimate, if we have 100,000 transistors each with a source and a drain 10<SPAN CLASS="Symbol">
m</SPAN>
m by 4 <SPAN CLASS="Symbol">
m</SPAN>
m, and half of them are biased at 3.3 V, then the total leakage current would be </P>
<TABLE>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnRight">
<A NAME="pgfId=203756">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=203758">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnLeft">
<A NAME="pgfId=203760">
</A>
(100 <SPAN CLASS="Symbol">
¥</SPAN>
10<SUP CLASS="Superscript">
5</SUP>
) (2) (0.5) (280 <SPAN CLASS="Symbol">
¥</SPAN>
10<SUP CLASS="Superscript">
–15</SUP>
)</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=203762">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqn">
<A NAME="pgfId=203764">
</A>
</P>
</TD>
</TR>
<TR>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnRight">
<A NAME="pgfId=203766">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=203768">
</A>
=</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnLeft">
<A NAME="pgfId=203770">
</A>
2.8 <SPAN CLASS="Symbol">
¥</SPAN>
10<SUP CLASS="Superscript">
–6</SUP>
A ,</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnCenter">
<A NAME="pgfId=203772">
</A>
</P>
</TD>
<TD ROWSPAN="1" COLSPAN="1">
<P CLASS="TableEqnNumber">
<A NAME="pgfId=203774">
</A>
(15.12)</P>
</TD>
</TR>
</TABLE>
<P CLASS="BodyAfterHead">
<A NAME="pgfId=161432">
</A>
or approximately 3 <SPAN CLASS="Symbol">
m</SPAN>
A. This is the same order of magnitude (a few microamperes) as the quiescent leakage current, <SPAN CLASS="EquationVariables">
I</SPAN>
<SUB CLASS="SubscriptVariable">
DDQ</SUB>
, that we expect to measure when we test an ASIC with power applied, but with no signal activity. A measurement of more current than this in a nonactive CMOS ASIC indicates a problem with the chip manufacture or the design. We use this measurement to test an ASIC using an <A NAME="marker=182785">
</A>
IDDQ test.</P>
</DIV>
<HR><P>[ <A HREF="CH15.htm">Chapter start</A> ] [ <A HREF="CH15.4.htm">Previous page</A> ] [ <A HREF="CH15.6.htm">Next page</A> ]</P></BODY>
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