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Subject: Design of Counters - Example 1.6
Date: Sat, 1 Jan 2011 15:05:26 +0200
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f>
<P><BIG><BIG><STRONG>Design of Counters</STRONG></BIG></BIG></P>
<P>This example is taken from M. M. Mano, <EM>Digital Design</EM>, =
Prentice=20
Hall, 1984, p.243.</P>
<P><FONT color=3D#0000ff><BIG><STRONG>Example 1.6</STRONG></BIG></FONT>=20
&nbsp;&nbsp;&nbsp; <FONT face=3DArial>Design a counter specified by the =
state=20
diagram in <A=20
href=3D"http://www.eelab.usyd.edu.au/digital_tutorial/part3/example1-5.ht=
m">Example=20
1.5</A> using T flip-flops. The state diagram is shown here again in =
Figure=20
22.</FONT></P>
<TABLE border=3D0 width=3D"62%">
  <TBODY>
  <TR>
    <TD width=3D"37%"><IMG alt=3D"fig-18.gif (5027 bytes)"=20
      =
src=3D"http://www.eelab.usyd.edu.au/digital_tutorial/part3/fig-18.gif"=20
      width=3D270 height=3D260></TD>
    <TD width=3D"63%"><BIG>Figure 22. State diagram of a 3-bit binary=20
      counter.</BIG></TD></TR></TBODY></TABLE>
<P><FONT face=3DArial>The state table will be the same as in Example=20
1.5.</FONT></P>
<P><FONT face=3DArial>Now derive the excitation table from the state =
table, which=20
is shown in Table 17.</FONT></P>
<P><BIG>Table 17. Excitation table.</BIG></P>
<TABLE border=3D1 width=3D"64%">
  <TBODY>
  <TR>
    <TD bgColor=3D#6feccd width=3D"62%" colSpan=3D2>
      <P align=3Dcenter>Output State Transitions</P></TD>
    <TD bgColor=3D#ff8040 rowSpan=3D2 width=3D"38%">
      <P align=3Dcenter>Flip-flop inputs</P>
      <TABLE border=3D0 width=3D"100%">
        <TBODY>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>T2&nbsp; T1&nbsp;=20
      T0</TD></TR></TBODY></TABLE></TD></TR>
  <TR>
    <TD bgColor=3D#ff8040 width=3D"29%" align=3Dmiddle>Present State
      <TABLE border=3D0 width=3D"100%">
        <TBODY>
        <TR>
          <TD width=3D"100%">
            <P align=3Dcenter>Q2 Q1 =
Q0</P></TD></TR></TBODY></TABLE></TD>
    <TD bgColor=3D#ff8040 width=3D"33%" align=3Dmiddle>Next State
      <TABLE border=3D0 width=3D"100%">
        <TBODY>
        <TR>
          <TD width=3D"100%">
            <P align=3Dcenter>Q2 Q1 =
Q0</P></TD></TR></TBODY></TABLE></TD></TR>
  <TR>
    <TD width=3D"29%">
      <TABLE border=3D0 width=3D"100%">
        <TBODY>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>0&nbsp;&nbsp; 0&nbsp;&nbsp; =
0</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>0&nbsp;&nbsp; 0&nbsp;&nbsp; =
1</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>0&nbsp;&nbsp; 1&nbsp;&nbsp; =
0</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>0&nbsp;&nbsp; 1&nbsp;&nbsp; =
1</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>1&nbsp;&nbsp; 0&nbsp;&nbsp; =
0</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>1&nbsp;&nbsp; 0&nbsp;&nbsp; =
1</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>1&nbsp;&nbsp; 1&nbsp;&nbsp; =
0</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>1&nbsp;&nbsp; 1 &nbsp;=20
      1</TD></TR></TBODY></TABLE></TD>
    <TD width=3D"33%">
      <TABLE border=3D0 width=3D"100%">
        <TBODY>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>0&nbsp;&nbsp; 0&nbsp;&nbsp; =
1</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>0&nbsp;&nbsp; 1&nbsp;&nbsp; =
0</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>0&nbsp;&nbsp; 1&nbsp;&nbsp; =
1</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>1&nbsp;&nbsp; 0&nbsp;&nbsp; =
0</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>1&nbsp;&nbsp; 0&nbsp;&nbsp; =
1</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>1&nbsp;&nbsp; 1&nbsp;&nbsp; =
0</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>1&nbsp;&nbsp; 1&nbsp;&nbsp; =
1</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>0&nbsp;&nbsp; 0 &nbsp;=20
      0</TD></TR></TBODY></TABLE></TD>
    <TD width=3D"38%">
      <TABLE border=3D0 width=3D"100%">
        <TBODY>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>0&nbsp;&nbsp;&nbsp;=20
            0&nbsp;&nbsp;&nbsp; 1</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>0&nbsp;&nbsp;&nbsp;=20
            1&nbsp;&nbsp;&nbsp; 1</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>0&nbsp;&nbsp;&nbsp;=20
            0&nbsp;&nbsp;&nbsp; 1</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>1&nbsp;&nbsp;&nbsp;=20
            1&nbsp;&nbsp;&nbsp; 1</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>0&nbsp;&nbsp;&nbsp;=20
            0&nbsp;&nbsp;&nbsp; 1</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>0&nbsp;&nbsp;&nbsp;=20
            1&nbsp;&nbsp;&nbsp; 1</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>0&nbsp;&nbsp;&nbsp;=20
            0&nbsp;&nbsp;&nbsp; 1</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>1&nbsp;&nbsp;&nbsp;=20
            1&nbsp;&nbsp;&nbsp; =
1</TD></TR></TBODY></TABLE></TD></TR></TBODY></TABLE>
<P><FONT face=3DArial>Next step is to transfer the flip-flop input =
functions to=20
Karnaugh maps to derive a simplified Boolean expressions, which is shown =
in=20
Figure 23.</FONT></P>
<TABLE border=3D0 width=3D"73%">
  <TBODY>
  <TR>
    <TD width=3D"60%"><IMG alt=3D"fig-23.gif (9767 bytes)"=20
      =
src=3D"http://www.eelab.usyd.edu.au/digital_tutorial/part3/fig-23.gif"=20
      width=3D424 height=3D194></TD>
    <TD width=3D"40%"><BIG>Figure 23. Karnaugh =
maps</BIG></TD></TR></TBODY></TABLE>
<P><FONT face=3DArial>The following expressions are obtained:</FONT></P>
<P><STRONG><BIG>T0 =3D 1;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; T1 =
=3D Q0;=20
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; T2 =3D =
Q1*Q0</BIG></STRONG></P>
<P><FONT face=3DArial>Finally, draw the logic diagram of the circuit =
from the=20
expressions obtained. The complete logic diagram of the counter is shown =
in=20
Figure 24.</FONT></P>
<TABLE border=3D0 width=3D"61%">
  <TBODY>
  <TR>
    <TD width=3D"45%"><IMG alt=3D"fig-24.gif (2994 bytes)"=20
      =
src=3D"http://www.eelab.usyd.edu.au/digital_tutorial/part3/fig-24.gif"=20
      width=3D271 height=3D135></TD>
    <TD width=3D"55%"><BIG>Figure 24. Logic diagram of 3-bit binary=20
      counter.</BIG></TD></TR></TBODY></TABLE>
<P><FONT face=3DArial>To see the timing and state transitions of the =
counter,=20
click on this image.</FONT><A=20
href=3D"javascript:loadSession('cnt-animate.htm')"><IMG=20
alt=3D"signal.gif (1039 bytes)"=20
src=3D"http://www.eelab.usyd.edu.au/digital_tutorial/part3/signal.gif" =
width=3D70=20
height=3D25></A></P>
<P>&nbsp;</P>
<P><FONT face=3DArial>Now that you have reached the end of the tutorial, =
you=20
should be able to understand the basic concept of sequential circuits. =
You=20
should be able to analyse and design a basic sequential circuit. Now you =
can=20
practice some of the exercises using the analysis and design procedures =
shown in=20
the examples.</FONT></P>
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