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📄 serial interface (rs-232) 2.baud generator.mht

📁 用FPGA实现RS232通信,此代码是用VHDL语言编写
💻 MHT
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Subject: FPGAs are fun!
Date: Wed, 19 Apr 2006 00:28:21 +0800
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    <TD vAlign=3Dtop>
      <H4>Baud generator</H4>Here we want to use the serial link at =
maximum=20
      speed, i.e. 115200 bauds. Other slower speeds would also be easy =
to=20
      generate.<BR><BR>FPGAs usually run at speed well above 115200Hz =
(RS-232 is=20
      pretty slow by today's standards). That means we use a high-speed =
clock=20
      and divide it down to generate a "tick" as close as possible to =
115200=20
      times a second.=20
      <H4>Synchronous tick from a 1.8432MHz clock</H4>Traditionally, =
RS-232=20
      chips use a 1.8432MHz clock, because that makes generating the =
standard=20
      baud frequencies very easy. So let's assume we have a 1.8432MHz =
clock=20
      available.<BR><BR>1.8432MHz divided by 16 gives 115200Hz, what a=20
      coincidence!<BR><BR>
      <TABLE cellPadding=3D10>
        <TBODY>
        <TR>
          <TD bgColor=3D#d0d0f0><B>reg</B> [3:0] =
BaudDivCnt;<BR><B>always</B>=20
            @(<B>posedge</B> clk) BaudDivCnt &lt;=3D BaudDivCnt +=20
            1;<BR><BR><B>wire</B> BaudTick =3D (BaudDivCnt=3D=3D15);=20
      </TD></TR></TBODY></TABLE><BR>So "BaudTick" is asserted once every =
16=20
      clocks, i.e. 115200 times a second when using a 1.8432MHz clock.=20
      <H4>Synchronous tick from any frequency</H4>The earlier generator =
was=20
      assuming the use of a 1.8432MHz clock. But what do you do if all =
your have=20
      is, say, a 2MHz clock? To generate 115200Hz from a 2MHz clock, you =
divide=20
      the clock by "17.361111111..." Not exactly a round number. The =
solution is=20
      to divide sometimes by 17, sometimes by 18, making sure the ratio =
stays=20
      "17.361111111". That's actually easy to do.<BR><BR>Look at the =
following=20
      "C" code:<BR>
      <TABLE cellPadding=3D10>
        <TBODY>
        <TR>
          <TD bgColor=3D#d0d0f0>while(1) // repeat =
forever<BR>{<BR>&nbsp; acc +=3D=20
            115200;<BR>&nbsp; if(acc&gt;=3D2000000) printf("*"); else =
printf("=20
            ");<BR><BR>&nbsp; acc %=3D=20
      2000000;<BR>}<BR></TD></TR></TBODY></TABLE><BR>That prints the "*" =
in the=20
      exact ratio, once every "17.361111111..." loops on =
average.<BR><BR>To=20
      obtain the same thing efficiently in an FPGA, we rely on the fact =
that the=20
      serial interface can tolerate a few % of error in the baud =
frequency=20
      generator. It really won't matter if we use "17.3" or "17.4".=20
      <H4>FPGA baud generator</H4>It is desirable that the 2000000 be a =
power of=20
      two. Obviously 2000000 is not a power of two. So we change the =
ratio.=20
      Instead of the ratio "2000000/115200", let's use "1024/59" =3D =
17.356.=20
      That's very close to our ideal ratio, and makes an efficient FPGA=20
      implementation.<BR><BR>
      <TABLE cellPadding=3D10>
        <TBODY>
        <TR>
          <TD bgColor=3D#d0d0f0>// 10 bits for the accumulator ([9:0]), =
and one=20
            extra bit for the accumulator carry-out ([10])<BR><B>reg</B> =
[10:0]=20
            acc; &nbsp; // 11 bits total!<BR><BR><B>always</B> =
@(<B>posedge</B>=20
            clk)<BR>&nbsp; acc &lt;=3D acc[9:0] + 59; // use only 10 =
bits from the=20
            previous result, but save the full 11 =
bits<BR><BR><B>wire</B>=20
            BaudTick =3D acc[10]; // so that the 11th bit is the =
carry-out=20
        </TD></TR></TBODY></TABLE><BR>Using our 2MHz clock, "BaudTick" =
is asserted=20
      115234 times a second, a 0.03% error from the ideal 115200.=20
      <H4>Parameterized FPGA baud generator</H4>The previous design was =
using a=20
      10 bits accumulator, but as the clock frequency increases, more =
bits are=20
      required.<BR><BR>Here's a design with a 25MHz clock and a 16 bits=20
      accumulator. The design is parameterized, so easy to =
customize.<BR><BR>
      <TABLE cellPadding=3D10>
        <TBODY>
        <TR>
          <TD bgColor=3D#d0d0f0><B>parameter</B> ClkFrequency =3D =
25000000; //=20
            25MHz<BR><B>parameter</B> Baud =3D =
115200;<BR><B>parameter</B>=20
            BaudGeneratorAccWidth =3D 16;<BR><B>parameter</B> =
BaudGeneratorInc =3D=20
            =
(Baud&lt;&lt;BaudGeneratorAccWidth)/ClkFrequency;<BR><BR><B>reg</B>=20
            [BaudGeneratorAccWidth:0] BaudGeneratorAcc;<BR><B>always</B> =

            @(<B>posedge</B> clk)<BR>&nbsp; BaudGeneratorAcc &lt;=3D=20
            BaudGeneratorAcc[BaudGeneratorAccWidth-1:0] +=20
            BaudGeneratorInc;<BR><BR><B>wire</B> BaudTick =3D=20
            BaudGeneratorAcc[BaudGeneratorAccWidth]; =
</TD></TR></TBODY></TABLE><BR>One=20
      last implementation issue: the "BaudGeneratorInc" calculation is =
wrong,=20
      due to the fact that Verilog uses 32 bits intermediate results, =
and the=20
      calculation exceeds that. Change the line as follow for a=20
      workaround.<BR><BR>
      <TABLE cellPadding=3D10>
        <TBODY>
        <TR>
          <TD bgColor=3D#d0d0f0><B>parameter</B> BaudGeneratorInc =3D=20
            =
((Baud&lt;&lt;(BaudGeneratorAccWidth-4))+(ClkFrequency&gt;&gt;5))/(ClkFre=
quency&gt;&gt;4);=20
          </TD></TR></TBODY></TABLE><BR>This line has also the added =
advantage to=20
      round the result instead of truncating.<BR><BR>That's it.<BR>Now =
that we=20
      have a precise enough Baud generator, we can go ahead with the =
RS-232=20
      transmitter and receiver modules. <BR><BR><A=20
      href=3D"http://www.fpga4fun.com/SerialInterface3.html"><FONT =
color=3Dblue=20
      size=3D2>&gt;&gt;&gt; NEXT: RS-232 transmitter module=20
      &gt;&gt;&gt;</FONT></A><BR><BR><BR>
      <HR>
      This page was last updated on December 01=20
2004.<BR><BR></TD></TR></TBODY></TABLE></BODY></HTML>

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