On Mar 1, 12:24=A0pm, "patilrohit103"
<patilrohit103@n_o_s_p_a_m.rediffmail.com> wrote:
> I am having a problem calculating the baud rate of 9600 bauds for ARM7
> (LPC2148) at 12MHZ crystal. I have figured out a value of :
> =A0 =A0 =A0 =A0U1DLM=3D0x00;
> =A0 =A0 =A0 =A0U1DLL=3D0x62;
>
> following is my UART1 initialization code:
>
> void init_UART1()
> {
>
> =A0PINSEL0=3D0x00050000; // FOR UART1
>
> =A0U1LCR=3D0x83;
> =A0U1DLM=3D0x00;
> =A0U1DLL=3D0x62;
> =A0U1LCR=3D0x03;
> =A0U1FCR=3D0x01;
>
> }
>
> please help me if there is any other solution.....
>
> --------------------------------------- =A0 =A0 =A0 =A0
> Posted throughhttp://www.EmbeddedRelated.com

this looks like a standard 16550, which needs a divisor to give you
16x the desired baud rate.  9600 x 16 =3D 153600.
12e6 / 153600 =3D 78.125, implies 0x4e would be the divisor.  Note that
async serial can usually tolerate some baud rate deviation as the
receiver should resync on the start bit of every character.

I am having a problem calculating the baud rate of 9600 bauds for ARM7
(LPC2148) at 12MHZ crystal. I have figured out a value of :
       U1DLM=0x00;
       U1DLL=0x62;


following is my UART1 initialization code:

void init_UART1()
{

 PINSEL0=0x00050000; // FOR UART1
 
 U1LCR=0x83;
 U1DLM=0x00;
 U1DLL=0x62;
 U1LCR=0x03;
 U1FCR=0x01;
   
}

please help me if there is any other solution.....

	   
					
---------------------------------------		
Posted through http://www.EmbeddedRelated.com