Input to the timerclk on the MPC8280

[Hans Schillstrom]

Hi,
I have the same problem on MPC8270,
and solved it two parts in u-boot:
- Setup you PC[26] to somtething else than TMCLK ex. g,p. I/O
    then you an automatic selecttion of BRG1 as source.
    
- Write a proper value in BRGC1 and make shure that 
    bit 16-17 is 0 for selection of CPM as clock source.

and then you have to "move around the brgs" in (2.6.10)
drivers/serial/cpm_uart/cpm_uart_cpm2.c

I moved smc 1&2 to brg7 & 8
scc1 to brg2 etc
scc4 ext clk (48MHz for USB)

void smc1_lineif(struct uart_cpm_port *pinfo)
{
	volatile iop_cpm2_t *io = &cpm2_immr->im_ioport;

	/* SMC1 is only on port D */
	io->iop_ppard |= 0x00c00000;
	io->iop_pdird |= 0x00400000;
	io->iop_pdird &= ~0x00800000;
	io->iop_psord &= ~0x00c00000;
#ifndef CONFIG_RCB8270
	/* Wire BRG1 to SMC1 */
	cpm2_immr->im_cpmux.cmx_smr &= 0x0f;
	pinfo->brg = 1;
#else
        /* Wire BRG7 to SMC1 */
        cpm2_immr->im_cpmux.cmx_smr &= 0x0f;
        cpm2_immr->im_cpmux.cmx_smr |= 0x10;  /* BRG 7 */
        pinfo->brg = 7;
#endif
}

I guess you can do it in many ways ...

Any one that knows why SMC1 & SMC2 gets the same BRG as SCC1 & SCC2

Regards
/Hans
 
On Wed, 2005-01-19 at 23:08, annamaya wrote:
> I've asked this question on this forum before but it
> did not get answered. I tried asking the same question
> in the U-Boot forum with no results. So, I am going to
> try this again.
> 
> According to section 4.1.2 on Page 4-4 in the MPC8280
> User's Manual, one should be able to use a combination
> of  external clocks on Port C pins 25, 26, 27 and 29,
> the CPM cloak and BRG1 to generate a 32KHz or a 4MHz
> clock. I am unable to find  any code in u-boot or any
> other place that initializes any of these pins to be
> external clocks. And as for BRG1, I believe it is
> being used for the SMC clock.
> 
> I am unable to understand how this is done. Can we
> have a timerclk input WITHOUT an external clock
> signal? Can I just use the CPM clock to generate this?
> How is this being done in U-Boot now?
> 
> Thanks much for your help. I really need an answer to
> this. 
> 
> 
> 		
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