On Fri, 12 Jul 2024 19:14:57 +0200 Michael Walle mwalle@kernel.org wrote:
Hi,
If the maximum frequency is requested, we still fall into the CDR2 handling. But there the minimal divider is 2. For the sun6i and sun8i we can do better with the CDR1 setting where the minimal divider is 1: SPI_CLK = MOD_CLK / 2 ^ cdr with cdr = 0
Thus, handle the div = 1 case specially.
Yes, this is correct: we lose half the performance without this patch, in the (common) full clock speed case. However ....
While at it, correct the comment above the calculation.
Signed-off-by: Michael Walle mwalle@kernel.org
drivers/spi/spi-sunxi.c | 7 ++++++- 1 file changed, 6 insertions(+), 1 deletion(-)
diff --git a/drivers/spi/spi-sunxi.c b/drivers/spi/spi-sunxi.c index bfb402902b8..3048ab0ecf7 100644 --- a/drivers/spi/spi-sunxi.c +++ b/drivers/spi/spi-sunxi.c @@ -249,6 +249,8 @@ static void sun4i_spi_set_speed_mode(struct udevice *dev) * We have two choices there. Either we can use the clock * divide rate 1, which is calculated thanks to this formula: * SPI_CLK = MOD_CLK / (2 ^ (cdr + 1))
* Or for sun6i/sun8i variants:* SPI_CLK = MOD_CLK / (2 ^ cdr)
- Or we can use CDR2, which is calculated with the formula:
- SPI_CLK = MOD_CLK / (2 * (cdr + 1))
- Whether we use the former or the latter is set through the
@@ -256,12 +258,15 @@ static void sun4i_spi_set_speed_mode(struct udevice *dev) * * First try CDR2, and if we can't reach the expected * frequency, fall back to CDR1.
* There is one exception if the requested clock is the input* clock. In that case we always use CDR1 because we'll get a* 1:1: ration for sun6i/sun8i variants.*/
div = DIV_ROUND_UP(SUNXI_INPUT_CLOCK, priv->freq); reg = readl(SPI_REG(priv, SPI_CCR));
- if ((div / 2) <= (SUN4I_CLK_CTL_CDR2_MASK + 1)) {
- if (div != 1 && ((div / 2) <= (SUN4I_CLK_CTL_CDR2_MASK + 1))) { div /= 2;
This is still not fully correct, is it? If I ask for 10 MHz, the algorithm should select 8 MHz (24/3) or actually 6 MHz (24/4), but it chooses 12 MHz (24/2), which is too much. So I think this division here should be either: div = (div + 1) / 2; or: div = DIV_ROUND_UP(div, 2);
Can someone confirm this?
Cheers, Andre
if (div > 0) div--;