[Intel-gfx] [PATCH 1/2] drm/dp/i915: Fix DP link rate math

[Ville Syrjälä]

On Wed, Nov 09, 2016 at 09:32:29PM -0800, Dhinakaran Pandiyan wrote:
> We store DP link rates as link clock frequencies in kHz, just like all
> other clock values. But, DP link rates in the DP Spec are expressed in
> Gbps/lane, which seems to have led to some confusion.
> 
> E.g., for HBR2
> Max. data rate = 5.4 Gbps/lane x 4 lane x 8/10 x 1/8 = 2160000 kBps
> where, 8/10 is for channel encoding and 1/8 is for bit to Byte conversion
> 
> Using link clock frequency, like we do
> Max. data rate = 540000 kHz * 4 lanes = 2160000 kSymbols/s
> Because, each symbol has 8 bit of data, this is 2160000 kBps
> and there is no need to account for channel encoding here.
> 
> But, currently we do 540000 kHz * 4 lanes * (8/10) = 1728000 kBps
> 
> Similarly, while computing the required link bandwidth for a mode,
> there is a mysterious 1/10 term.
> This should simply be pixel_clock kHz * bpp * 1/8  to give the final
> result in kBps
> 
> Signed-off-by: Dhinakaran Pandiyan <dhinakaran.pandiyan at intel.com>
> ---
>  drivers/gpu/drm/i915/intel_dp.c | 28 +++++++++-------------------
>  1 file changed, 9 insertions(+), 19 deletions(-)
> 
> diff --git a/drivers/gpu/drm/i915/intel_dp.c b/drivers/gpu/drm/i915/intel_dp.c
> index 8f313c1..7a9e122 100644
> --- a/drivers/gpu/drm/i915/intel_dp.c
> +++ b/drivers/gpu/drm/i915/intel_dp.c
> @@ -161,33 +161,23 @@ static u8 intel_dp_max_lane_count(struct intel_dp *intel_dp)
>  	return min(source_max, sink_max);
>  }
>  
> -/*
> - * The units on the numbers in the next two are... bizarre.  Examples will
> - * make it clearer; this one parallels an example in the eDP spec.
> - *
> - * intel_dp_max_data_rate for one lane of 2.7GHz evaluates as:
> - *
> - *     270000 * 1 * 8 / 10 == 216000
> - *
> - * The actual data capacity of that configuration is 2.16Gbit/s, so the
> - * units are decakilobits.  ->clock in a drm_display_mode is in kilohertz -
> - * or equivalently, kilopixels per second - so for 1680x1050R it'd be
> - * 119000.  At 18bpp that's 2142000 kilobits per second.
> - *
> - * Thus the strange-looking division by 10 in intel_dp_link_required, to
> - * get the result in decakilobits instead of kilobits.
> - */
> -
>  static int
>  intel_dp_link_required(int pixel_clock, int bpp)
>  {
> -	return (pixel_clock * bpp + 9) / 10;
> +	/* pixel_clock is in kHz, divide bpp by 8 to return the value in kBps*/

Probably best not to mix in the kBps unit here and instead just talk
in terms of the symbol clock.

> +	return (pixel_clock * bpp + 7) / 8;

DIV_ROUND_UP()

>  }
>  
>  static int
>  intel_dp_max_data_rate(int max_link_clock, int max_lanes)
>  {
> -	return (max_link_clock * max_lanes * 8) / 10;
> +	/* max_link_clock is the link symbol clock (LS_Clk) in kHz and not the
> +	 * link rate that is generally expressed in Gbps. Since, 8 bits data is
> +	 * transmitted every LS_Clk per lane, there is no need to account for
> +	 * the channel encoding that is done in the PHY layer here.
> +	 */
> +
> +	return (max_link_clock * max_lanes);

Useless parens.


>  }
>  
>  static int
> -- 
> 2.7.4

-- 
Ville Syrjälä
Intel OTC