Whenever we request a power domain it has to guarantee that all HW
resources are enabled that are needed to access a HW register associated
with that power domain. In case a register is on an always-on power well
this won't result in turning on a power well, but it may require
enabling some other HW resource. One such resource is the HSW/BDW device
D0 state that is required for all register accesses and thus for all
power wells/power domains.
So far the init power domain (guaranteeing access to all HW registers)
was part of the default i9xx always-on power well, but not the HSW/BDW
always-on power wells. Add the domain to the latter power wells too.
Atm, all the always-on power wells have noop handlers, so this doesn't
change the functionality.
v2:
- clarify semantics of always-on power wells (Paulo)
Signed-off-by: Imre Deak <imre.deak@intel.com>
Reviewed-by: Jesse Barnes <jbarnes@virtuousgeek.org>
Signed-off-by: Daniel Vetter <daniel.vetter@ffwll.ch>
#define HSW_ALWAYS_ON_POWER_DOMAINS ( \
BIT(POWER_DOMAIN_PIPE_A) | \
- BIT(POWER_DOMAIN_TRANSCODER_EDP))
+ BIT(POWER_DOMAIN_TRANSCODER_EDP) | \
+ BIT(POWER_DOMAIN_INIT))
#define HSW_DISPLAY_POWER_DOMAINS ( \
(POWER_DOMAIN_MASK & ~HSW_ALWAYS_ON_POWER_DOMAINS) | \
BIT(POWER_DOMAIN_INIT))