Re: [PATCH v2 1/2] regulator: pcf50633: Don't write to reserved bits of AUTO output voltage select register

[Lars-Peter Clausen <lars@metafoo.de>]

On 03/17/2012 01:07 AM, Axel Lin wrote:
> The datasheet says 00000000 to 00101110 are reserved, and the min value of the
> voltage setting is 1.8 V.
> Thus don't write 0 to AUTO output voltage select register (address 1Ah).
> 
> Table 50. AUTOOUT - AUTO output voltage select register (address 1Ah) bit description[1]
> Bit Symbol Access Description
> 7:0 auto_out R/W VO(prog) = 0.625 + auto_out × 0.025 V
> eg. 00000000 to 00101110: reserved
> 00101111: 1.8 V (min)
> 01010011: 2.7 V
> 01101010: 3.275 V
> 01101011: 3.300 V
> 01101100: 3.325 V
> 01111111 : 3.800 V (max)
> ..... .....
> 11111110 : 3.800 V
> 11111111 : 3.800 V
> 
> This patch also fixes a bug in pcf50633_regulator_list_voltage.
> It is wrong to do "index += 0x2f" for PCF50633_REGULATOR_AUTO in
> pcf50633_regulator_list_voltage. The purpose of adding 0x2f to index is because
> current code return 0 in auto_voltage_bits when millivolts < 1800.
> For millivolts > 1800, adding 0x2f to index is wrong.
> 

I think you misunderstood what the current code does. The first usable voltage
is 1.8V which is equal to a index of of 0x2f. So the driver adds 0x2f to the
index so that there is not a headroom of 0x2f unusable voltages. So a selector
of 0 translates to 1.8V, a selector of 1 translates to 1.825V and so on.

I can see why you'd want to change it to simplify the code, but you also have
to change the number of voltages for the AUTO regulator to 128.