On Thu, May 17, 2018 at 1:11 AM, Andy Shevchenko
<andy.shevchenko@xxxxxxxxx> wrote:
> On Wed, May 16, 2018 at 11:43 PM, Dan Murphy <dmurphy@xxxxxx> wrote:
>> On 05/15/2018 05:24 PM, Andy Shevchenko wrote:
>>> On Wed, May 16, 2018 at 1:08 AM, Dan Murphy <dmurphy@xxxxxx> wrote:
>>>> On 05/15/2018 04:56 PM, Andy Shevchenko wrote:
>>>>>> +static const struct lm3601x_max_timeouts strobe_timeouts[] = {
>>>>>> + { 40000, 0x00 },
>>>>>> + { 80000, 0x01 },
>>>>>> + { 120000, 0x02 },
>>>>>> + { 160000, 0x03 },
>>>>>> + { 200000, 0x04 },
>>>>>> + { 240000, 0x05 },
>>>>>> + { 280000, 0x06 },
>>>>>> + { 320000, 0x07 },
>>>>>> + { 360000, 0x08 },
>>>>>> + { 400000, 0x09 },
>>>>>> + { 600000, 0x0a },
>>>>>> + { 800000, 0x0b },
>>>>>> + { 1000000, 0x0c },
>>>>>> + { 1200000, 0x0d },
>>>>>> + { 1400000, 0x0e },
>>>>>> + { 1600000, 0x0f },
>>> if (x < 9)
>>> strobe_timeout = (x + 1) * 40 * MSECS_IN_SEC;
>>> else
>>> strobe_timeout = (x - 7) * 200 * MSECS_IN_SEC;
>>
>> Not sure if this will produce the register value I am looking for. I have to run through the
>> algorithim. The idea is to take in the timeout and get the reg value to program.
>>
>> If it yields the expected output, or with a modified equation, then I can pull it in.
>
> So, you need a reversed calculus in this case.
>
> u8 regval;
>
> /* To make it in range... */
> timeout = clamp_val(timeout, 40000, 1600000);
> /* ...other possibility to issue an error */
>
> if (timeout >= 400000)
> regval = timeout / 200000 + 0x07;
> else
> regval = timeout / 40000 - 0x01;
Forgot to mention that there are also three approaches (considering
division here):
- round up
- round down
- round to closest
>
> In any case, please double check it gives a correct values.
--
With Best Regards,
Andy Shevchenko
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