Messages in this thread | ![/](/images/icornerl.gif) | | Date | Tue, 15 Mar 2016 12:14:50 +0530 | From | Laxman Dewangan <> | Subject | Re: [PATCH 4/5] regulator: pwm: Add support for voltage linear equal steps |
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On Monday 14 March 2016 09:58 PM, Mark Brown wrote: > * PGP Signed by an unknown key > > On Sun, Mar 13, 2016 at 06:36:06PM +0530, Laxman Dewangan wrote: >> On Saturday 12 March 2016 11:39 AM, Mark Brown wrote: >>> I can't see any reason why this would ever be preferable to just using >>> the flat linear range (you certainly haven't articulated one, you're >>> just stating it). This seems like you are bodging around a limited >>> consumer driver, you should fix the consumer to cope with regulators >>> with lots of voltages - PWM regulators aren't the only ones with high >>> resolution steps. >> The requirement is to have perfect linear steps interms of the period/pulse >> time of PWM without loosing any voltage. >> Continuous mode is pretty much near to what you said but here we are loosing >> the perfect step as this divides the periods to 100 parts and then set >> voltage. > Could you be more specific about what the issue is? We've hopefully got > errors of less than 1% in the values here... > If I use the continuous mode of PWM regulator then the calculation for PWM pulse ON time(duty_pulse) done as: duty_cycle = ((requested - minimum) * 100) / voltage_range.
duty_pulse = (pwm_period/100) * duty_cycle
This leads to the calculation error if we have the requested voltage where accurate pulse time is possible. For example: Let's have following case voltage range is 800000uV to 1350000uV. pwm-period = 1550ns (1ns time is 1mV). Requested 900000uV.
duty_cycle = ((900000uV - 800000uV) * 100)/ 1550000 = 6.45 but we will get 6 due to integer division.
duty_pulse = (1550/100) * 6 = 90 pulse time.
90 pulse time is equivalent to 90mV and this gives us pulse time equivalent to 890000uV instead of 900000uV.
>> If new mode is not accpetable then need to enhance the existing continuous >> mode like before scaling for 100% of period, first look if we get the >> perfect pulse time of of PWM period and if it is there then use this direct >> instead of converting required voltage to 100% scale and then back >> calculating duty time. > That seems a lot better,
You mean using the continuous mode only.
If I add following logic then also it resolve the issue: if (((req_uV - min_uV) * pwm_period) % voltage_range == 0) duty_pulse = ((req_uV - min_uV) * pwm_period) / voltage_range; else existing_continuous mode calculation.
So on above example: duty_pulse = ((900000uV - 800000uV) * 1550)/1550000) = 100
and this is equivalent to 100mV and so final voltage is (800000 + 100000) = 900000uV which is same as requested,
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