Rafael Ignacio Zurita wrote: > But, for this question : If the battery needs replacement, I would > prefer to have to do that > like in a hand watch : never for several years.
Hmm, tricky. Let's use the original assumption of an average on-time of 5 minutes every day, standby for the rest of the day. For CR2032 we can then assume an average current of 15 mA while on. That's what it now draws from a 2.4 V supply with the display on but not doing much else. In real life, there will be a higher current, due to user activity, but the CR2032 will also provide a higher voltage, requiring a lower current. So 15 mA looks like a good estimate. Furthermore, let's conservatively assume that the usable charge of the battery is half its nominal capacity. Let's also assume the nominal capacity is 200 mAh. Then we can achieve with Anelok in its present state a battery life of 200 mAh * 0.5 / (15 mA / 12 h^-1 + 24 h * 0.055 mA) = 38.9 days If the standby current is reduced to 10 uA, the equation would be 200*.5/(15/12+24*.01) = 67.1 days If we could reduce the standby current to zero, we'd be at 80 days. If we let the device rest in standby (at 10 uA) all the time, we get 417 days. Let's run the numbers for AAA: let's assume the battery has 1000 mAh at an average voltage of 1.2 V. Let's assume we can use 70% of the nominal capacity (we should be able to drain an AAA deeper than a CR2032 because the AAA has a much lower internal resistance). The 15 mA in Active state would then become roughly 30 mA. Since the boost converter has to run also in Ready and Standby, the unboosted current at 2.4 V would get the following multipliers: - operating at 3.3 V instead of 2.4 V: 3.3 / 2.4 = 1.375 (assuming the MCU acts like a resistive load) - boost conversion from 1.2 V to 3.3 V = 3.3 / 1.2 = 2.75 - boost coverter efficiency: 50% - to this, add a quiescent current of 5.5 uA (TLV61220) So the battery current from AAA corresponding to 55 uA from CR2032 would be 421.4 uA, and for 10 uA it would be 81 uA. The battery life would thus be: - current design (power-guzzling RC oscillator): 1000*.7/(30/12+24*.42) = 55 days - using the crystal oscillator, 10 uA: 1000*.7/(30/12+24*.081) = 158 days - standby all the time: 1000*.7/(24*.081) = 360 days CR2032 even beats AAA in this case because CR2032 is great when things demand extremely little current but over a long time. Could we go even lower ? Well, yes. We could stop sampling the touch sensor. That should bring us down from the estimated 10 uA to 5.5 uA. The MCU could also enter a lower-current mode, from which it wakes up with a reset, saving about another 2 uA. To wake up, one could abuse rfkill as a wakeup source. We could also stop the crystal clock. That should save another 500 nA. If we assume such radical power-saving measures, then we'd be at 200*.5/(24*.003) = 1389 days for CR2032 and 1000*.7/(24*.041) = 711 days for AAA. That's for a sleep the device should still be able to wake up from. It would keep on running for some more time in Standby, but if you tried to turn it on, the battery would collapse. Of course, an easy means to avoid _any_ drain is to simply take out the battery if you don't plan to use Anelok for a few years :-) In the case of AAA, that's what you should do anyway, due to the risk of battery leakage. - Werner _______________________________________________ Qi Hardware Discussion List Mail to list (members only): [email protected] Subscribe or Unsubscribe: http://lists.en.qi-hardware.com/mailman/listinfo/discussion
