Hi Urs, unfortunately that is not the biggest problem and so the solution to force the MCU to stay ON during the sampling had no effect: the fluctuations with Duracells remain. Moreover I also found this problem with samplings of major periods (3000 ms), where the MCU should not have problems of state changes. I really think that the main cause depends on the batteries. However thank you very much for your idea, you are very helpful.
Greetings,
--
Francesco Ficarola <francesco.ficarola_at_gmail_dot_com>
[GPG KeyID: 0xDBA99D92]
Urs Hunkeler ha scritto:
> Hi Francesco,
>
> At a quick glance your calculations appear to be correct. If the effort
> is small to change the resistors, it's worth a try.
>
> I also have been thinking, and maybe the simple model of the battery is
> not enough to explain the phenomenon you see. I suspect that the
> microcontroller turning on and off while sampling the sensor could also
> be a problem. You can force the microcontroller to stay on during the
> sampling by overriding the default behavior of the power management.
> You'll have to wire an additional interface for this:
>
> components McuSleepC;
> McuSleepC.McuPowerOverride -> YourApplication;
>
> In your application, you need to implement the McuPowerOverride interface:
>
> module ... {
> // ...
> provides interface McuPowerOverride;
> // ...
> }
> implementation {
> uint8_t sensorSampling = 0;
> // ...
> async command mcu_power_t McuPowerOverride.lowestState() {
> // WARNING: this code is specific to the Atmel microcontrollers
> // Should run on Mica2, MicaZ, and Iris
> if(sensorSampling == 0) {
> // we can allow the MCU to sleep
> return ATM128_POWER_DOWN;
> } else {
> // MCU needs to remain active
> return ATM128_POWER_IDLE;
> }
> }
> // when starting to sample the sensor
> sensorSampling = 1;
> // sample sensor
> // when done sampling sensor
> sensorSampling = 0;
> // ...
> }
>
> Cheers,
> Urs
>
>
> On 12/4/10 11:14 AM, Francesco Ficarola wrote:
>> Urs, in the meantime I thought:
>>
>> Since I use a weathstone bridge
>> (http://en.wikipedia.org/wiki/Wheatstone_bridge) with:
>> - R1 and R3: 1.2 kOhm resistors
>> - R2: 470 Ohm potentiometer
>> - Rx: 354 Ohm strain gauge
>>
>> the equivalent resistance of the bridge is (with potentiometer in max
>> resistance):
>> [1/Re] = [1/(1200+470)] + [1/(1200+354)] --> Re = 804.95 Ohm, right?
>>
>> Now, since I = V / R, the current consumption of this bridge is:
>>
>> I = 3 V / 804.95 Ohm = 3.7 mA
>>
>> So, assuming that the battery resistance is 3 Ohm, the voltage
>> fluctuations are about 3.7 mA * 3 Ohm = 11.1 mV.
>>
>> Now, if I increase the values of resistors, for example:
>> - R1 and R3: 10 kOhm resistors
>> - R2: 10 kOhm potentiometer
>> - Rx: 354 Ohm strain gauge
>>
>> the equivalent resistance of this bridge will be (with potentiometer in
>> max resistance):
>>
>> [1/Re] = [1/(10000+10000)] + [1/(10000+354)] --> Re = 6822,17 Ohm
>>
>> So the current consumption is:
>>
>> I = 3 V / 6822,17 Ohm = 0.4 mA
>>
>> In this case the voltage fluctuations are about 0.4 mA * 3 Ohm = 1.2 mV.
>>
>> Thanks to this value I should solve the problem in my ADC readings
>> because a level of the quantization is greater then 2 mV.
>>
>>
>> Do you think the reasoning is correct?
>>
>> Greetings,
>
>
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