Yes indeed i need to conditionate the signal, i though there were pins tolerant to 5 V but i just read that is only for logic input. Until now i was measuring currents lower than 10 A so i didn't remarked the need of the conditioning. Thank you for your support.
Also i was already using the buffered version of AnalogInput, i did just like you said, i measured the peak-to-peak voltage and converted it to rms, then i did a simple conversion to current, that's when i got the accuracy of +-0.1A, but i need it to be less than that, that's why i'm looking for the option of doing it in the firmware. I wonder if that is possible? El miércoles, 15 de julio de 2015, 12:09:16 (UTC-5), Ytai escribió: > > I think I understand now. So: > > 1. If you're really intending to measure a signal as high as 30A, the > output voltage will be around 4.5V, which exceeds the range the ADC is > willing to accept (0-3.3V). > 2. Unless the sensor has a built-in amplifier, it is likely that it > has a fairly high output impedance. > 3. To address both of the above problems, I recommend conditioning the > signal using one of more opamp stages, so that it has about 1.5V bias and > about 1.5V amplitude @ 30A (or 50mV/A). > 4. The IOIO samples at 1kHz. There should be no problem sampling of a > 60Hz signal. Moreover, since you know it is a sine wave, all you really > need is measure the peak-to-peak voltage which you can then trivially > convert to RMS. I suggest that you look at the buffered version of > AnalogInput, so that you don't miss or depend on your application thread > loop frequency. > > > On Wed, Jul 15, 2015 at 7:15 AM, Julio Castellanos <[email protected] > <javascript:>> wrote: > >> Hi Ytai, thank you for your help. The signal is the output of an A.C. >> current sensor, the frequency is the same as the network, 60 hz, and the >> sensor can measure up to 30 A with a sensitivity of 66 mV/A. The signal has >> always a D.C. value of 2.5 V so if i measure 1 A the signal will have a max >> of 2.566 V and a min of 2.434 V. The absolute maximum with 30 A is 4.48 V >> and absolute minimum of 0.52 V, so the signal is always in the range of the >> analog input of the ioio. I want to know the value rms of the current in my >> smartphone, and i tried obtaining the rms of the voltage by calculating the >> max value of a buffer of 1000 values and the multiplying it by 0.707. I get >> the value rms of the current with an accuracy of +-0.1 A, but i need it to >> be less than that, that's why i though it would be better to compute the >> rms value of the signal in hardware so i dont have to send the whole buffer >> of the signals. >> >> El martes, 14 de julio de 2015, 18:44:30 (UTC-5), Ytai escribió: >>> >>> Before we begin, can you please specify the signal you're trying to >>> measure better? >>> Specifically, its min and max (instantaneous) voltage, and its bandwidth >>> would be important. >>> Based on that I can recommend a way of measuring it. >>> >>> On Tue, Jul 14, 2015 at 3:06 PM, Julio Castellanos <[email protected]> >>> wrote: >>> >>>> Hello, i'm currently working with the IOIO analog input with a sensor >>>> signal that comes out as an A.C. signal in the range of 2,5 V as mean >>>> value. I've tried to get the voltage through the analog input and get the >>>> rms value in software, but the value changes to much because of the >>>> bandwith limit of the IOIO:Android communication. I'm now looking to get >>>> the RMS directly in hardware and just send that value to the software but >>>> i >>>> gave a look at the hardware code for adc.c and i don't understand almost >>>> nothing, can someone point out to me a source where i can learn how does >>>> this code works? >>>> >>>> -- >>>> You received this message because you are subscribed to the Google >>>> Groups "ioio-users" group. >>>> To unsubscribe from this group and stop receiving emails from it, send >>>> an email to [email protected]. >>>> To post to this group, send email to [email protected]. >>>> Visit this group at http://groups.google.com/group/ioio-users. >>>> For more options, visit https://groups.google.com/d/optout. >>>> >>> >>> -- >> You received this message because you are subscribed to the Google Groups >> "ioio-users" group. >> To unsubscribe from this group and stop receiving emails from it, send an >> email to [email protected] <javascript:>. >> To post to this group, send email to [email protected] >> <javascript:>. >> Visit this group at http://groups.google.com/group/ioio-users. >> For more options, visit https://groups.google.com/d/optout. >> > > -- You received this message because you are subscribed to the Google Groups "ioio-users" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To post to this group, send email to [email protected]. Visit this group at http://groups.google.com/group/ioio-users. For more options, visit https://groups.google.com/d/optout.
