Post by thread:[PIC] Question about a/d converter

Rochester, 25 februari 2006. Dear Tia, It is likely that you'll have to "massage" your input signal to comply with the requirements of the A/D input pin: scale down the voltage, and possibly use an OPAMP to increase impedance (see a previous discussion about this on this same list). What is more relevant is the frequency of your signal: if it is too high, the A/D converter will be too slow to convert it. With the AC transformer I assume you're measuring a 50 Hz signal, which should be no problem for any PICmicro with A/D converter. Displaying the result on an LCD is a different story, but at 50 Hz should certainly be possible. Most graphical LCD don't have a fixed response time, so they might "hang" for a while occasionally as you write to them. An interrupt buffering the A/D conversions is a solution there. Greetings, Maarten Hofman. \

> It is likely that you'll have to "massage" your input signal to comply > with the requirements of the A/D input pin: scale down the voltage, > and possibly use an OPAMP to increase impedance (see a previous > discussion about this on this same list). What is more relevant is the > frequency of your signal: if it is too high, the A/D converter will be > too slow to convert it. With the AC transformer I assume you're > measuring a 50 Hz signal, which should be no problem for any PICmicro > with A/D converter. > > Displaying the result on an LCD is a different story, but at 50 Hz > should certainly be possible. Most graphical LCD don't have a fixed > response time, so they might "hang" for a while occasionally as you > write to them. An interrupt buffering the A/D conversions is a > solution there. Hi, I can use a pc to display the results, with uart or usb to transfer the samples. But do I need to think about any other problems. For example, what about eliminating noise, and also what is the best sampling period, since I can choose the a/d sampling time. I.e. if I choose a long sampling period will I be more exposed to the noise problem. Should I use an RC circuit to filter out noises? Tia > > Greetings, > Maarten Hofman.

> Rochester, 25 februari 2006. > > Dear Tia, > > It is likely that you'll have to "massage" your input signal to comply > with the requirements of the A/D input pin: scale down the voltage, > and possibly use an OPAMP to increase impedance (see a previous > discussion about this on this same list). What is more relevant is the > frequency of your signal: if it is too high, the A/D converter will be > too slow to convert it. With the AC transformer I assume you're > measuring a 50 Hz signal, which should be no problem for any PICmicro > with A/D converter. It's not "no problem", it will be ok, but there is not a lot of margin, here is why: Tacq= Aquisition time= 20usec n= Number of samples for good waveform display = 100 f= Frequency= 50hz, but for rectified ac it is actually double = 100 1/100 = 0.01 (T= 1/f) 0.01/Tacq = 0.01/20u = 10/20m= 0.5k= 500 namely, we should be able to obtain 500 samples per waveform on screen, which is very good quality. But you can see that if you increase your frequency to 500Hz, you are approaching the limit. > > Displaying the result on an LCD is a different story, but at 50 Hz > should certainly be possible. Most graphical LCD don't have a fixed > response time, so they might "hang" for a while occasionally as you > write to them. An interrupt buffering the A/D conversions is a > solution there. > > Greetings, > Maarten Hofman.

Will wrote: > Tacq= Aquisition time= 20usec > n= Number of samples for good waveform display = 100 > f= Frequency= 50hz, but for rectified ac it is actually double = 100 > > 1/100 = 0.01 (T= 1/f) > 0.01/Tacq = 0.01/20u = 10/20m= 0.5k= 500 > > namely, we should be able to obtain 500 samples per waveform on > screen, [...] Maybe you should read up again the ADC section of the manual and pay attention to the difference between acquisition time and conversion time. The conversion time is what you always have (and you seem not to be considering it here). The acquisition time is what you only need after you switched channels. If you are reading only one ADC channel, you usually can use an acquisition time of 0. Gerhard

-----Original Message----- From: .....piclist-bouncesKILLspam@spam@mit.edu [piclist-bouncesKILLspammit.edu] On Behalf Of Gerhard Fiedler Sent: Wednesday, March 01, 2006 1:54 PM To: .....piclistKILLspam.....mit.edu Subject: Re: [PIC] Question about a/d converter >The conversion time is what you always have (and you seem not to be >considering it here). The acquisition time is what you only need after you >switched channels. If you are reading only one ADC channel, you usually can >use an acquisition time of 0. >Gerhard I disagree. Selection of charge source is NOT the sampling switch! The sampling switch and series impedance dictate a phase/time delay in combination with the internal charge hold capacitor. (120 pF) The holding capacitor is NOT continuously refreshed. Read the example, you will see microseconds for the minimum charging time. Tom * | __O Thomas C. Sefranek EraseMEtcsspam_OUTTakeThisOuTcmcorp.com |_-\<,_ Amateur Radio Operator: WA1RHP (*)/ (*) Bicycle mobile on 145.41MHz PL74.4 ARRL Instructor, Technical Specialist, VE Contact. hamradio.cmcorp.com/inventory/Inventory.html http://www.harvardrepeater.org

{Quote hidden}>-----Original Message----- >From: piclist-bouncesspam_OUTmit.edu [@spam@piclist-bouncesKILLspammit.edu] >Sent: 01 March 2006 20:02 >To: 'Microcontroller discussion list - Public.' >Subject: RE: [PIC] Question about a/d converter > > > > >-----Original Message----- >From: KILLspampiclist-bouncesKILLspammit.edu [RemoveMEpiclist-bouncesTakeThisOuTmit.edu] >On Behalf Of Gerhard Fiedler >Sent: Wednesday, March 01, 2006 1:54 PM >To: spamBeGonepiclistspamBeGonemit.edu >Subject: Re: [PIC] Question about a/d converter > >>The conversion time is what you always have (and you seem not to be >>considering it here). The acquisition time is what you only >need after you >>switched channels. If you are reading only one ADC channel, >you usually can > >>use an acquisition time of 0. > >>Gerhard > > >I disagree. Selection of charge source is NOT the sampling >switch! The sampling switch and series impedance dictate a >phase/time delay in combination with the internal charge hold >capacitor. (120 pF) The holding capacitor is NOT continuously >refreshed. >From the datasheet diagrams, the capacitor would appear to be continuously connected to the A/D pin once a channel has been selected (the "sampling switch" refers to the analog mux according to my FAE). In fact common sense tells us this must be the case, as you don't have to do anything to initiate an acquisiation once a channel has been selected for the first time. There is some impedance in the analog switch though, so the cap would undoubtedly need some recovery time after a sample has been taken. Regards Mike ======================================================================= This e-mail is intended for the person it is addressed to only. The information contained in it may be confidential and/or protected by law. If you are not the intended recipient of this message, you must not make any use of this information, or copy or show it to any person. Please contact us immediately to tell us that you have received this e-mail, and return the original to us. Any use, forwarding, printing or copying of this message is strictly prohibited. No part of this message can be considered a request for goods or services. =======================================================================

> No, if the next sample is a complete random value. This is correct... The example I read (in the manual for the 16F630) shows that the hold capacitor charging/decharging is the largest part of the time. However, it also talks about the maximum allowed error. This all brings me to another point, which is the actual LCD display. When I read the original question, I was thinking of the usual type of display I would attach to a PICmicro. Obviously it would need to be graphical, so a T6963C of 64*128 ($25 on futurlec) pixels seemed like an appropriate choice. Obviously you can go more expensive, but at some point a dedicated external A/D converter would become preferable (as the price/size of the display would outweigh the cost/size of the A/D converter). Given a 64*128 bit display you only need 128 samples, and not 500. Also, you only need 6-7 bits of resolution, which means you can skimp on the acquisition time (note that the 10F220 manual, with its 8-bit A/D converter, doesn't even mention acquisition times, and assumes all conversions can take place within 7us). Greetings, Maarten Hofman.

Jan-Erik Soderholm wrote: > The "holding cap" is disconnected from the input during the conversion > time. After the conversion it is automaticly connected to the input pin > again. Now, if the voltage on the input pin has changed conciderably > during the conversion, I'd guess that you need a new acquisition time to > let the holding cap re-charge to the new input voltage, before starting > a new conversion, not ? Yes. That's why I said "usually". This of course means that there /are/ (IMO rare) conditions where this is not the case. I'd venture a guess and say that in most cases when only one input is read there is a suitable combination of slow enough input changes and the program's reaction time to the end of conversion signal that the hold cap has enough time to charge before the next conversion. Especially when we are talking about mains wave forms. Gerhard

> Especially when we are talking about mains wave forms. Yes, in our current scenario there is no reason to assume that there would be a sudden change between samples, unless something odd happens on the input signal (which brings me to the question: if we are designing this to just display a 50 Hz or 60 Hz curve, why are we even bothering sampling? We could just make a table, display it and be done with it... When we are sampling, we would even have to make intelligent decisions on when to restart at the left of the screen...). > Yes. That's why I said "usually". This of course means that there /are/ > (IMO rare) conditions where this is not the case. Why rare? Wouldn't a proper design of sampling an analog signal (not for display on a screen by a PICmicro, but for other purposes) use the Nyquist–Shannon sampling theorem, and therefore be designed such that at the maximum frequency slightly more than two samples/period are taken? With only two samples/period the values could be anywhere within the range. Greetings, Maarten Hofman.

part 1 2751 bytes content-type:text/plain; charset="big5" (decoded quoted-printable) Maarten Hofman wrote: >> Yes. That's why I said "usually". This of course means that there /are/ >> (IMO rare) conditions where this is not the case. > > Why rare? Wouldn't a proper design of sampling an analog signal (not for > display on a screen by a PICmicro, but for other purposes) use the > Nyquist�VShannon sampling theorem, and therefore be designed such that at > the maximum frequency slightly more than two samples/period are taken? > With only two samples/period the values could be anywhere within the > range. This is correct. I said rare, because I think that in most cases, you need many more than two samples per period to fulfill your application requirements. The theorem only says that if you take less than two samples per period of the highest frequency present in the signal, you get folding in the frequency domain (that is, you measure frequencies that don't exist in the signal). It does /not/ say that taking two samples per period gives you any useful data (for a given purpose). So in most cases, you would measure many more than two samples per period. The most frequent case probably is arbitrarily changing signals that do not actually have a period as such (process signals like temperature, acceleration, etc.). They have usually components of higher frequencies, that also usually get filtered out, with filter cut-offs often way below the maximum conversion frequency. Which then of course means that the signal changes much slower than one sample at maximum and the next sample at minimum. I think it is rare that you get to the point where charging the holding cap becomes a problem (remember, that's all in the case of measuring one single analog signal). Has anybody had such an application, where exactly this was a limiting factor? At 5V Vdd, the internal resistance that charges Chold (25pF) is about 3k. Assuming a small source resistance (op amp), that gives a time constant tRC of 75ns. A full scale jump in the input takes about 7*tRC or 525ns to dis/charge the cap to within 1/1023th (10 bit resolution) of the final value. With a PIC running at 4MHz, that's less than a single instruction cycle, with a PIC running at 20MHz, that's less than three instruction cycles. But you're not really measuring any useful data in this situation (in most applications), because you're too close to the frequency limit. If you want to measure anything, you need more points, the jumps between points won't be full-scale anymore, and the time to charge Chold will become smaller. I don't say it doesn't happen... you can have a bigger source resistance, for example. But I still think it's rare. Very rare. Gerhard part 2 35 bytes content-type:text/plain; charset="us-ascii" (decoded 7bit)

> This is correct. I said rare, because I think that in most cases, you need > many more than two samples per period to fulfill your application > requirements. The theorem only says that if you take less than two samples > per period of the highest frequency present in the signal, you get folding > in the frequency domain (that is, you measure frequencies that don't exist > in the signal). It does /not/ say that taking two samples per period gives > you any useful data (for a given purpose). As far as I understood the theorem (and I agree that it has been a few years since I studied it) it is that if the signal is limited to a certain maximum frequency f then if you take slightly more than than two times this number in samples ("slightly more" because of the > instead of the >=, so just one more is fine, i.e. 40001 samples/second for a 20000 Hz signal) then, based on those two samples, you can reproduce the signal perfectly (this is assuming that each sample is taken with an infinite number of bits, digitizing the sample introduces an amount of noise equal to the resolution gap). Note that to do this perfect reproducing, you'll have to do something more than just convert them back to an analog level: you will need to multiply each sample with sin(x)/x (1 for x=0). However, the resulting signal will be identical (except for possible noise caused by digitizing the signal). Obviously, for all these calculations you probably will need a dsPIC to actually process anything, but for merely gathering the data other PICmicro could suffice. I, however, do agree that I have not seen these things done with PICmicros, and that therefore it would indeed be quite rare, and just like you would like to hear from people that were bothered by the Chold time (the example in one of the manuals should it being around 20 microseconds, so it could be a significant time). Greetings, Maarten Hofman.

Maarten Hofman wrote: >> This is correct. I said rare, because I think that in most cases, you >> need many more than two samples per period to fulfill your application >> requirements. The theorem only says that if you take less than two >> samples per period of the highest frequency present in the signal, you >> get folding in the frequency domain (that is, you measure frequencies >> that don't exist in the signal). It does /not/ say that taking two >> samples per period gives you any useful data (for a given purpose). > > As far as I understood the theorem (and I agree that it has been a few > years since I studied it) it is that if the signal is limited to a > certain maximum frequency f then if you take slightly more than than two > times this number in samples ("slightly more" because of the > instead > of the >=, so just one more is fine, i.e. 40001 samples/second for a > 20000 Hz signal) then, based on those two samples, you can reproduce the > signal perfectly (this is assuming that each sample is taken with an > infinite number of bits, digitizing the sample introduces an amount of > noise equal to the resolution gap). This is correct, but only if the signal is time-invariant in the frequency domain and you can take enough samples (two won't do it :). Which again is rare in a typical PIC application. Imagine a pure 20kHz sine wave sampled with 40001 samples/s. You'd start with a 0. The next sample would be just a bit above 0. The 3rd sample would be just a bit below 0. There's any number of time domain curves that would fit through those three points, especially since they are not points, but small rectangles (or at least lines). The theorem considers the points to be points, and it considers the signal to be invariant in the frequency domain. Then it works. But that's rare in a real-world application. > I, however, do agree that I have not seen these things done with > PICmicros, and that therefore it would indeed be quite rare, and just > like you would like to hear from people that were bothered by the Chold > time (the example in one of the manuals should it being around 20 > microseconds, so it could be a significant time). I've been bothered by the charging time of Chold (as it can be significant, especially with higher source resistances), but only when switching between channels. Then the situation is different, because you may have arbitrary differences up to full-scale between channels, and you have to wait after you switch channels. When measuring only one channel, the charging time starts automatically at the end of the conversion without intervention by firmware; when switching between channels, the acquisition time only starts after the firmware switched the input channel. That's why the programmable acquisition time is such a convenient thing: you don't have to take care of the acquisition time between switching channels and the start of the conversion in firmware; it's now taken care of by the ADC hardware. But that's also why you can set the programmable acquisition time to 0: if you don't switch channels, usually you don't need it. IMO, of course :) Gerhard