5
\$\begingroup\$

I am making use of the dsPIC33FJ64GP802's ADC to do Undersampling and for that, I need to know the analog bandwidth of the ADC of this Microcontroller. My signal is around a few MHz. I have already contacted Microchip for the specs but no reply.

So would anyone know about the typical analog bandwidth of this tier of ADCs?

Note that I am using the 4 simultaneous channels option with the 10-bit ADC.

I am making a circuit to measure the analog bandwidth anyway. I just wanted to know if anyone knows beforehand maybe there's no need to go through the trouble.

Let me stress out again that I am Undersampling the few MHz signal which has a bandwidth that satisfies the Nyquist–Shannon sampling theorem.

And yes, I am aware that the sampling rate is 1.1 MS/s so a max signal BW of 550 kHz.

\$\endgroup\$
10
  • 2
    \$\begingroup\$ Look at the datasheet(s) \$\endgroup\$
    – PlasmaHH
    May 2, 2015 at 9:14
  • 2
    \$\begingroup\$ Duh, I did look at the datasheet. Nothing there about the analog bandwidth.@PlasmaHH \$\endgroup\$
    – br4him
    May 2, 2015 at 10:17
  • 1
    \$\begingroup\$ Analog Peripherals 10-bit, 1.1 Msps. Don't worry about the analog bandwidth if your maximum sample rate is already at least an order magnitude out. \$\endgroup\$
    – jippie
    May 2, 2015 at 10:46
  • 1
    \$\begingroup\$ @jippie, He specifically mentions undersampling. That means he wants to do down-mixing in the same step as sampling. There are many ADCs out there that would allow this. \$\endgroup\$
    – The Photon
    May 2, 2015 at 14:27
  • 2
    \$\begingroup\$ That said, if the datasheet doesn't specify analog bandwidth, it's not likely the ADC was designed for undersampling, and I wouldn't expect it to work well above Nyquist without an external S/H. \$\endgroup\$
    – The Photon
    May 2, 2015 at 15:27

5 Answers 5

3
\$\begingroup\$

A typical general-purpose ADC will have an input frequency response which is essentially flat up to frequencies at least up to the Nyquist rate, and in many cases significantly beyond [such frequency content, if present, will be aliased down to lower frequencies]. Given a typical ADC that can process 100,000 samples/second, feeding in a 101,000Hz signal while the device is sampling at that rate would likely yield a 1,000Hz signal with an amplitude close to that of the original (when taking 100,000 samples/second of such a signal, each sample would be advanced 1% further along the input waveform than the previous one).

Most general-purpose ADCs capture the state of the input during a small but non-trivial fraction of the overall sampling period, but aren't particularly intended to be used with signals that will change significantly during a capture, so one shouldn't rely upon the converter to have flat frequency response above Nyquist, but in most cases the capture time will be short enough that using the full range of frequencies up to Nyquist shouldn't be a problem.

\$\endgroup\$
3
\$\begingroup\$

Disclaimer: the answer below is about definition of the analog input bandwidth (AIB) of an ADC; not the particular value of AIB of a general purpose ADC.

An input analog signal, before being digitized, on its way between input pins of the ADC chip and a digitizing unit (usually a comparator), is attenuated (due to intentional or parasitic RC circuits present or not enough current to charge capacitors). For instance, signal attenuation may happen due to large input capacitance of sample and hold unit (assuming it is embedded into the ADC chip) or parasitic capacitances of comparators. The frequency at which the input signal is attenuated by 3 dB is called analog input bandwidth.

If one sends an input sinewave at the AIB frequency, before being digitized it is attenuated by 30% (which is unlikely acceptable). Thus, it’s preferable that maximum input frequency should be about 1/3 – 1/5 of the AIB. For instance, if AIB=500 MHz and input signal is 100 MHz, attenuation (measured by amplitude) is 2% (vs 30% for an input signal at 500 MHz).

Below is a chart from the datasheet of ADS54J40 showing how input circuitry attenuates the input signal (as seen from the picture, the AIB is about 1.2 GHz). Also see the discussion on the topic (pages 23 and 24 of the datasheet).

enter image description here

PS: Drop of SNR by 3 dB (equivalent to 0.5 LSB decrease in ENOB) is a consequence of analog input bandwidth limitation, but not definition of it. SNR (and ENOB) are driven by many factors, not only by AIB.

Useful reading on the topic (paragraph “Bandwidth”).

\$\endgroup\$
2
  • 1
    \$\begingroup\$ Thanks for your input. This gives an idea about the order of magnitude the ADC's analog bandwidth. \$\endgroup\$
    – br4him
    Sep 22, 2016 at 8:28
  • \$\begingroup\$ @br4him my pleasure \$\endgroup\$ Sep 22, 2016 at 8:30
2
\$\begingroup\$

The bandwidth of the ADC is described in the AC Electrical Specification section and it's different in 10 bit and 12 bit mode. Sorry for the pictures I'm from my phone:

enter image description here

enter image description here

See in Input Signal Bandwidth!

\$\endgroup\$
2
  • \$\begingroup\$ Thanks, but that (F_NYQ) is the Nyquist frequency (1.1 MHz/2 = 550 kHz ) not the actual analog BW. There is a difference... the analog BW should be higher than this. \$\endgroup\$
    – br4him
    May 2, 2015 at 15:40
  • \$\begingroup\$ It's different because of the different Sampling rates depending on the resolution. It can provide 500 kS/s in 12 bits or 1.1 MS/s in 10 bits. So the max signal BWs are -as the datasheet indicates- 250 kHz and 550 kHz respectively. \$\endgroup\$
    – br4him
    May 2, 2015 at 15:47
2
\$\begingroup\$

First, most people don't seem to understand the question. For a typical ADC, the effective number of bits resolution, N = (SN - 1.72)/6.02, where SN is the signal to noise ratio in DB. So, a drop in 3dB would be N = (3 - 1.72)/6.02 = 0.2 bits... You want it to be less than 0.5 bits. At 4.73db, you'd get 0.5 bits and your ADC's LS bit would be flawed. 3db is generally used as the point before which we'd be assured our LS bit would start to be inaccurate. So, the analog input bandwidth is the frequency at which the SN ratio has fallen by 3db. Anyway, I looked and couldn't find it in the specs, but I think that that could be because if you use the device as specified, there shouldn't be attenuation that high. I did find a few app notes that might help you. http://ww1.microchip.com/downloads/en/AppNotes/00546e.pdf and http://ww1.microchip.com/downloads/en/AppNotes/00546e.pdf ....Typically that really isn't a problem. Is there a particular reason why you need that info? The max sampling rate is in the spec (I'm assuming you already know that), and if you are under that there really shouldn't be a problem. I've used the ADCs inside many processors (including PICs), and I've never had to worry about the analog bandwidth.... Let us know what you find out....I'm curious :-) Good hunting.

\$\endgroup\$
5
  • \$\begingroup\$ Thanks for your input, that is the problem, I was not trying to "use the device as specified" I was trying to sample a signal that is centered around a few MHz while the ADC could only go for hundred kHz. I was trying to implement an undersampling scheme. So I had to worry about the analog bandwidth, I didn't want my signal to be attenuated by the sample & hold circuitry which may act as a filter. Ultimately I changed the IF frequency to be within the ADCs specs. \$\endgroup\$
    – br4him
    Feb 10, 2016 at 8:42
  • \$\begingroup\$ What do you mean by "undersampling scheme"? Do you mean it's a repetitive signal and by sampling at a non-harmonic of the fundamental frequency, you are trying to reconstruct the signal? \$\endgroup\$ Feb 11, 2016 at 9:44
  • \$\begingroup\$ I meant that I am sampling the signal way below its Nyquist frequency (taken to be double the highest frequency component) and I was worried that the signal would be attenuated by the ADC since the analog bandwidth of the ADC wasn't specified and there's no guarantee that it'll allow signals way above the ADC's sampling rate without attenuation. \$\endgroup\$
    – br4him
    Feb 11, 2016 at 10:28
  • \$\begingroup\$ I guess my question is why are you undersampling? What are you trying to achieve by doing so? There might be another way to achieve what you want if I knew what you were trying to do. the only application that I'm aware of for undersampling might be to reconstruct a higher frequency repetitive signal, which coul be done by undersampling. Is it something like that? \$\endgroup\$ Feb 13, 2016 at 5:07
  • \$\begingroup\$ I was trying to sample an IF signal without having to downconvert it to baseband. It was a low IF signal ~ a few Megs. Anyway, the project is over now, I was able to use a professional digitizer that could go to 60 MHz instead. Thanks again :) \$\endgroup\$
    – br4him
    Feb 13, 2016 at 8:06
0
\$\begingroup\$

You will likely not find a direct specification for the analog bandwidth in the data sheet for an A/D input of a microprocessor. Bandwidth to the analog input pin is dependent on the source impedance of your analog signal, capacitance of the connection circuitry and the input impedance of the MCU pin.

Beyond that it becomes far more important for you to look at the "Throughput Rate" that Microchip will show for various classes of input pins. Throughput will be largely affected by the sample and hold characteristics of the part. And as you can see the sample time is dependent on the source impedance and the selected ADC clock frequency.

enter image description here

This data taken from PIC32MZ family data sheet.

\$\endgroup\$
2
  • 2
    \$\begingroup\$ The OP wants to use undersampling (not oversampling) and therefore the analogue bandwidth is all-important for this. \$\endgroup\$
    – Andy aka
    May 2, 2015 at 11:14
  • \$\begingroup\$ Like @Andyaka said, I need to know the analog bandwidth because I am undersampling. \$\endgroup\$
    – br4him
    May 2, 2015 at 12:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.