2

For an ADC evaluation board like this one, where signals go in through exposed pins, I'd like to have the signals come from a BNC cable (or SMA). However, I'm concerned about noise. What's the right way to do this connection? There doesn't seem to be commercial solutions that go from jumper cables/pins to BNC/SMA cables.

EDIT:

So to be clearer, here's what I need to connect to (which is another board for the same ADC). I need to connect to the J2 part on the left, I need to get two adjacent pins (AIN0 + AIN1, for example) connected to BNC.

enter image description here

The Quantum Physicist
  • 225
  • 2
  • 12
  • You are more likely to run into problems due to grounding and shielding than the actual connections. Keep the latter short and surround the board with a metal case (or at least mount it above a metal plate) connected to *analog* ground, if possible. – Spehro Pefhany Aug 22 '17 at 17:54
  • There's no ground. I'm going for [differential](http://www.ti.com/lit/ds/symlink/ads1256.pdf). – The Quantum Physicist Aug 22 '17 at 18:09
  • 1
    Its unusual to be trying to use a BNC for a differential signal. It's usually signal+ground. Are you sure you have a differential signal coming in? – brhans Aug 22 '17 at 18:13
  • @brhans We do high-precision measurements in the atomic magnetometry sector, and all our measurements are differential to minimize noise. My doctoral dissertation had a magnetometer with relative precision of 7x10^-8. You can't reach that with a common ground. – The Quantum Physicist Aug 22 '17 at 18:16
  • 1
    Ok - then I would've expected a shielded twisted pair. Differential signals over coax/BNC just seems very strange. – brhans Aug 22 '17 at 18:25
  • 2
    @TheQuantumPhysicist You will still need to reference the inputs to a voltage related to the ADC analog ground (typically something like +1.25V) to maintain both inputs in the common mode range. You can use two BNCs. – Spehro Pefhany Aug 22 '17 at 19:32
  • @SpehroPefhany I'm sorry, I don't understand. Correct me if I'm wrong: There's the ground of the whole circuit, that's a given, but doesn't matter. For a single channel, only 2 poles suffice, which is why it's "differential". It measures the difference between two voltages independent of the other parts of the ADC. – The Quantum Physicist Aug 22 '17 at 19:40
  • 1
    @TheQuantumPhysicist It will only work if BOTH input voltages are within the "common mode range" of the ADC. If you don't know what I mean, read up on it, it's a bit much for a comment. – Spehro Pefhany Aug 22 '17 at 19:44
  • 1
    @SpehroPefhany I definitely will. Thanks for the tip :) – The Quantum Physicist Aug 22 '17 at 19:50

3 Answers3

2

Solution with highest signal integrity is:

  • Cut BNC-BNC cable in half, strip it down
  • Solder the braids to the boards' ground plane on the back (remove soldermask with blade first)
  • Solder coax center conductor to input pin

Less annoying solution:

Since half the pins on the analog input connector seem to be ground, I would use a 16 pin 0.1" dual row female header soldered on the edge of a bit of blank PCB. Only solder the ground pins, let the other ones hang in the air. Now drill 2 holes in the PCB, install BNC connectors which will be properly grounded, and connect the signal conductors to the header with short wires.

EDIT

The enormous CMRR of "110dB typical" rejects differences in voltage between "local ADC ground" and "signal common mode voltage or remote source ground" but the signals have to be between 0V and AVCC, that's not negotiable. Most likely the 8 unused pins are connected to the ground plane. We've been having a little bit of a misunderstanding because to me it is obvious that the pins should be grounded, so I had trouble understanding why you think they are not.

For example, someone would plug a ribbon cable on the connector, in this case it is nice to have one wire for ground between each signal line, as this prevents crosstalk.

If the cable is shielded (like shielded twisted pair) then the shield becomes useless except for very low frequencies unless it is grounded at both ends. The shield can be grounded through a capacitor at one end though, to block low frequency currents in it while allowing it to shield at HF.

So, I wonder what you really mean...

bobflux
  • 70,433
  • 3
  • 83
  • 203
  • Btw, it's a differential signal (which is why it's 2 pins). So no ground. But I see your point. – The Quantum Physicist Aug 22 '17 at 17:52
  • Are you sure? ADS1256 has only 8 input pins, there are 16 pins on the connector... – bobflux Aug 22 '17 at 18:02
  • Yes. Otherwise I wouldn't have picked it. [Check this](http://www.ti.com/lit/ds/symlink/ads1256.pdf). It's either 8 single, or 4 differential. – The Quantum Physicist Aug 22 '17 at 18:08
  • Yes, I suspected that (a 24 bit ADC without differential inputs would be really weird). But in both cases these are 8 signal lines, the connector has 16 pins, thus the rest should be ground... Do you plan to not connect the ground at all? Could lead to problems, what keeps the signal voltages inside the ADC's acceptable input range? – bobflux Aug 22 '17 at 18:53
  • There will be ground, but the ground of the circuit, not the ground of the signals. The ground for the signals is not common, and that's very important for high precision. If you [check here](http://www.ti.com/lit/ds/symlink/ads1256.pdf), at page 14, the right column, you'll see that the same 8 pins for single ended measurement are used for differential. Apparently the extra pins on the board are superfluous with no function, or just copies of the same pins on the left of each. – The Quantum Physicist Aug 22 '17 at 19:12
  • Yes, the input multiplexing is quite flexible! This chip seems to be pretty nice. But that's not what I'm talking about... The chip can only measure the difference between voltages which lie inside the range 0V-5V (if AVCC is 5V) relative to its local ground. Outside of this range, the input protection diodes will shunt the signal to AVCC or ground. Thus whatever generates the voltages to be measured (sensors, I presume) must have some sort of common ground to ensure voltages stays within acceptable range. – bobflux Aug 22 '17 at 20:24
  • (see edits to my answer, it was too long for a comment) – bobflux Aug 22 '17 at 20:31
  • I'm a little confused. On page 16 of the datasheet it says the range is +/-, not 0 to something, depending on the configuration. Actually I haven't thought about crosstalk that way. I thought twisted pair should be sufficient for this. I really have to think about this more deeply. Perhaps I should create another question for this very issue to learn how to do this best. – The Quantum Physicist Aug 22 '17 at 20:51
  • Yeah it would be better you give all the info unless there is a NDA you signed in blood ;) Look at datasheet page 3, "absolute input voltage" this gives you the input voltage range where the ADC will work, and it is roughly between GND and AVCC. This is normal, as having a chip handle signals outside of its supply voltage range is either impossible or comes with compromises... So, this ADC will accept a differential voltage range which depends on the internal gain setting, but both wires of the differential pair must be within the 0-5V range. – bobflux Aug 22 '17 at 20:58
  • Isn't that "absolute input voltage" for single ended ADC, not for differential? – The Quantum Physicist Aug 22 '17 at 21:04
  • Nope, check page 2 "absolute maximum rating" => "Analog inputs to AGND −0.3 to AVDD + 0.3" this is valid for all inputs no matter how they are configured, single ended or differential. Also, I searched for "diodes" in the datasheet, which jumped to page 14 "ESD diodes protect the analog inputs. To keep these diodes from turning on, make sure the voltages on the input pins do not go below AGND by more than 100mV, and likewise do not exceed AVDD by more than 100mV: −100mV < (AIN0 − 7 and AINCOM) < AVDD + 100mV." – bobflux Aug 22 '17 at 21:07
  • This dialogue is extremely helpful for myself as well -- consider the case @TheQuantumPhysicist where (In+ - In- = 0.2V), but each with respect to external ground is over a million volts. Yikes, none of the transistors in our lil amplifier will be able to operate on them w/any reasonable circuitry. – bunkerdive Oct 09 '22 at 02:40
1

If you're looking for a clean solution, this board might be a decent solution.

enter image description here

This gives you two options. In the first, you can use these jumper wires, which you might have gotten with your Arduino. Either solder the male ends of the jumpers to the through-holes in the BNC board or clamp them in the screw terminal. Slide the female ends on J1.1 and J1.2 per this table from the reference manual for your board:

enter image description here

You'll probably want to twist the wires about each other to help with noise immunity. There's also a possible solution without cables. Pins 1 and 2 of J1 are on the short edge of the connector, so you can solder a female header like this (any 0.1" pitch header would work) on the BNC breakout, and then slide it on. Both of these solutions will work fine since your ADC is fairly slow, the slide-on method will have better noise immunity. Both will probably require you to epoxy/superglue/tape things down so nothing shakes loose.

Edit: I just saw your edit with the exact board you're using. I'd have to look at the schematic but so long as the adjacent pins in the Ain0 row are Ain+ and Ain-/GND, this still applies.

jalalipop
  • 835
  • 5
  • 13
0

Try these. There are also varieties with crocodile clip ends.

enter image description here

pipe
  • 13,748
  • 5
  • 42
  • 72
Trevor_G
  • 46,364
  • 8
  • 68
  • 151
  • These seem to be OK for testing and probing. I'm looking for a solution for reliable and continuous operation. – The Quantum Physicist Aug 22 '17 at 16:44
  • @TheQuantumPhysicist they also sell them with just wire ends that you can add pretty much any connector to or solder directly to your board or whatever. For a permanent solution you want to keep the unshielded wires as short as possible though. – Trevor_G Aug 22 '17 at 16:46
  • Actually, I was hoping there's a solution that doesn't involve soldering. I'm a little traumatized from soldering and am worried it would create additional noise in the results. – The Quantum Physicist Aug 22 '17 at 16:53
  • Solder shouldn't affect signal integrity at low speeds but soldering to pins can definitely get messy – jalalipop Aug 22 '17 at 17:10
  • @jalalipop My signal's sample rate is ~10 kHz. – The Quantum Physicist Aug 22 '17 at 17:14
  • In my world low speed is under a couple MHz haha. I have an idea I'm formulating into a answer now, but it still involves very light soldering unfortunately – jalalipop Aug 22 '17 at 17:16
  • @jalalipop Soldering by itself is not the problem. The problem is a solution that's vulnerable to doing it wrong. The cable solution suggested above can easily get messy. If you have a clean solution, please post it. – The Quantum Physicist Aug 22 '17 at 17:17
  • @TheQuantumPhysicist the real issue here is you have not really shown what you are trying to connect to. Your image of the whole board does not help. Perhaps a close-up image of the connection points might help. – Trevor_G Aug 22 '17 at 17:19
  • @Trevor Please check the edit. Does that clear it up? – The Quantum Physicist Aug 22 '17 at 17:26
  • 1
    @TheQuantumPhysicist buy a connector to the pins, buy a BNC connector, wire them together, plug onto the pins, keep the assembly for when you need to do it for the next board. I am afraid that an EE (whether hobby or for paid employment) that's traumatized by soldering is an EE that's not going to last long. I don't often say 'man up', but in this case, do some soldering, practice soldering, until it doesn't phase you. An EE that can't solder is like a poster that can't think of the right, you know, whatever I'm looking for, dammit ... See, it doesn't work if you don't solder. – Neil_UK Aug 22 '17 at 17:28
  • @TheQuantumPhysicist there are numerous crimp style connectors you could use to attach your BNC cable to those headers. As I said, just make sure you keep the unshielded part as short as you can. – Trevor_G Aug 22 '17 at 17:31
  • @Neil_UK Thanks for the advice. I'll try to do that. – The Quantum Physicist Aug 22 '17 at 17:33
  • @Trevor Absolutely. Being as short as possible is very important in my book to avoid having them act like antennas. – The Quantum Physicist Aug 22 '17 at 17:34