Designing an amplification circuit (amplifying small sensor signals)

Discussion in 'Analog & Mixed-Signal Design' started by Henry603, Nov 19, 2018.

  1. Henry603

    Thread Starter New Member

    Nov 19, 2018
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    Hi,

    I'm new to circuit design and I need your help to design an amplification circuit. I did a lot of research so far but still have quite some misunderstandings I suppose. I hope I can change that with the help of some experienced engineers! :)

    My Problem:
    I have a sensor that outputs a signal between 0 - 20mV.
    It is a passive sensor and only has two wires: HI (+) and LO (-). So I guess the best way is to amplify it differentially.
    In the datasheet it says:
    "To accurately measure changes the data logger or data acquisition system requires a total input measurement uncertainty (error) of 5 µV, or less; including noise, offsets, resolution, temperature effects, etc."

    I have chosen a 16Bit ADC that will be powered from 5V and has an internal reference voltage of 2.048V (V_ref). Because of the accuracy requirement mentioned above (after amplification of the signal by a factor of ~100, I need a resolution of 500uV) I need at least a 14Bit ADC (I chose a 16 Bit to have some air). I have posted the datasheet of the chosen ADC below.

    So before I can feed my signal to the ADC I need to amplify it. I have chosen an Instrumentation Amplifier (INA) with low offset voltage and low drift and rail-to-rail output capability. The datasheet is again linked below.
    I would attach the HI (+) wire from the sensor to the non-inverting input and the LO (-) wire to the inverting input of the INA. Then set a gain resistor to generate a total gain of x102 (to amplify the sensor signal to match V_ref of 2.048V of the chosen ADC) and feed it directly to the V_in (+) of the chosen ADC. To make clear what I want to do: the max. output signal I get from the sensor is 20mV. By amplifying it with ~102 gain, the signal will be amplified to ~2.048V. This will match the internal reference voltage of the chosen ADC. So I will accurately be able to convert the full range of my analog signal to digital.

    The simplified circuit would look like this:
    [​IMG]

    [​IMG]
    [​IMG]

    Potential issues:
    • Even the INA is specified as rail-to-rail it will not be fully able to swing to GND and VDD so I guess that might be a problem for my amplified signal as it will not cover the entire range of my sensor signal? Is that smth. I have to consider and how do you handle this kind of issue?
    • There should not be any impedance issues as the sensor needs a load that is >1M Ohm and the INA fulfills this requirement. I guess there should not be any problem feeding the amplified signal into the ADC directly either?
    • Can I attach my sensor to the INA as described above? Do you see other potential issues with my circuit idea?
    • The chosen ICs should be pretty accurate or do you see any issues for my application?
    I hope you can give me some hints/feedback as i'm pretty new to this. I highly appreciate any advise! :)
    Thank you.

    INA:
    Datasheet of Instrumentation Amplifier (AD8553)
    ADC:
    Datascheet of ADC (ADS1110)



     
    Last edited: Nov 19, 2018
  2. BR-549

    AAC Fanatic!

    Sep 22, 2013
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    Frequency of sensor signal?
     
  3. Henry603

    Thread Starter New Member

    Nov 19, 2018
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    It is a pyranometer (measurement of sun radiation) and changes very slowly. I only need to take one reading per second.
    Thank you! :)
     
  4. BR-549

    AAC Fanatic!

    Sep 22, 2013
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    May I ask what character you are detecting with that resolution? And field of interest? Have installed many pyranometers but not at that resolution. That was years ago.....just curious.
     
  5. Henry603

    Thread Starter New Member

    Nov 19, 2018
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    This is the pyranometer: Kipp and Zonen CMP10
    This are some specs, the one marked in red is the one i'm using (its for a weather station, recording the daily/yearly trend):
    [​IMG]
     
  6. BR-549

    AAC Fanatic!

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    May I ask at what latitude you are measuring sunlight?
     
  7. Henry603

    Thread Starter New Member

    Nov 19, 2018
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    Latitude and longitude coordinates are: 48.137154, 11.576124.
    Any thoughts on the circuit I proposed? :)
     
  8. BR-549

    AAC Fanatic!

    Sep 22, 2013
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    It's been almost twenty yrs......I was just curious. Have you ever seen a glass sphere burn a line on a strip chart?

    We replaced a lot of these....and even installed some.

    https://www.amusingplanet.com/2016/10/campbellstokes-recorder-simple-device.html

    I noticed that the data sheet has link to sensor amplifier by same company. Probably expensive.

    And I will quit bothering you and let the experts advise you on your circuitry.
     
  9. Henry603

    Thread Starter New Member

    Nov 19, 2018
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    No, I have never seen something like this, I do not know much about pyranometers to be honest, I'm just interested in the technical circuit design stuff to amplify the signal properly so I can use it with a microcontroller :).
    The sensor amplifier of the company you mentioned is not an option. I have to develop an amplification circuit myself and I hope some of you experienced people will help me with that.
    I would really highly appreciate any feedback on the proposed design in the first post.
     
  10. BR-549

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    Can you find any more information on that sensor? I am not familiar with that type. And from the information....not clear on needed house keeping conditions. Have you a data sheet with operating plots and device under test circuit?

    If the senor self outputs millivolts......then millivolts into what impedance? I would need a clear understanding of the sensor before any consideration.

    Is it a "light" version of a thermocouple?
     
  11. danadak

    Well-Known Member

    Mar 10, 2018
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    Note there are processors out there with DelSig ADC onboard.

    This family even has DSP and DMA so that you can create a HW process
    on chip that acquires/filters/transmits out the result w/o code intervention.

    Example shown is 20 bit, resolution 4 uV, Vref (onboard) set to +/- 2.048 V

    As you can see in resources window most of the chip free to do other stuff.

    Board is $ 10 for basic board with debug.

    upload_2018-11-19_20-0-36.png

    Regards, Dana.
     
  12. Henry603

    Thread Starter New Member

    Nov 19, 2018
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    Ok, thank you to both of you! :)
    @BR-549:
    I only know the data from the datasheets I posted. In the datasheet is also written that if the driven device has a larger impedance than 1M Ohm the it is fine. The proposed INA I posted has even a way larger input impedance so that should be ok.

    @danadak:
    Ok, I will check it out. But do you have circuit design experience? Could you give me some feedback on the proposed combination of ICs/circuit I described in the first post?

    If the signal amplification I wanna do works with the porposed ICs that would be great! But as i'm inexperienced I would need some kind of feedback if this is going to work or not. Will the amplification work as described? Will the top and bottom of the signal range give me a headache due to the rail-to-rail limits of the INA (even if it is an rail-to-rail).
    Do you see potential issues with the proposed method?

    Best. henry.
     
  13. KeepItSimpleStupid

    Distinguished Member

    Mar 4, 2014
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    A couple of comments:

    1. You always need a place for Ib to go
    2. Vref most of the time has to be supplied by a low Z source
    3. Vref is many times Vcc/2; Note Vcc/2 is not 2.5 V if the supply is 5V. This is called a ratiometric measurement.
    If Vref was 2.5V then that's an absolute measurement.
    4. Sensors, especially Automotive use the Vcc/2. The sensors can't output 0V, nor can output the supply voltage. They get close.
    5. OP amps may blow up if the input current is higher that allowed. When the supply is 0V, you effectively are operating into max of +-0.3V diode that needs to be current limited. This depends on the OP amp. Analog Devices/Linear Technology has some "Over the Top" op amps which allow the inputs to exceed the supply.
    6. The output of an IA direclty into another OP-amp without a resistor will cause you undue grief, if you don;t know what's going on.
    7. Bypass caps are often neglected. Use manufacturer's recommendation.
     
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  14. danadak

    Well-Known Member

    Mar 10, 2018
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    Yes, the CM range of the IA both in and out looks OK. Note
    the IA calls for a low noise ref to minimize offset. You should
    do an end to end error analysis to insure you meet design
    goals. Easiest way to do this is convert all errors to a common
    base, like ppm, and then add to get total error. Noise of course
    handled RMS.

    The G error in the ADC is .4% which translates to ~ 8 bits absolute
    accuracy. So you have to decide if goal is absolute vs relative accuracy.

    The CM range ADC input is excellent.

    Board layout critical to minimize errors due to coupling into hi-z nodes,
    and noise. 16 bits over T and V and Layout very challenging. Do the
    error budgets.


    Regards, Dana.
     
  15. Henry603

    Thread Starter New Member

    Nov 19, 2018
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    @KeepItSimpleStupid:
    Thank you I will reply to your advise, would be nice of you could give some feedback on it :)

    1.
    Do you mean the curernt return path? Could I just add a ground like I did in the picture below?
    [​IMG]
    But I guess this would cause my measurements to be single ended and not differential anymore?

    2.
    For my IC V_ref is internal and should be fine I guess.

    3. and 4. (important, please give me your thoughts on this one)
    I have chosen a rail-to-rail instrumentation amplifier and I would supply it with 2.048V. Therefore it will output the amplified signal in the range of 0V up to 2.048V. I chose the supply like this so the internal reference voltage of the ADC (where V_ref is 2.048V) matches my INA output voltage range. So now I should be able to use the full range of the 16 Bit ADC from going from 0 up to 2^16-1=65535, right?
    So in my example, if the sensors measures 0mV it will amplify to 0V, so my ADC will output 0. If the sensor measures 20mV, the INA will amplify it to 2.048V (by setting the gain resistor of the INA for x102 gain) and my ADC will output the maximum value 65535. I know in practice this will not be that ideal behaviour but will it do what I just described roughly? (so I can still get a decent accuracy out of the described system?)

    5.
    Would I use diodes for blocking a beyond rail voltage? Would the resulting voltage drop not be a problem for my system?

    6. and 7.
    Thanks I will. I guess in the datasheet the manufacturer will state how to drive an ADC, like e.g. in my case for the INA:
    [​IMG]
    Simple 100Ohm resistor and a capacitor should be enough?


    @danadak:
    Thank you.
    But the .4% error only is there when using the internal ADC PGA gain I assume? I want to use an INA in front of the ADC for amplification like statet above so this error should be way smaller for me?
    Do you still think I could only achieve 8Bit accuracy with the proposed system? I do not understand how you came up with that bad accuracy. That accuracy for me is useless :/


    Thank you for the help so far.
    I hope you can answer my questions and maybe also some other users can give some feedback on my undertaking.
    Also it would be awesome if you could link me to some thread/project/tutorial that might help me with my problem.
     
  16. danadak

    Well-Known Member

    Mar 10, 2018
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    Look at the G error in the datasheet, its is sum of PGA and onboard Vref error.
    Thats why I asked you do you care about absolute accuracy or relative. If later
    this is largely not a problem. In a cascade of signal processing elements the
    product of all errors is the overall G error. So the fact the INA is excellent
    does not obviate the elephant in the signal chain, if you are seeking absolute
    accuracy. Suppose you have a sig chain of 3 components, each with a G
    accuracy of .99. The Goverall = .99 x .99 x .99 = ~ .97. Now suppose one of
    those is replaced by an element with a G accuracy of .8, Goverall = .99 x .99 x
    .8 = ~ .78. As you can see the good parts did not obviate the lousy performer.

    With a G error of .4 % thats ~ 1 part in 200 (rounding .4% up to .5%) . An 8 bit
    converter, good to 1/2 LSB, is 1 part in 256, that's why I called it effectively 8 bit
    performance.

    Then look at your sensor, total of all errors approaches 1%, which effectively is
    a 7 bit converter.

    There is a way of taking a "pig" part and making it look high performance.
    Its done in production test.

    upload_2018-11-20_19-39-16.png

    But your sensor has a long term drift problem of .5%, so that has to be modeled
    and compensated for as well.

    Note the PGA in the A/D is always present in the signal chain, so its
    poor specs limit the system, for absolute measurements. Note G error
    in a PGA is largely driven by R ratio accuracy in the feedback loop,
    and if Aol of the OpAmp used in the PGA is low, that as well. You can
    do a simple matlab or algebra analysis to show this. Very instructive.

    The method I showed above to achieve high accuracy is used to laser trim
    the Analog Devices IA on the production line, probably at final test.


    Regards, Dana.
     
    Last edited: Nov 20, 2018
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  17. KeepItSimpleStupid

    Distinguished Member

    Mar 4, 2014
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    Your understanding is WAY OFF. Its been a long day and it's bed time for me.

    The input bias current needs a place to go. (transistors, FETS, OP-amps etc.) it's effectively a leakage current and it has to go somewhere. It will develop an offset voltage across some resistor. Even if it's 1 pA, it has to go somewhere. A 10 M resistor from each input to ground is fine. It can generate (1 pA * 10 M) or not a lot of input voltage and you have the same, but opposite on the other input. Now, you see where matching comes into play. These devices are not ideal.

    You can't reach the supply rails. Look at figure 3 and figure 9 here: https://www.analog.com/media/en/technical-documentation/application-notes/an18f.pdf there is a transistor and a resistor between the rail and the output. So, at saturation, you have some diode drop that's dependent on temperature and some IR drop across the resistor that's dependent on current.

    The output CANNOT equal the supply rails. FETS get you a LOT closer, but even in the limit, you can never reach the rail.
     
  18. Henry603

    Thread Starter New Member

    Nov 19, 2018
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    @danadak: Ok, thanks, that is very informative.

    @KeepItSimpleStupid:
    Ok thank you! So that means by implementing resistors as shown in the following image it should be fine (current return path is handled properly)? Is that what you meant?
    [​IMG]

    Ok, if I can not get up to the rails, how is this handled properly then? Every Tutorial I found was describing how to implement an INA to match the V_ref range of the ADC. However they never mentioned that it can not get up/down to the rail completly.
    How would you make it like I can properly amplify my signal and still use the whole ADC range/resolution steps.
    If I power the INA with VDD_INA = V_ref_ADC my INA will not be able to completly go to rails as you just said.
    But I can not power the INA with VDD_INA > V_ref_ADC as then my INA might output a higher signal than V_ref_ADC right?
    So how can I match this? Amplify my sensor signal so it matches the ADC range?
    Would really highly appreciate it if you could help me out here, as most project/tutorials only say that one has to match the AMP to the ADC range/V_ref but they do not go into detail at all. Im pretty much clueless at the moment.

    Thank you so much! :)
     
  19. danadak

    Well-Known Member

    Mar 10, 2018
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    The INA you chose will do almost R-R on output. Spec is 75 mV off the rail with a 10K
    ohm load, room temp.


    Regards, Dana.
     
  20. KeepItSimpleStupid

    Distinguished Member

    Mar 4, 2014
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    Dear clueless.

    the datasheet for the A/d converter basically says:

    What does that mean? The datasheet basically says, you loose a bit. So you have 15 bits instead of 16. The A/D has a PGA so for low levels, you can increase the resolution. Quantization error is sometimes a problem.

    Suppose you had an A/D that went 0-100 (this is totally ficticious). they usually measure +- 1 bit, so at low levels like 4 bits, the % error is LARGE. 100 just makes percentages easy.

    This means that you have to apply your own Vref of ~2.048 V to the Vref input of the IA. This IA does not require a low Z Vref. The A/D allows some adjustment of the 2.048 V internal reference. That's basically your 0V.

    At the other extreme, you won't quite be able to get to 5V. With a 5V supply, you can get to 4.92 V max, NOT 5V. Notice that 4.92 is 2.46 Volts.

    So a 4.92 V output from the IA is +2.46 V Full scale of your sensor. So, you have to loose a little bit on your upper end too.
    So, you might scale the sensor to 2V. The IA will output Vref+(Sensor voltage)* (IA gain). Your A/D will "see" 2V.

    So, although you can get 2.048 V max, your going to have to guarantee the 5V supply is a little tiny bit bigger than 5V. 4.92/5 is NOT >= 2.048.
    It's close.

    This is part of the design process where you look at tolerances and your "error budget" like resistor tolerances, power supply tolerances, accuracy, resolution, temperature effects etc.

    The other part is are the inputs within range including power up where the rails are 0V,

    My brain isn't the best right now (migraine), but maybe you get the idea.
     
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