Electrochemical Sensor & Op Amp Baseline understanding

Discussion in 'General Electronics Chat' started by Sam Matthews, May 6, 2016.

  1. Sam Matthews

    Thread Starter Member

    Jan 16, 2016
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    I'm having trouble understanding how to 'set' a baseline output on a Op Amp, this may be something very simple to you guys or even to explain, but for some reason i can't seem to find information regarding this to educate myself.

    My current circuit here is one to detect Carbon Monoxide, specifically using a Figaro TGS5042 electrochemical sensor. The schematic for my basic circuit is:
    cir1.JPG

    This works pretty well and i can monitor the uA changes in the sensor when exposed to burning paper. However, i start reading through the Datasheet and the Appnotes and they recommend to use a baseline value of >1V in my proposed setup, and this is where i become stuck. For some reason i seem to keep assuming that the baseline would be a voltage that i put onto the +Input of the Op Amp. But then i read in the datasheet that any voltages on the sensor terminals over 10mV can cause damage to the sensor itself.

    Baseline voltage is mentioned on page 4, 6 and 10 of the Appnote document. Pages 4 and 6 touch on the use of a FET as the short circuit precautions, so this may not be of any interest. I currently plan to use the basic resistor short circuit method, but i may switch to using a FET maybe. But I diversify, the baseline is what i really need information regarding at the moment.

    In the Appnote there is a "Change baseline voltage circuit" but i have no idea how/where i'm supposed to connect this into the current circuit. Could someone please explain how i go about creating a baseline voltage please?
     
  2. BR-549

    Well-Known Member

    Sep 22, 2013
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    The offset voltage is not located at the sensor.

    Edit......The degree of concern will depend on op-amp selection.
     
  3. jpanhalt

    AAC Fanatic!

    Jan 18, 2008
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    I fond this rather old article helpful in understanding biasing (pdf attached).

    John
     
  4. Sam Matthews

    Thread Starter Member

    Jan 16, 2016
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    Am i wrong in thinking that it is as simple as feeding 1V into the input of the OP AMP to then cause the OP AMP's output to sit naturally at 1v to compensate for the input voltage?
     
  5. dannyf

    Well-Known Member

    Sep 13, 2015
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    It is a current sensor. So you want to offset the output signal a little so that the opamp doesn't have to work near extremes (in this case, ground).

    I would simply use a resistive divider + capacitor.
     
  6. Sam Matthews

    Thread Starter Member

    Jan 16, 2016
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    My current schematic for this setup is:

    [​IMG]

    Is this voltage divide in the location that you would put it?
     
  7. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    Why do you want to shift the baseline, and to what value? Anything you do to the input is going to be multiplied by 1001, so it won't take much.

    Also, even with the matched impedances, input offset voltage and bias current errors will be multiplied by 1001. A +1 mV offset error will appear as 1 V at the output. A -1 mV offset error (with no other input) will cause the opamp output to saturate against the negative rail.

    ak
     
  8. Sam Matthews

    Thread Starter Member

    Jan 16, 2016
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    I would like to offset the voltage on the output of the OP AMP by 1V-2V simply because as the application notes of my sensor states, "the sensor output may have a negative value due to the offset voltage of the OP AMP, the baseline sensor output should be set >1V".

    Am i miss-interpreting the information that the app notes is giving me? For what reason would the op amp go negative on the output, only if the sensor itself goes negative, right? I think that the only situation that this could actually happen is as the Electrochemical sensor ages, it loses water from its internal cavity due to evaporation. This would, in my opinion, cause the sensor to move away from its 0PPM reading negatively. There is also effects from temperature that may drop it negatively. Again, those are just my opinions and assumptions. I'm currently searching the datasheet and appnotes to find any more information out regarding why there should be a baseline of >1V.

    Also, figure 21a on the app notes document shows a "change baseline circuit", would you apply the two resistors to the left of the OP AMP in this imagine to my circuit?
     
  9. KeepItSimpleStupid

    Well-Known Member

    Mar 4, 2014
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    Are you doing current conversion by amplifying the voltage across the shunt or the I-V converter method?
    It looks like your amplifying a voltage.

    The one thing they "kinda hid" is that the I-V converter used doesn't used the "standard" terminology.
    They did illustrate with arrows. I've done a few I-V converter designs but I used a dual supply.

    In a standard I-V converter, the + input of the OP-amp is connected to ground. This can be connected to any low Z source to BIAS the sensor or add offset. lets call the voltage applied to the (+) input as Vbias.
    the standard way of writing the output is Vb-I*Rf. The currents are summed into the (-) input.
    So, that (-I*rf) is equal to Vb. Vb happens to be imposed across the sensor.

    With an I-V converter Ib and Vos are the important parameters. Vos likes to change with temperature.
    In some systems a zero check function can be implemented which effectively amplifies Vos.

    Vos can go either way positive or negative.

    You may have a choice of auto-zero OP-amps.

    They did mention recovery, hence the multiple diode-clamp.

    With a J-fet the unbiased state is ON, so with no power the sensor is shorted.

    I need to read the app note more carefully to understand baseline. What I think they are doing is level shifting the output. This can do a lot for you. With a microprocessor, quantization error is reduced and resolution can be improved for low concentrations.

    e.g. a 10-bit converter still has a hard time resolving zero, but when zero is 1/2 full scale, it's a lot better.
    That's partially why you see 1-5V outputs and not 0-5 sometimes.

    With a rail to rail OP-amp, again zero is a problem. the Rail isn't quite the rail. it's close, but not the rail.
    So, with a single supply, you need to offset the output.

    Another concept to introduce is "Ratiometric". e.g zero can be 1/2 the supply voltage whatever it is. "Ratiometric" sensors are used a lot in automobiles. The engineering unit output is a ratio to the current value of the supply voltage. 2.5 V on a 5V system is not always 0 for instance. If the 5V supply is 5.1V, zero shifts to 1/2*5.1.

    The capacitance across the I-V converter resistor restricts the bandwidth and makes the I-V converter faster settling and more stable.

    It's an interesting app note, BTW.
     
  10. BR-549

    Well-Known Member

    Sep 22, 2013
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    Sam.....I am assuming you have a resistive load for the op amp.

    What voltage do you want to output to represent 0.0 CO?

    Well, whatever it is..............................another circuit builder might want a different output voltage to represent 0.0 CO.

    Or another builder might use a junky op amp.

    These two conditions MAY (that means might not certainly will) cause a negative voltage on the sensor.

    If this happens....they recommend changing the 0.0 CO level to a little over 1 volt. OK?

    This is why they recommend what amps to use. They even give an op circuit to calibrate and measure CO, not just detect it.

    Also.....always consult the op-amp data sheet and app notes for same, that you are using, in which you have not mentioned.

    As said earlier and in sensor note, most of this depends on the op-amp chosen.
     
  11. crutschow

    Expert

    Mar 14, 2008
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    My understanding is that the reason for the 1V offset is to take care of op amp input offsets.
    If you use an op amp with an offset adjustment pot, then you shouldn't need the added 1V offset.
     
  12. KeepItSimpleStupid

    Well-Known Member

    Mar 4, 2014
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    1.827 nA/ppm is a pretty small current. Vos * Rf gets amplified too. As I said, usually a ZERO check function is employed for really low currents. If I remember, if you essentially short the sensor and add a resistor from the - terminal to ground in an I-V converter, you essentially now have an amplifier that is amplifying Vos which you can correct for and/or adjust. Vos is a strong function of temperature.

    Note also that the app note mentions the ability to check the health of the sensor.
     
  13. Sam Matthews

    Thread Starter Member

    Jan 16, 2016
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    OP AMP: TLC272
    Datasheet: http://www.ti.com/lit/ds/symlink/tlc272.pdf

    Sensor: Figaro TGS5042
    Datasheet: http://www.figarosensor.com/products/5042Dtl.pdf
    App Note: http://www.figarosensor.com/products/5042app.pdf

    I have included the documents for the OP AMP and the Carbon Monoxide sensor above. Now, to answer your questions;

    As far as I'm aware of from reading information regarding this sensor, an extract from the datasheet reads "The sensor generates a minute current which is converted into sensor output voltage (Vout) by an OP-AMP/Resistor combination.". This is what their basic suggested circuit does. Then I'm adding in polarization safety in the form of a fixed resistor (this is the simplest way, they claim).

    I was wanting 1V, that should give plenty of room for movement, from my tests i only need around 2-3V movement for a huge scale all the way to >1000PPM. I'm very tempted to dial this in to be more accurate but only cover 0PPM - 400PPM. Humans can't withstand 400PPM for anything over 30 minutes anyway. So I will not care if its above 400PPM, i will be well away from the area before it could even tell me.
     
  14. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    No. Either the input offset voltage, input bias current, or both can cause errors that drive the output negative enough to clip against the negative rail. Something to consider is a chopper-stabilized amplifier, sometimes called a CAZ (commutating auto-zero) amp. Intersil, Linear Tech, Maxim.

    ak
     
  15. KeepItSimpleStupid

    Well-Known Member

    Mar 4, 2014
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    AnalogKid likes this.
  16. Sam Matthews

    Thread Starter Member

    Jan 16, 2016
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    @AnalogKid - Okay, well to be honest i had to re-read what you put 3 times over and then check back to what i said. I can't quite believe i said that the only reason the OP AMP will go negative is if the sensor itself goes negative. Sorry.

    While i will definitely read the linked article, could you tell me what i'm supposed to be getting from it to apply to my situation? I'm a little confused on what it has to do with my proposed use of an OP AMP.
     
  17. KeepItSimpleStupid

    Well-Known Member

    Mar 4, 2014
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    The article uses a supply splitter (Rail splitter) and an auto-zero amplifier with a CR2032 battery for simple amplification. Stuff that's been talked about.
     
  18. Sam Matthews

    Thread Starter Member

    Jan 16, 2016
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    I'm still confused though. Are you suggesting that i build this circuit into mine to change the way its amplifying the sensor output?

    I'm tempted to just ignore any negative readings as these are no use to me anyway, the only time i want this sensor to be of any use is when there is CO in the ambient vicinity. So, unless not allowing the opamp output to show these negative readings from the sensor effects its validity in the positive range then i'm thinking of allowing it to not show these readings.

    Do you see this causing issues do you think?
     
  19. Sam Matthews

    Thread Starter Member

    Jan 16, 2016
    178
    3
    I've been reading through some documents regarding the use of a voltage follower to generate voltage bias onto the input of the OP AMP for the sensor. The main document that i've read through is: http://ocw.mit.edu/courses/media-ar...g-2011/readings/MITMAS_836S11_read02_bias.pdf

    At the bottom of page 3 and the top of page 4 it shows how to configure the voltage divider, i assume this is connected to the sensors OP AMP input. However, every circuit using this method that i have seen has it connected to the non-inverting input, to connect it to my current circuit i would need to provide -1V as the bias wouldn't I, or to connect it to the non-inverting input? [​IMG]

    The above is my current setup. How would you connect this voltage follower into this layout?

    Does this sound like what i need to do, am I even on the right lines?
     
  20. KeepItSimpleStupid

    Well-Known Member

    Mar 4, 2014
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    If you shorted R3, Vos or the voltage (because of not being a perfect OP amp) between the + and - gets amplified.

    The current some 2 pA per ppm gets converted to a voltage e.g. (2e-12 Amps) and multiplied by a big number.

    Vos has to be small, so it doesn't swamp out the measured value.
     
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