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.

K.I.SS. is totally correct about the input return current connection being a potential issue. I can tell you that it can be a very real issue as an input without any ground reference will drift around until one input reaches saturation So rather than a direct ground, a 1Megohm resistor from each input side to the analog ground buss should solve your problems completely.

 

@MisterBill2 :
Ok thank you very much. So to connect the sensor to the instrumentation amplifier like I did in the schematic in post #18 is the right way to do it? It should not introduce any extra issues (noise, voltage drops etc.) into my circuit then?

@KeepItSimpleStupid :
Thank you for the explanation and I hope your head is better by now

This means that you have to apply your own Vref of ~2.048 V to the Vref input of the IA.
The A/D allows some adjustment of the 2.048 V internal reference. That's basically your 0V.

Why would I apply my own v_ref to the IA? Are you suggesting that I should supply a ref voltage to the IA of 2.048V (from a voltage reference chip I guess?) and shift the sensor signal to a level of [signal in mV range]+[v_ref of 2.048]?
Then the gain I had to apply would only have to be set about x2?

With a 5V supply, you can get to 4.92 V max, NOT 5V. Notice that 4.92 is 2.46 Volts.

Could you please elaborate more on the last sentence? I do not really get the "Notice that 4.92 is 2.46 Volts" part.
I'm a bit confused reading your previous paragraph again.
If I apply ~2.048V v_ref to my IA like you said before, how would I get 2.46V then (if my sensor signal is only in 0-20mV range)?

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.

Because of my misunderstanding of the previous quoted paragraphs I'm confused.
Could you please try to explain that to me? I would really appreciate it!

Thank you a lot for your help so far!
Regards, the clueless .

 

The only purpose of adding an offset voltage is to bring the operation into a better part of the response curve. Even most "rail to rail" amplifiers work better away from the extremes of their input range. One other reason is that it is quite challenging to see "the difference between Zero and Nothing." That is why many current loops use 4 to 20 mA as their active range. A broken connection then outputs Nothing, while a "zero" output from the sensor is 4mA.

 

You have two parts that DON'T QUITE work together. The IA can ALMOST make 2.048+2.048 AND your going to have to make a zero adjustment using the A/D.

The output of the A/D says it's twos complement, so Vref is the 0 point. If Vref = 2.048 and you apply 0 V relative to 2.048V, the A/D's output is 0V.

So, yes, you have to provide your own 2.048 V reference to the I/A

The voltages CANNOT be exact because of tolerances. The A/D, I believe, allows you to adjust it's 2.048 V reference to match your external one to get a zero.

You'll have 15 bits 0 to +2.048, but you need to loose some on the upper end too. So, you might want to amplify your 20 mV with a gain of 2 for a maximum read of 2 V.

Numbers are NOT ABSOLUTE. They have tolerances. They run into brick walls (the power supply rails). They vary with temperature.

If the A/D can read up to 2.048V, but with a 5V supply the IA can only output 2.048+2.040, you have to use the IA's limits. If your 5V supply was say 5.1 V, you might be able to actually read 2.048 V, BUT what about the zeroing of the IA which looks like it can be done by mucking with the A/D's 2.048V reference.

 

thank you!

You have two parts that DON'T QUITE work together. The IA can ALMOST make 2.048+2.048 AND your going to have to make a zero adjustment using the A/D.

2.048+2.048? The IA outputs only a single ended signal?

Also as you mentioned these two components are not a good match for this application, could you maybe push me into a better direction?
Are you able to suggest some IAs/ADCs combinations that will work better/easier together? (as you seem to have quite some experience in this field I assume you have your best practices and favourite components figured out)

So, you might want to amplify your 20 mV with a gain of 2 for a maximum read of 2 V.

Still confused by this one: If I do like you suggested I supply v_rev = 2.048V to the IA, then I apply a gain of 2 to my sensor signal.
That will leave me with an max. IA-output of: 2.048V + 20mV*2?

 

Scratch some of my answers - head bad.

QUOTE="Henry603, post: 1328158, member: 586190"]thank you!

2.048+2.048? The IA outputs only a single ended signal?
[/quote]
The IA inputs a differential signal and outputs a differential signal. 2.048 on the refeference pin, A=1, you get 2.048+2.048 out.

Also as you mentioned these two components are not a good match for this application, could you maybe push me into a better direction?
Are you able to suggest some IAs/ADCs combinations that will work better/easier together? (as you seem to have quite some experience in this field I assume you have your best practices and favourite components figured out)

You have a misconception, that you can use everything and that all numbers are absolute. Rail to Rail is very dependent on the datasheet.
For a 0 V, 5V powered Op-amp; rail to rail is not likely 0V and 5V. The common mode voltage at the inputs may not be 0-5V; they could be restricted.

Still confused by this one: If I do like you suggested I supply v_rev = 2.048V to the IA, then I apply a gain of 2 to my sensor signal.
That will leave me with an max. IA-output of: 2.048V + 20mV*2?

Yep, purely wrong on may part. Try this again: 20 mV * 10 = 200 mV; 200 mV*10 = 2Volts.

So a gain of 1000 makes the 0-20 mV sensor read 2V. The negative range is gone. You lost 1 bit of 15. A negative number may show upin calibration. You have your PGA to increase the resolution of low value signals.

My math was really wrong due to migraine. I haven't yet put money in a parking meter and expected bubble-gum to come out. Migrainers have.
I miss interstate exits when driving or I might put ketchup where the dishes go and not in the fridge.

You need "wiggle room" and you can't treat numbers as absolutes.

I did a system for my first time. I needed to convert dividers to divide 0-40 V to 0-5V. No trimmers. 1% resistors. All were initally set for naminally 5V nominal using the best ratios.

The software had a fudge factor that was close to 1. I did not set FS = 4.930. I set it to 5V * 0.986

I had 40 V (power supply output), 5V (Nominal D/A) value and a (Calibration factor ~1)

In your system there is zero and it's the difference of the 2.048 V references. You need to verify 2V FS works which it probably will.
Panel meters are often 0-1.9999 volts, so that's a good thing.

Your choice of IA and A/D is good, but they are not LEGO's.

 

Here https://www.allaboutcircuits.com/ne...-operational-amplifier-from-texas-instrument/ is a little blurb about rail to rail OP amps.

 

@KeepItSimpleStupid : Thats's great info thanks.

I did some more research and came up with the conclusion that picking an ADC IC that has an intergrated PGA (and low drift reference voltage and oscillator) might be the best choice regarding complexity and error/noise sources.
What do you think about this one?:

http://www.ti.com/lit/ds/symlink/ads122c04.pdf

Its a delta sigma with a 24Bit ADC and internal PGA.
The new idea is to use the internal PGA to use the amplification of x64 to get my initial sensor signal of max. 20mA to about 1.3V.
The ADC has an internal reference voltage of 2.048V (and ADC has 24Bits).
So this should be enough to get my required 0.5mV resolution of the amplified signal?
Input offset voltage is very low, so is the ofset drift and gain error, so do you see problems with this idea?

I would attach the sensors differentially. Sensor outputs a signal between 0-20mV.
Would amplify it by 64 and after amplification i need a resolution of at least 0.5mV.

Do you see problems with this idea?
This way I do not have to match an instrumentation amplifier with an ADC but just have only one chip for amplification and A/D conversion. Seems like this chip is made for my kind of purposes.

Thank you!
Regards.

 

That datasheet is a bit messy for me to deal with. It's a bit more complicated than the other one. I remembered somewhere that it said effective 20 bits.

You might be missing something else too.

20 mV /64 =0.3125 mV. That's like a 5 bit converter. 2^6=64.

Are you going WAY OVERBOARD an overthinking this?

I did mention quantization errors where values close to zero are hard to measure. There you can add gain, but you have to refer the resolution to the input.

I do think you understand the OP-amp rails better. Make sure your not treating the 0-20 mV signal having a hard 0V. That almost never happens and can get you in trouble. It did me once. I set a D/A to zero volts for a calibration routine. Unfortunately it may have been a few 100 uV and not zero. I really didn't need it anyway because the AC performance is what I was after. 40 pA of offset current and 1 mV of offset voltage didn;t matter, but I could not get 0V in and 0V out nor could I do the calibration with 100 uV. I needed a hard 0 like the shorted inputs this datasheet talked about.

 

@KeepItSimpleStupid :

20 mV /64 =0.3125 mV. That's like a 5 bit converter. 2^6=64.

What are you talking about? That has nothing to do with my resolution steps??! This is a PGA (Programmable Gain Amplifier).
I would apply a GAIN of 64 to the sensor signal (initially ranging from 0 to 20mV) so I get a stronger signal up to max. 1.3V (1.3V corresponding to 20mV without amplification, before the PGA).
Then the amplified signal will be forwarded to the ADC of 24Bits and a 2.048V reference where the conversion magic happens.
This IC contains a PGA and an ADC (+voltage reference and TXO).
Check the first page of the datasheet if you did not see that before.
I should get way more accuracy out of this than 5-6Bits...

Are you going WAY OVERBOARD an overthinking this?

I dont think so, using an IC that is a complete analog front-end and does amplification and conversion in one chip is simple and basically can not get any simpler... In case you have a simpler idea, I would love to hear that

I did mention quantization errors where values close to zero are hard to measure. There you can add gain, but you have to refer the resolution to the input.
I do think you understand the OP-amp rails better. Make sure your not treating the 0-20 mV signal having a hard 0V. That almost never happens and can get you in trouble.

The voltage offset is very low + I can calibrate that out on the software side by shorting the input and check how much offset there already is.

Any thoughts?

 

20 mV /64 =0.3125 mV. That's like a 5 bit converter. 2^6=64.

The point being made is a chain in signal path of processes each with its own G
will have a composite G of the overall product of each element in the chain.

Let this be an absolute accuracy discussion.

So if the PGA can only resolve to 6 bits accurately, the fact you have a 2000 bit A/D
after that is irrelevant to the absolute accuracy discussion. In fact you may also no longer
necessarily have a monotonic transfer function measurement system. The dominant
element of accuracy is the 6 bit PGA.

Now if discussion is relative accuracy then digitizing the PGA output range to 20 bits will
result in high accuracy of that range, even if that range is inherently inaccurate (6 bits).
So its relatively accurate, but not absolute accurate.

Do the error budget, converting all errors to a common base, like ppm or LSBs,
then algebraically add (except for noise, that has to be handled as RMS).
Errors are offsets, PSRR, CMRR, G, INL, DNL......

The voltage offset is very low + I can calibrate that out on the software side by shorting the input and check how much offset there already is.

That's classic auto zero (of sorts) and yes it works. But it should be done over T & V and
using a least squares or geometric power equation curve fitted for use in interp-
rolating values under all conditions and values.

A real true high resolution absolute accuracy approach looks like this -

Attached.


Regards, Dana.

 

@danadak :
Thank you for your input!
I dont really get it. I thought the PGA only amplifies the analiog signal and then sends it to the ADC for analog -> digital conversion?
Why should this PGA only be accurate like a 6 Bit ADC?
If thats true, why should they put a PGA on board of a high precision 24 BIt ADC if one can not even use the PGA because it would lower the accuracy down to 8 Bit max. (as max gain of the PGA is 128)? That does not really make sense to me.
The PGA is working with the analog signal and sending it to the ADC then to digitalize it?
In that case a amplification with that internal PGA of x2 would mean I only can get 1 precision out of it?
Doesnt really make sense to me.
And yes, absolute accuracy is important to me.

Why should they make internal PGAs if one can not use them?
This ADCs (like I suggested above) with internal PGAs (with PGA gain up to 128) are complete analog front-ends meant for precise data acquisition and supposed to simplify the design by containing most needed parts (amplifier, oscillator, ref_voltage etc.).
What would be the reason to make them if one can not use them cause one would get about 6 bit accuracy out of a 24Bit ADC?

Please help.
Do you think that this is not a good solution for my case? (to successfully amplify my max. 20mV sensor signal to a 0.5mV accurate signal that can be used by a µC?)

 

A PGA can be trimmed, like an IA, to achieve hi accuracy, you have to
looks at its G accuracy spec. It can have N bits of G setting, and if its G
accuracy spec is 20 bit accurate (for each G setting) then you are good to
go. But if it only has a +/- 1% G accuracy, then that's 2 parts out of
100 which is just slightly higher than equivalent 6 bits accurate. Not to
confuse its accuracy with the number of its possible G settins.

Often you will see a PGA inside a A/D and its G accuracy very low compared
to A/D accuracy, especially if A/D DelSig type. That's because either you bypass
using it, or you are doing relative accuracy, and the V of focus is output of PGA,
and you want high res of that PGA output range, but don't care that absolute
accuracy is not so great.

Regards, Dana.

Regards, Dana.

 

A lot of folks do not realize just how difficult it is to make high precision
instrumentation, like Keithly, Agilent/Keysight make. Taking out all the
non linearity's in a design challenging.

Doing a 12 bit design, +/- 1 LSB, absolute accurate, over T and V, if you do
it right not easy.

That's why the configure at test method is widely used. Even O Scopes are
doing this now to correct signal path gremlins. Both timing and G and T and V.

Analog Devices has tons of ap notes on A/D, PGA, Amplifiers, errors.....


Regards, Dana.

 

The specs of the ADS122C04 say ±0.015% gain error only?
That would make me only about 6 Bit accurate?

 

PGA G error is -



A/D Vref error is



So PGA is accurate to .1 % (G < 64) which is ~ 10 bits. And I do not see a T drift error for it. And I did not add
in Vref error or drift or long term or PSRR or INL or DNL....

And this is just some of the errors....


Regards, Dana.

 

That is good info thank you! Could you provide me the calculation how you end up with ~ 10 bits? As i right now do not get how you end up with 10 bit? Could you please show me the formula?

Also I did not see any much more accurate ADCs/PGAs out there so how can i even get more accurate than this? Or do you know some devices I might check out? Hints on that would help me so much (as you seem very experienced I assume you came accross some low error devices and could suggest me some for my case that have less error)?

thank you!

 

https://www.analog.com/media/en/tra...ndbooks/Data-Conversion-Handbook/Chapter2.pdf



Regards, Dana.

 

.1% is 1 part out of 1000. A 10 bit converter has 1024 parts, or values.

So .1% is ~ equivalent to 10 bit converter.

Keithly ultra precision handbook.

https://download.tek.com/document/LowLevelHandbook_7Ed.pdf

Excellent ap note on basics -

https://www.analog.com/media/en/tec...294542582256114777959693992461771205AN280.pdf

Regards, Dana.

 

Correlated Double Sampling one approach -

http://www.cypress.com/file/119666/download

http://www.google.com/url?sa=t&rct=...666/download&usg=AOvVaw3lNuREWgnMLOtl3_j7btBY

Regards, Dana.