Load Cell Application - An ADC decision

[carl]

Just a quick one,

I am doing a new project - which is a wireless loadcell.
The wireless part is already done - which is reusable from another project.

The electronic design path from load cell to PIC has not been done and this is my question.

Is it better to use a stand alone ADC or use the intergrated ones within a PIC. I have never used the ADC in a PIC before, but am reasonably familiar with PIC's.

The design has to be:

minimum power
excited from DC 5V
no external clock - use the internal 4MHz clock.
analogue output from load cell is 1.5mV/V, 10Kg load.
Considering using AD623 as the instrumentation amplifier.




Any ideas would be thankfully accepted.

[arunb]

you could try using a bridge transducer ADC such as CS5520/CS5516, they are quite accurate and fairly fast. These are SPI based both 3 and 4 wire.

The device also contains facilities for calibration auto taring etc.

Datasheet link

http://www.chipcatalog.com/Datasheet/0CF8B76D7E3A26C589EEA499D84BF946.htm

thanks
arunb

[carl]

Hi Arunb,

Thanks for the reply. On initial inspection of the datasheet - it looks complicated. It also looks power hungry. I think it is powered from +/- 5V.

Although I didn't say, it needs to be DC.

Thanks for your suggestion, but I was just really after the pro's and con's of using a stand alone ADC compared to a PIC intergrated one. With power and simplicity being the key elements.

Thanks
Carl

[John P]

I've built the exact setup you describe, and it worked fine.

Note that PIC ADC's are 10 bit resolution. Is this enough?

The AD623 seems to be a good choice for a single-supply low voltage amp. But you have to think about setting an offset; in general the output has to be biased to give you a midrange (2.5V) output when there is no load on the transducer, unless your mechanical setup only receives load in one direction. I ended up using a voltage divider and a precision op-amp to generate this, though you could do it by biasing the positive input to the AD623 and just grounding the offset pin.

[Ttelmah]

On the ADC, will depend totally on resolution/accuracy you need.

You talk about a 'cell'. Hopefully this is a bridge design. Otherwise you will never be able to reliably correct for temperature. Even if your gauge is a temperature compensated type, the metal to which it is attached will expand/contract, and the wires themselves will show changes in resistance as well.

Most of the compexity in the CS5516, is allowing it to give 0.0015% linearity error, and 20bit reolution with no missing codes. You need to quantify the resolution, linearity, and repeatability you need. To get to the same sort of level as this chip, you probably wil not do better in terms of power. Remember you can leave it in low power mode, and only switch it 'up' when reading, and that you can derive a -5v rail very easily with a charge pump, and you really will need to be using +/- rails otherwise the performance near zero, will suffer. You could even use a FET, and turn the whole sircuit 'off' when not in use.

Best Wishes

[carl]

Thanks for your replies Ttelmah and John P. The load cell is an off centre load cell (bridge type) - and will only give out a differential DC positive voltage.

1) the resolution required is 10-12 bit.
2) I cannot use any negative voltage rails because the design has to meet hazardous area criteria.
3) Performance and offsetting is not a criteria at the moment.
4) An up-converter is used to change the power from battery 3.6V to 5V.

In reference to the AD623, With +Vs = 5V and -Vs = 0V, this gives a bias Voltage of 2.5V automatically with no load applied.

With the maximum 7.5mV pk/pk (this is at 10kg load = 1.5mV/V @ 5V) the +signal is measured at 2.50375V
the -signal is measured at 2.49625V

Now I have tied the reef pin to ground - the reason being is because the load cell is unipolar (I thijnk this means that a neg line is not used) - so I cannot tie the ref pin to 2V say (as it explains in the datasheet) so that I get a 2V +/-1V swing or whatever gain is chosen.

What I get is a 0V to 2V with the same gain. Do you agree with me John that this is the way to go?

so if I changed the gain and had a 0V (at no load) and 4.096V (at full load), then this is how I am thinking about generating the input innto the ADC.

Now onto the ADC. The whole design is battery powered - minimal power consumption is paramount. The stand alone ADC's that i have seen so far all need a clock - but doesn't this itself use more power.

Wouldn't be easier just to take the analogue signal directly into the PIC?
wouldn't this be the most power conservative way?

Thanks for your help
Carl

[FvM]

For low resolutions up to 16 bit, it's no problem to operate the load cell at 5 V or even lower voltage to my opinion, if a low offset voltage amplifier is available. You have to check, that the load cell is intended for constant voltage operation.

The PIC internal ADC's accuracy won't be sufficient normally. You can at best expect 9 bit accuracy from the chip. In my opinion, a low speed external ADC with at least 16 Bit resolution would be the best solution. Although some ADC have internal programmable amplifiers, zero stabilility and noise performance would be much better with an external instrumentation amplifier as AD623.

Regarding "unipolar" range you should consider, that without a small negative range, you are unable to do a zero calibration.

[carl]

Thanks FvM,

An informative response. I didn't realise that the ADC module on PIC's was only 'at best 9 bit'.

Just on this point alone, I think it would be better to get a stand alone module.

Some further questions if I went for the stand alone option:

1) Would a parallel (16 lines) type be easier to interface (C code understanding) from the ADC to the PIC. I am not the best at writing code.

Or would you only consider Serial like SPI. Would a particular option have an effect on the power consumption.


2) After testing the AD623 - it definitely does not seem to be linear. I adjusted the gain to give out 4.096V at full load - 10kg. Therefore you would expect it to give out 2.048V at 5kg - but it is a lot lower - around 1.8V. At 0Kg it is about 18mV. Any ideas??


Thanks for your help
Carl

[mkuang]

[FvM]

[carl]

Sorry FvM,

I don't quiet understand what you mean.

If I measure the mV/V signal coming out of the load cell (difference between +sig and -sig) - it is linear. i.e. at 0Kg - it is 0, at 5Kg it is 3.75mV, and at 10Kg it is 7.5mV.

So you would assume that this would be amplified in a linear way when connecting to the AD623. But when i measure the output of the AD623 it is not linear.

Can you explain further.

Thanks
Carl

[bkamen]

[FvM]

You calculated a gain from two points, one of them outside the linear amplifier range. Your calculation would be correct, if 1. the load cell offset and 2. the amp offset can be neglected. I don't know, if this is the case.

As said before, you can't achieve a meaningfull scale operation without having zero inside the linear range.

[SherpaDoug]

Is your load cell going to be used with a scale platform, or anything else that will cause a mechanical force offset. If so then it gets you away from having to accurately work with 0V with is difficult even for "rail to rail" amps.
_________________
The search for better is endless. Instead simply find very good and get the job done.

[carl]

Thanks everyone for your help.

I am going to have a play around with it. At the moment, I dont know if it will have a mechanical offset - but it would help if it did.

As for the ADC, i think my question has been answered - above 9 bit and it has to be a stand alone unit (unipolar).

If I can live with 9 bit, then I will use a PIC.

Thanks again
Carl