+/- 10v adc interface with pic micro

[raman00084]

I am having a load cell unit that can output a analog voltage of +/- 10v
Positive voltage for push and negative voltage for pull. It is a push pull load cell. I am using dspic33ep512mu810 ic.
How to interface +/-10v in this microcontroller.

 

[KlausST]

Hi,

the first question is: Do you like to use the PIC interneal ADC or an external one? Both is possible.

There may be external ADCs which provide the input voltage range you need.

Or - in any case - you can use an external analog circuit to adjust Sensor_output_voltage_range to ADC_input_voltage_range.

It depends on what YOU need. And on the specifications for that ... like any anlog/digital circuit does.
If you want detailed answers you need to provide complete and detailed specifications first.


Klaus

 

[raman00084]

My first choice is to use the internal adc. How can I do this kindly explain

 

[D.A.(Tony)Stewart]

Link your 10V output load cell first.

I suspect it is only10 mV/V FS.

The approach is simply to shift bipolar 0V to Vref/2 and choose gain = full scale {ADC/source} often equal to Vref or Vdd for ADC.

 

[raman00084]

Is there ant external adc ic can I interface directly and communicate with i2c
The load cell unit has 2 terminals marked + and - 10v

 

[D.A.(Tony)Stewart]

Is there ant external adc ic can I interface directly and communicate with i2c
The load cell unit has 2 terminals marked + and - 10v

Is there a model number?
Those are usually supply inputs.

 

[betwixt]

If +/-10V really is the output do this:
Find half the PIC supply voltage (or half Vref if you are using an external reference),
Scale the entire -10V to +10V (in other words 20V) to full PIC supply or full Vref, whichever you have to configured to use with a potential divider,
Add on the half voltage you first calculated.

For example if the PIC supply is 3.3V and you select VDD as Vref:
1. half of 3.3V is 1.65V
2. use resistors to scale 20V to 3.3V, the scaled voltage will now be -1.65V to +1.65V
3. add the 1.65 from step 1 and the ADC will now see 0V to 3.3V as the input goes from -10V to +10V.

You can do it entirely with resistors or you could use a combination of resistors to drop the voltage and an op-amp to do the adding.

Brian.

 

[danadakk]

Your DSP only has a 10 bit resolution A/D, seems low for a load cell application.

What range of load and resolution do you want ?


Regards, Dana.

 

[KlausST]

Hi,

My first choice is to use the internal adc. How can I do this kindly explain

Is there ant external adc ic can I interface directly and communicate with i2c

Wow, did not last long ...
Just go to the electronics distributor of your choice and use the ADC selection tool.

If you want detailed answers you need to provide complete and detailed specifications first.

Is still valid.
You give no details .. but ask for solutions...

Klaus

 

[D.A.(Tony)Stewart]

Is there a model number?
Those are usually supply inputs.

 

[BradtheRad]

Resistor divider lifts bipolar AC up into positive polarity. Incoming signal must source and sink tiny amount of current:

 

[D.A.(Tony)Stewart]

Using a non-inverting (unity gain) Op Amp, the R divider is 50% with an offset of 5V. Both AC and DC offset are attenuated by 50% thus the output gives the necessary linear gain = 0.25 and offset = +2.5V)

Both OpAmp inputs are the same with a null difference in theory. In practice, it is the output reduced by the open loop gain for the virtual ground (which means ~0V input.)

Any questions?