vne147
Member
Hello everyone. Thanks in advance for your help. I’m seeking some advice for a project I’m working on.
What I’m trying to do is build a force measuring device. The force will be sensed by a 0 – 50 lb load cell (data sheet at bottom of post). The outputs from the load cell will feed the inputs of an instrumentation amplifier circuit. I’ll be building that circuit and it will be similar to the one in this thread:
Instrumentation Amplifier Circuit Question
The output from the IA will in turn feed into a PIC’s onboard 10-bit ADC. I’ll probably be using an 18F4620 but I’m not 100% on that yet.
So far there’s nothing really special but the twist is I want to have 2 user selectable output ranges and both need to have roughly the same 10 bit resolution. The two output ranges I need are 0 – 50 lbs and 0 – 5 lbs. For the smaller range without changing anything else, I’d need to multiply the IA gain by a factor of 10 to achieve this. I know that squeezing the smaller range out of the larger load cell isn’t the best approach but my hands are kind of tied on that one.
Moving on…with a 15 VDC input to the load cell, the full scale range of 0-50 lbs will equate to about 0-45 mV. So, if I want the IA output range to be from 0 – 5 VDC then I need a gain of about 111. That means in the 0-5 lb range, I’d need a gain of roughly 1110 to maintain the 0 – 5 VDC output of the IA. Unfortunately, the AMP04 IC I’m using only goes up to a gain of 1000.
So, the idea I came up with to make it all work is this:
The gain of the AMP04 IC is set using a single external resistor. Therefore, I can switch it back and forth by using digital potentiometer IC controlled by the PIC. Not sure yet which specific chip I’ll use. Suggestions are welcome.
For the 2 ranges:
0 – 50 lbs
IA input = 0 – 45 mV
IA output ≈ 0 – 5VDC
Gain ≈ 111
ADC VREF+ = VDD = 5V
ADC VREF- = VSS = 0V
0 – 5 lbs
IA input = 0 – 4.5 mV
IA output ≈ 0 – 2.5 VDC
Gain ≈ 555
ADC VREF+ = VDD/2 = 2.5V (obtained from the comparator voltage reference module)
ADC VREF- = VSS = 0V
So here are my questions:
Thanks again for the help. I welcome any and all comments/suggestions. Please let me know if I need to supply any additional information.
Load Cell Datasheet
AMP04 Datasheet
PIC 18F4620 Datasheet
What I’m trying to do is build a force measuring device. The force will be sensed by a 0 – 50 lb load cell (data sheet at bottom of post). The outputs from the load cell will feed the inputs of an instrumentation amplifier circuit. I’ll be building that circuit and it will be similar to the one in this thread:
Instrumentation Amplifier Circuit Question
The output from the IA will in turn feed into a PIC’s onboard 10-bit ADC. I’ll probably be using an 18F4620 but I’m not 100% on that yet.
So far there’s nothing really special but the twist is I want to have 2 user selectable output ranges and both need to have roughly the same 10 bit resolution. The two output ranges I need are 0 – 50 lbs and 0 – 5 lbs. For the smaller range without changing anything else, I’d need to multiply the IA gain by a factor of 10 to achieve this. I know that squeezing the smaller range out of the larger load cell isn’t the best approach but my hands are kind of tied on that one.
Moving on…with a 15 VDC input to the load cell, the full scale range of 0-50 lbs will equate to about 0-45 mV. So, if I want the IA output range to be from 0 – 5 VDC then I need a gain of about 111. That means in the 0-5 lb range, I’d need a gain of roughly 1110 to maintain the 0 – 5 VDC output of the IA. Unfortunately, the AMP04 IC I’m using only goes up to a gain of 1000.
So, the idea I came up with to make it all work is this:
The gain of the AMP04 IC is set using a single external resistor. Therefore, I can switch it back and forth by using digital potentiometer IC controlled by the PIC. Not sure yet which specific chip I’ll use. Suggestions are welcome.
For the 2 ranges:
0 – 50 lbs
IA input = 0 – 45 mV
IA output ≈ 0 – 5VDC
Gain ≈ 111
ADC VREF+ = VDD = 5V
ADC VREF- = VSS = 0V
0 – 5 lbs
IA input = 0 – 4.5 mV
IA output ≈ 0 – 2.5 VDC
Gain ≈ 555
ADC VREF+ = VDD/2 = 2.5V (obtained from the comparator voltage reference module)
ADC VREF- = VSS = 0V
So here are my questions:
- Is there anything wrong with using the comparator reference voltage module for the ADC VREF+ input? I didn’t seem to find anything in the data sheet that indicated this would cause problems.
- Should I buffer the output of the reference voltage module before feeding it back into the ADC VREF+ pin?
- Does anyone have any better ideas for how to go about doing this?
Thanks again for the help. I welcome any and all comments/suggestions. Please let me know if I need to supply any additional information.
Load Cell Datasheet
AMP04 Datasheet
PIC 18F4620 Datasheet