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super-d
02-21-2008, 09:30 PM
Can someone explain what the voltage divider does with the FlexiForce? How do you adjust it? I've watched the video but I'm sort of confused on how / when you adjust it.

How I see this working (novice thinking here) is if I had a 25 pound FlexiForce I would apply 25 pounds of pressure and then adjust the voltage divider for 5 volts. Of course 0 volts would mean no pressure. In my mind this would create a baseline for the sensor. I don't quite understand what they mean by "compensate" in the video. Are you just adjusting it to get max voltage so instead of possibly having 0 to 2 volts for 0 to 25 pounds you would get 0 to 5 for more precision on the input?

I was on another site that used / sold an Op-Amp for the FlixiForce and for each model they had a different resistor combination for each one since the sensors themselves vary between models. Will the Phidgets voltage divider work for each model of the FlexiForce?

Last but not least using my example with the 25 pound FlexiForce is there a formula to figure out how much voltage would be equal to 5 pounds or 22 pounds?

Thanks in advance
David

Dave
02-22-2008, 05:33 PM
Ok, here's the deal:
An op-amp is the ideal solution for interfacing with a Flexiforce sensor. With one simple circuit, you can shift and scale your output to give you a perfect 0-5V range. The problem is that this requires a bi-polar (negative and positive voltage) power supply of at least +-6.3V (because the op-amp can amplify the signal to within 1.3V of the supply voltage.

Op-amps are great, but they're a major pain when you're trying to use something like the 8/8/8, which limits you to +5V to drive your sensor. Powering an op-amp off a 0-5V supply will only give you a range of 1.3V to 3.7V. That's pretty much useless.

The voltage divider serves several purposes. Mainly, it gives you a simple, passive way to attach pretty much any resistive sensor to the 8/8/8. With a voltage divider, the voltage present at the analog input is a ratio of the sensor resistance and the potentiometer resistance, so you can compensate if the sensor has a high resistance by adjusting the potentiometer. In your case, you'll be tuning to to be most sensitive to the range of pressure you'll be exerting on the forse sensor.

As for a pressure-to-voltage equation, we don't really have one. The output isn't linear, so you need to test it with a series of known weights to calibrate it.

super-d
02-22-2008, 06:56 PM
Dave,

I appreciate the info. That helps. The one thing I don't understand about the op-amp is the bi-polar power supply. I was looking at this site http://www.imagesco.com/sensors/opamp.html
and their op-amp uses 5 volts only other than the 9v for the regulator. So now I am confused :confused: although when I was looking at op-amp circuits I did find some with a negative and positive voltage. The above op-amp was designed (I'm assuming) for the FSR and FlexiForce because they list a chart for the type of sensor and the resistive values required for each model. Quite a bit more than the Phidget voltage divider but I thought if I could get the amp and a voltage divider I could do a little comparison.

Thanks
David


Ok, here's the deal:
An op-amp is the ideal solution for interfacing with a Flexiforce sensor. With one simple circuit, you can shift and scale your output to give you a perfect 0-5V range. The problem is that this requires a bi-polar (negative and positive voltage) power supply of at least +-6.3V (because the op-amp can amplify the signal to within 1.3V of the supply voltage.

Dave
02-27-2008, 05:44 PM
Ok, I was just referring to the drive circuit that the manufacturer recommends, which uses a different op-amp chip (MC34071) and a different configuration than the Images board. The Images op-amp board doesn't actually amplify the signal, it just buffers it. It looks like the Images board incorporates a voltage divider with an op-amp to isolate the sensor circuit from the device that's reading it. I'm going to have to get one of those to play around with to see how much difference it makes.

auralius
04-24-2009, 06:45 AM
please note that recommended maximum current is only 2,5 mA...
so better use opamp since opamp has a very high input impedance...
meaning that it will give a no load condition to the sensor so the current flowing in the sensor will be very small...
so it is safe...
even tough it is only for buffer....

MikeG
04-24-2009, 11:10 PM
The circuit on http://www.imagesco.com/sensors/opamp.html is a voltage follower with unity gain.

The reason to use the operational amplifier in this configuration is to isolate the source and load as stated above.

The voltage divider connected to v+ divides the voltage between two variable resistors. One variable resistor is the sensor and the other is a pot which is connected through a fixed resistance (R1) to ground. The pot is used for tuning. The fixed resistor helps the POT because the FlexiForce has a relatively large resistance value. See Ohm's law.

Anyway, Op-Amps like it when (v+) + (v-) = 0 so as the voltage across the FlexiForce changes the output will follow due to the feedback loop on v-.

I couldn’t find the tekscan recommended driver so I can’t comment on that but there’s a bazillion ways to integrate a variable resistor sensor. Like using a simple RC circuit. Count the ticks it takes to charge or discharge the cap before/after the TTL or CMOS voltage threshold is reached. Probably not the best method as these devices have varied thresholds even among production siblings.