Non-linear voltage adder?

Thread Starter

relativity

Joined Dec 12, 2006
4
First of all, I'm new to this sort of thing but I have a project in mind that I'd like to try to create and I need help.

I have a non-linear sensor with a 0-5v output that I'm trying to combine with a second, identical sensor's output to measure a combined total. The sensors may have significantly different values at times, so I can't average their voltages. I need to add them using a lookup table.

For example, the sensor's output voltage is the square root of the measured quantity (lbf for example). So a measurement of 4 lbs yields a sensor output of 2 volts. 9 lbs = 3 volts, 16 lbs = 4 volts, etc.

So if sensor A reads 2 volts (4 lbs), and sensor B reads 3 volts (9 lbs), the combined voltage output would be 3.6 volts (13 lbs).

Logic:
  1. Compare sensor A's voltage to the lookup table (2v = 4)
  2. Compare sensor B's voltage to the lookup table (3v = 9)
  3. Add the table values (4 + 9) and, using the table, determine the combined voltage (13 = 3.6v)
The actual circuit is not a convenient equation, so it would need to interpolate between adjacent values (linear interpolation is fine).

How would you recommend creating such an adder? If you can at least point me in the right direction, I'm willing to do the legwork. I'm hoping you can at least save me from going down the wrong path(s) and wasting time. Thanks!

Rel.
 

kubeek

Joined Sep 20, 2005
5,733
what about using opamp logarithmical converter for each signal, amplify it by 2,
then use exponential converter. This will make the signals powered by two.

then you add them, and use the same method to make a square root if you want (use gain 0.5).

hint: 2*log(x) = log(x^2)
 

Thread Starter

relativity

Joined Dec 12, 2006
4
hint: 2*log(x) = log(x^2)
Thanks for the reply! It turns out the actual table isn't a convenient X^2, I just used that as an example. However I found a way to simplify this. I can represent this whole operation as a single equation as follows:

Vout = 0.8054*(VA^3.2 + VB^3.2)^0.3125

For example, let's say Sensor A is 2 volts (VA) and Sensor B is 3 volts (VB). Using the equation above my combined Vout = 2.61 volts.

This is an approximation of the actual table, but it's accurate to within 10%. How do I create a circuit which replicates the equation above?

To clarify, there are two input voltages (VA and VB) and one combined output voltage (VOUT) with a ground reference of course.
 

beenthere

Joined Apr 20, 2004
15,819
Hi,

Look up a data sheet for a four-quadrant multiplier ic. As long as you're dealing with constants, then it should be possible to crank out a function. The four-quadrant device will act as a mixer for your two sensor signals, and may also be able to tailor the response of each to fit the function.
 

kubeek

Joined Sep 20, 2005
5,733
And what about using A/D converter and a PIC to count the equation? It might be easier than creating this quite complex equation discretely.
 

Thread Starter

relativity

Joined Dec 12, 2006
4
Hi,

Look up a data sheet for a four-quadrant multiplier ic. As long as you're dealing with constants, then it should be possible to crank out a function. The four-quadrant device will act as a mixer for your two sensor signals, and may also be able to tailor the response of each to fit the function.
If I'm following correctly, you're suggesting a selection of these arranged in order to reproduce the equation?

Analog Multipliers/Dividers

AD538 looks like it will perform the exponential and scalar calculations, but I'm not sure about the addition.

And what about using A/D converter and a PIC to count the equation? It might be easier than creating this quite complex equation discretely.
I like easy, this sounds interesting?
 

beenthere

Joined Apr 20, 2004
15,819
Hi,

The addition can be done in an op amp inverter. The inputs each put current through resistors feeding the inverting input, which will add them together.
 
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