Can't get simple op amp circuit to work

[BobM2]

I'm using an MC34071 op amp with a "FlexiForce" load sensor connected per the drawing supplied by the maker of the load sensor (Tekscan):

http://www.tekscan.com/flexiforce/specs_flexiforce.html.

The output of the op amp goes to about +8.5 volts and won't change much (changes about .2 volts) as I change the load on the sensor. I can get the voltage at the inverting input to change from positive to negative by varying the load. It seems like I have something not connected correctly but I can't find any errors. I've tried 3 different op amps, 2 breadboards, and different feedback resistances but still no luck. Any suggestions?

 

[felixc]

As per the theory shown on the page you should see a voltage going from near zero (no load) to +5 volts at the output of the opamp. If it goes to 8.5 volts, it means that the opamp is probably saturated at the upper voltage rail, and that the voltage at the negative input does not balance with the positive one, at zero volts. Can you measure the voltage at the negative input of the opamp? You should read zero volts.

How do you generate the -5v? If you put this voltage at ground, do you get 0 volts at the output? It should be.

 

[ScottyUK]

Bob,

An op-amp should never have any significant voltage differential between its inverting and non-inverting inputs. Consider the gain of the op-amp: typically 10^6 or greater, open loop. For any non-saturated output voltage, the differential between the inputs must be very small indeed. You can measure it, but you need laboratory class instruments. The op-amp gain remains the same when the loop is closed, but the circuit gain is modified by the feedback network. In a negative FB circuit, the feedback acts to reduce the differential input voltage to almost zero. In the special case where the non-inverting leg is grounded, the op-amp creates a 'virtual earth' at the inverting input because the ungrounded leg is 'virtually' at earth potential as a result of feedback reducing the input differential. If you are measuring a significant input differential, there is something wrong with the feedback, or the circuit gain is far too high for the sensor (or open loop).

A last daft thought - you did spot that the upper power connection (pin 4) on the example circuit is -9V and the lower connection (pin 7) is the positive? The drawing is correct, it just breaks with convention.



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If we learn from our mistakes,
I'm getting a great education!

 

[IRstuff]

The pin connections match ONLY the single device 8-pin SOIC or DIP. The dual device model in the same packages use different power connections:

http://www.onsemi.com/pub/Collateral/MC34071-D.PDF

TTFN

 

[BobM2]

Well, this is a bit embarassing. I was counting the pins from the wrong direction on the right side so I had pins 6 and 7 mixed up. I should stick with gears and bearings. Thanks to all for the suggestions!