Hello,
Please view the attachment below called "Circuit".
I think that, initially, this op amp is in inversion mode since the non-inverting input senses 1.5VDC and the inverting input senses 2VDC... I could be wrong!!! But this is how I see it. The op amp will stabilize its input voltages so the are equal to each other which will be at 1.5VDC as shown in the "Circuit" attachment below.
At RA, we have a VRA of 0.5VDC voltage drop with an IRA of 50ua. At RF, we also have a 0.5VDC voltage drop with an IRA of 50ua. If we begin at v1, we can say, 2VDC - 0.5VDC (VRA) - 0.5VDC(VRF) equals an output of approximately 1.0VDC which is what I measured when I built the circuit.
But as a text book formula (see TBFormula attachement), we can see the following formula:
Vout = Rf/Ra (v2 - v1)
If I plug in my measurements into this formula, I get 3.0VDC as an output instead of 1VDC??
Vout = 10K/10K (5 - 2) = 3VDC
Why doesn't the formula work?
confused!
All help very appreciated!
thanks
Please view the attachment below called "Circuit".
I think that, initially, this op amp is in inversion mode since the non-inverting input senses 1.5VDC and the inverting input senses 2VDC... I could be wrong!!! But this is how I see it. The op amp will stabilize its input voltages so the are equal to each other which will be at 1.5VDC as shown in the "Circuit" attachment below.
At RA, we have a VRA of 0.5VDC voltage drop with an IRA of 50ua. At RF, we also have a 0.5VDC voltage drop with an IRA of 50ua. If we begin at v1, we can say, 2VDC - 0.5VDC (VRA) - 0.5VDC(VRF) equals an output of approximately 1.0VDC which is what I measured when I built the circuit.
But as a text book formula (see TBFormula attachement), we can see the following formula:
Vout = Rf/Ra (v2 - v1)
If I plug in my measurements into this formula, I get 3.0VDC as an output instead of 1VDC??
Vout = 10K/10K (5 - 2) = 3VDC
Why doesn't the formula work?
confused!
All help very appreciated!
thanks
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