Hello,

Intuitive deductions in reference to attachment:

1) Initial op amp assumption:

Output = 0V

Va will sense: va = v1(Ri/Rf+Ri) = 2(10K/30K) = 0.6VDC.... Therefore: 2-0.6 = aprox 1.3VDC

Vb will sense: vb = v2(R2/R1+R2) = 5(1.5K/5K) = 1.5VDC

Therefore, since the Non inverting input > inverting input, we know that Vout will be a positive voltage.

2) We also know that increasing Va (Inverting input) will cause Vout to ramp down!

3) If Vb = 1.5VDC, Then, Va should also equal to 1.5VDC

4) Iri = (v1-va)/Ri = (2v-1.5V)/10K = 50ua

5) We can see that Vri is 0.5VDC... and now we can solve for Vrf.

If Iri = 50ua, then Irf also equals 50ua, so then Vrf is:

50ua x 20K = 1VDC

6) Vout = v1 - Vri - Vrf = 2 - 0.5 - 1 = +0.5VDC

7) Op amp gain = Rf/Ri = 20K/10K = 2

8) Proof with Differential op amp formula:

Vout = V2 (R2/(R1+R2)) ((Ri+Rf)/Ri) - V1(Rf/Ri)

Vout = 5(1.5K/5K) (30K/10K) - 2(20K/10K)

Vout = 4.5 - 4 = +0.5VDC

My question is, in reference to the attachment, if the differential op amp amplifies the voltage difference between va and vb by a gain factor of 2, why is the amount divided by 2 instead of multiplied by 2. In other words, a quintillianth of a second after we power up the op amp we have a difference of 0.16666 VDC at the op amp's inputs:

1.5VDC - (2-0.6666VDC) = 0.16666VDC

Therefore, 0.16666VDC multiplied by a gain of 2:

should give 0.166666 x 2 = 0.3333VDC as vout instead of +0.5VDC ??

I am a little confused on how the gain applies here.... We know we have a gain of 2 right! So what exactly is multipied by 2??

Thanks for all the help!

Intuitive deductions in reference to attachment:

1) Initial op amp assumption:

Output = 0V

Va will sense: va = v1(Ri/Rf+Ri) = 2(10K/30K) = 0.6VDC.... Therefore: 2-0.6 = aprox 1.3VDC

Vb will sense: vb = v2(R2/R1+R2) = 5(1.5K/5K) = 1.5VDC

Therefore, since the Non inverting input > inverting input, we know that Vout will be a positive voltage.

2) We also know that increasing Va (Inverting input) will cause Vout to ramp down!

3) If Vb = 1.5VDC, Then, Va should also equal to 1.5VDC

4) Iri = (v1-va)/Ri = (2v-1.5V)/10K = 50ua

5) We can see that Vri is 0.5VDC... and now we can solve for Vrf.

If Iri = 50ua, then Irf also equals 50ua, so then Vrf is:

50ua x 20K = 1VDC

6) Vout = v1 - Vri - Vrf = 2 - 0.5 - 1 = +0.5VDC

7) Op amp gain = Rf/Ri = 20K/10K = 2

8) Proof with Differential op amp formula:

Vout = V2 (R2/(R1+R2)) ((Ri+Rf)/Ri) - V1(Rf/Ri)

Vout = 5(1.5K/5K) (30K/10K) - 2(20K/10K)

Vout = 4.5 - 4 = +0.5VDC

My question is, in reference to the attachment, if the differential op amp amplifies the voltage difference between va and vb by a gain factor of 2, why is the amount divided by 2 instead of multiplied by 2. In other words, a quintillianth of a second after we power up the op amp we have a difference of 0.16666 VDC at the op amp's inputs:

1.5VDC - (2-0.6666VDC) = 0.16666VDC

Therefore, 0.16666VDC multiplied by a gain of 2:

should give 0.166666 x 2 = 0.3333VDC as vout instead of +0.5VDC ??

I am a little confused on how the gain applies here.... We know we have a gain of 2 right! So what exactly is multipied by 2??

Thanks for all the help!

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