Deriving Vout in an inverting operational amplifier.

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To derive Vout1 and Vout3 for an inverting operational amplifier, the discussion clarifies that the relevant input voltages are Vin1 and Vin2, not Va and Vb. The total output voltage Vout1 is expressed as Vout1 = -[(R3/R1)Vin1 + (R3/R2)Vin2]. The contributions from each input voltage are specified, emphasizing the negative feedback characteristic of the inverting amplifier. The participants confirm that understanding the relationship between the resistances and input voltages is crucial for accurate derivation. The clarification provided was appreciated and deemed helpful.
Sammyboss
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Pls does anyone know how to derive for Vout1 and Vout3 in terms of Va and Vb for an inverting operational amplifier. Thank you.
 
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Sammyboss said:
Pls does anyone know how to derive for Vout1 and Vout3 in terms of Va and Vb for an inverting operational amplifier. Thank you.
See inverting amplifier section in http://physics.usask.ca/~angie/ep316/lab7/theory.htm
 
There is no Va and Vb! I see two input voltage. I only see Vin1 and Vin2. So you look at Vin1 and Vin2 contribute. Vout1 part due to Vin1 is -(R3/R1)Vin1. Vout1 due to Vin2 is -(R3/R2)Vin2 so the total Vout1:

V_{out1}=-\left[ \frac{R_3}{R_1}V_{in1}+\frac{R_3}{R_2}V_{in2} \right]

You try to do the next one.
 
yungman said:
There is no Va and Vb! I see two input voltage. I only see Vin1 and Vin2. So you look at Vin1 and Vin2 contribute. Vout1 part due to Vin1 is -(R3/R1)Vin1. Vout1 due to Vin2 is -(R3/R2)Vin2 so the total Vout1:

V_{out1}=-\left[ \frac{R_3}{R_1}V_{in1}+\frac{R_3}{R_2}V_{in2} \right]

You try to do the next one.
Thanks a lot. that really helped
 
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