A non-inverting amplifier

[Clari]

hi there,

i have two questions concerning operational amplifier which i can't figure out. i have tried them for hours, and finally get on a bit, yet i am not sure whether i have the answer right. :(

1a. What is the value of the feedback resistance in figure 1?

My ans: V_out/ V_in = 1+ R_f/ R_in
30/ 6 = 1+ R_f/ 10 x 10^3
R_f = 40000 ohm

2. In figure 2, Assume the op amp does not saturate, find the potentials at A and B.
At A, the current = 3.2 x 5k = 0.64 mA
At B, the current = 2.8 x (5k+15K) = 0.14 mA

Total current = 0.64 + 0.14 = 0.78 mA

Potential at A = (0.78 -0.64) x 15k = 2.1V
Potential at B = 0.14mA x 15k = 2.1V

please help me. ^^

 

[BobG]

hi there,

i have two questions concerning operational amplifier which i can't figure out. i have tried them for hours, and finally get on a bit, yet i am not sure whether i have the answer right. :(

1a. What is the value of the feedback resistance in figure 1?

My ans: V_out/ V_in = 1+ R_f/ R_in
30/ 6 = 1+ R_f/ 10 x 10^3
R_f = 40000 ohm

2. In figure 2, Assume the op amp does not saturate, find the potentials at A and B.
At A, the current = 3.2 x 5k = 0.64 mA
At B, the current = 2.8 x (5k+15K) = 0.14 mA

Total current = 0.64 + 0.14 = 0.78 mA

Potential at A = (0.78 -0.64) x 15k = 2.1V
Potential at B = 0.14mA x 15k = 2.1V

please help me. ^^

Question 1 is correct.

You also got the right answer for the second one, but I'm not sure why you did what you did.

If your resistances were labeled from left to right, with the top resistors first, you'd have R1=5K, R2=15K, R3=5K, R4=15K.

For an ideal op amp, the inverting input (-) and the non-inverting input (+) have the same voltage. Once you solve for the potential at B, you've solved for the potential at A.

Your currents are wrong, however. B is correct, but A is wrong. To find the current at A, you need (Vin - Va)/R1.

(3.2V-2.1V)/5k = .22mA

The rationale is that 1.2 volts were dropped across R1 (2.1 is 1.2 Volts lower than the input voltage). Per Ohm's law, the current through the resistor is V/R.

It's the current at A that will give you your output voltage:

Va-(.22mA*R2) = 2.1 - (.22mA*15K) = 2.1 - 3.2 = -1.2

Rationale is the same as before. You're still using Ohm's law: V=RI. You're dropping 3.2 volts, so the output voltage has to 3.2 volts lower than point A.

If you subtracted your input voltages (3.2 - 2.8) and multiplied by the inverting gain (-R2/R1), you'd get the same answer for the output voltage (but might not understand why). Or if you subtracted your input voltages (2.8 - 3.2) and multiplied by the non-inverting gain {(R1+R2)/R1}*{R4/(R3+R4)}, you'd get the same answer for the output voltage.

 

[Clari]

If you subtracted your input voltages (3.2 - 2.8) and multiplied by the inverting gain (-R2/R1), you'd get the same answer for the output voltage (but might not understand why). Or if you subtracted your input voltages (2.8 - 3.2) and multiplied by the non-inverting gain {(R1+R2)/R1}*{R4/(R3+R4)}, you'd get the same answer for the output voltage.

Hello BobG, i can figure it out now. Thank you very much. Your help is greatly appreciated. well, i have also tried the alternative method before but i don't know why the two voltages have to subtracted from each other. :rofl:

 

[BobG]

Hello BobG, i can figure it out now. Thank you very much. Your help is greatly appreciated. well, i have also tried the alternative method before but i don't know why the two voltages have to subtracted from each other. :rofl:

That's the goal of the circuit. It was designed so the output would depend on the difference between the two inputs. You can also design opamp circuits to sum two or more voltages, integrators, differentiators, etc.