6. Below is a model of a differential amplifier. v, is a small signal from a sen

Question

6. Below is a model of a differential amplifier. v, is a small signal from a sensor, it's perhaps a few millivolts. The signal represents noise that might contaminate the output; it could be a few tens of volts. The purpose of this differential amplifier circuit is to amplify the sensor voltage while rejecting the noise voltage, R1 = R2 = 2 RF = R3-240 kQ. Label the circuit as appropriate. Carefully show your analysis of the circuit using the assumptions for an ideal op-amp. Find yo as a function of ya and v. Be careful not to round off values until your final calculation. RF R1 Vo V. R2 Vx R3

Explanation / Answer

Let's first calculate output Vo1 due to signal Vs

Vo1 = -(Rf/R1)*Vs                                     ( gain of inverting amplifier = -Rf/R1)

Vo1 = -(240/2)*Vs

Vo1 = -120Vs                         ..........1

Let's calculate output Vo2 due to signal Vx

Vx1 = (R3 /R2+R3)Vx = input voltage at noninverting input

Vx1 = (240 / 240 +2)Vx = (240/242)Vx

Vo2 = -(Rf/R1)Vx + ( 1 + Rf/R1 )Vx1                                ( gain of non inverting amplifier = 1 + Rf/R1 )

Vo2 = -(240/2)*Vx + ( 1 + 240/2 )(240/242)Vx

Vo2 = -120Vx + (121*240/242)Vx

Vo2 = -120Vx + 120 Vx

Vo2 = 0 V                             ....................2

Therefore by superposition theorem output Vo is

Vo = Vo1 + Vo2

Vo = -120Vs + 0

Vo = -120Vs                                   .................1

This is the required output voltage.

We can see that output due to input noise voltage Vx is eliminated by differential amplifier configuration.

And we have output only due to signal input.

interference due to noise is eliminated

Related Questions

Navigate

• Browse (All)
• Browse 6
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.