CHAPTER 8 IDEAL OPERATIONAL AMPLIFIER AND OPAMP CIRCUITS

CHAPTER 8 IDEAL OPERATIONAL AMPLIFIER AND OP-AMP CIRCUITS

inverting output non-inverting Op-amp circuit symbol • Open loop mode • Vo = Aod ( v 2 – v 1) – Aod is referred to as the open loop gain. – Notice that is v 2 = v 1, the open loop gain equals to

• Two main characteristics: • We want the open loop gain to be equal to which means that v 2 = v 1 0 0 • the input resistance to be equal to , hence there is no current going into the op-amp

• Op amp can be configured to be used for different type of circuit applications: – Inverting Amplifier – Non – inverting Amplifier – Summing Amplifier – Integrator – Differentiator

Inverting Amplifier Op-amp as an inverting amplifier Voltage at node 1 (inverting) = voltage at node 2 (non-inverting ) KCL at node 1: I 1 – I 2 – Iin = 0 (Vi – 0) / R 1 = (0 – Vo) / R 2 Vi / R 1 = - V o / R 2 Vo = - R 2 Vi R 1

Example 1 Gain = - (R 2 / R 1) = -(150/12) = -12. 5

Example 2 Answers: (a) - 6 (b) - 0. 27 V

Non - Inverting Amplifier Voltage at node 1 (inverting) = voltage at node 2 (non-inverting ) KCL at node 1: i 1 – i 2 = 0 (0– Vi) / R 1 = (Vi – Vo) / R 2 -(Vi / R 1) = (Vi / R 2) – (Vo / R 2) Vo / R 2 = (Vi / R 2) + (Vi / R 1) = Vi 1 + 1 R 2 R 1 Vo / V i = R 2 1 + 1 R 2 R 1 Noninverting amplifier

Voltage Follower / Buffer Amplifier vo = v. I Hence, gain = 1

Example 1 Answers: Vo = 5 V, Current = 1 m. A Vo = 10 V, current = 2 m. A

Summing Amplifier Similarly, i 1 + i 2 + i 3 – i 4 – 0 = 0 Example 8. 2 Design a summing amplifier as shown in figure to produce a specific output signal, such that vo = 1. 25 – 2. 5 cos t volt. Assume the input signals are v. I 1 = -1. 0 V, v. I 2 = 0. 5 cos t volt. Assume the feedback resistance RF = 10 k

Solution: output voltage

Other Op-Amp Applications

Integrator When the feedback resistor of an inverter circuit is replaced by a capacitor the circuit is worked as an integrator circuit -cause the output to respond to changes in the input voltage over time Integrator circuit

Example 1 Solution: The output voltage

Differentiator When the inverting input terminal resistor of an op-amp inverter circuit is replaced by a capacitor the circuit is worked as a differentiator circuit. Differentiator circuit Because Q = CVS

Example 1

Calculating Gain and Design Questions NON - INVERTING Calculating Output and Design Questions SUMMING AMPLIFIER DIFFERENTIATOR AMPLIFIER INTEGRATOR AMPLIFIER

Va NON - INVERTING Vb INVERTING Calculate the input voltage if the final output, VO is 10. 08 V. Finally: Va = (1 + 10/5) V 1 0. 504 = 3 V 1 = 0. 168 V Then: Vb = -(5/5) Va -0. 504 = - Va Va = 0. 504 V Have to work backwards: Vo = -(100/5) Vb 10. 08 = -20 Vb Vb = -0. 504 V

Va INVERTING SUMMING Calculate the output voltage, VO if V 1 = V 2 = 700 m. V Va = -(500/250) 0. 7 Va = -1. 4 V Then: Vo = - 500 [ Va / 100 + V 2 / 50 ] Vo = - 500 [ -1. 4 / 100 + 0. 7 / 50 ] Vo = 0 V

Calculate the output voltage VO of the operational amplifier circuit as shown in the figure. Answer: -3 V