CHAPTER 8 IDEAL OPERATIONAL AMPLIFIER AND OPAMP CIRCUITS inverting noninverting output Open loop mode V o A od v 2 v 1 A od is referred to as the open loop gain Notice that is v 2 v ID: 769778
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CHAPTER 8 IDEAL OPERATIONAL AMPLIFIER AND OP-AMP CIRCUITS
inverting non-inverting output Open loop mode V o = A od ( v2 – v1)Aod is referred to as the open loop gain.Notice that is v2 = v1, the open loop gain equals to Op-amp circuit symbol
Two main characteristics:We want the open loop gain to be equal to which means that v 2 = v1 the input resistance to be equal to , hence there is no current going into the op-amp 0 0
Op amp can be configured to be used for different type of circuit applications: Inverting Amplifier Non – inverting Amplifier Summing AmplifierIntegratorDifferentiator
Op-amp as an inverting amplifier Inverting Amplifier Voltage at node 1 (inverting) = voltage at node 2 (non-inverting ) KCL at node 1: I 1 – I2 – Iin = 0(Vi – 0) / R1 = (0 – V o) / R2Vi / R1 = - Vo / R2Vo = - R2 V i R 1
Example 1 Gain = - (R 2 / R 1 ) = -(150/12) = -12.5
Example 2 Answers: - 6 - 0.27 V A voltage source with source resistance is connected to the input of an op-amp inverting amplifier circuit(a) If the and then calculate the voltage gain, VO / VS(b) Determine the output voltage for the source voltage
Noninverting amplifier 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) / R1 = (V i – V o ) / R 2 -(V i / R 1 ) = (V i / R 2 ) – (V o / R 2 ) V o / R 2 = (V i / R 2 ) + (V i / R 1 ) = V i 1 + 1 Vo / Vi = R2 1 + 1 R2 R1 R 2 R 1
Voltage Follower / Buffer Amplifier v o = v IHence, gain = 1
Example 1 A voltage source with source resistance is connected to a load resistance input of directly and through a buffer amplifier as shown in the figures below. Determine the output voltage across the load and the current in the resistance RL for each circuit configuration Answers: Vo = 5 V, Current = 1 mAVo = 10 V, current = 2 mA
Summing Amplifier i 1 + i 2 + i 3 – i4 – 0 = 0 Similarly, Example 8.2 Design a summing amplifier as shown in figure to produce a specific output signal, such that v o = 1.25 – 2.5 cos t volt. Assume the input signals are v I1 = -1.0 V, v I2 = 0.5 cos t volt. Assume the feedback resistance R F = 10 k
Solution: output voltage
Other Op-Amp Applications
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 Integrator circuit
Example 1 The integrator circuit as shown in figure has an initial voltage V across the capacitor at time. A step input voltage V is applied at time. Determine the RC time constant necessary such that the output voltage reaches +10.2 V at time 5.0 ms. 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 = CV S
Example 1 Solution : The output voltage 1.65 volt
Calculating Gain and Design Questions INVERTING NON - INVERTING Calculating Output and Design Questions SUMMING AMPLIFIER DIFFERENTIATOR AMPLIFIERINTEGRATOR AMPLIFIER
Calculate the input voltage if the final output, V O is 10.08 V. NON - INVERTING INVERTING INVERTING Va Vb Have to work backwards: Vo = -(100/5) Vb 10.08 = -20 Vb Vb = -0.504 V Then: Vb = -(5/5) Va -0.504 = - Va Va = 0.504 V Finally: Va = (1 + 10/5) V 1 0.504 = 3V 1 V 1 = 0.168 V
Calculate the output voltage, V O if V 1 = V2 = 700 mV INVERTING SUMMING Va Va = -(500/250) 0.7 Va = -1.4 V Then: Vo = - 500 [ V a / 100 + V 2 / 50 ] Vo = - 500 [ -1.4 / 100 + 0.7 / 50 ] Vo = 0 V
Calculate the output voltage V O of the operational amplifier circuit as shown in the figure. Answer: -3 V