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Op Amps Lecture 30.

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Presentation on theme: "Op Amps Lecture 30."— Presentation transcript:

1 Op Amps Lecture 30

2 Introduction Op Amp is short for operational amplifier.
An operational amplifier is modeled as a voltage controlled voltage source. An operational amplifier has a very high input impedance and a very high gain. Lecture 30

3 Use of Op Amps Op amps can be configured in many different ways using resistors and other components. Most configurations use feedback. Lecture 30

4 Applications of Op Amps
Amplifiers provide gains in voltage or current. Op amps can convert current to voltage. Op amps can provide a buffer between two circuits. Op amps can be used to implement integrators and differentiators. Lecture 30

5 More Applications Lowpass and bandpass filters. Lecture 30

6 The Op Amp Symbol High Supply Non-inverting input + Output -
Ground Low Supply Lecture 30

7 The Op Amp Model v+ Non-inverting input + vo Rin + - v-
A(v+ -v- ) Lecture 30

8 Typical Op Amp The input resistance Rin is very large (practically infinite). The voltage gain A is very large (practically infinite). Lecture 30

9 “Ideal” Op Amp The input resistance is infinite. The gain is infinite.
The op amp is in a negative feedback configuration. Lecture 30

10 The Basic Inverting Amplifier
- + Vin Vout R1 R2 Lecture 30

11 Consequences of the Ideal
Infinite input resistance means the current into the inverting input is zero: i- = 0 Infinite gain means the difference between v+ and v- is zero: v+ - v- = 0 Lecture 30

12 Solving the Amplifier Circuit
Apply KCL at the inverting input: i1 + i2 + i-=0 R2 i2 R1 - i1 i- Lecture 30

13 KCL Lecture 30

14 Solve for vout Amplifier gain: Lecture 30

15 Recap The ideal op amp model leads to the following conditions: i- = 0
v+ = v- These conditions are used, along with KCL and other analysis techniques, to solve for the output voltage in terms of the input(s). Lecture 30

16 Where is the Feedback? - + Vin Vout R1 R2 Lecture 30

17 Review To solve an op amp circuit, we usually apply KCL at one or both of the inputs. We then invoke the consequences of the ideal model. The op amp will provide whatever output voltage is necessary to make both input voltages equal. We solve for the op amp output voltage. Lecture 30

18 The Non-Inverting Amplifier
+ - vin vout R1 R2 Lecture 30

19 KCL at the Inverting Input
+ + - + i- vin - vout i1 i2 R2 R1 - Lecture 30

20 KCL Lecture 30

21 Solve for Vout Lecture 30

22 A Mixer Circuit - + v2 vout R2 Rf R1 v1 Lecture 30

23 KCL at the Inverting Input
Rf i1 if + R2 v1 i2 - - i- v2 + - + + vout - Lecture 30

24 KCL Lecture 30

25 KCL Lecture 30

26 Solve for Vout Lecture 30

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