lecture 91 Loop Analysis (3.2) Circuits with Op-Amps (3.3) Prof. Phillips February 19, 2003

lecture 92 Op Amps 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 93 Use of Op Amps Op amps can be configured in many different ways using resistors and other components. Most configurations use feedback.

lecture 94 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. Lowpass and bandpass filters.

lecture 95 The Op Amp Symbol + - Non-inverting input Inverting input Ground High Supply Low Supply Output

lecture 96 The Op Amp Model + –Inverting input Non-inverting input R in v+v+ v-v- +–+– A(v + -v - ) vovo

lecture 97 Typical Op Amp The input resistance R in is very large (practically infinite). The voltage gain A is very large (practically infinite).

lecture 98 “Ideal” Op Amp The input resistance is infinite. The gain is infinite. The op amp is in a negative feedback configuration.

lecture 99 The Basic Inverting Amplifier – + V in + – V out R1R1 R2R2 +–+–

lecture 910 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 911 Solving the Amplifier Circuit Apply KCL at the inverting input: i 1 + i 2 + i - =0 – R1R1 R2R2 i1i1 i-i- i2i2

lecture 912 KCL

lecture 913 Solve for v out Amplifier gain:

lecture 914 Recap The ideal op-amp model leads to the following conditions: i - = 0 = i + 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 915 Where is the Feedback? – + V in + – V out R1R1 R2R2 +–+–

lecture 916 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 917 The Non-Inverting Amplifier + – v in + – v out R1R1 R2R2 +–+–

lecture 918 KCL at the Inverting Input + – v in + – v out R1R1 R2R2 i-i- i1i1 i2i2 +–+–

lecture 919 KCL

lecture 920 Solve for V out

lecture 921 A Mixer Circuit – + v2v2 + – v out R2R2 RfRf R1R1 v1v1 +–+– +–+–

lecture 922 KCL at the Inverting Input – + v2v2 + – v out R2R2 RfRf R1R1 v1v1 i1i1 i2i2 ifif i-i- +–+– +–+–

lecture 923 KCL

lecture 924 KCL

lecture 925 Solve for V out

lecture 926 Class Example