ECE201 Lect-131 Loop Analysis (7.8) Circuits with Op-Amps (3.3) Dr. Holbert October 9, 2001

ECE201 Lect-132 Another Analysis Example We will re-analyze the possible implementation of an AM Radio IF amplifier. We will solve for output voltages using mesh analysis this time. This circuit is a bandpass filter with center frequency 455kHz and bandwidth 40kHz.

ECE201 Lect-133 IF Amplifier 4k  1V  0  + – V out 100pF 160  100pF 80k  – + VxVx 100V x +–+– +–+–

ECE201 Lect-134 Mesh AC Analysis Use AC steady-state analysis. Use a frequency of  =2  100,000.

ECE201 Lect-135 Impedances 4k  1V  0  + – V out 160  80k  – + VxVx 100V x -j15.9k  +–+– +–+–

ECE201 Lect-136 Mesh Analysis I1I1 I2I2 I3I3 4k  1V  0  + – V out 160  80k  – + VxVx 100V x -j15.9k  +–+– +–+–

ECE201 Lect-137 KVL Around Loop 1

ECE201 Lect-138 KVL Around Loop 2

ECE201 Lect-139 KVL Around Loop 2 (Cont)

ECE201 Lect-1310 KVL Around Loop 3

ECE201 Lect-1311 Solve Equations I 1 = j0.23  A I 2 = j5.98  A I 3 = j3.46  A

ECE201 Lect-1312 Solve for V out

ECE201 Lect-1313 |V out | as a function of 

ECE201 Lect-1314 Class Example Learning Extension E7.14(a)

ECE201 Lect-1315 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.

ECE201 Lect-1316 Use of Op Amps Op amps can be configured in many different ways using resistors and other components. Most configurations use feedback.

ECE201 Lect-1317 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.

ECE201 Lect-1318 The Op Amp Symbol + - Non-inverting input Inverting input Ground High Supply Low Supply Output

ECE201 Lect-1319 The Op Amp Model + –Inverting input Non-inverting input R in v+v+ v-v- +–+– A(v + -v - ) vovo

ECE201 Lect-1320 Typical Op Amp The input resistance R in is very large (practically infinite). The voltage gain A is very large (practically infinite).

ECE201 Lect-1321 “Ideal” Op Amp The input resistance is infinite. The gain is infinite. The op amp is in a negative feedback configuration.

ECE201 Lect-1322 The Basic Inverting Amplifier – + V in + – V out R1R1 R2R2 +–+–

ECE201 Lect-1323 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

ECE201 Lect-1324 Solving the Amplifier Circuit Apply KCL at the inverting input: i 1 + i 2 + i - =0 – R1R1 R2R2 i1i1 i-i- i2i2

ECE201 Lect-1325 KCL

ECE201 Lect-1326 Solve for v out Amplifier gain:

ECE201 Lect-1327 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).

ECE201 Lect-1328 Where is the Feedback? – + V in + – V out R1R1 R2R2 +–+–

ECE201 Lect-1329 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.

ECE201 Lect-1330 The Non-Inverting Amplifier + – v in + – v out R1R1 R2R2 +–+–

ECE201 Lect-1331 KCL at the Inverting Input + – v in + – v out R1R1 R2R2 i-i- i1i1 i2i2 +–+–

ECE201 Lect-1332 KCL

ECE201 Lect-1333 Solve for V out

ECE201 Lect-1334 A Mixer Circuit – + v2v2 + – v out R2R2 RfRf R1R1 v1v1 +–+– +–+–

ECE201 Lect-1335 KCL at the Inverting Input – + v2v2 + – v out R2R2 RfRf R1R1 v1v1 i1i1 i2i2 ifif i-i- +–+– +–+–

ECE201 Lect-1336 KCL

ECE201 Lect-1337 KCL

ECE201 Lect-1338 Solve for V out

ECE201 Lect-1339 Class Example Learning Extension E3.16