Presentation on theme: "ECE201 Lect-161 Operational Amplifiers (4.1-4.3) Dr. Holbert April 3, 2006."— Presentation transcript:
ECE201 Lect-161 Operational Amplifiers ( ) Dr. Holbert April 3, 2006
ECE201 Lect-162 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-163 Use of Op Amps Op amps can be configured in many different ways using resistors and other components. Most configurations use feedback.
ECE201 Lect-164 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-165 The Op Amp Symbol + - Non-inverting input Inverting input Ground High Supply Low Supply Output
ECE201 Lect-166 The Op Amp Model + –Inverting input Non-inverting input R in v+v+ v-v- +–+– A(v + -v - ) vovo
ECE201 Lect-167 Typical Op Amp The input resistance R in is very large (practically infinite). The voltage gain A is very large (practically infinite).
ECE201 Lect-168 “Ideal” Op Amp The input resistance is infinite. The gain is infinite. The op amp is in a negative feedback configuration.
ECE201 Lect-169 The Basic Inverting Amplifier – + V in + – V out R1R1 R2R2 +–+–
ECE201 Lect-1610 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-1611 Solving the Amplifier Circuit Apply KCL at the inverting input: i 1 + i 2 + i - =0 – R1R1 R2R2 i1i1 i-i- i2i2
ECE201 Lect-1612 KCL
ECE201 Lect-1613 Solve for v out Amplifier gain:
ECE201 Lect-1614 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-1615 Where is the Feedback? – + V in + – V out R1R1 R2R2 +–+–
ECE201 Lect-1616 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-1617 The Non-Inverting Amplifier + – v in + – v out R1R1 R2R2 +–+–
ECE201 Lect-1618 KCL at the Inverting Input + – v in + – v out R1R1 R2R2 i-i- i1i1 i2i2 +–+–
ECE201 Lect-1619 KCL
ECE201 Lect-1620 Solve for V out
ECE201 Lect-1621 A Mixer Circuit – + v2v2 + – v out R2R2 RfRf R1R1 v1v1 +–+– +–+–
ECE201 Lect-1622 KCL at the Inverting Input – + v2v2 + – v out R2R2 RfRf R1R1 v1v1 i1i1 i2i2 ifif i-i- +–+– +–+–
ECE201 Lect-1623 KCL
ECE201 Lect-1624 KCL
ECE201 Lect-1625 Solve for V out
ECE201 Lect-1626 Class Example Learning Extension E4.1 Learning Extension E4.2 Learning Extension E4.3