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Operational Amplifiers (4.1-4.3) Dr. Holbert April 3, 2006 ECE201 Lect-16 1 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-16 2 Use of Op Amps • Op amps can be configured in many different ways using resistors and other components. • Most configurations use feedback. ECE201 Lect-16 3 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-16 4 The Op Amp Symbol High Supply Non-inverting input Inverting input + Output Ground Low Supply ECE201 Lect-16 5 The Op Amp Model Non-inverting input v+ + Rin Inverting input v- vo + – – ECE201 Lect-16 A(v+ -v- ) 6 Typical Op Amp • The input resistance Rin is very large (practically infinite). • The voltage gain A is very large (practically infinite). ECE201 Lect-16 7 “Ideal” Op Amp • The input resistance is infinite. • The gain is infinite. • The op amp is in a negative feedback configuration. ECE201 Lect-16 8 The Basic Inverting Amplifier R2 R1 Vin + – – + ECE201 Lect-16 + Vout – 9 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-16 10 Solving the Amplifier Circuit Apply KCL at the inverting input: R2 i2 R1 – i1 i- i1 + i2 + i-=0 ECE201 Lect-16 11 KCL i 0 vin v vin i1 R1 R1 vout v vout i2 R2 R2 ECE201 Lect-16 12 Solve for vout vin vout R1 R2 Amplifier gain: vout R2 vin R1 ECE201 Lect-16 13 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-16 14 Where is the Feedback? R2 R1 Vin + – – + ECE201 Lect-16 + Vout – 15 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-16 16 The Non-Inverting Amplifier + + – vin + – R2 vout R1 – ECE201 Lect-16 17 KCL at the Inverting Input + + – vin + – ii1 i2 R2 vout R1 – ECE201 Lect-16 18 KCL i 0 v vin i1 R1 R1 vout v vout vin i2 R2 R2 ECE201 Lect-16 19 Solve for Vout vin vout vin 0 R1 R2 vout R2 vin 1 R1 ECE201 Lect-16 20 A Mixer Circuit R1 v1 R2 + – v2 + – Rf – + ECE201 Lect-16 + vout – 21 KCL at the Inverting Input R1 i1 v1 R2 i 2 + – iv2 Rf if + – – + ECE201 Lect-16 + vout – 22 KCL v1 v v1 i1 R1 R1 v2 v v2 i2 R2 R2 ECE201 Lect-16 23 KCL i 0 vout v vout if Rf Rf ECE201 Lect-16 24 Solve for Vout v1 v 2 vout 0 R1 R2 R f vout Rf R1 v1 ECE201 Lect-16 Rf R2 v2 25 Class Example • Learning Extension E4.1 • Learning Extension E4.2 • Learning Extension E4.3 ECE201 Lect-16 26