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Operational Amplifiers Dr. Holbert February 11, 2008 Lect8 EEE 202 1 Op Amps • Op Amp is short for operational amplifier • Amplifiers provide gains in voltage or current • Op amps can convert current to voltage Lect8 EEE 202 2 Applications of Op Amps • Op amps can be configured in many different ways using resistors and other components • Most configurations use feedback • Op amps can provide a buffer between two circuits • Op amps can be used to implement integrators and differentiators • Lowpass and bandpass filters Lect8 EEE 202 3 The Op Amp Symbol High Supply Non-inverting input Inverting input + Output – Ground Low Supply Lect8 EEE 202 4 The Op Amp Model • An operational amplifier is modeled as a voltage-controlled voltage source. Non-inverting input v+ + Rin Inverting input Lect8 v– vo + – – EEE 202 A(v+ – v– ) 5 Typical vs. Ideal Op Amps Typical Op Amp: • The input resistance (impedance) Rin is very large (practically infinite). • The voltage gain A is very large (practically infinite). Lect8 Ideal Op Amp: • The input resistance is infinite. • The gain is infinite. • The op amp is in a negative feedback configuration. EEE 202 6 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 Lect8 EEE 202 7 The Basic Inverting Amplifier R2 R1 Vin Lect8 + – – + EEE 202 + Vout – 8 Solving the Amplifier Circuit Apply KCL at the inverting (–) input: R2 Vin Vout i2 R1 V– i1 Vin V Vin i1 R1 R1 – i– Vout V Vout i2 R2 R2 i1 + i2 + i– =0 Lect8 i 0 EEE 202 9 Solve for Vout • From KCL i1 i2 i 0 • Thus, the amplifier gain is Vout R2 Vin R1 Vin Vout 00 R1 R2 Vin Vout R1 R2 Lect8 EEE 202 10 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 (e.g., nodal), to solve for the output voltage in terms of the input(s) Lect8 EEE 202 11 Where is the Feedback? R2 R1 Vin Lect8 + – – + EEE 202 + Vout – 12 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 Lect8 EEE 202 13 The Non-Inverting Amplifier + + – vin + – R2 vout R1 – Lect8 EEE 202 14 KCL at the Inverting Input i 0 + vin + – – + i2 vout i– i1 R2 R1 – Lect8 EEE 202 v vin i1 R1 R1 vout v i2 R2 vout vin R2 15 Solve for vout i1 i2 i 0 vin vout vin 0 R1 R2 R2 vout vin 1 R1 • Hence, the non-inverting amplifier has a gained output (> unity) relative to the resistance ratio Lect8 EEE 202 16 A Mixer Circuit R1 v1 R2 + – v2 Lect8 + – Rf – + EEE 202 + vout – 17 KCL at the Inverting Input v1 v v1 i1 R1 R1 v1 i 0 R1 i1 R2 i 2 + – i– v2 + – – + vout v vout if Rf Rf Lect8 Rf if + vout – v2 v v2 i2 R2 R2 EEE 202 18 Solve for vout i1 i2 i f i 0 v1 v2 vout 0 R1 R2 R f vout Rf R1 v1 Rf R2 v2 • So, the mixer circuit output is a (negative) combination of the input voltages Lect8 EEE 202 19 Class Examples • Drill Problem P4-1 Lect8 EEE 202 20