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**Operational Amplifiers**

Dr. Holbert Lecture 8 Operational Amplifiers Dr. Holbert February 11, 2008 Lect8 EEE 202 EEE 202

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**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

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**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

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**The Op Amp Symbol High Supply Non-inverting input + Output –**

Ground Low Supply Lect8 EEE 202

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The Op Amp Model An operational amplifier is modeled as a voltage-controlled voltage source. + – Inverting input Non-inverting input Rin v+ v– A(v+ – v– ) vo Lect8 EEE 202

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**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). Ideal Op Amp: The input resistance is infinite. The gain is infinite. The op amp is in a negative feedback configuration. Lect8 EEE 202

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**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

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**The Basic Inverting Amplifier**

– + – + + Vin Vout – Lect8 EEE 202

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**Solving the Amplifier Circuit**

Apply KCL at the inverting (–) input: i1 + i2 + i– =0 – R1 R2 i1 i– i2 Vin Vout V– Lect8 EEE 202

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Solve for Vout From KCL Thus, the amplifier gain is Lect8 EEE 202

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**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

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Where is the Feedback? R2 R1 – + – + + Vin Vout – Lect8 EEE 202

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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

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**The Non-Inverting Amplifier**

+ + – + – vin vout R2 R1 – Lect8 EEE 202

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**KCL at the Inverting Input**

+ – vin vout R1 R2 i– i1 i2 Lect8 EEE 202

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Solve for vout Hence, the non-inverting amplifier has a gained output (> unity) relative to the resistance ratio Lect8 EEE 202

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A Mixer Circuit R1 Rf + – R2 v1 – + – + + v2 vout – Lect8 EEE 202

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**KCL at the Inverting Input**

– + v2 vout R2 Rf R1 v1 i1 i2 if i– Lect8 EEE 202

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Solve for vout So, the mixer circuit output is a (negative) combination of the input voltages Lect8 EEE 202

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Class Examples Drill Problem P4-1 Lect8 EEE 202

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