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. Op amps can be configured in many different ways using resistors and other components. Most configurations use feedback.
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. Can be used to built low-pass and bandpass filters
The Op Amp Symbol + - Non-inverting input Inverting input Ground High Supply Low Supply Output
The Op Amp Model + -Inverting input Non-inverting input R in v+v+ v-v- + - A(v + -v - ) vovo
Typical Op Amp The input resistance R in is very large (practically infinite). The voltage gain A is very large (practically infinite).
CIRCUIT AND MODEL FOR UNITY GAIN BUFFER WHY UNIT GAIN BUFFER? BUFFER GAIN PERFORMANCE OF REAL OP-AMPS
“Ideal” Op Amp The input resistance is infinite. The gain is infinite. The op amp is in a negative feedback configuration.
THE IDEAL OP-AMP
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 These conditions are used, along with KCL and other analysis techniques, to solve for the output voltage in terms of the input(s).
THE VOLTAGE FOLLOWER OR UNITY GAIN BUFFER CONNECTION WITHOUT BUFFER CONNECTION WITH BUFFER THE VOLTAGE FOLLOWER ACTS AS BUFFER AMPLIFIER THE SOURCE SUPPLIES POWER THE SOURCE SUPPLIES NO POWER THE VOLTAGE FOLLOWER ISOLATES ONE CIRCUIT FROM ANOTHER ESPECIALLY USEFUL IF THE SOURCE HAS VERY LITTLE POWER
LEARNING EXTENSION
“INVERSE VOLTAGE DIVIDER” LEARNING EXTENSION
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.