Teachers Name - Suman SarkerSubject Name – Industrial Electronics (6832)

Department – Computer (3rd)

IDEAL INSTITUTE OF SCIENCE & TECHNOLOGY

Lecture no- 08 CH-08

THE FEATURES OF OPERATIONAL AMPLIFIER

Operational amplifier

A μA741 integrated circuit, one of the most successful operational amplifiers.

An operational amplifier ("op-amp") is a DC-coupled high-gainelectronic voltage amplifier with a differential input and, usually, a single-ended output.[1]

 In this configuration, an op-amp produces an output potential (relative to circuit ground) that is typically hundreds of thousands of times larger than the potential difference between its input terminals

Operational amplifiers had their origins in analog computers, where they were used to do mathematical operations in many linear, non-linear and frequency-dependent circuits. The popularity of the op-amp as a building block in analog circuits is due to its versatility. Due to negative feedback, the characteristics of an op-amp circuit, its gain, input and output impedance, bandwidth etc. are determined by external components and have little dependence on temperature coefficients or manufacturing variations in the op-amp itself.

Op-amps are among the most widely used electronic devices today, being used in a vast array of consumer, industrial, and scientific devices. Many standard IC op-amps cost only a few cents in moderate production volume; however some integrated or hybrid operational amplifiers with special performance specifications may cost over $100 US in small quantities.[3] Op-amps may be packaged as components, or used as elements of more complex integrated circuits.

An op-amp without negative feedback (a comparator)The amplifier's differential inputs consist of a non-inverting input (+) with voltage V+ and an inverting input (–) with voltage V−; ideally the op-amp

amplifies only the difference in voltage between the two, which is called thedifferential input voltage. The output voltage of the op-amp Vout is given by

the equation:

An op-amp with negative feedback (a non-inverting amplifier)If predictable operation is desired, negative feedback is used, by applying a portion of the output voltage to the inverting input. The closed loopfeedback greatly reduces the gain of the circuit. When negative feedback is used, the circuit's overall gain and response becomes determined mostly by the feedback network, rather than by the op-amp characteristics. If the feedback network

An equivalent circuit of an operational amplifier that models some resistive non-ideal parameters.

The non-inverting op-amp configuration of slide 2-4 has an apparent input resistance of infinity, since iIN = 0 and RIN = vIN/iIN = vIN/0 = infinityThe inverting op-amp configuration, however, has an apparent input resistance of R1

since RIN = vIN/iIN = vIN/[(vIN – 0)/R1] = R1

op-amps into one circuitUsing superposition of the results from the two previous cases, we can writevOUT = [(R1 + R2)/R1]v1 – (R2/R1)v2

The gain factors for both inputs are different, howeverWe can obtain the same gain factors for both v1 and v2 by using the modified circuit belowHere the attenuation network at v1 delivers a reduced input v+ = v1(R2/(R1 + R2))Replacing v1 in the expression above by the attenuation factor, gives us

vOUT = (R2/R1)(v1 – v2)The difference amplifier will work properly if the attenuation network resistors (call them R3 & R4) are related to the feedback resistors R1 & R2 by the relation R3/R4 = R1/R2 (i.e. same ratio)

The example of Fig’s 2.14 and 2.15 in the text shows a difference amplifier used with a bridge circuit and strain gauge to measure strain.

A summation op-amp (shown at left) can be used to obtain a weighted sum of inputs v1…vN

The gain for any input k is given by RF/Rk

If any input goes positive, vOUT goes negative just enough to force the input v- to zero, due to the virtual short nature of the op-ampCombining all inputs, we have

vOUT = -RF(v1/R1 + v2/R2 + .. + vN/RN)The input resistance for any input k is given by Rk due to the virtual short between v- and v+Example 2.5 – use as an audio preamp with individual adjustable gain controlsNote effect of microphone’s internal resistance

Given an input signal of 4V square wave for 10 ms duration, what is the integrator output versus time for the integrator circuit at the left?

Using the integral expression from the previous chart, the capacitor voltage will increase linearly in time (1/R1C) 4t = 0.8t V/ms during the square wave durationThe output will therefore reduce linearly in time by – 0.8t V/ms during the pulse duration, falling from 0 to –8 volts, as shown in the figure at leftSince at 10 ms the output will be –8 V > VNEG, the op-amp will not saturate during the 10 ms input pulse

Next Lecture

POWER SWITCHING DEVICE