Operational Amplifiers part1

Operational Amplifiers

Introduction

Op amps have been circuit building blocks of universal importance for a long time. Early op amps were constructed from discrete components (vacuum tubes, transistors and resistors) with very high cost. With the advent of IC op amps in 1960, within a few years, high quality op amps became available at very low prices from a large number of suppliers

Importance of Op amp

• Used in all modern Communication, Instrumentation and computation systems

• IC op amp has characteristics that closely approximate the ideal – High Zin, Low Zo and very high Av, i.e. Op amp circuits work at performance levels that are quite close to those predicted theoretically

• They are made up of large number of transistors, resistors and one capacitor – all put together considered as a block

The IDEAL OPAMP

From a signal point-of-view the op amp has three terminals: Two input terminals (1 and 2) and one output terminal (3). Amplifiers require dc power to operate - Most IC op amps require two dc power supplies. Two terminals, 4 and 5, are brought out of the op-amp package and connected to a positive voltage +VCC and a negative voltage -VEE respectively. It also has other terminals for frequency compensation, offset nulling etc.

Figure 5.1 Circuit symbol for the op amp.

Figure 5.2 The op amp shown connected to dc power supplies.

Function and Characteristics of Ideal Op amp

• An ideal op amp senses the difference between the two input voltages and multiplies it with a gain factor A, causing A (v2-v1) to appear at its output terminal.

• An ideal op amp should not draw any input current – Hence is said to have Infinite Input Impedance.

• The output terminal should act like an ideal voltage source – causing Output voltage to remain at A (v2-v1) independent of the current.

The Equivalent Circuit of Ideal Op amp with its characteristics is as shown below:

Figure 5.3 Equivalent circuit and characteristics of an ideal op amp

Differential and Common-Mode signals

The difference between two input signals gives the differential input, and the average of the two input signals gives the common mode input, as given below:

Expressing inputs in terms of differential and common mode gains, we get

Exercise 1

1. Consider an opamp that is ideal except that its open loop gain is 103.The opamp is used in a feedback circuit, and the voltages appearing at two of its three signal terminals are measured. In each case use the measured value to find the expected value of voltage at the third terminal. Also give the differential and Common mode input signals in each case. a) v2=0V and v3=2V b) v2=+5V and v3=-10V c) v1=1.002V and v2=0.998V d) v1=-3.6V and v3=-3.6V.

Solution:

a) Given v2=0V and v3=2V in an op amp

0.002V

Basic Op amp Configurations

Op amps are not used alone; rather, the op amp is connected to passive components in a feedback circuit. There are two such basic circuit configurations employing an op amp and two resistors: The inverting configuration, and the non-inverting configuration.

THE INVERTING CONFIGURATION

Inverting Configuration consists of one op amp and two resistors R1 and R2. Resistor R2 is connected from the output terminal of the op amp, terminal 3, back to the inverting or negative input terminal, terminal 1, we speak of R2 as applying negative feed-back; if R2 were connected between terminals 3 and 2 we would have called this positive feed-back.

In addition to adding R2, we have grounded terminal 2 and connected a resistor between terminal 1 and an input signal source with a voltage

Figure 5.5 The inverting closed-loop configuration.

The output of the overall circuit is taken at terminal 3 (i.e., between terminal 3 and ground). Terminal 3 is, of course, a convenient point to take the output, since the impedance level there is ideally zero. Thus the voltage will not depend on the value of the current that might be supplied to a load impedance connected between terminal 3 and ground.

The Closed-Loop Gain

Analyzing the circuit in Fig. 5.5 to determine the closed-loop gain G, defined as

We will do so assuming the op amp to be ideal. Figure 5.6(a) shows the equivalent circuit, and the analysis proceeds as follows: The gain A is very large (ideally infinite). If we assume that the circuit is "working" and producing a finite output voltage at terminal 3, then the voltage between the op amp input terminals should be negligibly small and ideally zero.

Specifically, if we call the output voltage , then, by definition,

It follows that the voltage at the inverting input terminal is given by

.That is, because the gain A approaches infinity, the voltage approaches and ideally equals . We speak of this as the two input terminals "tracking each other in potential." We also speak of a "virtual short circuit" that exists between the two input terminals. Here the word virtual should be emphasized, and one should not make the mistake of physically shorting terminals 1 and 2 together while analyzing a circuit. A virtual short circuit means that whatever voltage is at 2 will automatically appear at 1 because of the infinite gain A, But terminal 2 happens to be connected to ground; thus, = 0 and = 0. We speak of terminal 1 as being a virtual ground—that is, having zero voltage but not physically connected to ground.

The closed-loop gain is simply the ratio of the two resistances and. The minus sign means that the closed-loop amplifier provides signal inversion. Thus if and we apply at the input a sine-wave signal of 1V peak-to-peak, then the output will be a sine wave of 10V peak-to-peak and phase-shifted 180° with respect to the input sine wave. Because of the minus sign associated with the closed-loop gain, this configuration is called the inverting configuration.

Note:

i. Since the closed-loop gain depends entirely on external passive components (resistors and) , we can make the closed-loop gain as accurate as we want by selecting passive components of appropriate accuracy.

ii. It also means that the closed-loop gain is (ideally) independent of the op-amp gain.

We stalled out with an amplifier having very large gain A, and by applying negative feedback we have obtained a closed-loop gain that is much smaller than A but is stable and predictable. That is, we are trading gain for accuracy.

Effect of Finite Open-Loop Gain

The points just made are more clearly illustrated by deriving an expression for the closed-loop gain under the assumption that the op-amp open-loop gain A is finite. Figure 5.7shows the analysis. If we denote the output voltage , then the voltage

between the two input terminals of the op amp will be . Since the positive input terminal is grounded, the voltage at the negative input terminal must be .

The infinite input impedance of the op amp forces the current to flow entirely through . The output voltage can thus be determined from

Collecting terms, the closed loop gain G is found as,

Note:

i. As A approaches ∞, G approaches the ideal value of -R2/R1.

ii. As A approaches ∞, the voltage at the inverting input terminal approaches zero.

This is the virtual-ground assumption we used in our earlier analysis when the op amp was assumed to be ideal.

iii. Eq. (5.5), in fact indicates that to minimize the dependence of the closed- loop gain G on the value of the open-loop gain A, we should make

Exercise 2

Consider the inverting configuration with = 1 kΩ and = 100 kΩ.

(a) Find the closed-loop gain for the cases A = 103, 104, and 105. In each case determine the percentage error in the magnitude of G relative to the ideal value of (obtained with A =∞).Also determine the voltage that appears at the inverting input terminal when.

(b) If the open-loop gain A changes from 100,000 to 50,000 (i.e., drops by 50%), what is the corresponding percentage change in the magnitude of the closed-loop gain G?

Solution

(a) Substituting the given values in Eq. (5.5), we obtain the values given in the following table where the percentage error ε is defined as