FREQUENCY RESPONSE Electronics Help

Stability

When Ilw wm! “!dhiltly ‘,! used ill “/,1 ncction with. a high-gain amplifier. it usually means the property of behaving like an amplifier (ather than like an os .illau»: An osci lutor is H, device that .pontaneously generates an ac signal because of positive. feedback. We will st udy oscillator theory in Chapter 14, but for now it is sufficient to know that oscillations are easily induced in high-gain.we’ r  amplifiers, due to positive feedback. that occurs through reactive clements, As WI! ri.llUW. an operational amplifier has very high gain, so precautions must be taken in its design to ensure that it does not oscillate, i.e., to ensure that it remains stable. Large gains at high frequencies tend to make an amplifier unstable because those properties enable positive feedback through stray capacitance.

To ensure stable operation, most operational amplifiers have internal contpensalion circuitry that causes the open-loop gain to diminish with increasing frequency. This reduction in gain is called “rolling-off” the amplifier. Sometimes it is necessary to connect external roll-off networks to reduce high-frequency gain even more , rapidly. Because the de and low-frequency open-loop gain of an operational amplifier is so great. the gain roll-off must begin at relatively low frequencies. As a consequence, the open-loop bandwidth of an operational amplifier is generally rather small.In I1l()S~operational amplifiers. the gain over the usable frequent’} range rolls off at the rate of -20 dBidecaue, or -6 dll/octave. Recall, from Chapter  10 that this rate of gain reduction is the same as that of a single low-pass RC network.

Any device whose gain falls Off like that of a single RC network is said to have a single-pole frequency response, a name derived from advanced mathematical theory. Beyond a certain very high frequency, the frequency response of an operational amplifier exhibits further break frequencies. meaning that the gain falls off at greater rates. but for most practical uualysis purposes we can treat the umplilicr as if it had a single-pole response,

The Gain-Bandwidth Product

Figure 13-15 shows a typical frequency response characteristic for the open-loop gain of an operational amplifier having a single-pole frequency response, plotted on log-log scales. We use f~to denote the cut off frequency, which, as usual.

Frequency at which the gain A falls to Y2/2 times its low-frequency, or de, value (An). Recall that the slope of the single-pole response is -I. In ‘Chapter 10. we studied a very similar frequency response: that of the beta of a bipolar junction transistor. Using the fact that the slope is -1, we showed that the frequency Ir at which the f3 falls to the value I (unity) is given by fr = f3″,J where f3,,>is the 10wfrequency f3 and 1;, is the f3 cutoff frequency. Using exactly the same approach, we can deduce that the frequency at which the amplifier gain falls to value J equals the product Of the cutoff frequency and the low-frequency gain

Since the amplifier is de (lower cutoff frequency ~ 0). the bandwidth equals .t:. The term An!: in equation 13-37 is called the gain-bandwidth product. In specitications. a value may be quoted either for the gain-bandwidth product, or for its equivalent, the unity-gain frequency.

The significance of the gain-bandwidth product is that it makes it possible for us to compute the bandwidth of an amplifier when it is operated in one of the more useful closcu-loop conllgurutions. Obviously, a knowledge of the upper-frequency -lirniuuion of a certain amplifier configuration is vital information when designing it for a particular application. The relationship between closed-loop bandwidth (8W(I.) and the gain-bandwidth product is closely approximated

Each of the amplifiers shown in Figure 13-·16 has an open-loop, gain-bandwidth product equal to I x lOC’. Find the cutoff frequencies in the closed-loop configurations shown .

SOLUTION

1. In Figure 13-16(a), (3 = RI/(R, + Rf) = (10 kIl)/r(1O kn) + (240 kO)] = 0.04.

From equation 13-38, BWCl = f,{3 = (101′)(0.04) = 411 k l lz.

Since the amplifier is dc, the closed-loop cutoff Frequency is the same as the closed-loop bandwidth, 40 kHz.
2. In Figure l3-16(b). (3 = RI/(R, + Rf) = (10 k!1)/[(1O kfl) + (15 k!1)] = 0.4. Then I3Wn. == HY'(O.4) == 400 kHz.

It is worthwhile noting that in the case of the  amplifier, the fact that the ideal dosed-loop gain is 1/{3 makes equation 13-33 equivalent

or (ideal closed-loop gain) X (closed-loop bandwidth) == gain-bandwidth product. To illustrate the validity of this expression, refer to part I of Example 13-7. Here, the ideal closed-loop gain is (R, + Rf)IRI == (250 k!1)/(1O k!1) == 25, so 25 X (closedloop bandwidth) == 106, which yields

Although some authors interpret the gain-bandwidth product to be the product of closed-loop gain and closed-loop bandwidth regardless of configuration, we have seen that this interpretation yields a bandwidth for the inverting amplifier that is larger than its actual value. At large values of closed-loop gain, the bandwidths of the inverting and noninverting amplifiers are comparable, but at low gains the noninverting amplifier has a larger bandwidth. For example, when the closed-loop gain is l. the bandwidth of the non inverting amplifier is twice that 01 the inverting amplifier.

Figure 13-17 shows a typical set of frequency response plots for a non inverting amplifier, as the gain ranges from its open-loop value of lOs to a closed-loop value of 1. This figure clearly shows how the bandwidth decreases as the closed-looped gain increases. Notice that the bandwidth at maximum (open-loop) gain is only 10 Hz.

W-ith reference to the amplifier whose frequency response is shown in Figure 13-17, find
Example

1. the ,mil y-gain frequency:
2. the gain-bandwidth product:
3. the bandwidth when the feedback ratio i1,0.U2; and
4. the closed-loop gain at 0.4 MHz whe~ the feedback ratio is 0.04.

Solution

1. In Figure 13-17. it is apparent that the open-loop gain equals when the frequency is I MHz. Thus, f, == 1 MHz.
2. Gain-bandwidth product = Au!.. = f, == 10″:
3. From equation 13-38, BWn. = f,f3 = 101>(0.02) = 20 kHz.
4. BW. = f,f3 = 10/'(0.04) = 40 kHz. Thus, the closed-loop cutoff frequency is 40 kHz. Since Ihe amplifier is noninverting. the closed-loop gain is 1/f3 = 25. Since 0.4 MHz is I decade above the cutoff frequency, the gain is down 20 dB from 25, or down by a factor of 0: 0.1 (25) = 2.5

Posted on November 19, 2015 in Operational Ampiifier Theory

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