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AMPLIFIERS AND OPERATIONAL AMPLIFIER CIRCUITS

[CHAP. 5

Fig. 5-4

EXAMPLE 5.3 Find v2 =vs in Fig. 5-4 and express it as a function of the ratio b ¼ R1 =ðR1 þ R2 Þ. From the amplifier we know that or

v2 ¼ kv1

v1 ¼ v2 =k

ð3Þ

Applying KCL at node A, v1  vs v1  v2 þ ¼0 R1 R2

ð4Þ

Substitute v1 in (3) into (4) to obtain v2 R2 k k ¼ ð1  bÞ ¼ 1  bk vs R2 þ R1  R1 k

where b ¼

R1 R1 þ R2

ð5Þ

EXAMPLE 5.4 In Fig. 5-5, R1 ¼ 1 k and R2 ¼ 5 k. (a) Find v2 =vs as a function of the open-loop gain k. (b) Compute v2 =vs for k ¼ 100 and 1000 and discuss the results.

Fig. 5-5 (a) Figures 5-4 and 5-5 differ only in the polarity of the dependent voltage source. Example 5.3 and change k to k in (5). v2 k ¼ ð1  bÞ 1 þ bk vs

where b ¼

To find v2 =vs , use the results of

R1 1 ¼ R1 þ R2 6

v2 5k ¼ vs 6 þ k (b) At k ¼ 100, v2 =vs ¼ 4:72; at k ¼ 1000, v2 =vs ¼ 4:97. Thus, a tenfold increase in k produces only a 5.3 percent change in v2 =vs ; i.e., ð4:97  4:72Þ=4:72 ¼ 5:3 percent. Note that for very large values of k, v2 =vs approaches R2 =R1 which is independent of k.

5.3

OPERATIONAL AMPLIFIERS

The operational amplifier (op amp) is a device with two input terminals, labeled þ and  or noninverting and inverting, respectively. The device is also connected to dc power supplies (þVcc and

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