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After rearranging, (3) and the given approximations lead to hre By (1.18), hfe ic ib vce 0 6 hrb 1 hfb hib hob h h % ib ob hrb hib hob hrb 1 1 hfb 1 hfb

In the same or equal circles, the greater minor arc has the greater chord. (This is the converse of Principle 10.)

Comparing the two columns in Table 5-6 we see that the smaller Ri reduces the overall gain G . However, as the open-loop gain k increases, the e ect of Ri is diminished. As k becomes very large, v2 =v1 approaches 5 unless Ri 0.

By KCL at node B of Fig. 6-1(b), with vce 0 (and thus vcb veb vbe , ib 1 hfb ie hob vcb 1 hfb ie hob vbe Solving (2) for vbe with ie ib ic and substituting now give ib 1 hfb ib ic hib hob i ic 1 hrb b

in the circuit of Fig. 5-33. Replace the circuit to the left of node A including vs , R1 , and Ri by its Thevenin equivalent. Then use (5) to derive (36).

PRINCIPLE 12:

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After rearranging, (6) and the given approximations lead to hfe By (1.19), hoe hfb 1 hrb hib hob hfb % 1 hfb 1 hrb hib hob 1 hfb ic vce ib 0

: From (5), v2 1 b where Therefore, v2 5 1 Ri k Ri 5Ri k v v 6Ri 5 1 Ri k= 6Ri 5 1 Ri s 6Ri 5 Ri k s b RTh Ri RTh R2 6Ri 5 k v 1 bk Th and 1 b 5 1 Ri 6Ri 5 R i vs Rv i s R1 Ri 1 Ri R1 Ri Ri R1 Ri 1 Ri

Fig. 13-7

which is identical with (36).

PRINCIPLE 13:

If ib 0, then ic ie . Replacing ie with ic in (4) and (5), solving (4) for vcb , and substituting into (5) give hob vce hib ic ic 1 hfb 1 hrb 1 hrb After rearranging, (7) and the given approximations lead to hoe hob hob % 1 hfb 1 hrb hib hob 1 hfb

Find the output voltage of an op amp with A 105 and Vcc 10 V for v 0 and v sin t (V). Refer to Figs. 5-7 and 5-8.

In the same or equal circles, the chord at the smaller distance from the center is the greater chord. (This is the converse of Principle 12.)

Apply the de nitions of the z parameters given by (1.10) through (1.13) to the CB h-parameter circuit of Fig. 6-1(b) to nd values for the z parameters in terms of the CB h parameters.

CHAP. 5]

Selecting inequality symbols Determine which inequality symbol, (a) 5 3 (b) 6 9

Because of high gain, saturation occurs quickly at jv2 j 105 jvd j 10 V We may ignore the linear interval and write v2 where vd v v sin t (V). 10 V 10 V vd > 0 vd < 0 or jvd j 10 4 V

The circuit of Fig. 6-1(b) is described by the linear system of equations hib hrb ie v eb hfb hob vcb ic By (1.10) and Fig. 1-8, z11 veb ie ic 0

(a) (b) (c) (d)

For a more exact v2 , we use the transfer characteristic of the op amp in Fig. 5-7. 8 < 10 vd < 10 4 V 5 4 v2 10 vd 10 < vd < 10 4 V : 10 vd > 10 4 V Saturation begins at jvd j j sin tj 10 4 V. output v2 is then given by v2 v2 v2 v2 Since this is a very small range, we may replace sin t by t. 10 4 10 4 10 4 10 4 < t < 10 4 s < t < 10 4 s < t < 10 4 s < t < 2 10 4 s The

(c) (d)

CHAP. 6]

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