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In a triangle, the measure of an exterior angle is larger than the measure of either nonadjacent interior angle.

Fig. 5-31

PRINCIPLE 3:

hrc Lec _

. Find (a) the Thevenin equivalent of the circuit seen by Rl and (b) v2 and the power dissipated in Rl for Rl 0:5, 1, 3, 5, 10, 100, and 1000

If the lengths of two sides of a triangle are unequal, the measures of the angles opposite these sides are unequal, the larger angle being opposite the longer side. (Corollary: The largest angle of a triangle is opposite the longest side.)

(a) The open-circuit voltage and short-circuit current at A B terminal are vo:c: 5v1 and is:c: 5v1 =3, respectively. We nd v1 by dividing vs between Rs and Ri . Thus, v1 Ri 990 20 19:8 V v Rs Ri s 10 990

hoc (S)

PRINCIPLE 4:

Fig. 5-32

Redraw the CE small-signal equivalent circuit of Fig. 6-1(a) so that the collector C is common to the input and output ports. Then apply KVL at the input port and KCL at the output port to nd a set of equations that can be compared with (3) and (4) of Problem 6.13 to determine the CC h parameters in terms of the CE h parameters.

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If the measures of two angles of a triangle are unequal, the lengths of the sides opposite these angles are unequal, the longer side being opposite the larger angle. (Corollary: The longest side of a triangle is opposite the largest angle.)

Therefore, vo:c: 5 19:8 99 V is:c: 99=3 33 A The Thevenin equivalent is shown in Fig. 5-32. (b) With the load Rl connected, we have v2 Rl 99Rl v Rl RTh Th Rl 3 and p v2 2 Rl vTh vo:c: 99 V RTh vo:c: =is:c: 3

Figure 6-1(a) is rearranged, to make the collector common, in Fig. 6-17. B; C loop, with vce vec , results in vbc hie ib hre vce vec hie ib 1 hre vec Applying KVL around the

PRINCIPLE 5:

Table 5-3 shows the voltage across the load and the power dissipated in it for the given seven values of Rl . The load voltage is at its maximum when Rl 1. However, power delivered to Rl 1 is zero. Power delivered to Rl is maximum at Rl 3

, which is equal to the output resistance of the ampli er. Table 5-3 Rl ;

Fig. 13-3

Fig. 6-17

0.5 1 3 5 10 100 1000 v2 ; V 14.14 24.75 49.50 61.88 76.15 96.12 98.70 p; W 400.04 612.56 816.75 765.70 579.94 92.38 9.74

PRINCIPLE 6:

CHAP. 6]

. Find the gains G v2 =vs in Fig. 5-4 and G v2 =vs in Fig. 5-5 for k 1, 2, 4, 6, 8, 10, 100, 1000, and 1. Compare the results.

Fig. 13-4

The gains G and G are calculated for nine values of k in Table 5-4. As k becomes very large, G and G approach the limit gain of 5, which is the negative of the ratio R2 =R1 and is independent of k. The circuit of Fig. 5-5 (with negative feedback) is always stable and its gain monotonically approaches the limit gain. However, the circuit of Fig. 5-4 (with positive feedback) is unstable. The gain G becomes very large as k approaches six. At k 6, G 1. Table 5-4 k 1 2 4 6 8 10 100 1000 1 G 1:0 2:5 10:0 1 20:0 12:5 5:32 5:03 5:00 G 0:71 1:25 2:00 2:50 2:86 3:12 4:72 4:97 5:00

Applying KCL at node E gives ie ib hfe ib hoe vec hfe 1 ib hoe vec Comparison of (1) and (2) above with (3) and (4) of Problem 6.13 yields, by direct analogy, hic hie hrc 1 hre hfc hfe 1 hoc hoe 3 2

If two sides of a triangle are congruent to two sides of another triangle, the triangle having the greater included angle has the greater third side.

CHAP. 5]

PRINCIPLE 7:

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