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Use the CC transistor model of Fig. 6-16 to nd the Thevenin equivalent for the circuit to the right of terminals B; C in Fig. 6-2(b), assuming hrc % 1 and hoc % 0. Compare the results with (6.16) to determine relationships between hie and hic , and between hfe and hfc .

in the circuit of Fig. 5-33. Find v2 =vs for k 1, 10, 100, 1000, 1 and compare the results with the values of G in Table 5-4.

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

The circuit to be analyzed is Fig. 6-16 with a resistor RE connected from E to C. With terminal pair B; C open, the voltage across terminals C; E is zero; thus, the Thevenin equivalent circuit consists only of ZTh RTh . Now consider vbc as a driving-point source, and apply KVL around the B; C loop to obtain vdp vbc hic ib hrc vec % hic ib vec Use KCL at node E to obtain vec ie RE hfc ib hoc vce RE % hfc RE ib Substitute (2) into (1), and solve for the driving-point impedance: RTh vbc hic hfc RE ib 3 2 1

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Fig. 5-33 This problem is solved by application of KCL at node A (another approach which uses the Thevenin equivalent is suggested in Problem 5.30). Thus, v1 vs v1 v2 v1 0 1 5 50 From the ampli er we obtain v2 kv1 or v1 v2 =k 34 33

13.1E

Replacing v1 in (34) into (33) and rearranging terms, we obtain v2 50k 5k vs 61 10k 6:1 k 35

Now (3) is compared with (6.16), it becomes apparent that hic hie and hfc hfe 1 , as given in (3) of Problem 6.14.

PRINCIPLE 8:

input resistance of the ampli er reduces the overall gain very slightly, as seen by comparing Tables 5-4 and 5-5. The feedback has made the input resistance of the ampli er less e ective in changing the overall gain. Table 5-5 k 1 10 100 1000 1 v2 =vs 0:704 3:106 4:713 4:97 5:00

PRINCIPLE 9:

Apply the de nitions of the general h parameters given by (1.16) to (1.19) to the circuit of Fig. 6-1(b) to determine the CE h parameters in terms of the CB h parameters. Use the typically good approximations hrb ( 1 and hob hib ( 1 hfb to simplify the results.

. Find v2 =v1 for k 1; 10; 100; 1000; 1. Repeat for Ri 1: Discuss the effects of Ri and k on the overall gain. Show that, for k 1 and Ri 6 0; the gain of the amplifier is independent of Ri and is equal to R2 =R1 :

In the same or equal circles, the greater arc has the greater central angle. (This is the converse of Principle 8.)

(a) Apply KCL to currents leaving node A to obtain v1 vs v1 v2 v1 0 1 5 Ri From the ampli er we get v2 kv1 or v1 v2 =k. rearranging terms we get Substituting for v1 in the KCL equation and

If vce 0 (short-circuited) in the network of Fig. 6-1(b), then vcb vbe , so that, by KVL around the E; B loop, vbe hib ie hrb vcb hib ie hrb vbe h 1 ie rb vbe hib 1 hfb 1 hrb hob vbe hib

Fig. 13-5

36

ib 1 hfb ie hob vcb Now, (1) and the given approximations, hie By (1.17), hre If ib 0, then ic ie in Fig. 6-1(b). By KVL,

PRINCIPLE 10:

, c 1=11 which, substituted into (36), gives v2 5k vs 11 k For Ri 1 we get c 1=6 and so v2 5k vs 6 k 38 37

KCL at node C then gives ic ie hfb ie hob vcb h ie ob 1 hfb hib hob v vbe 1 hfb ce

Fig. 13-6

Table 5-6 gives values of v2 =vs in (37) and (38) versus k. Note that (38) is identical with (32).

PRINCIPLE 11:

0:31 2:38 4:51 4:95 5:00 Ri 1 0:71 3:12 4:72 4:97 5:00

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