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Finally, we de ne @vBC % vBC @iB Q iB Q @v vBC Reverse voltage ratio hrc BC % @vEC Q vEC Q @i i Forward current gain hfc E % E @iB Q iB Q @i iE Output admittance hoc E % @v v Input resistance hic

5 6 7 8

(a) Move R vertically upward by one unit to form a new region R . The line y = 1 moves up to become the x-axis. R is bounded above by y = x 2 + 1 and below by the line y = 1. The volume we want is obtained by revolving R

A circuit that satis es (3) and (4) with de nitions (5) to (8) is displayed by Fig. 6-16.

hrc Lec _

Note!

hoc (S)

Fig. 33-16

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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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

1 2 + 5 3

Fig. 6-17

CHAP. 6]

The amount of goods sold during the period is shown as cost of goods sold in the income state ment.

(b) Move R two units to the right to form a new region R # [see Fig. 33-16(b)]. The line x = 2 moves over to become the y-axis. R # is bounded above by y = (x 2)2 and below by the x-axis and lies between x = 2 and x = 3. The volume we want is obtained by revolving R # about the y-axis. The cylindrical shell formula applies, V = 2 = 2

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

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 .

1 4 4 3 (3) (3) + 2(3)2 4 3

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