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7 = sin + + cos sin = sin cos 12 2 12 2 12 2 12 2+ = (1) 4 2+ 6 = 4 6 + (0) sin 12 [by example (a)]

6:34

6:35

6:36

Substituting u = /2 and v = in the difference formulas of Theorem 26.6 yields cos = cos cos + sin sin = 0 cos + 1 sin = sin 2 2 2 = sin cos cos sin = 1 cos 0 sin = cos sin 2 2 2

6:37

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Comparing (6.34) with (6.9) and (6.37) with (6.10), we see that hib hrb hie hfe 1 hie hoe hre hfe 1 hfe hfe 1 6:38 6:39 6:40 6:41

Thus we have: Theorem 26.7: cos = sin and sin = cos ; that is, the sine of an angle is the cosine of its 2 2 complement.

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Ampli ers are usually designed to emphasize one or more of the following interrelated performance characteristics, whose quantitative measures of goodness are de ned in terms of the quantities of Fig. 6-4: 1. 2. 3. 4. 5. 6. Current ampli cation, measured by the current-gain ratio Ai io =iin . Voltage ampli cation, measured by the voltage-gain ratio Av vo =vin . Power ampli cation, measured by the ratio Ap Av Ai vo io =io iin . Phase shift of signals, measured by the phase angle of the frequency-domain ratio Av j! or Ai j!). Impedance match or change, measured by the input impedance Zin (the driving-point impedance looking into the input port). Power transfer ability, measured by the output impedance Zo (the driving-point impedance looking into the output port with the load removed). If Zo ZL , the maximum power transfer occurs.

Fig. 6-4

In This :

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We also have the useful double-angle formulas: Theorem 26.8: cos 2 = cos2 sin2 = 2 cos2 1 = 1 2 sin2 and sin 2 = 2 sin cos . For cos 2 , note that: (i) cos 2 = cos ( + ) = cos cos sin sin (ii) = cos2 sin2 = cos2 (1 cos2 ) = 2 cos2 1

A simpli ed (bias network omitted) CE ampli er is shown in Fig. 6-5(a), and the associated smallsignal equivalent circuit in Fig. 6-5(b).

Example 6.2. In the CE ampli er of Fig. 6-5(b), let hie 1 k ; hre 10 4 ; hfe 100; hoe 12 S, and RL 2 k . (These are typical CE ampli er values.) Find expressions for the (a) current-gain ratio Ai , (b) voltage-gain ratio Av , (c) input impedance Zin , and (d) output impedance Zo . (e) Evaluate this typical CE ampli er. (a) By current division at node C, iL 1=hoe hfe ib 1=hoe RL 6:42

[by Theorem 26.6(a)] [by Theorem 26.1] [by (ii) and Theorem 26.1]

CHAP. 6]

RL LL _

For sin 2 , sin 2 = sin ( + ) = sin cos + cos sin = 2 sin cos [by Theorem 26.6(c)]

_ (a)

Summary

Fig. 6-5

CHAP. 26]

6:43

Note that Ai % hfe , where the minus sign indicates a 1808 phase shift between input and output currents. (b) By KVL around B; E mesh, vs vbe hie ib hre vce Ohm s law applied to the output network requires that hfe RL ib 1 kRL vce hfe ib hoe 1 hoe RL Solving (6.45) for ib , substituting the result into (6.44), and rearranging yield Av hfe RL vs vce hie RL hie hoe hfe hre 100 2 103 199:2 1 103 2 103 1 103 12 10 6 100 1 10 4 6:46 6:44

6:45

From Theorem 26.8, we deduce: Theorem 26.9 (Half-Angle Formulas): 1 + cos = 2 2 1 cos 2 (b) sin = 2 2 In fact, replace by /2 in Theorem 26.8: (a) cos2 (a) cos = cos 2 (b) cos = cos 2 EXAMPLES

Observe that Av % hfe RL =hie , where the minus sign indicates a 1808 phase shift between input and output voltages. (c) Substituting (6.45) into (6.44) and rearranging yield Zin hre hfe RL vs 1 10 4 100 2 103 hie 1 103 980:5 ib 1 hoe RL 1 12 10 6 2 103 6:47

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