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FREQUENCY EFFECTS IN AMPLIFIERS
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Poles or zeros of multiplicity n, ( j! n , whose amplitude ratio is Mdb 20n log !, where the plus sign corresponds to zeros and the minus sign to poles of the transfer function. (See Example 8.3.) First-order lead and lag factors, 1 j! n , as discussed in Examples 8.1 and 8.2. They are usually approximated with asymptotic Bode plots; if greater accuracy is needed, the asymptotic plots are corrected using Table 8-1:
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The circuit of Fig. 8-15(a) is driven by a sinusoidal source vS . (a) Sketch the asymptotic Bode plot Mdb only) associated with the Laplace-domain transfer function T s Vo =VS . (b) Use Table 8-1 to correct asymptotic plot, so as to show the exact Bode plot.
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20 log
RL RL + RS Exact Bode plot 20 log RLC
_ 20 log |1 + jMC (R + R )| L S
_ 20
Fig. 8-15 (a) By voltage division, Vo RL V RL RS 1=sC S 1
so that
Vo sRL C Kb s VS 1 sC RL RS 1 s
Using the result of Problem 8.2, we recognize (2) as the combination of a rst-order lag, a constant gain, and a zero of multiplicity 1. The components of the asymptotic Bode plot are shown dashed in Fig. 8-15(b), and the composite is solid. For purposes of illustration, it was assumed that 1= C RL RS > 1, which is true in most cases. (b) The correction factors of Table 8-1 lead to the exact Bode plot as drawn in Fig. 8-15(b).
CHAP. 8]
FREQUENCY EFFECTS IN AMPLIFIERS
Sketch the asymptotic Bode plot (Mdb only) associated with the output-to-input voltage ratio of the circuit in Fig. 8-16(a).
i1 +
R1 + R2 C2
R2 20 log R1 + R2
20 Decade
_ (a)
_ (b)
Fig. 8-16 By voltage division, R2 sR2 C2 1 Vs R2 R1 sR2 C2 1
R2 k 1=sC2 V2 V R1 R2 1=sC2 s
and the Laplace-domain transfer function is T s V2 R = R1 R2 Kb   2 R1 R2 VS sReq C2 1 s C 1 R1 R2
From T s , it is apparent that the circuit forms a low-pass lter with low-frequency gain T 0 R2 = R1 R2 and a corner frequency at !1 1=1 1=Req C2 . Its Bode plot is sketched in Fig. 8-16(b).
For the ampli er of Fig. 3-10, assume that CC ! 1; hre hoe 0, and Ri 0. The bypass capacitor CE cannot be neglected. Find expressions for (a) the current-gain ratio Ai s , (b) the current-gain ratio at low frequencies, and (c) the midfrequency current-gain ratio. (d) Determine the low-frequency cuto point, and sketch the asymptotic Bode plot (Mdb only).
(a) The small-signal low-frequency equivalent circuit is given in Fig. 8-4. By current division for Laplacedomain quantities, Ib RB I RB hie ZE i 1
where Also
ZE RE k IL
1 RE sCE sRE CE 1
(2) (3)
RC h I RC RL fe b
FREQUENCY EFFECTS IN AMPLIFIERS
[CHAP. 8
Substitution of (1) into (3) gives the current-gain ratio as hfe RB IL RC Ii RC RL RB hie ZE Using (2) in (4) and rearranging leads to the desired current-gain ratio: hfe RC RB sRE CE 1 IL RC RL RE RB hie Ai s R C R hie Ii 1 s E E B RE RB hie (b) The low-frequency current-gain ratio follows from letting s ! 0 in (5): Ai 0 lim (c) hfe RC RB IL Ii RC RL RE RB hie 6 4
The midfrequency current-gain ratio is obtained by letting s ! 1 in (5): Ai 1 lim hfe RC RB IL Ii RC RL RB hie 7
(d) Inspection of (5) shows that the Laplace-domain transfer function is of the form Ai s Kb 1 s 1 2 s 1 1 R RB hie E 2 RE CE RB hie
where
!1
1 1 1 RE CE
!2
With !1 and !2 as given by (8) and with MdbL 20 log Ai 0 and MdbM 20 log Ai 1
the Bode plot is identical to that of Fig. 8-5. Since !2 > !1 , !2 is closer to the midfrequency region and thus is the low-frequency cuto point.
In the ampli er of Fig. 3-10, CC ! 1; Ri 0; RE 1 k; R1 3:2 k; R2 17 k; RL 10 k; and hoe hre 0. The transistors used are characterized by 75 hfe 100 and 300 hie 1000 . (a) By proper selection of RC and CE , design an ampli er with low-frequency cuto fL 200 Hz and high-frequency voltage gain jAv j ! 50. (b) For the nished design, determine the low-frequency voltage-gain ratio if hie and hfe have median values.
(a) According to (8.17), the worst-case transistor parameters for high Av 1 are minimum hfe and maximum hie . Using those parameter values allows us to determine a value for the parallel combination of RC and RL : Req RC kRL ! jAv 1 j hie 1000 50 666:7  hfe 75
Then Now, from (8.20), for fL CE !
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