barcode fonts for ssrs Substitution of (1) into (3) and rearrangement then give Av s in Software

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Substitution of (1) into (3) and rearrangement then give Av s
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2 Vo s2 gm RG RD RL rds CC = RD rds     RD rds Vi RL 1 sCC RG 1 sCC RD rds
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(b) Since high-frequency capacitances have not been modeled, the midfrequency gain follows from letting s ! 1 in (4): Amid Av 1 gm RD RL rds RD rds RL RD rds
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FREQUENCY EFFECTS IN AMPLIFIERS
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For the CS JFET ampli er of Fig. 7-2, (a) nd an expression for the voltage-gain ratio Av s and (b) determine the low-frequency cuto point.
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(a) The low-frequency equivalent circuit is shown in Fig. 8-22. Id By KVL, 1 Vgs  sRS CS 1 Vgs RS k1=sCS rds RD sCS RS RD rds RS RD rds   1 R S ID Vgs Vi ID RS k Vi sCS sRS CS 1 sCS RS RD rds RS RD rds V sCS RS RD rds RD rds  1 RS i RD sRS CS 1 Vgs sCS RS RD rds RS RD rds sRS CS 1 CS RS RD rds s 1 RD rds  1 RS
But KVL requires that
Substituting (1) into (2) and solving for Vgs give Vgs Now, by Ohm s law and (1), Vo RD ID 4 3
Substituting Vgs as given by (3) into (4) and rearranging yield, nally, Av s Vo RD Vi RD rds  1 RS 5
(b) It is apparent that the low-frequency cuto is the larger of the two break frequencies; from (5), it is !L RD rds  1 RS CS RS RD rds
mLgs + iD RS CS
4 + Lo _
RG _
Fig. 8-22
The hybrid- equivalent circuit for the CE ampli er of Fig. 3-10 with the output shorted is shown in Fig. 8-23. (a) Find an expression for the so-called cuto frequency f , which is simply the high-frequency current-gain cuto point of the transistor with the collector and emitter terminals shorted. (b) Evaluate f if rx 100 ; r 1 k; C 3 pF; and C 100 pF.
(a) Ohm s law gives Vb 0 e Ib g s C C where g 1 r 1
But with the collector and emitter terminals shorted, IL gm Vb 0 e 2
FREQUENCY EFFECTS IN AMPLIFIERS
[CHAP. 8
ib +
C iL Cp + Cm gm Lb e
Lb e
Fig. 8-23
Substituting (1) into (2) and rearranging give the current-gain ratio IL gm gm r Ib g s C C sr C C 1 From (3), the cuto frequency is seen to be f ! 1 2 2r C C 4 3
(b) Substituting the given high-frequency parameters in (4) yields f 1 1:545 MHz 2 1000 103 10 12
Apply the hybrid- high-frequency model to the CB ampli er of Fig. 6-15(b): (a) Find an expression for the high-frequency voltage-gain ratio. (b) Describe the high-frequency behavior of the CB ampli er.
(a) Use of the hybrid- model of Fig. 8-8 results in the high-frequency small-signal equivalent circuit of Fig. 8-24. The coupling capacitors are assumed to be short circuits at high frequency. For typical values, rx ( 1=sC ; r ; 1=sC for frequencies near the break frequencies; thus, letting rx 0 introduces little error (but considerable simplicity).
gm Lb e E _ +
C Cp rp B rx B
Lb e
Cm RC || RL
RE _
+ LL _
Fig. 8-24 A Thevenin equivalent can be found for the network to the left of terminal pair a; a 0 . With rx 0, current from the dependent source ows only through C, so VTh By the method of node voltages, VS Vb 0 e gm Vb 0 e Vb 0 e sC GE g 0 RS 2 1 g V 0 sC m b e 1
CHAP. 8]
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Solving (2) for Vb 0 e and substituting the result into (1) yield VTh gm VS sC 1 RS gm RS sC GE g 3
Deactivating (shorting) VS also shorts E to B 0 . Consequently, Vb 0 e 0, the dependent current source is open-circuited, and ZTh 1=sC . Now, the Thevenin equivalent and voltage division lead to VL RC kRL V RC kRL ZTh Th 4
Substitution of (3) into (4) and rearrangement give the desired voltage-gain ratio: Av VL gm RC kRL VS sC RC kRL 1 sC RS RS gm g GE 1 5
(b) Since (5) involves the upper frequency range, it describes the ampli er as a low-pass (midfrequency) lter with break frequencies at !1 1 C RC kRL and !2 RS gm g GE 1 C RS 6
(a) Apply the results of Section 8.6 to the small-signal equivalent circuit of Fig. 8-9(a) to determine the Miller capacitance. (b) Using the Miller capacitance, draw the associated equivalent circuit and from it nd an expression for the high-frequency voltage-gain ratio.
(a) First, the gain KF must be found with capacitor C and load resistor RL removed. VL gm Vb 0 e RC the desired gain is KF VL gm RC Vb 0 e YF 1 gm RC C s 1 Since
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