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6.1 For the CB ampli er of Fig. 3-23, nd the voltage-gain ratio Av vL =vS using the tee-equivalent small-signal circuit of Fig. 6-3.
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The small-signal circuit for the ampli er is given by Fig. 6-9. ic vcb R RL vL C RC kRL RC RL By Ohm s law, 1
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iL + LL _
Fig. 6-9 Substituting (1) into (6.27) and (6.28) gives, respectively, RC RL vL RC RL R RL vL vL vcb rc rb ie rb rc C RC RL vS veb re rb ie rb where we also made use of (6.29). Solving (2) for ie and substituting the result into (3) yield vS rb RC RL vL R RL RC RL rb rc C v re rb RC RL L 2 3
vL rc rb
The voltage-gain ratio follows directly from (4) as Av vL rc rb RC RL vS RC RL re rb RC RL 1 rc rb re rb rc
CHAP. 6]
SMALL-SIGNAL MIDFREQUENCY BJT AMPLIFIERS
Assume that rc is large enough so that ic % ie for the CB ampli er of Fig. 3-23, whose smallsignal circuit is given by Fig. 6-9. Find an expression for the current-gain ratio Ai iL =is and evaluate it if re 30 ; rb 300 ; rc 1 M; RE 5 k; RC RL 4 k, and 0:99.
Letting ic % ie in (6.27) allows us to determine the input resistance Rin : veb re rb ie rb ie v Rin eb re 1 rb ie
from which By current division at node E,
ie Solving for is gives is
RE i RE Rin s
RE Rin R re 1 rb ie E ie RE RE RC RC ie i RC RL c RC RL
Current division at node C, again with ic % ie , yields iL 2
The current gain is now the ratio of (2) to (1): Ai iL RC = RC RL RC RE is RE re 1 rb =RE RC RL RE re 1 rb
Substituting the given values results in Ai 0:99 4 103 5 103 0:492 4 103 4 103 5 103 30 1 0:99 300
The transistor of a CE ampli er can be modeled with the tee-equivalent circuit of Fig. 6-3 if the base and emitter terminals are interchanged, as shown by Fig. 6-10(a); however, the controlled source is no longer given in terms of a port current an analytical disadvantage. Show that the circuits of Fig. 6-10(b) and (c), where the controlled variable of the dependent source is the input current ib , can be obtained by application of Thevenin s and Norton s theorems to the circuit of Fig. 6-10(a).
The Thevenin equivalent for the circuit above terminals 1,2 of Fig. 6-10(a) has vth rc ie By KCL, ie ic ib , so that vth rc ic rc ib 1 Zth rc
We recognize that if the Thevenin elements are placed in the network, the rst term on the right side of (1) must be modeled by using a negative resistance. The second term represents a controlled voltage source. Thus, a modi ed Thevenin equivalent can be introduced, in which the negative resistance is combined with Zth to give
0 vth rc ib rm ib 0 Zth 1 rc
With the modi ed Thevenin elements of (2) in position, we obtain Fig. 6-10(b). The elements of the Norton equivalent circuit can be determined directly from (2) as ZN 1 0 Zth 1 rc YN IN
0 vth rc ib ib 0 Zth 1 rc
The elements of (3) give the circuit of Fig. 6-10(c).
SMALL-SIGNAL MIDFREQUENCY BJT AMPLIFIERS
[CHAP. 6
= ie
ib B + rb re ie E (a) ib B + rb 2 re ie E (b) rmib + (1 _ =) rc 1 + C 2 rc 1 ic + C
> ib
ib B + rb re ie E (c) 2 (1 _ =) rc 1 + C
Fig. 6-10
Utilize the r-parameter equivalent circuit of Fig. 6-10(b) to nd the voltage gain ratio Av vL =vi for the CE ampli er circuit of Fig. 3-10.
The small-signal equivalent circuit for the ampli er is drawn in Fig. 6-11. After nding the Thevenin equivalent for the network to the left of terminals B; E, we may write vbe
ii +
RB RB Ri v i RB Ri i RB Ri b
_ rm ib + ic (1 _ =) rc C + re ie E
Lce _
ib B + RB
iL + LL _
Fig. 6-11 Ohm s law at the output requires that vce vL RC RL i RC RL c 2
CHAP. 6]
SMALL-SIGNAL MIDFREQUENCY BJT AMPLIFIERS
Applying KVL around the B; E mesh and around the C; E mesh while noting that ie ic ib yields, respectively, vbe rb ib re ie rb re ib re ic and vce re ie rm ib 1 rc ic re rm ib 1 rc re ic 3 (4)
Equating (1) to (3) and (2) to (4) allows formulation of the system of linear equations 2  3  RB Ri RB Ri r R Ri rb re e B 6 7 i   v  RB RB Ri RB i 6 7 b  7 6 4 RC RL 5 ic 0 re rm 1 rc re RC RL from which, by Cramer s rule, ic 2 = , where     R Ri RB Ri RC RL B rb re 1 rc re re re rm RB RB Ri RC RL 2 re rm vi Then Av vL RL kRC ic R L R C re rm vi vi RL RC
The CE tee-equivalent circuit of Fig. 6-10(b) is suitable for use in the analysis of an EF ampli er if the collector and emitter branches are interchanged. Use this technique to calculate (a) the voltage-gain ratio Av vL =vB and (b) the input impedance for the ampli er of Fig. 3-26(a).
(a) The appropriate small-signal equivalent circuit is given in Fig. 6-12. By KVL around the B; C loop, with rm rc (from Problem 6.3), vB rb ib rm ib 1 rc ib ie rb rc ib 1 rc ie
B rb ib + iS + LB _ rm ib RB _ (1 _ ) rc C ic RE RL ie re E iL + LL _
Fig. 6-12 Application of KVL around the C; E loop, again with rm rc , gives   RE RL RE RL ie rc ib re 1 rc i RE RL RE RL e
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