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+ RB _ hie hfe ib RC
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Example 8.6. In the circuit of Fig. 3-20, battery VS is replaced with a sinusoidal source vS . The impedance of the coupling capacitor is not negligibly small. (a) Find an expression for the voltage-gain ratio M jAv j! j jvo =vS j. (b) Determine the midfrequency gain of this ampli er. (c) Determine the low-frequency cuto point !L , and sketch an asymptotic Bode plot. (a) The small-signal low-frequency equivalent circuit is shown in Fig. 8-6. VS IS RS hie kRB 1=sC By Ohm s law, 8:24
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FREQUENCY EFFECTS IN AMPLIFIERS
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[CHAP. 8
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Then current division gives Ib But Ohm s law requires that Vo hfe RC Ib Substituting (8.25) into (8.26) and rearranging give A s hfe RC RB Cs Vo VS RB hie 1 sC RS hie kRB hfe RC RB C! q RB hie 1 !C 2 RS hie kRB 2 8:27 8:26 RB RB VS I RB hie S RB hie RS hie kRB 1=sC 8:25
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Now, with s j! in (8.27), its magnitude is M jA j! j 8:28
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(b) The midfrequency gain follows from letting s j! ! 1 in (8.27). We may do so because reactances associated with inherent capacitances have been assumed in nitely large (neglected) in the equivalent circuit. We have, then, Amid (c) From (8.27), !L 1= 1 RB hie C RS hie kRB C RS hie RB hie RB 8:30 hfe RC RB RB hie RS hie kRB 8:29
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The asymptotic Bode plot is sketched in Fig. 8-7.
Mdb 20 log |Amid| 20 Decade 0 L B ib rp (rb e) Cp (Cb e) + _
Lb e
rx (rbb )
Cm (Ccb ) ic gm Lb e
Fig. 8-7
Fig. 8-8 Hybrid-p model for the BJT
HIGH-FREQUENCY HYBRID- BJT MODEL
Because of capacitance that is inherent within the transistor, ampli er current- and voltage-gain ratios decrease in magnitude as the frequency of the input signal increases beyond the midfrequency p range. The high-frequency cuto point !H is the frequency at which the gain ratio equals 1= 2 times its midfrequency value [see Fig. 8-1(a)], or at which Mdb has decreased by 3 db from its midfrequency value. The range of frequencies above !H is called the high-frequency region. Like !L , !H is a break frequency. The most useful high-frequency model for the BJT is called the hybrid- equivalent circuit (see Fig. 8-8). In this model, the reverse voltage ratio hre and output admittance hoe are assumed negligible. The base ohmic resistance rbb 0 , assumed to be located between the base terminal B and the base junction B 0 ,
CHAP. 8]
FREQUENCY EFFECTS IN AMPLIFIERS
has a constant value (typically 10 to 50 ) that depends directly on the base width. The base-emitterjunction resistance rb 0 e is usually much larger than rbb 0 and can be calculated as rb 0 e VT 1 VT IEQ ICQ 8:31
(see Problem 6.9). Capacitance C is the depletion capacitance (see Section 2.3) associated with the reverse-biased collector-base junction; its value is a function of VBCQ . Capacitance C () C ) is the di usion capacitance associated with the forward-biased base-emitter junction; its value is a function of IEQ .
Example 8.7. Apply the hybrid- model of Fig. 8-8 to the ampli er of Fig. 3-10 to nd an expression for its voltage-gain ratio Av s valid at high frequencies. Assume Ri 0. The high-frequency hybrid-, small-signal equivalent circuit is drawn in Fig. 8-9(a). To simplify the analysis, a Thevenin equivalent circuit may be found for the network to the left of terminal pair B 0 ; E, with VTh r V r rx S r rx r rx
Cm B + +
Ls Lb e
8:32
RTh r krx
(8.33)
RB _
gm Lb e
+ LL _
_ E (a) Cm
B +
+ VTh _ _ E (b)
Lb e
gm Lb e
+ LL _
Fig. 8-9 Figure 8-9(b) shows the circuit with the Thevenin equivalent in position. Using vb 0 e and vL as node voltages and working in the Laplace domain, we may write the following two equations: Vb 0 e VTh V 0 V 0 VL be be 0 RTh 1=sC 1=sC VL V Vb 0 e gm Vb 0 e L 0 RC kRL 1=sC 8:34 8:35
The latter equation can be solved for Vb 0 e , then substituted into (8.34), and the result rearranged to give the voltage ratio VTh =VL :
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