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barcode fonts for ssrs Mdb MdbM iS MdbL 0 RS C B ib C ic in Software
Mdb MdbM iS MdbL 0 RS C B ib C ic Code 3 Of 9 Decoder In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. Code 3 Of 9 Creator In None Using Barcode drawer for Software Control to generate, create Code 3/9 image in Software applications. + RB _ hie hfe ib RC
Decode Code 39 Full ASCII In None Using Barcode recognizer for Software Control to read, scan read, scan image in Software applications. Paint Code 39 Extended In Visual C#.NET Using Barcode generation for .NET Control to generate, create Code 3/9 image in .NET applications. Fig. 8-5 Code 39 Generator In .NET Framework Using Barcode creator for ASP.NET Control to generate, create Code 39 Extended image in ASP.NET applications. Making Code 3 Of 9 In .NET Using Barcode generator for Visual Studio .NET Control to generate, create Code39 image in .NET framework applications. Fig. 8-6 ANSI/AIM Code 39 Creation In VB.NET Using Barcode maker for .NET framework Control to generate, create USS Code 39 image in VS .NET applications. Encoding EAN 128 In None Using Barcode generator for Software Control to generate, create GTIN - 128 image in Software applications. 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 Creating DataMatrix In None Using Barcode generation for Software Control to generate, create Data Matrix 2d barcode image in Software applications. Painting Code 39 In None Using Barcode creator for Software Control to generate, create Code 3/9 image in Software applications. FREQUENCY EFFECTS IN AMPLIFIERS
Encode Code 128B In None Using Barcode creator for Software Control to generate, create Code 128A image in Software applications. Generate Barcode In None Using Barcode generator for Software Control to generate, create bar code image in Software applications. [CHAP. 8
Rationalized Codabar Maker In None Using Barcode generation for Software Control to generate, create ANSI/AIM Codabar image in Software applications. Barcode Reader In Visual Studio .NET Using Barcode scanner for .NET framework Control to read, scan read, scan image in .NET framework applications. 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 Generating Bar Code In None Using Barcode creation for Font Control to generate, create bar code image in Font applications. Making Code39 In C# Using Barcode maker for Visual Studio .NET Control to generate, create USS Code 39 image in .NET applications. Now, with s j! in (8.27), its magnitude is M jA j! j 8:28 Making Bar Code In Visual Studio .NET Using Barcode printer for Visual Studio .NET Control to generate, create bar code image in .NET applications. Decode EAN / UCC - 13 In Visual C#.NET Using Barcode decoder for .NET framework Control to read, scan read, scan image in .NET applications. (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 ANSI/AIM Code 128 Creator In None Using Barcode drawer for Microsoft Word Control to generate, create Code 128C image in Word applications. EAN128 Printer In .NET Using Barcode creation for Visual Studio .NET Control to generate, create GS1 128 image in Visual Studio .NET applications. 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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