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which can be solved to obtain 1 IB = VB RB + hie + (hf e + 1)RE and VC = IC RC = hf e IB RC = hf e RC RB + hie + (hf e + 1)RE VB (107) (106)
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Figure 105 DC equivalent circuit for the BJT ampli er of Figure 104
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Then, the AC/open-loop voltage gain of the ampli er is given by the expression = hf e RC VC = VB RB + hie + (hf e + 1)RE (108)
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You may recall that the open-loop voltage gain was introduced in Section 91 and Example 91 The small-signal model for the BJT will be further explored in the next section
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Transconductance In addition to the h parameters described above, another useful small-signal transistor parameter is the transfer conductance, or transconductance The transconductance of a bipolar transistor is de ned as the local slope of the collector current base-emitter voltage curve: iC gm = vBE (109)
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E RE
and it can be expressed in terms of the h parameters if we observe that we can write hf e iC iC iB gm = = = vBE iB vBE hie (1010)
AC equivalent circuit RB B IB hie
+ ~ VB
C hfe IB
It can be shown that the expression for the transconductance can be approximated by mA ICQ ICQ gm = = 39ICQ = kT 0026 V q (1011)
E RE
Small-signal model
at room temperature, where k is Boltzmann s constant, T is the temperature in degrees Kelvin, and q the electron charge (see 8 for a review of the pn junction equation) Now, the transconductance is an important measure of the voltage ampli cation of a BJT ampli er, because it relates small oscillations in the base-emitter junction voltage to the corresponding oscillations in the collector current We shall see in the next sections that this parameter can be related to the voltage gain of the transistor Note that the approximation of equation 1011 suggests that the transconductance parameter is dependent on the operating point
Figure 106 AC equivalent circuit and small-signal model for the ampli er of Figure 104
EXAMPLE 101 Determining the AC Open-Loop Voltage Gain of a Common-Emitter Ampli er
Problem
Determine the Q point and AC open-loop voltage gain of the ampli er of Figure 104; the ampli er employs a 2N5088 npn transistor
Solution
Known Quantities: Ampli er supply voltages; base, collector, and emitter resistances; h parameters; AC circuit model of Figure 103 Find: Quiescent values of IB , IC , and VCE ; open-loop AC voltage gain,
RB = 100 k ; RC = 500 hoe = 150 S
Schematics, Diagrams, Circuits, and Given Data: VBB = 6 V; VCC = 12 V;
; RE = 100
; V = 06 V hf e = 350; hie = 14 k ;
Assumptions: Use the linear small-signal h-parameter model of the BJT
10
Transistor Ampli ers and Switches
Analysis:
1 Q-point calculation (also see Example 94) We rst write the collector circuit equation by applying KVL: VCC = VCE + RC IC + RE IE = VCE + RC IC + RE (IB + IC ) VCE + (RC + RE )IC where the emitter current has been approximately set equal to the collector current since the current gain is large and IC IB Next, we write the base circuit equation, also via KVL: VBB = RB IB + VBE + RE IE = (RB + ( + 1)RE )IB + V The above equation can be solved numerically (with hf e = ) to obtain: IB = Then, IC = IB = 350 40 10 6 = 14 mA and VCE = VCC (RC + RE )IC = 12 600 14 10 3 = 36 V Thus, the Q point for the ampli er is: IBQ = 40 A ICQ = 14 mA VCEQ = 36 V VBB V 6 06 = = 40 A RB + ( + 1) RE 100 103 + 351 100
con rming that the transistor is indeed in the active region 2 AC open-loop gain calculation The AC open-loop gain for the ampli er of Figure 94 can be computed by using equation 107: = 350 500 103 hf e RC = 100 103 + 14 103 + 351 100 RB + hie + hf e + 1 RE V V
= 128
Comments: You may wish to examine the data sheets for the 2N5088 npn transistor They are available in electronic form in the accompanying CD-ROM Look for the values of the parameters used in this example Note that if the parameter hie were neglected in the expression for , the answer would not change signi cantly This is true in this particular case because of the large values of the base resistor and of hf e , but is not true in the general case
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