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Note the negative sign, due to the direction of the controlled current source This sign reversal is a typical characteristic of the common-emitter ampli er Knowing the open-circuit output voltage and the short-circuit output current, we can nd the output resistance of the ampli er as the ratio of these two quantities: ro = vout 1 = RC iout hoe (1024)
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Next, if we de ne the AC open-circuit voltage gain of the ampli er, , by the expression = vout vin (1025)
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(1026)
Figure 1021 Simpli ed equivalent circuit for the common-emitter ampli er
At this point, it is possible to take advantage of the equivalent-circuit representation of Figure 108, with the expressions for , ro , and ri just given Figure 1021 illustrates the equivalent circuit for the common-emitter ampli er This circuit replaces the transistor ampli er and enables us to calculate the actual voltage and current gain of the ampli er for any given load-and-source pair Referring to
Part II
Electronics
the circuit of Figure 108 and to equations 1012 and 1013, we can compute these gains to be AV = ri vL RL = vS ro + R L R S + r i RC hoe RL ri hie ro + RL RS + ri (1027)
= hfe = gm and AI =
RL ri ro + R L R S + r i
ri + R S iout vout /RL = = AV iin vS /(ri + RS ) RL
(1028)
Example 105 provides numerical values for a typical small-signal common-emitter ampli er
EXAMPLE 105 Common-Emitter Ampli er Analysis
Problem
Compute the input and output resistances and the voltage, current and power gains of the common-emitter ampli er of Figure 109
Solution
Known Quantities: Ampli er supply voltages; base, collector, and emitter resistances; source and load resistances; transistor parameters Find: ri ; ro ; ;
AV =
vout vS
AI =
iout iS
AP =
Pout PS
Schematics, Diagrams, Circuits, and Given Data: hie = 1,400 ; hfe = 100; hoe = 125 S R1 = 20 k ; R2 = 5 k ; RC = 4 k ; RE = 1 k ; RL = 500 ; RS = 50 Assumptions: A suitable Q point has already been established The coupling capacitors have appropriately been selected to separate the AC circuit from the DC bias circuit Analysis: We replace the BJT in the circuit of Figure 109 with the BJT h-parameter circuit model of Figure 1020 The resulting AC equivalent circuit is shown in Figure 1022, where RB = R1 ||R2 = 4 k The input resistance is given by the parallel combination of hie and RB , as stated in equation 1021:
ri = hie ||RB = 4,000||1,400 = 104 k To determine the output resistance, we short circuit the voltage source, vS , leading to vin = 0 as well Thus, ro = RC || 1 4 8 = 267 k = hoe 4+8
10
Transistor Ampli ers and Switches
RS + vS vin _
IB hie h fe IB
C + 1 hoe RC RL
vout
_ E ri ro
The open-circuit voltage gain of the ampli er is found by using equation 1026: = hfe V 4,000 RC = 286 = 100 hie 1,400 V
RS + vS vin _ ri
ro +
With the above three parameters in place, we can build the equivalent AC circuit model of the ampli er shown in Figure 1023 Then we can use equations 1027 and 1028 to compute the actual ampli er gains: AV = vout RL 05 ri 104 V = = 286 = 43 vS ro + R L RS + r i 267 + 05 005 + 104 V vout iout RS + ri A R AI = = vL = AV = 94 S iin RL A R +r
in v + _ RL
vout _
Pout iout vout W AP = = = AI AV = 4,042 PS iin vS W
Comments: Although the open-loop voltage gain, , is very large, the actual gain of the
ampli er is signi cantly smaller You may have already noted that it is the small value of load resistance as compared to the AC output resistance of the ampli er that causes AV to be smaller than The current gain is actually quite large because of the relatively large input resistance of the ampli er A word of warning is appropriate with regard to the power gain While such a large power gain may seem a desirable feature, one always has to contend with the allowable power dissipation of the transistor For example, the data sheets for the 2N3904 general purpose npn transistor show a maximum power dissipation of 350 mW at room temperature for the TO-236 package, and of 600 mW for the TO-92 package Thus, a power gain of 4,000 can be meaningful only if the input signal power is very low (less than 150 W), or else the transistor will exceed its thermal rating and will be destroyed
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