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IC (mA) 50 IB = 230 A 190 A
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286 22 153 0 Q 150 A 110 A 75 A 0 5 10 15 V CE (V)
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Figure 916 Ampli cation of sinusoidal oscillations in a BJT
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The i-v characteristic of Figure 916 illustrates how an increase in collector current follows the same sinusoidal pattern of the base current but is greatly ampli ed Thus, the BJT acts as a current ampli er, in the sense that any oscillations in the base current appear ampli ed in the collector current Since the voltage across the collector resistance, RC , is proportional to the collector current, one can see how the collector voltage is also affected by the ampli cation process Example 95 illustrates numerically the effective ampli cation of the small AC signal that takes place in the circuit of Figure 915
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EXAMPLE 95 A BJT Small-Signal Ampli er
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With reference to the BJT ampli er of Figure 917 and to the collector characteristic curves of Figure 913, determine: (1) the DC operating point of the BJT; (2) the nominal current gain, , at the operating point; (3) the AC voltage gain AV = Vo / VB
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Known Quantities: Base, collector, and emitter resistances; base and collector supply voltages; collector characteristic curves; BE junction offset voltage
iB (t)
iC (t) + C + vo (t) vCE (t)
voltage, VCEQ ; (2) =
Find: (1) DC (quiescent) base and collector currents, IBQ and ICQ , and collector-emitter
IC / IB ; (3) AV =
Vo / V B
+ ~ _
Schematics, Diagrams, Circuits, and Given Data: RB = 10 k ; RC = 375 ; VBB = 21 V; VCC = 15 V; V = 06 V The collector characteristic curves are shown in Figure 913 Assumptions: Assume that the BE junction resistance is negligible when compared to the base resistance Assume that each voltage and current can be represented by the superposition of a DC (quiescent) value and an AC component For example: v0 (t) = V0Q + V0 (t) Analysis:
VB VBB
vBE (t)
VCC E
1 DC operating point On the assumption the BE junction resistance is much smaller than RB , we can state that the junction voltage is constant: vBE (t) = VBEQ = V , and plays a role only in the DC circuit The DC equivalent circuit for the base is shown in Figure 918 and described by the equation VBB = RB IBQ + VBEQ from which we compute the quiescent base current: IBQ = VBB VBEQ VBB V 21 06 = 150 A = = RB RB 10,000
21 V
10 k IB 06 V
To determine the DC operating point, we write the load line equation for the collector circuit: VCE = VCC RC IC = 15 375IC The load line is shown in Figure 919 The intersection of the load line with the 150 A base curve is the DC operating or quiescent point of the transistor ampli er, de ned below by the three values: VCEQ = 72 V; ICQ = 22 mA; IBQ = 150 A
IC (mA) 50 IB = 230 A
190 A
286 22 153 0
150 A 110 A 75 A 35 A
15 VCE (V)
Figure 919 Operating point on the characteristic curve
2 AC current gain To determine the current gain, we resort, again, to the collector curves Figure 919 indicates that if we consider the values corresponding to base currents of 190 and 110 A, the collector will see currents of 286 and 153 mA,
9
Transistor Fundamentals
respectively We can think of these collector current excursions, IC , from the Q point as corresponding to the effects of an oscillation IB in the base current, and calculate the current gain of the BJT ampli er according to: = IC 286 10 3 153 10 3 = = 16625 IB 190 10 6 110 10 6
Thus, the nominal current gain of the transistor is approximately = 166 3 AC voltage gain To determine the AC voltage gain, AV = Vo / VB , we need to express Vo as a function of VB Observe that vo (t) = RC iC (t) = RC ICQ + RC IC (t) Thus we can write: Vo (t) = RC IC (t) = RC IB (t)
Using the principle of superposition in considering the base circuit, we observe that IB (t) can be computed from the KVL base equation VB (t) = RB IB (t) + VBE (t)
but we had stated in part 1 that, since the BE junction resistance is negligible relative to RB , VBE (t) is also negligible Thus, IB = VB RB Vo (t), we can write RC VB (t) RB
Substituting this result into the expression for Vo (t) = RC IB (t) = or RC Vo (t) = AV = = 623 VB RB
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