c° Copyright 2008. W. Marshall Leach, Jr., Professor, Georgia Institute of Tec hnology, School of Electricaland Computer Engineering.Complemen tary CC AmplifierFigure 1 shows a complementary common-collector stage. This is commonly used as the final output stagein op amps and audio power amplifiers. Compared to a non-complemen tary CC amplifier, it can supplylarge positive and negative load curren ts with low power dissipation in the absence of a signal. The npn BJTsupplies positive load current while the pnp BJT supplies negative load current.Figure 1: Complementary common-collector amplifier.Let us first examine the performance of the circuit with VB=0.ForvI=0, both transistors are cut off.In order to obtain a positive output voltage, vImust be increased until Q1turns on. Denote the turn-onvoltage for Q1by Vγ. Similarly, denote the turn-on voltage for Q2by −Vγ.When−Vγ<vI<Vγ,bothtransistors are off and there is very little output voltage. For vI>Vγ, Q1turns on and vOgoes positive.For vI< −Vγ, Q2turns on and vOgoes negative. The plot of vOversus vIwould resemble curve a in Fig.2. A sine wave applied to the circuit would exhibit distortion in the crossover range for −Vγ<vI<Vγasis illustrated by curve a in Fig. 3. The distortion in the wav eform is called crossover distortion or centerclipping.For VB> 0, a positive bias voltage is applied to the base of Q1and a negative bias voltage is applied tothebaseofQ2.AsVBis increased, both transistors turn on and emitter currents flow that are given byIE1= IE2=2VB− VBE1− VBE2RE1+ RE2The bias voltage causes the portion of curve a in Fig. 2 for vI> 0 to be shifted to the left and the portionfor vI< 0 to be shifted to the right. The effective sum curve changes into approximately a straight line asshownincurveb in Fig. 2. This eliminates the crossover distortion in the output waveform in shown bycurve b in Fig. 3.Once the transistors are turned on, the emitter currents are extremely sensitive to the value of VB.Toreduce this sensitivity, resistors are often used in series with the emitters as shown in Fig. 1. If an excessiveemitter current flows, the voltage drops across RE1and RE2cause VBE1and VEB2to decrease, causing thecurrent to decrease. For minimum power dissipation in these resistors, their value must be much smallerthan that of RL. In the design of op amps, the emitter resistors are usually omitted. In this case, the value1Figure 2: Plots of vOversus vI.Curvea-VB=0.Curveb-VBadjusted to eliminate the deadband region.Figure 3: Sine wave (a)withand(b) without crossover distortion.2of VBis chosen to bias the transistors just below cutoff. Although a small amount of crossover remains, itis minimized by the negative feedback that is used in the application of the op amps.The circuit that is commonly used to bias a common-collector stage is the VBEmultiplier. Fig. 4 showsasimpleVBEmultiplier consisting of transistor Q3connected between the bases of Q1and Q2. The v oltageacross the multiplier is given byVB= I1R1+ VBE3IfthebasecurrentinQ3is small compared to I1,wecanwriteI1= VBE3/R2. When this is substituted intothe equation for VB,weobtainVB=VBE3R2R1+ VBE3= VBE3µ1+R1R2¶It follows that the voltage VBcan be set by proper c hoice of the ratio R1/R2.Figure 4: Complementary common-collector amplifier with a VBEmultiplier bias circuit.The exact equation for VBcan be obtained by writing a node equation at the base node of Q3.Theequation isVB− VBER1=VBER2+1βµI −VB− VBER1¶ThiscanbesolvedforVBto obtainVB= VBEµα +R1R2¶+IR11+βwhere α = β/ (1 + β) has been used. This agrees with the approximate solution above if β is large andIR1¿
View Full Document
Unlocking...