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ANALOG ELECTRONIC CIRCUITS 1. EKT 204 Basic BJT Amplifiers (Part 2). Basic Common-Emitter Amplifier. The basic common-emitter circuit used in previous analysis causes a serious defect : If BJT with V BE =0.7 V is used, I B =9.5 μA & I C =0.95 mA
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ANALOG ELECTRONIC CIRCUITS 1 EKT 204 Basic BJT Amplifiers (Part 2)
Basic Common-Emitter Amplifier • The basic common-emitter circuit used in previous analysis causes a serious defect : • If BJT with VBE=0.7 V is used, IB=9.5 μA & IC=0.95 mA • But, if new BJT with VBE=0.6 V is used, IB=26 μA & BJT goes into saturation; which is not acceptable Previous circuit is not practical • So, the emitter resistor is included: Q-point is stabilized against variations in β, as will the voltage gain, AV • Assumptions • CC acts as a short circuit • Early voltage = ∞ ==> ro neglected due to open circuit
Common-Emitter Amplifier with Emitter Resistor inside transistor CE amplifier with emitter resistor Small-signal equivalent circuit (with current gain parameter, β)
Common-Emitter Amplifier with Emitter Resistor • ac output voltage • Input voltage loop • Input resistance, Rib • Input resistance to amplifier, Ri • Voltage divider equation of Vin to Vs Remember: Assume VA is infinite, ro is neglected
Common-Emitter Amplifier with Emitter Resistor Cont.. • So, small-signal voltage gain, AV • If Ri >> Rs and (1 + β)RE >> rπ Remember: Assume VA is infinite, ro is neglected
VCC RC R1 vO RS CC vs R2 RE CE RS Vo Vs R1|| R2 r gmV ro RC Common-Emitter Amplifier with Emitter Bypass Capacitor Emitter bypass capacitor, CE provides a short circuit to ground for the ac signals Emitter bypass capacitor is used to short out a portion or all of emitter resistance by the ac signal. Hence no RE appear in the hybrid-π equivalent circuit Small-signal hybrid-π equivalent circuit
DC & AC LOAD LINE ANALYSIS • DC load line • Visualized the relationship between Q-point & transistor characteristics • AC load line • Visualized the relationship between small-signal response & transistor characteristics • Occurs when capacitors added in transistor circuit
Common Emitter Amplifier with emitter bypass capacitor Example 1 Common-emitter amplifier with emitter bypass capacitor
DC Load Line Solution... • KVL on C-E loop
AC Load Line Solution... • KVL on C-E loop AC equivalent circuit
DC & AC Load Lines Full solution
AC LOAD LINE ANALYSIS Determine the dc and ac load line. VBE=0.7V, β=150, VA=∞ Example 2
DC Load Line • To determine dc Q-point, KVL around B-E loop
AC Load Line Small signal hybrid-π equivalent circuit
DC & AC Load lines Full solution
Maximum Symmetrical Swing • When symmetrical sinusoidal signal applied to the input of an amplifier, the output generated is also a symmetrical sinusoidal signal • AC load line is used to determine maximum output symmetrical swing • If output is out of limit, portion of the output signal will be clipped & signal distortion will occur
Maximum Symmetrical Swing • Steps to design a BJT amplifier for maximum symmetrical swing: • Write DC load line equation (relates of ICQ & VCEQ) • Write AC load line equation (relates ic, vce ; vce = - icReq, Req = effective ac resistance in C-E circuit) • Generally, ic = ICQ – IC(min), where IC(min) = 0 or some other specified min collector current • Generally, vce = VCEQ – VCE(min), where VCE(min) is some specified min C-E voltage • Combination of the above equations produce optimum ICQ & VCEQ values to obtain maximum symmetrical swing in output signal
Maximum Symmetrical Swing Example 3 Determine the maximum symmetrical swing in the output voltage of the circuit given in Example 2. Solution: • From the dc & ac load line, the maximum negative swing in the Ic is from 0.894 mA to zero (ICQ). So, the maximum possible peak-to-peak ac collector current: • The max. symmetrical peak-to-peak output voltage: • Maximum instantaneous collector current:
Self-Reading • Textbook: Donald A. Neamen, ‘MICROELECTRONICS Circuit Analysis & Design’,3rd Edition’, McGraw Hill International Edition, 2007 • Chapter 6: Basic BJT Amplifiers • Page: 397-413, 415-424.
Exercise • Textbook: Donald A. Neamen, ‘MICROELECTRONICS Circuit Analysis & Design’,3rd Edition’, McGraw Hill International Edition, 2007 • Exercise 6.5, 6.6, 6.7,6.9 • Exercise 6.10 , 6.11