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ANALOG ELECTRONIC CIRCUITS 1 EKT 104 Basic BJT Amplifiers (Part 2) 1 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 2 Common-Emitter Amplifier with Emitter Resistor inside transistor CE amplifier with emitter resistor Small-signal equivalent circuit (with current gain parameter, β) 3 Common-Emitter Amplifier with Emitter Resistor ac output voltage Input voltage loop Input resistance, Rib Vo I b RC Vin I b r I b I b RE Vin Rib r 1 RE Ib Input resistance to amplifier, Ri Ri R1 R2 Rib Remember: Assume VA is infinite, ro is neglected Voltage divider equation of Vin to Vs Ri Vin Ri RS Vs 4 Common-Emitter Amplifier with Emitter Resistor Cont.. So, small-signal voltage gain, AV Av Vo I b RC Vs Vs Vin 1 RC Rib Vs RC Av r 1 RE Remember: Assume VA is infinite, ro is neglected Ri Ri RS If Ri >> Rs and (1 + β)RE >> rπ RC RC Av `1 1 RE RE 5 6 7 Common-Emitter Amplifier with Emitter Bypass Capacitor Sometimes for the purpose of dc biased, emitter resistor set to be large. Hence degrades the small-signal voltage gain too severely. Apply an emitter bypass capacitor to effectively short out a portion or all of the emitter resistance as seen by the ac signals. 8 Common-Emitter Amplifier with Emitter Bypass Capacitor VCC Emitter bypass capacitor, CE provides a short circuit to ground for the ac signals RC R1 vO RS 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 CC vs R2 RS Vs R1|| R2 RE CE B C r gmV ro Vo RC E Small-signal hybrid-π equivalent circuit 9 Common-Emitter Amplifier with Emitter Bypass Capacitor 10 Common-Emitter Amplifier with Emitter Bypass Capacitor 11 Common-Emitter Amplifier with Emitter Bypass Capacitor 12 Common-Emitter Amplifier with Emitter Bypass Capacitor 13 Common-Emitter Amplifier with Emitter Bypass Capacitor 14 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 15 Common Emitter Amplifier with emitter bypass capacitor Example 1 Common-emitter amplifier with emitter bypass capacitor 16 DC Load Line Solution... KVL on C-E loop V I C RC VCE I E ( RE1 RE 2 ) V I C RC VCE 1 1 I C ( RE1 RE 2 ) V , when I E I C 1 I C ( RE1 RE 2 ) V V VCE I C RC 1 1 For Q - point, when 1, So, V V VCEQ I CQ ( RC RE1 RE 2 ) -1 Slope RC RE1 RE 2 17 AC Load Line Solution... KVL on C-E loop ic RC vce ie RE1 0 Assuming ic ie vce ic RC ic RE1 ic ( RC RE1 ) -1 Slope RC RE1 AC equivalent circuit 18 DC & AC Load Lines Full solution 19 AC LOAD LINE ANALYSIS Determine the dc and ac load line. VBE=0.7V, β=150, VA=∞ Example 2 20 DC Load Line To determine dc Q-point, KVL around B-E loop V I BQ RB VEB I E RE I BQ RB VEB (1 ) I BQ RE I BQ V VEB 5.96 A RB (1 ) RE Then I CQ I BQ 0.894mA & I EQ (1 ) I BQ 0.9mA For Q - point, VCEQ (V V ) I CQ RC I EQ RE 6.53 1 -1 Slope RC RE 15k 21 AC Load Line Small signal hybrid-π equivalent circuit I CQ 0.894mA ;VECQ 6.53V r VT gm I CQ I CQ VT 4.36k 34.4mA / V VA ro I CQ vo vec ( g m v )( RC // RL ) ic ( RC // RL ) 22 DC & AC Load lines Full solution 23 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 24 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 25 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-topeak ac collector current: ic 2( I CQ I C (min)) 2(0.894) 1.79 mA The max. symmetrical peak-to-peak output voltage: | vce || ic | Req | ic | ( RC || RL ) (1.79)(5 || 2) 2.56 V Maximum instantaneous collector current: iC I CQ 1 | ic | 0.894 0.894 1.79 mA 2 26 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. 27 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 28