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EKT104 ANALOG ELECTRONIC CIRCUITS [LITAR ELEKTRONIK ANALOG] BASIC BJT AMPLIFIER (PART II) DR NIK ADILAH HANIN BINTI ZAHRI [email protected] 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 Vo I b RC • Input voltage loop • Input resistance, Rib Vin I b r I b I b RE Rib • Remember: Assume VA is infinite, ro is neglected Vin r 1 RE Ib Input resistance to amplifier, Ri Ri R1 R2 Rib • Voltage divider equation of Vin to Vs Ri Vin Ri RS Vs 4 COMMON-EMITTER AMPLIFIER WITH EMITTER RESISTOR • So, small-signal voltage gain, AV Av Vo I b RC Vs Vs V 1 RC in Rib Vs Av Remember: Assume VA is infinite, ro is neglected • RC r 1 RE Ri Ri RS If Ri >> Rs and (1 + β)RE >> rπ Av RC R C 1 RE RE 5 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. CC vs R2 RS Vs R1|| R2 RE CE Hence no RE appear in the hybrid-π equivalent circuit B C r gmV ro Vo RC E Small-signal hybrid-π 6 equivalent circuit VOLTAGE GAIN WITH AND WITHOUT BYPASS CAPACITOR VCC=10V R1=56k vO RS=0.5 k CC vs Compare the Voltage gain value with and without Bypass Capacitor of the following circuit. RC=2k R2=12.2k RE= CE 0.4k 100,VA ,VBE 0.7 7 VOLTAGE GAIN WITH AND WITHOUT BYPASS CAPACITOR Voltage gain Measurement Without Bypass Capacitor VCC=10V I CQ 2.16mA, VCEQ 4.81V r R1=56k RC=2k I CQ vO I CQ RS=0.5 k CC vs VT R2=12.2k RE= 0.4k gm CE VT (100)(0.026) 1.2k 2.16 2.16 83.1mA / V 0.026 VA ro I CQ Rib r (1 ) RE 1.2 (101)(0.4) 41.6k Ri R1 || R2 || Rib 56 || 12.2 || 4.16 8.06k 100,VA ,VBE 0.7 Av (100)( 2) 8.06 4.53 8 1.2 (101)(0.4) 8.06 0.5 VOLTAGE GAIN WITH AND WITHOUT BYPASS CAPACITOR Voltage gain Measurement With Bypass Capacitor VCC=10V I CQ 2.16mA, VCEQ 4.81V R1=56k RC=2k vO RS=0.5 k CC vs r R2=12.2k RE= 0.4k VT I CQ gm ro CE I CQ VT (100)(0.026) 1.2k 2.16 2.16 83.1mA / V 0.026 VA I CQ Rib r 1.2k Ri R1 || R2 || Rib 56 || 12.2 || 1.2 1.07 k 100,VA ,VBE 0.7 Av (100)( 2) 1.07 113.6 1.2 1 . 07 0 . 5 9 STABILITY OF VOLTAGE GAIN • Stability : Measure of how well an amplifier maintains its design values over changes • Bypassing external RE does produce maximum voltage gain, however there is stability problem because ac voltage depends internal ac emitter resistance, re • Where re depends on IE and on temperature 10 COMMON-EMITTER AMPLIFIER WITH EMITTER BYPASS CAPACITOR Swamping Method • Method to minimize effect of re without reducing the voltage gain to minimum value • • RE is partially bypassed so that reasonable gain can be achieved and the effect of re on the gain can be greatly eliminated RE is formed with two separate emitter resistor, where RE2 is bypassed and RE1 is not Common-emitter amplifier with emitter bypass capacitor 11 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 12 COMMON-EMITTER AMPLIFIER WITH EMITTER BYPASS CAPACITOR Example 1: Determine the Q-point (VBE=0.7V, β=150, VA=∞) and DC & AC Load Line. Then plot the graph. 13 DC LOAD LINE SOLUTION... • KVL on C-E loop V I C RC VCE I E ( RE1 RE 2 ) V 1 1 I C ( RE1 RE 2 ) V , when I E I C I C RC VCE 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 ) Slope -1 RC RE1 RE 2 14 DC & AC LOAD LINES FULL SOLUTION 15 AC LOAD LINE ANALYSIS Example 2 Determine the dc and ac load line. VBE=0.7V, β=150, VA=∞ 16 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 17 AC LOAD LINE Small signal hybrid-π equivalent circuit I CQ 0.894mA ;VECQ 6.53V r VT gm I CQ I CQ VT VA I CQ 4.36k ro 34.4mA / V vo vec ( g m v )( RC // RL ) ic ( RC // RL ) 18 DC & AC LOAD LINES FULL SOLUTION 19 MAXIMUM SYMMETRICAL SWING • Symmetrical sinusoidal signal applied to the input of an amplifier produces an output of 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 20 MAXIMUM SYMMETRICAL SWING • Steps to design a BJT amplifier for maximum symmetrical swing: 1. Write DC load line equation (relates of ICQ & VCEQ) 2. Write AC load line equation (relates ic, vce ; vce = icReq, Req = effective ac resistance in C-E circuit) 3. Generally, ic = ICQ – IC(min), where IC(min) = 0 or some other specified min collector current 4. Generally, vce = VCEQ – VCE(min), where VCE(min) is some specified min C-E voltage 5. Combination of the above equations produce optimum ICQ & VCEQ values to obtain maximum symmetrical swing in output signal 21 MAXIMUM SYMMETRICAL SWING Example 3 Determine the maximum symmetrical swing in the output voltage of the following circuit (same as Example 2). 22 MAXIMUM SYMMETRICAL SWING 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: ic 2( I CQ I C (min)) 2(0.894) 1.79 mA • Maximum instantaneous collector current: | vce || ic | Req | ic | ( RC || RL ) (1.79)(5 || 2) 2.56 V iC I CQ 1 | ic | 0.894 0.894 1.79 mA 2 23 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. 24 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 25