Coupling and Bypass Capacitors Coupling capacitors (or dc blocking capacitors) are use to decouple ac and dc signals so as not to disturb the quiescent point of the circuit when ac signals are injected at the input. Bypass capacitors are used to force signal currents around elements by providing a low impedance path at the frequency. V =12V C1 and C3 are coupling capacitors CC C2 is a bypass capacitor Ω R2 R 4.3 k C → ∞ 30 kΩ C3 v RS C R3 + Q Ω 1kΩ C → ∞ 100 k vO 1 - + v - s R1 Ω R 10 k E C → ∞ 1.3 kΩ 2 Common emitter amplifier stage - Complete ac coupled circuit. Lecture 30 30 - 1 Circuit Analysis - dc & ac Equivalent Circuits DC Analysis 1. Find dc equivalent circuit. C’s replaced by open circuits and L’s replaced by short circuits. 2. Find Q-point from dc equivalent circuit using appropriate large-signal model for transis- tor. AC Analysis 3. Find ac equivalent circuit. C’s replaced by short circuits and L’s replaced by open circuits. DC voltage sources are replaced by ground connections and dc current sources by open circuits in ac equivalent circuit. 4. Replace transistor by small-signal model. 5. Analyze ac characteristics from small-signal ac equivalent circuit. 6. Combine results from #2 and #5 to get total voltages and currents in complete network. Lecture 30 30 - 2 BJT Common-Emitter Amplifier VCC=12V Ω R2 R 4.3 k C → ∞ 30 kΩ C3 v RS C R3 + Q Ω 1kΩ C → ∞ 100 k vO 1 - + v - s R1 Ω R 10 k E C → ∞ 1.3 kΩ 2 Lecture 30 30 - 3 DC Analysis VCC=12V R 4.3 kΩ 2 RC disconnected disconnected 30 kΩ v RS C R3 + Q Ω 1kΩ 100 k vO - + v - s R1 10 kΩ RE 1.3 kΩ Simplified equivalent circuit for dc analysis µ The Q-point is IC = 201.5 A, VCE = 4.30V. Lecture 30 30 - 4 AC Analysis R 4.3 kΩ 2 RC 30 kΩ vC RS Q 1kΩ R3 + Ω 100 k vO + v R - s 1 - 10 kΩ RE 1.3 kΩ Simplified equivalent circuit for ac analysis Lecture 30 30 - 5 vC RS Q 1kΩ + v + RC R3 O vs R1 R2 Ω - - 4.3 kΩ 100 k 10 kΩ 30 kΩ Thevenin’s Equivalent v Rth C Q 882 Ω + R R3 vO + v = 0.88v C - th s 4.3 kΩ 100 kΩ - Simplified ac equivalent circuit. Lecture 30 30 - 6 Rth B C Ω 882 ib + + r R R + vbe rπ o C 3 vO g v - m be 4.3 kΩ 100 kΩ - vth = 0.88vs E AC Equivalent Circuit with BJT replaced with its SS model. Final equivalent circuit of common-emitter Rth amplifier. Ω i 882 b + + R = R || R ;;R = R || R B 1 2 th S B + vbe rπ RL v - g v O v = 0.88v m be - RB th s vth = ---------------------vs RS + RB r • π vbe = vth -------------------- rπ + Rth || || The load resistance is RL = ro RC R3 Lecture 30 30 - 7 Final equivalent circuit of common-emitter Rth amplifier. Ω i 882 b + + R = R || R ;;R = R || R B 1 2 th S B + vbe rπ RL v - g v O v = 0.88v m be - RB th s vth = ---------------------vs RS + RB r • π vbe = vth -------------------- rπ + Rth || || The load resistance is RL = ro RC R3 vo gmvbeRL Voltage Gain AVth, ==------- – ----------------------- . vth vth β vo –gmrπRLRB – oRL RB AV ==----- ----------------------- --------------------- =----------------------------------------- . vs rπ + Rth RS + RB rπ + RthRS + RB v –g R 1 A ==-----o -----------------m L- • ----------------- . V v R R s 1 + --------th 1 + -------S rπ RB Lecture 30 30 - 8 Model Simplifications ()|| || Generally RS « RB and Rth « rπ , so AV ==–gmRL –gm ro RC R3 . This is the mismatched condition. Design Guide for CE BJT Amplifier || || ≈ Typically, ro » R3 and in design, R3 » RC , so ro RC R3 RC . –I R ≈ C C Therefore, AV ==–gmRL –gmRC ----------------- . VT –ζV ζ ≤≤ζ ≈≈CC ζ Putting ICRC =0VCC with 1 , we get AV ----------------- –40 VCC. VT ζ ≈ For = 0.25, we get, AV –10VCC Upper Bound for CE Voltage Gain → , → ∞ Letting RL ro and in design, R3 RC , we get µ AV ==–gmro – f . Lecture 30 30 - 9 Input Resistance of CE BJT Amplifier VCC=12V Ω R2 R 4.3 k C → ∞ 30 kΩ C3 R → ∞ vC S C1 R3 + Q Ω 1kΩ 100 k vO - + v R - s RIN 1 Ω R 10 k E C → ∞ signal 1.3 kΩ 2 source Lecture 30 30 - 10 ix ib + + + R r R R v vx B vbe rπ o C 3 O - gmvbe - R1//R2 v ()|| x || || || vx = ix RB rπ and RIN ==----- RB rπ =R1 R2 rπ ix Lecture 30 30 - 11 Output Resistance of CE BJT Amplifier VCC=12V Ω R2 R 4.3 k C → ∞ 30 kΩ C3 v RS C → ∞ C 1 R Q 3 + ROUT 1kΩ vO 100 kΩ - + v - s R1 Ω R 10 k E C → ∞ signal 1.3 kΩ 2 source Lecture 30 30 - 12 ix + R R r R + S B vbe rπ o C vx g v - R1//R2 m be vx vx ix =v------- ++----- gmvbe . No excitation at the base node, so be = 0. RC r0 v V x || ≈ ≈ CC ROUT ==----- RC r0 . ICr0 VA and ICRC ----------- . ix 2 Lecture 30 30 - 13 PSPICE EXAMPLE Vcc 5v 10k RC C3 10u R3 220k Q1 R1 C1 10u 330 RB C2 100k 10u VOFF = 0 RE VAMPL = 1m Vin FREQ = 5k 16K VEE 5v *Libraries: * Local Libraries : .LIB ".\example12.lib" * From [PSPICE NETLIST] section of C:\Program Files\OrcadLite\PSpice\PSpice.ini file: .lib "nom.lib" *Analysis directives: .TRAN05ms01u .PROBE V(*) I(*) W(*) D(*) NOISE(*) .INC ".\example12-SCHEMATIC1.net" Lecture 30 30 - 14 PSPICE EXAMPLE (Cont’d) **** INCLUDING example12-SCHEMATIC1.net **** * source EXAMPLE12 R_RC N00280 N00349 10k R_RE N00334 N00307 16K R_R3 0 N00527 220k V_Vin N00822 0 +SIN01m5k000 R_R1 N00726 N00822 330 R_RB N00656 0 100k C_C1 N00726 N00656 10u V_Vcc N00349 0 5v C_C3 N00280 N00527 10u V_VEE 0 N00334 5v C_C2 N00307 0 10u Q_Q1 N00280 N00656 N00307 Qbreakn **** RESUMING example12-SCHEMATIC1-Example12Profile.sim.cir **** .END **** BJT MODEL PARAMETERS ****************************************************************************** Qbreakn NPN IS 100.000000E-18 BF 65 NF 1 VAF 1.000000E+03 BR 1 NR 1 CN 2.42 Lecture 30 30 - 15 PSPICE EXAMPLE (Cont’d) D.87 **** INITIAL TRANSIENT SOLUTION TEMPERATURE = 27.000 DEG C ****************************************************************************** NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE (N00280) 2.6024 (N00307) -1.1050 (N00334) -5.0000 (N00349) 5.0000 (N00527) 0.0000 (N00656) -.3678 (N00726) 0.0000 (N00822) 0.0000 VOLTAGE SOURCE CURRENTS NAME CURRENT V_Vin 0.000E+00 V_Vcc -2.398E-04 V_VEE -2.434E-04 100mV 50mV 0V -50mV -100mV 0s 0.5ms 1.0ms 1.5ms 2.0ms 2.5ms 3.0ms 3.5ms 4.0ms 4.5ms 5.0ms V(R3:2) V(Vin:+) Time Lecture 30 30 - 16.