Chap 4 BJT transistors Widely used in amplifier

Chap. 4 BJT transistors • Widely used in amplifier circuits • Formed by junction of 3 materials • npn or pnp structure ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 0

ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 1

pnp transistor ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 2

ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 3

ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 4

Operation of npn transistor Large current ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 5

Modes of operation of a BJT transistor Mode BE junction cutoff reverse biased linear(active) forward biased reverse biased saturation forward biased ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 BC junction 6

Summary of npn transistor behavior npn IC base IB collector large current + VBE small current - emitter IE ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 7

Summary of pnp transistor behavior pnp IC base IB collector large current + VBE small current - emitter IE ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 8

Summary of equations for a BJT IE IC IC = b. IB b is the current gain of the transistor 100 VBE = 0. 7 V(npn) VBE = -0. 7 V(pnp) ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 9

4. 5 Graphical representation of transistor characteristics IC IB Output circuit Input circuit IE ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 10

Input characteristics IB IB VBE 0. 7 V • Acts as a diode • VBE 0. 7 V ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 11

Output characteristics IC IC IB = 40 m. A IB = 30 m. A IB = 20 m. A IB = 10 m. A VCE Early voltage Cutoff region • At a fixed IB, IC is not dependent on VCE • Slope of output characteristics in linear region is near 0 (scale exaggerated) ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 12

Biasing a transistor • We must operate the transistor in the linear region. • A transistor’s operating point (Q-point) is defined by IC, VCE, and IB. ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 13

4. 6 Analysis of transistor circuits at DC For all circuits: assume transistor operates in linear region write B-E voltage loop write C-E voltage loop Example 4. 2 B-E junction acts like a diode VE = VB - VBE = 4 V - 0. 7 V = 3. 3 V IC IE = (VE - 0)/RE = 3. 3/3. 3 K = 1 m. A IC IE = 1 m. A IE VC = 10 - ICRC = 10 - 1(4. 7) = 5. 3 V ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 14

Example 4. 6 B-E Voltage loop b = 100 5 = IBRB + VBE, solve for IB IB = (5 - VBE)/RB = (5 -. 7)/100 k = 0. 043 m. A IC IB IE IC = b. IB = (100)0. 043 m. A = 4. 3 m. A VC = 10 - ICRC = 10 - 4. 3(2) = 1. 4 V ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 15

Exercise 4. 8 VE = 0 -. 7 = - 0. 7 V b = 50 IE = (VE - -10)/RE = (-. 7 +10)/10 K = 0. 93 m. A IC IC IE = 0. 93 m. A IB = IC/b =. 93 m. A/50 = 18. 6 m. A IB IE VC = 10 - ICRC = 10 -. 93(5) = 5. 35 V VCE = 5. 35 - -0. 7 = 6. 05 V ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 16

Summary of npn transistor behavior npn IC base IB collector large current + VBE small current - emitter IE ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 17

Summary of equations for a BJT IE IC IC = b. IB b is the current gain of the transistor 100 VBE = 0. 7 V(npn) VBE = -0. 7 V(pnp) ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 18

Prob. 4. 32 • Use a voltage divider, RB 1 and RB 2 to bias VB to avoid two power supplies. • Make the current in the voltage divider about 10 times IB to simplify the analysis. Use VB = 3 V and I = 0. 2 m. A. (a) RB 1 and RB 2 form a voltage divider. Assume I >> IB I = VCC/(RB 1 + RB 2) I . 2 m. A = 9 /(RB 1 + RB 2) IB AND VB = VCC[RB 2/(RB 1 + RB 2)] 3 = 9 [RB 2/(RB 1 + RB 2)], Solve for RB 1 and RB 2. RB 1 = 30 KW, and RB 2 = 15 KW. ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 19

Prob. 4. 32 Find the operating point • Use the Thevenin equivalent circuit for the base • Makes the circuit simpler • VBB = VB = 3 V • RBB is measured with voltage sources grounded • RBB = RB 1|| RB 2 = 30 KW || 15 KW =. 10 KW ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 20

Prob. 4. 32 Write B-E loop and C-E loop B-E loop VBB = IBRBB + VBE +IERE C-E loop B-E loop C-E loop VCC = ICRC + VCE +IERE Solve for, IC, VCE, and IB. This is how all DC circuits are analyzed and designed! ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 21

Example 4. 8 • 2 -stage amplifier, 1 st stage has an npn transistor; 2 nd stage has an pnp transistor. IC = b. IB IC IE VBE = 0. 7(npn) = -0. 7(pnp) b = 100 Find IC 1, IC 2, VCE 1, VCE 2 • Use Thevenin circuits. ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 22

Example 4. 8 • RBB 1 = RB 1||RB 2 = 33 K • VBB 1 = VCC[RB 2/(RB 1+RB 2)] VBB 1 = 15[50 K/150 K] = 5 V Stage 1 IB 1 IE 1 • B-E loop VBB 1 = IB 1 RBB 1 + VBE +IE 1 RE 1 Use IB 1 IE 1/ b 5 = IE 133 K / 100 +. 7 + IE 13 K IE 1 = 1. 3 m. A ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 23

Example 4. 8 C-E loop neglect IB 2 because it is IB 2 << IC 1 IE 1 VCC = IC 1 RC 1 + VCE 1 +IE 1 RE 1 15 = 1. 3(5) + VCE 1 +1. 3(3) VCE 1= 4. 87 V ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 24

Example 4. 8 Stage 2 • B-E loop IE 2 VCC = IE 2 RE 2 + VEB +IB 2 RBB 2 + VBB 2 15 = IE 2(2 K) +. 7 +IB 2 (5 K) + 4. 87 + 1. 3(3) Use IB 2 IE 2/ b, solve for IE 2 IB 2 IE 2 = 2. 8 m. A ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 25

Example 4. 8 Stage 2 • C-E loop IE 2 VCC = IE 2 RE 2 + VEC 2 +IC 2 RC 2 15 = 2. 8(2) + VEC 2 + 2. 8 (2. 7) solve for VEC 2 IC 2 VCE 2 = 1. 84 V ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 26

Summary of DC problem • Bias transistors so that they operate in the linear region B-E junction forward biased, C-E junction reversed biased • Use VBE = 0. 7 (npn), IC IE, IC = b. IB • Represent base portion of circuit by the Thevenin circuit • Write B-E, and C-E voltage loops. • For analysis, solve for IC, and VCE. • For design, solve for resistor values (IC and VCE specified). ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 27

4. 7 Transistor as an amplifier • Transistor circuits are analyzed and designed in terms of DC and ac versions of the same circuit. • An ac signal is usually superimposed on the DC circuit. • The location of the operating point (values of IC and VCE) of the transistor affects the ac operation of the circuit. • There at least two ac parameters determined from DC quantities. ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 28

Transconductance IB ac output signal DC output signal A small ac signal vbe is superimposed on the DC voltage VBE. It gives rise to a collector signal current ic, superimposed on the dc current IC. The slope of the ic - v. BE curve at the bias point Q is the transconductance gm: the amount of ac current produced by an ac voltage. ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 (DC input signal 0. 7 V) ac input signal 29

Transconductance ac output signal Transconductance = slope at Q point gm = dic/dv. BE|ic = ICQ DC output signal where IC = IS[exp(-VBE/VT)-1]; the equation for a diode. gm = ISexp(-VBE/VT) (1/VT) gm IC/VT (A/V) DC input signal (0. 7 V) ac input signal ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 30

ac input resistance of transistor ac output signal IB DC output signal ac input resistance 1/slope at Q point rp = dv. BE/dib|ic = ICQ rp VT /IB DC input signal (0. 7 V) ac input signal re VT /IE ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 31

4. 8 Small-signal equivalent circuit models • ac model • Hybrid-p model • They are equivalent • Works in linear region only ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 32

Steps to analyze a transistor circuit 1 DC problem Set ac sources to zero, solve for DC quantities, IC and VCE. 2 Determine ac quantities from DC parameters Find gm, rp, and re. 3 ac problem Set DC sources to zero, replace transistor by hybrid-p model, find ac quantites, Rin, Rout, Av, and Ai. ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 33

Example 4. 9 Find vout/vin, (b = 100) DC problem Short vi, determine IC and VCE B-E voltage loop 3 = IBRB + VBE IB = (3 -. 7)/RB = 0. 023 m. A C-E voltage loop VCE = 10 - ICRC VCE = 10 - (2. 3)(3) VCE = 3. 1 V Q point: VCE = 3. 1 V, IC = 2. 3 m. A ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 34

Example 4. 9 b c + vbe - e + vout - ac problem Short DC sources, input and output circuits are separate, only coupled mathematically gm = IC/VT = 2. 3 m. A/25 m. V = 92 m. A/V rp = VT/ IB = 25 m. V/. 023 m. A = 1. 1 K vbe= vi [rp / (100 K + rp)] = 0. 011 vi vout = - gm vbe. RC vout = - 92 (0. 011 vi)3 K vout/vi = -3. 04 ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 35

Exercise 4. 24 (a) Find VC, VB, and VE, given: b = 100, VA = 100 V IE = 1 m. A IB IE/b = 0. 01 m. A VB = 0 - IB 10 K = -0. 1 V VB VE = VB - VBE = -0. 1 - 0. 7 = -0. 8 V VC = 10 V - IC 8 K = 10 - 1(8) = 2 V ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 36

Exercise 4. 24 (b) Find gm, rp, and r 0, given: b = 100, VA = 100 V gm = IC/VT = 1 m. A/25 m. V = 40 m. A/V rp = VT/ IB = 25 m. V/. 01 m. A = 2. 5 K r 0 = output resistance of transistor r 0 = 1/slope of transistor output characteristics r 0 = | VA|/IC = 100 K ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 37

Summary of transistor analysis • Transistor circuits are analyzed and designed in terms of DC and ac versions of the same circuit. • An ac signal is usually superimposed on the DC circuit. • The location of the operating point (values of IC and VCE) of the transistor affects the ac operation of the circuit. • There at least two ac parameters determined from DC quantities. ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 38

Steps to analyze a transistor circuit 1 DC problem Set ac sources to zero, solve for DC quantities, IC and VCE. 2 Determine ac quantities from DC parameters Find gm, rp, and ro. 3 ac problem Set DC sources to zero, replace transistor by hybrid-p model, find ac quantites, Rin, Rout, Av, and Ai. ro ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 39

Circuit from Exercise 4. 24 + Vout IE = 1 m. A IB IE/b = 0. 01 m. A VB = 0 - IB 10 K = -0. 1 V VE = VB - VBE = -0. 1 - 0. 7 = -0. 8 V VC = 10 V - IC 8 K = 10 - 1(8) = 2 V gm = IC/VT = 1 m. A/25 m. V = 40 m. A/V rp = VT/ IB = 25 m. V/. 01 m. A = 2. 5 K ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 40

ac equivalent circuit b c e vbe = (Rb||Rpi)/ [(Rb||Rpi) +Rs]vi vbe = 0. 5 vi + vout - Neglecting Ro vout = -(gmvbe)(Rc ||RL) Av = vout/vi = - 80 vout = -(gmvbe)(Ro||Rc ||RL) vout = -154 vbe Av = vout/vi = - 77 ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 41

Prob. 4. 76 b=100 + Vout - ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 42

Prob. 4. 76 b c + ib e Vout = bib - Rin (a) Find Rin = Rpi = VT/IB = (25 m. V)100/. 1 = 2. 5 KW (c) Find Rout = Rc = 47 KW Rout (b) Find Av = vout/vin vout = - bib Rc vin = ib (R + Rpi) Av = vout/vin = - bib Rc/ ib (R + Rpi ) = - b. Rc/(R + Rpi) = - 100(47 K)/(100 K + 2. 5 K) = - 37. 6 ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 43

4. 9 Graphical analysis Input circuit B-E voltage loop VBB = IBRB +VBE IB = (VBB - VBE)/RB ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 44

Graphical construction of IB and VBE IB = (VBB - VBE)/RB VBB/RB If VBE = 0, IB = VBB/RB If IB = 0, VBE = VBB ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 45

Load line Output circuit C-E voltage loop VCC = ICRC +VCE IC = (VCC - VCE)/RC ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 46

Graphical construction of IC and VCE IC = (VCC - VCE)/RC VCC/RC If VCE = 0, IC = VCC/RC If IC = 0, VCE = VCC ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 47

Graphical analysis Input signal ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 Output signal 48

Bias point location effects • Load-line A results in bias point QA which is too close to VCC and thus limits the positive swing of v. CE. • Load-line B results in an operating point too close to the saturation region, thus limiting the negative swing of v. CE. ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 49

4. 11 Basic single-stage BJT amplifier configurations We will study 3 types of BJT amplifiers • CE - common emitter, used for AV, Ai, and general purpose • CE with RE - common emitter with RE, same as CE but more stable • CC common collector, used for Ai, low output resistance, used as an output stage CB common base (not covered) ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 50

Common emitter amplifier ac equivalent circuit ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 51

Common emitter amplifier ib Rin (Does not include source) Rin = Rpi Rout (Does not include load) Rout = RC AV = Vout/Vin Vout = - bib. RC Vin = ib(Rs + Rpi) AV = - b. RC/ (Rs + Rpi) iout + Vout Rin Ai = iout/iin iout = - bib iin = ib Ai = - b ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 52

Common emitter with RE amplifier ac equivalent circuit ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 53

Common emitter with RE amplifier ib Rin = V/ib iout + + bib)RE Rin = Rpi + (1 + b)RE V = ib Rpi + (ib (usually large) Vout V Rin ib + bib Rout = RC AV = Vout/Vin Vout = - bib. RC + bib)RE AV = - b. RC/ (Rs + Rpi + (1 + b)RE) + Vin = ib Rs + ib Rpi + (ib Rout - Ai = iout/iin iout = - bib iin = ib Ai = - b (less than CE, but less sensitive to b variations) ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 54

Common collector (emitter follower) amplifier b c e + vout - (vout at emitter) ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 ac equivalent circuit 55

Common collector amplifier ib + Rin = V/ib + bib)RL Rin = Rpi + (1 + b)RL V = ib Rpi + (ib V ib + bib Rin + vout - AV = vout/vs vout = (ib + bib)RL AV = (1+ b)RL/ (Rs + Rpi + (1 + b)RL) vs = ib Rs + ib Rpi + (ib (always < 1) ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 56

Common collector amplifier ib Ai = iout/iin iout = ib + bib iin = ib Ai = b + 1 ib + bib + vout Rout (don’t include RL, set Vs = 0) Rout = vout /- (ib + bib) vout = -ib Rpi + -ib. Rs Rout = (Rpi + Rs) / (1+ b) (usually low) ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 57

Prob. 4. 84 Given b = 50 + vout ac circuit Rpi =VT/IB = 25 m. V(50)/. 2 m. A = 6. 6 K CE with RE amp, because RE is in ac circuit ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 58

Prob. 4. 84 ib + V ib Rin + bib (a) Find Rin = V/ib V = ib Rpi + (ib + bib)RE Rin = Rpi + (1 + b)RE Rin = 6. 6 K + (1 + 50)125 Rin 13 K ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 59

Prob. 4. 84 ib -bib + vout ib + bib - (b) Find AV = vout/vs vout = - bib(RC||RL) vs = ib Rs + ib Rpi + (ib + bib)RE AV = - b (RC||RL) / (Rs + Rpi + (1 + b)RE) AV = - 50 (10 K||10 K) /(10 K + 6. 6 Ki + (1 + 50)125) AV - 11 ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 60

Prob. 4. 84 ib + vbe - + vout ib + bib - (c) If vbe is limited to 5 m. V, what is the largest signal at input and output? vbe = ib Rpi = 5 m. V ib = vbe /Rpi = 5 m. V/6. 6 K = 0. 76 m. A (ac value) vs = ib Rs + ib Rpi + (ib + bib)RE vs = (0. 76 m. A)10 K + (0. 76 m. A) 6. 6 K + (0. 76 m. A + (50)0. 76 m. A )125 vs 17. 4 m. V ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 61

Prob. 4. 84 ib + vbe - + vout ib + bib - (c) If vbe is limited to 5 m. V, what is the largest signal at input and output? vout = vs AV vout = 17. 4 m. V(-11) vout -191 m. V (ac value) ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 62

Prob. 4. 79 b = 100 Using this circuit, design an amp with: IE = 2 m. A AV = -8 current in voltage divider I = 0. 2 m. A (CE amp because RE is not in ac circuit) Voltage divider Vcc/I = 9/0. 2 m. A = 45 K = R 1 + R 2 Choose VB 1/3 Vcc to put operating point near the center of the transistor characteristics R 2/(R 1 + R 2) = 3 V Combining gives, R 1 = 30 K, R 2 = 15 K ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 63

Prob. 4. 79 Find RE (input circuit) Use Thevenin equivalent b = 100 B-E loop VBB=IBRBB+VBE+IERE using IB IE/b RE = [VBB - VBE - (IE/b)RBB]/IE IB + VBE - IE RE = [3 -. 7 - (2 m. A/100)10 K]/2 m. A RE = 1. 05 KW ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 64

Prob. 4. 79 + vout Find Rc (ac circuit) Rpi = VT/IB = 25 m. V(100)/2 m. A = 1. 25 K Ro = VA/IC = 100/2 m. A = 50 K Av = vout/vin vout = -gmvbe (Ro||Rc||RL) vbe = 10 K||1. 2 K / [10 K+ 10 K||1. 2 K]vi Av = -gm(Ro||Rc||RL)(10 K||1. 2 K) / [10 K||1. 2 K +Rs] Set Av = -8, and solve for Rc, Rc 2 K ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 65

CE amplifier ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 66

CE amplifier Av -12. 2 ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 67

CE amplifier FOURIER COMPONENTS OF TRANSIENT RESPONSE V($N_0009) DC COMPONENT = -1. 226074 E-01 HARMONIC FREQUENCY FOURIER NORMALIZED NO (HZ) COMPONENT 1 2 3 4 1. 000 E+03 2. 000 E+03 3. 000 E+03 4. 000 E+03 1. 581 E+00 1. 992 E-01 2. 171 E-02 3. 376 E-03 1. 000 E+00 1. 260 E-01 1. 374 E-02 2. 136 E-03 PHASE NORMALIZED (DEG) PHASE (DEG) -1. 795 E+02 0. 000 E+00 9. 111 E+01 2. 706 E+02 -1. 778 E+02 1. 668 E+00 -1. 441 E+02 3. 533 E+01 TOTAL HARMONIC DISTORTION = 1. 267478 E+01 PERCENT ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 68

CE amplifier with RE ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 69

CE amplifier with RE Av - 7. 5 ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 70

FOURIER COMPONENTS OF TRANSIENT RESPONSE V($N_0009) DC COMPONENT = -1. 353568 E-02 HARMONIC FREQUENCY FOURIER NO (HZ) COMPONENT 1 2 3 4 1. 000 E+03 2. 000 E+03 3. 000 E+03 4. 000 E+03 7. 879 E-01 1. 604 E-02 5. 210 E-03 3. 824 E-03 NORMALIZED PHASE COMPONENT (DEG) 1. 000 E+00 2. 036 E-02 6. 612 E-03 4. 854 E-03 NORMALIZED PHASE (DEG) -1. 794 E+02 0. 000 E+00 9. 400 E+01 2. 734 E+02 -1. 389 E+02 4. 056 E+01 -1. 171 E+02 6. 231 E+01 TOTAL HARMONIC DISTORTION = 2. 194882 E+00 PERCENT ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 71

Summary CE Av -12. 2 THD 12. 7% CE w/RE (RE = 100) -7. 5 2. 19% ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 72

Prob. 4. 83 b = 100 Rc=6. 8 K + vout RL=2 K - • 2 stage amplifier • Both stages are the same • Capacitively coupled (a) Find IC and VC of each transistor (same for each stage) ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 73

Prob. 4. 83 (a) Find IC and VC of each transistor (same for each stage) B-E voltage loop + VBB = IBRBB + VBE + IERE VC where RBB = R 1||R 2 = 32 K VBB = VCCR 2/(R 1+R 2) = 4. 8 V, and IB IE/b IE = [VBB - VBE ]/[RBB/b + RE] IE = 0. 97 m. A VC = VCC - ICRC VC = 15 -. 97(6. 8) VC = 8. 39 V ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 74

Prob. 4. 83 (b) find ac circuit b c b e c e + RL=2 K vout - RBB = R 1||R 2 = 100 K||47 K = 32 KW Rpi = VT/IB = 25 m. V(100)/. 97 m. A 2. 6 KW gm = IC/VT =. 97 m. A/25 m. V 39 m. A/V ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 75

Prob. 4. 83 b Rin 1 + vb 1 - b c e Rin 2 + vb 2 - (c) find Rin 1 = RBB||Rpi = 32 K||2. 6 K = 2. 4 KW find vb 1/vi = Rin 1/ [Rin 1 + RS] = 2. 4 K/[2. 4 K + 5 K] = 0. 32 ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 c + RL=2 K e vout - (d) find Rin 2 = RBB||Rpi = 2. 4 KW find vb 2/vb 1 vb 2 = -gmvbe 1[RC||RBB||Rpi] vb 2/vbe 1 = -gm[RC||RBB||Rpi] vb 2/vb 1 = -(39 m. A/V)[6. 8||32 K||2. 6 K] = -69. 1 76

Prob. 4. 83 b + vb 1 - c e b + vb 2 - (e) find vout/vb 2 vout = -gmvbe 2[RC||RL] vout/vbe 2 = -gm[RC||RL] vb 2/vb 1 = -(39 m. A/V)[6. 8 K||2 K] = -60. 3 ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 c + RL=2 K e vout - (f) find overall voltage gain vout/vi = (vb 1/vi) (vb 2/vb 1) (vout/vb 2) vout/vi = (0. 32) (-69. 1) (-60. 3) vout/vi = 1332 77

Prob. 4. 96 Q 1 has b 1 = 20 Q 2 has b 2 = 200 Q 1 IB 2 IE 1 Q 2 IE 2 Find IE 1, IE 2, VB 1, and VB 2 IE 2 = 2 m. A IE 1 = I 20 m. A + IB 2 IE 1 = I 20 m. A + IE 2/b 2 IE 1 = 20 m. A + 10 m. A IE 1 = 30 m. A ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 78

Prob. 4. 96 Find VB 1, and VB 2 Use Thevenin equivalent Q 1 has b 1 = 20 Q 2 has b 2 = 200 VB 1 = VBB 1 - IB 1(RBB 1) = 4. 5 - (30 m. A/20)500 K = 3. 8 V VB 2 = VB 1 - VBE = 3. 8 V - 0. 7 = 3. 1 V ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 IB 1 + v. B 1 - IB 2 + v. B 2 - 79

Prob. 4. 96 b 1 c 1 e 1 + v. B 2 b 2 c 2 e 2 - Rpi 2 = VT/IB 2 = VT b 2/IE 2 = 25 m. V(200)/2 m. A = 2. 5 KW + vout - (b) find vout/vb 2 vout = (ib 2 + b 2 ib 2)RL vb 2 = (ib 2 + b 2 ib 2)RL + ib 2 Rpi 2 vout/vb 2 = (1 + b 2)RL/[(1 + b 2)RL + Rpi 2] = (1 + 200)1 K/[(1 + 200)1 K + 2. 5 K] 0. 988 ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 80

Prob. 4. 96 b 1 c 1 e 1 + b 2 v. B 2 c 2 e 2 Rin 2 (b) find Rin 2 = vb 2/ib 2 vb 2 = (ib 2 + b 2 ib 2)RL + ib 2 Rpi 2 Rin 2 = vb 2/ib 2 = (1 + b 2)RL + Rpi 2 = (1 + 200)1 K + 2. 5 K 204 K ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 81

Prob. 4. 96 b 1 c 1 + i. B 1 e 1 v. B 1 b 2 c 2 e 2 Rin 1 (c) find Rin 1 = RBB 1||(vb 1/ib 1) = RBB 1|| [ib 1 Rpi 1 + (ib 1 + b 1 ib 1)Rin 2]/ib 1 = RBB 1|| [Rpi 1 + (1+ b 1)Rin 2], where Rpi 1 = VT b 1/IE 1 = 25 m. V(20)/30 m. A = 16. 7 K = 500 K||[16. 7 K + (1+ 20)204 K] 500 KW ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 82

Prob. 4. 96 b 1 c 1 + i. B 1 e 1 + v. B 1 b 2 c 2 ve 1 e 2 - - (c) find ve 1/vb 1 ve 1 = (ib 1 + b 1 ib 1)Rin 2 vb 1 = (ib 1 + b 1 ib 1)Rin 2 + ib 1 Rpi 1 ve 1/vb 1 = (1 + b 1) Rin 2 /[(1 + b 1) Rin 2 + Rpi 1] = (1 + 20)204 K/[(1 + 20)204 K + 16. 7 K] 0. 996 ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 83

Prob. 4. 96 b 1 + vb 1 c 1 e 1 b 2 c 2 e 2 (d) find vb 1/vi = Rin 1/[RS + Rin 1] = 0. 82 (e) find overall voltage gain vout/vi = (vb 1/vi) (ve 1/vb 1) (vout/ve 1) vout/vi = (0. 82) (0. 99) vout/vi = 0. 81 ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 84

(Prob. 4. 96) Voltage outputs at each stage Output of stage 2 Output of stage 1 Input ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 85

(Prob. 4. 96) Current Input to stage 2 (ib 2) Input current ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 86

(Prob. 4. 96) Current Input to stage 2 (ib 2) output current ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 87

(Prob. 4. 96) Current Input to stage 2 (ib 2) Input current output current ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 88

(Prob. 4. 96) Power and current gain Input current = (Vi)/Rin = 1/500 K = 2. 0 m. A output current= (Vout)/RL = (0. 81 V)/1 K= 0. 81 m. A current gain = 0. 81 m. A/ 2. 0 m. A = 405 Input power = (Vi)/Rin = 1 x 1/500 K = 2. 0 m. W output power = (Vout)/RL = (0. 81 V)/1 K= 656 m. W power gain = 656 m. W/2 m. W = 329 ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 89

BJT Output Characteristics • Plot Ic vs. Vce for multiple values of Vce and Ib • From Analysis menu use DC Sweep • Use Nested sweep in DC Sweep section ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 90

Probe: BJT Output Characteristics 2 Add plot (plot menu) -> Add trace (trace menu) -> IC(Q 1) 1 Result of probe 3 Delete plot (plot menu) ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 91

BJT Output Characteristics: current gain b at Vce = 4 V, and Ib = 45 m. A b = 8 m. A/45 m. A = 178 ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 Ib = 5 m. A (Each plot 10 m. A difference) 92

BJT Output Characteristics: transistor output resistance Ro = 1/slope At Ib = 45 m. A, 1/slope = (8. 0 - 1. 6)V/(8. 5 - 7. 8)m. A Rout = 9. 1 KW ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 Ib = 5 m. A (Each plot 10 m. A difference) 93

CE Amplifier: Measurements with Spice Rin ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 Rout 94

Input Resistance Measurement Using SPICE • Replace source, Vs and Rs with Vin, measure Rin = Vin/Iin • Do not change DC problem: keep capacitive coupling if present • Source (Vin) should be a high enough frequency so that capacitors act as shorts: Rcap = |1/w. C|. For C = 100 m. F, w = 1 KHz, Rcap = 1/2 p(1 K)(100 E-6) 1. 6 W • Vin should have a small value so operating point does not change Vin 1 m. V ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 95

Rin Measurement • Transient analysis ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 96

Probe results Rin = 1 m. V/204 n. A = 4. 9 KW I(C 2) ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 97

Output Resistance Measurement Using SPICE • Replace load, RL with Vin, measure Rin = Vin/Iin • Set Vs = 0 • Do not change DC problem: keep capacitive coupling if present • Source (Vin) should be a high enough frequency so that capacitors act as shorts: Rcap = |1/w. C|. For C = 100 m. F, w = 1 KHz, Rcap = 1/2 p(1 K)(100 E-6) 1. 6 W • Vin should have a small value so operating point does not change Vin 1 m. V ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 98

Rout Measurement • Transient analysis ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 99

Probe results Rout = 1 m. V/111 n. A = 9 KW -I(C 1) is current in Vin flowing out of + terminal ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 100

DC Power measurements Power delivered by + 10 sources: (10)(872 m. A) + (10)(877 m. A) = 8. 72 m. W + 8. 77 m. W = 17. 4 m. W ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 101

ac Power Measurements of Load average power instantaneous power Vout Vin ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 102