Chap. 4 BJT transistors
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1 Chap. 4 BJT transistors Widely used in amplifier circuits Formed by junction of 3 materials npn or pnp structure ECE Electronics - Dr. S. Kozaitis- 1
2 ECE Electronics - Dr. S. Kozaitis- 2
3 pnp transistor ECE Electronics - Dr. S. Kozaitis- 3
4 ECE Electronics - Dr. S. Kozaitis- 4
5 ECE Electronics - Dr. S. Kozaitis- 5
6 Operation of npn transistor Large current ECE Electronics - Dr. S. Kozaitis- 6
7 Modes of operation of a BJT transistor Mode BE junction BC junction cutoff reverse biased reverse biased linear(active) forward biased reverse biased saturation forward biased forward biased ECE Electronics - Dr. S. Kozaitis- 7
8 Summary of npn transistor behavior npn IC IB base + small current VBE - collector emitter IE large current ECE Electronics - Dr. S. Kozaitis- 8
9 Summary of pnp transistor behavior pnp IC IB base + small current VBE - collector emitter IE large current ECE Electronics - Dr. S. Kozaitis- 9
10 Summary of equations for a BJT IE IC IC = IB is the current gain of the transistor 100 VBE = 0.7V(npn) VBE = -0.7V(pnp) ECE Electronics - Dr. S. Kozaitis- 10
11 4.5 Graphical representation of transistor characteristics IC IB Output circuit Input circuit IE ECE Electronics - Dr. S. Kozaitis- 11
12 Input characteristics IB IB 0.7V VBE Acts as a diode VBE 0.7V ECE Electronics - Dr. S. Kozaitis- 12
13 Output characteristics IC IC IB = 40 A IB = 30 A IB = 20 A IB = 10 A Early voltage ECE Electronics - Dr. S. Kozaitis- 13 Cutoff region At a fixed IB, IC is not dependent on VCE Slope of output characteristics in linear region is near 0 (scale exaggerated) VCE
14 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 Electronics - Dr. S. Kozaitis- 14
15 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 = 4V - 0.7V = 3.3V IC IE IE = (VE - 0)/RE = 3.3/3.3K = 1mA IC IE = 1mA VC = 10 - ICRC = 10-1(4.7) = 5.3V ECE Electronics - Dr. S. Kozaitis- 15
16 Example 4.6 = 100 B-E Voltage loop 5 = IBRB + VBE, solve for IB IB = (5 - VBE)/RB = (5-.7)/100k = 0.043mA IC IC = IB = (100)0.043mA = 4.3mA IB IE VC = 10 - ICRC = (2) = 1.4V ECE Electronics - Dr. S. Kozaitis- 16
17 Exercise 4.8 VE = = - 0.7V = 50 IE = (VE - -10)/RE = ( )/10K = 0.93mA IC IC IE = 0.93mA IB IB = IC/ m IE VC = 10 - ICRC = (5) = 5.35V VCE = = 6.05V ECE Electronics - Dr. S. Kozaitis- 17
18 Summary of npn transistor behavior npn IC IB base + small current VBE - collector emitter IE large current ECE Electronics - Dr. S. Kozaitis- 18
19 Summary of equations for a BJT IE IC IC = IB is the current gain of the transistor 100 VBE = 0.7V(npn) VBE = -0.7V(pnp) ECE Electronics - Dr. S. Kozaitis- 19
20 I IB Prob Use a voltage divider, RB1 and RB2 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 = 3V and I = 0.2mA. (a) RB1 and RB2 form a voltage divider. Assume I >> IB AND.2mA = 9 /(RB1 + RB2) VB = VCC[RB2/(RB1 + RB2)] I = VCC/(RB1 + RB2) 3 = 9 [RB2/(RB1 + RB2)], Solve for RB1 and RB2. RB1 = 30K, and RB2 = 15K. ECE Electronics - Dr. S. Kozaitis- 20
21 Prob Find the operating point Use the Thevenin equivalent circuit for the base Makes the circuit simpler VBB = VB = 3V RBB is measured with voltage sources grounded RBB = RB1 RB2 = 30K 15K. 10K ECE Electronics - Dr. S. Kozaitis- 21
22 Prob Write B-E loop and C-E loop B-E loop VBB = IBRBB + VBE +IERE C-E loop C-E loop VCC = ICRC + VCE +IERE B-E loop Solve for, IC, VCE, and IB. This is how all DC circuits are analyzed and designed! ECE Electronics - Dr. S. Kozaitis- 22
23 Example stage amplifier, 1st stage has an npn transistor; 2nd stage has an pnp transistor. IC = IB IC IE VBE = 0.7(npn) = -0.7(pnp) = 100 Find IC1, IC2, VCE1, VCE2 Use Thevenin circuits. ECE Electronics - Dr. S. Kozaitis- 23
24 Example 4.8 IB1 IE1 RBB1 = RB1 RB2 = 33K VBB1 = VCC[RB2/(RB1+RB2)] VBB1 = 15[50K/150K] = 5V Stage 1 B-E loop VBB1 = IB1RBB1 + VBE +IE1RE1 Use IB1 IE1/ 5 = IE133K / IE13K IE1 = 1.3mA ECE Electronics - Dr. S. Kozaitis- 24
25 Example 4.8 C-E loop neglect IB2 because it is IB2 << IC1 IC1 VCC = IC1RC1 + VCE1 +IE1RE1 15 = 1.3(5) + VCE1 +1.3(3) IE1 VCE1= 4.87V ECE Electronics - Dr. S. Kozaitis- 25
26 Example 4.8 Stage 2 B-E loop IE2 VCC = IE2RE2 + VEB +IB2RBB2 + VBB2 15 = IE2(2K) +.7 +IB2 (5K) (3) Use IB2 IE2/ solve for IE2 IB2 IE2 = 2.8mA ECE Electronics - Dr. S. Kozaitis- 26
27 Example 4.8 IE2 Stage 2 C-E loop VCC = IE2RE2 + VEC2 +IC2RC2 15 = 2.8(2) + VEC (2.7) solve for VEC2 IC2 VCE2 = 1.84V ECE Electronics - Dr. S. Kozaitis- 27
28 Summary of DC problem Bias transistors so that they operate in the linear region B-E junction forward biased, C-B junction reversed biased Use VBE = 0.7 (npn), IC IE, IC = 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 Electronics - Dr. S. Kozaitis- 28
29 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 are at least two ac parameters determined from DC quantities. ECE Electronics - Dr. S. Kozaitis- 29
30 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 - vbe curve at the bias point Q is the transconductance gm: the amount of ac current produced by an ac voltage. ECE Electronics - Dr. S. Kozaitis- 30 (DC input signal 0.7V) ac input signal
31 Transconductance Transconductance = slope at Q point ac output signal gm = dic/dvbe 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 (0.7V) ac input signal ECE Electronics - Dr. S. Kozaitis- 31
32 ac input resistance of transistor IB ac output signal DC output signal ac input resistance 1/slope at Q point r = dvbe/dib ic = ICQ r VT /IB DC input signal (0.7V) re VT /IE ac input signal ECE Electronics - Dr. S. Kozaitis- 32
33 4.8 Small-signal equivalent circuit models ac model Hybrid- model They are equivalent Works in linear region only ECE Electronics - Dr. S. Kozaitis- 33
34 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, r and re. 3 ac problem Set DC sources to zero, replace transistor by hybrid- ac quantites, Rin, Rout, Av, and Ai. model, find ECE Electronics - Dr. S. Kozaitis- 34
35 Example 4.9 Find vout/vin, ( = 100) DC problem Short vi, determine IC and VCE B-E voltage loop 3 = IBRB + VBE IB = (3 -.7)/RB = 0.023mA C-E voltage loop VCE = 10 - ICRC VCE = 10 - (2.3)(3) VCE = 3.1V Q point: VCE = 3.1V, IC = 2.3mA ECE Electronics - Dr. S. Kozaitis- 35
36 Example 4.9 b + vbe - e c + vout - ac problem Short DC sources, input and output circuits are separate, only coupled mathematically gm = IC/VT = 2.3mA/25mV = 92mA/V r = VT/ IB = 25mV/.023mA = 1.1K vbe= vi [r / (100K + r 011vi vout = - gm vberc vout = - 92 ( 011vi)3K vout/vi = ECE Electronics - Dr. S. Kozaitis- 36
37 Exercise 4.24 (a) Find VC, VB, and VE, given: = 100, VA = 100V IE = 1 ma IB IE/ = 0.01mA VB = 0 - IB10K = -0.1V VB VE = VB - VBE = = -0.8V VC = 10V - IC8K = 10-1(8) = 2V ECE Electronics - Dr. S. Kozaitis- 37
38 Exercise 4.24 (b) Find gm, r, and r, given: = 100, VA = 100V gm = IC/VT = 1 ma/25mv = 40 ma/v r = VT/ IB = 25mV/.01mA = 2.5K r0 = output resistance of transistor r0 = 1/slope of transistor output characteristics r0 = VA /IC = 100K ECE Electronics - Dr. S. Kozaitis- 38
39 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 are at least two ac parameters determined from DC quantities. ECE Electronics - Dr. S. Kozaitis- 39
40 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, r and ro. 3 ac problem Set DC sources to zero, replace transistor by hybrid- ac quantites, Rin, Rout, Av, and Ai. model, find ro ECE Electronics - Dr. S. Kozaitis- 40
41 Circuit from Exercise Vout IE = 1 ma VC = 10V - IC8K = 10-1(8) = 2V IB IE/ = 0.01mA gm = IC/VT = 1 ma/25mv = 40 ma/v VB = 0 - IB10K = -0.1V r = VT/ IB = 25mV/.01mA = 2.5K VE = VB - VBE = = -0.8V ECE Electronics - Dr. S. Kozaitis- 41 -
42 ac equivalent circuit b e c + vout - vbe = (Rb Rpi)/ [(Rb Rpi) +Rs]vi vbe = 0.5vi Neglecting Ro vout = -(gmvbe)(rc RL) Av = vout/vi = - 80 vout = -(gmvbe)(ro Rc RL) vout = -154vbe Av = vout/vi = - 77 ECE Electronics - Dr. S. Kozaitis- 42
43 Prob =100 + Vout - ECE Electronics - Dr. S. Kozaitis- 43
44 Prob b ib c + Rin e = ib Rout Vout - (a) Find Rin Rin = Rpi = VT/IB = (25mV)100/.1 = 2.5K (c) Find Rout Rout = Rc = 47K (b) Find Av = vout/vin vout = - ib Rc vin = ib (R + Rpi) Av = vout/vin = - ib Rc/ ib (R + Rpi ) = - Rc/(R + Rpi) = - (47K)/(100K + 2.5K) = - ECE Electronics - Dr. S. Kozaitis- 44
45 4.9 Graphical analysis Input circuit B-E voltage loop VBB = IBRB +VBE IB = (VBB - VBE)/RB ECE Electronics - Dr. S. Kozaitis- 45
46 Graphical construction of IB and VBE IB = (VBB - VBE)/RB VBB/RB If VBE = 0, IB = VBB/RB If IB = 0, VBE = VBB ECE Electronics - Dr. S. Kozaitis- 46
47 Load line Output circuit C-E voltage loop VCC = ICRC +VCE IC = (VCC - VCE)/RC ECE Electronics - Dr. S. Kozaitis- 47
48 Graphical construction of IC and VCE IC = (VCC - VCE)/RC VCC/RC If VCE = 0, IC = VCC/RC If IC = 0, VCE = VCC ECE Electronics - Dr. S. Kozaitis- 48
49 Graphical analysis Input signal Output signal ECE Electronics - Dr. S. Kozaitis- 49
50 Bias point location effects Load-line A results in bias point Q A which is too close to V CC 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 Electronics - Dr. S. Kozaitis- 50
51 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 Electronics - Dr. S. Kozaitis- 51
52 Common emitter amplifier ac equivalent circuit ECE Electronics - Dr. S. Kozaitis- 52
53 Common emitter amplifier Rin (Does not include source) ib iout + Rin = Rpi Vout - Rout (Does not include load) Rin Rout Rout = RC AV = Vout/Vin Vout = - ibrc Vin = ib(rs + Rpi) Ai = iout/iin iout = - ib iin = ib AV = - RC/ (Rs + Rpi) Ai = - ECE Electronics - Dr. S. Kozaitis- 53
54 Common emitter with RE amplifier ac equivalent circuit ECE Electronics - Dr. S. Kozaitis- 54
55 Common emitter with RE amplifier Rin Rin = V/ib + ib iout + V = ib Rpi + (ib + ib)re Rin = Rpi + (1 + )RE (usually large) V Rin - ib + ib Rout Vout - Rout Rout = RC AV = Vout/Vin Vout = - ibrc Vin = ib Rs + ib Rpi + (ib + ib)re AV = - RC/ (Rs + Rpi + (1 + )RE) (less than CE, but less sensitive to variations) Ai = iout/iin iout = - ib iin = ib Ai = - ECE Electronics - Dr. S. Kozaitis- 55
56 Common collector (emitter follower) amplifier b c e + vout - (vout at emitter) ac equivalent circuit ECE Electronics - Dr. S. Kozaitis- 56
57 Common collector amplifier + ib Rin Rin = V/ib V = ib Rpi + (ib + ib)rl Rin = Rpi + (1 + )RL V - Rin ib + ib + vout - AV = vout/vs vout = (ib + ib)rl vs = ib Rs + ib Rpi + (ib + ib)rl AV = (1+ RL/ (Rs + Rpi + (1 + )RL) (always < 1) ECE Electronics - Dr. S. Kozaitis- 57
58 Common collector amplifier ib Ai = iout/iin iout = ib + ib iin = ib Ai = ib + ib + vout - Rout Rout (don t include RL, set Vs = 0) Rout = vout /- (ib + ib) vout = -ib Rpi + -ibrs Rout = (Rpi + Rs) / (1+ (usually low) ECE Electronics - Dr. S. Kozaitis- 58
59 Prob Given = 50 + vout - ac circuit Rpi =VT/IB = 25mV(50)/.2mA = 6.6K CE with RE amp, because RE is in ac circuit ECE Electronics - Dr. S. Kozaitis- 59
60 Prob ib + V - Rin ib + ib (a) Find Rin Rin = V/ib V = ib Rpi + (ib + ib)re Rin = Rpi + (1 + )RE Rin = 6.6K + (1 + )125 Rin 13K ECE Electronics - Dr. S. Kozaitis- 60
61 ib Prob ib + vout ib + ib - (b) Find AV = vout/vs vout = - ib(rc RL) vs = ib Rs + ib Rpi + (ib + ib)re AV = - (RC RL) / (Rs + Rpi + (1 + )RE) AV = - (10K 10K) /(10K + 6.6Ki + (1 + )125) AV - ECE Electronics - Dr. S. Kozaitis- 61
62 ib Prob vbe - + vout ib + ib - (c) If vbe is limited to 5mV, what is the largest signal at input and output? vbe = ib Rpi = 5mV ib = vbe /Rpi = 5mV/6.6K = 0.76 A (ac value) vs = ib Rs + ib Rpi + (ib + ib)re vs = (0.76 A)10K + (0.76 A) 6.6K + (0.76 A + ( 0.76 A )125 vs 17.4mV ECE Electronics - Dr. S. Kozaitis- 62
63 ib Prob vbe - + vout ib + ib - (c) If vbe is limited to 5mV, what is the largest signal at input and output? vout = vs AV vout = 17.4mV(-11) vout -191mV (ac value) ECE Electronics - Dr. S. Kozaitis- 63
64 Prob = 100 Using this circuit, design an amp with: IE = 2mA AV = -8 current in voltage divider I = 0.2mA (CE amp because RE is not in ac circuit) Voltage divider Vcc/I = 9/0.2mA = 45K 45K = R1 + R2 Choose VB 1/3 Vcc to put operating point near the center of the transistor characteristics R2/(R1 + R2) = 1/3 Combining gives, R1 = 30K, R2 = 15K ECE Electronics - Dr. S. Kozaitis- 64
65 Prob Find RE (input circuit) Use Thevenin equivalent = 100 B-E loop VBB=IBRBB+VBE+IERE using IB IE/ RE = [VBB - VBE - (IE/ )RBB]/IE RE = [ (2mA/ )10K]/2mA RE = 1.05K IB + VBE - IE ECE Electronics - Dr. S. Kozaitis- 65
66 Prob vout - Find Rc (ac circuit) Rpi = VT/IB = 25mV(100)/2mA = 1.25K Ro = VA/IC = 100/2mA = 50K Av = vout/vin vout = -gmvbe (Ro Rc RL) vbe = 10K 1.2K / [10K+ 10K 1.2K]vi Av = -gm(ro Rc RL)(10K 1.2K) / [10K 1.2K +Rs] Set Av = -8, and solve for Rc, Rc 2K ECE Electronics - Dr. S. Kozaitis- 66
67 CE amplifier ECE Electronics - Dr. S. Kozaitis- 67
68 CE amplifier Av ECE Electronics - Dr. S. Kozaitis- 68
69 CE amplifier FOURIER COMPONENTS OF TRANSIENT RESPONSE V($N_0009) DC COMPONENT = E-01 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG) E E E E E E E E E E E E E E E E E E E E+01 TOTAL HARMONIC DISTORTION = E+01 PERCENT ECE Electronics - Dr. S. Kozaitis- 69
70 CE amplifier with RE ECE Electronics - Dr. S. Kozaitis- 70
71 CE amplifier with RE Av ECE Electronics - Dr. S. Kozaitis- 71
72 FOURIER COMPONENTS OF TRANSIENT RESPONSE V($N_0009) DC COMPONENT = E-02 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG) E E E E E E E E E E E E E E E E E E E E+01 TOTAL HARMONIC DISTORTION = E+00 PERCENT ECE Electronics - Dr. S. Kozaitis- 72
73 Summary Av THD CE % CE w/re (RE = 100) % ECE Electronics - Dr. S. Kozaitis- 73
74 Prob Rc=6.8K = vout RL=2K - 2 stage amplifier (a) Find IC and VC of each transistor Both stages are the same (same for each stage) Capacitively coupled ECE Electronics - Dr. S. Kozaitis- 74
75 (a) Find IC and VC of each transistor (same for each stage) Prob B-E voltage loop VBB = IBRBB + VBE + IERE where RBB = R1 R2 = 32K VBB = VCCR2/(R1+R2) = 4.8V, and IB IE/ + VC - IE = [VBB - VBE ]/[RBB/ + RE] IE = 0.97mA VC = VCC - ICRC VC = (6.8) VC = 8.39V ECE Electronics - Dr. S. Kozaitis- 75
76 Prob (b) find ac circuit b c b c + e e RL=2K vout - RBB = R1 R2 = 100K 47K = 32K Rpi = VT/IB = 25mV(100)/.97mA 2.6K gm = IC/VT =.97mA/25mV 39mA/V ECE Electronics - Dr. S. Kozaitis- 76
77 Prob Rin1 + b vb1 - e c Rin2 b + vb2 - e c + RL=2K vout - (c) find Rin1 Rin1 = RBB Rpi = 32K 2.6K = 2.4K find vb1/vi = Rin1/ [Rin1 + RS] = 2.4K/[2.4K + 5K] = 0.32 (d) find Rin2 Rin2 = RBB Rpi = 2.4K find vb2/vb1 vb2 = -gmvbe1[rc RBB Rpi] vb2/vbe1 = -gm[rc RBB Rpi] vb2/vb1 = -(39mA/V)[6.8 32K 2.6K] = ECE Electronics - Dr. S. Kozaitis- 77
78 Prob b vb1 - e c b + vb2 - e c + RL=2K vout - (e) find vout/vb2 vout = -gmvbe2[rc RL] vout/vbe2 = -gm[rc RL] vb2/vb1 = -(39mA/V)[6.8K 2K] = (f) find overall voltage gain vout/vi = (vb1/vi) (vb2/vb1) (vout/vb2) vout/vi = (0.32) (-69.1) (-60.3) vout/vi = 1332 ECE Electronics - Dr. S. Kozaitis- 78
79 Prob Q1 IE1 IB2 Q1 has = 20 Q2 has = 200 Q2 IE2 Find IE1, IE2, VB1, and VB2 IE2 = 2mA IE1 = I20 A + IB2 IE1 = I20 A + IE2/ 2 IE1 = 20 A + 10 A IE1 = 30 A ECE Electronics - Dr. S. Kozaitis- 79
80 Prob Find VB1, and VB2 Use Thevenin equivalent Q1 has = 20 Q2 has = 200 VB1 = VBB1 - IB1(RBB1) = (30 A/20)500K = 3.8V IB1 + vb1 IB2 + VB2 = VB1 - VBE = 3.8V = 3.1V - vb2 - ECE Electronics - Dr. S. Kozaitis- 80
81 Prob b1 c1 e1 + vb2 b2 c2 - e2 + vout - Rpi2 = VT/IB2 = VT 2/IE2 = 25mV(200)/2mA = 2.5K (b) find vout/vb2 vout = (ib2 + ib2)rl vb2 = (ib2 + ib2)rl + ib2rpi2 vout/vb2 = (1 + 2)RL/[(1 + 2)RL + Rpi2] = (1 + )1K/[(1 + )1K + 2.5K] ECE Electronics - Dr. S. Kozaitis- 81
82 Prob b1 c1 e1 + vb2 b2 c2 e2 Rin2 - (b) find Rin2 = vb2/ib2 vb2 = (ib2 + ib2)rl + ib2rpi2 Rin2 = vb2/ib2 = (1 + )RL + Rpi2 = (1 + )1K + 2.5K 204K ECE Electronics - Dr. S. Kozaitis- 82
83 Prob b1 ib1 e1 c1 vb1 b2 c2 e2 Rin1 - (c) find Rin1 = RBB1 (vb1/ib1) = RBB1 [ib1rpi1 + (ib1 + ib1)rin2]/ib1 = RBB1 [Rpi1 + (1+ )Rin2], where Rpi1 = VT 1/IE1 = 25mV(20)/30 A = 16.7K = 500K [16.7K + (1+ )204K] 500K ECE Electronics - Dr. S. Kozaitis- 83
84 Prob b1 ib1 c1 e1 + vb1 ve1 b2 c2 e2 - - (c) find ve1/vb1 ve1 = (ib1 + ib1)rin2 vb1 = (ib1 + ib1)rin2 + ib1rpi1 ve1/vb1 = (1 + 1) Rin2 /[(1 + 1) Rin2 + Rpi1] = (1 + )204K/[(1 + )204K K] ECE Electronics - Dr. S. Kozaitis- 84
85 Prob b1 e1 c1 vb1 b2 c2 e2 - (d) find vb1/vi vb1/vi = Rin1/[RS + Rin1] = 0.82 (e) find overall voltage gain vout/vi = (vb1/vi) (ve1/vb1) (vout/ve1) vout/vi = (0.82) (0.99) (0.99) vout/vi = 0.81 ECE Electronics - Dr. S. Kozaitis- 85
86 (Prob. 4.96) Voltage outputs at each stage Output of stage 2 Output of stage 1 Input ECE Electronics - Dr. S. Kozaitis- 86
87 (Prob. 4.96) Current Input to stage 2 (ib2) Input current ECE Electronics - Dr. S. Kozaitis- 87
88 (Prob. 4.96) Current Input to stage 2 (ib2) output current ECE Electronics - Dr. S. Kozaitis- 88
89 (Prob. 4.96) Current Input to stage 2 (ib2) Input current output current ECE Electronics - Dr. S. Kozaitis- 89
90 (Prob. 4.96) Power and current gain Input current = (Vi)/Rin = 1/500K = 2.0 A output current= (Vout)/RL = (0.81V)/1K= 0.81mA current gain = 0.81mA/ 2.0 A = 405 Input power = (Vi) (Vi)/Rin = 1 x 1/500K = 2.0 W output power = (Vout) (Vout)/RL = (0.81V) (0.81V)/1K= 656 W power gain = 656 W/2 W = 329 ECE Electronics - Dr. S. Kozaitis- 90
91 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 Electronics - Dr. S. Kozaitis- 91
92 Probe: BJT Output Characteristics 2 Add plot (plot menu) -> Add trace (trace menu) -> IC(Q1) 1 Result of probe 3 Delete plot (plot menu) ECE Electronics - Dr. S. Kozaitis- 92
93 BJT Output Characteristics: current gain at Vce = 4V, and Ib = 45 A = 8mA/45 A = 178 Ib = 5 A (Each plot 10 A difference) ECE Electronics - Dr. S. Kozaitis- 93
94 BJT Output Characteristics: transistor output resistance Ro = 1/slope At Ib = 45 A, 1/slope = ( )V/( )mA Rout = 9.1K Ib = 5 A (Each plot 10 A difference) ECE Electronics - Dr. S. Kozaitis- 94
95 CE Amplifier: Measurements with Spice Rin Rout ECE Electronics - Dr. S. Kozaitis- 95
96 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/ C. For C = 100 F, KHz, Rcap = 1/2 (1K)(100E-6) 1.6 Vin should have a small value so operating point does not change Vin 1mV ECE Electronics - Dr. S. Kozaitis- 96
97 Rin Measurement Transient analysis ECE Electronics - Dr. S. Kozaitis- 97
98 Probe results I(C2) Rin = 1mV/204nA = 4.9K ECE Electronics - Dr. S. Kozaitis- 98
99 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/ C. For C = 100 F, KHz, Rcap = 1/2 (1K)(100E-6) 1.6 Vin should have a small value so operating point does not change Vin 1mV ECE Electronics - Dr. S. Kozaitis- 99
100 Rout Measurement Transient analysis ECE Electronics - Dr. S. Kozaitis- 100
101 Probe results -I(C1) Rout = 1mV/111nA = 9K -I(C1) is current in Vin flowing out of + terminal ECE Electronics - Dr. S. Kozaitis- 101
102 DC Power measurements Power delivered by + 10 sources: (10)(872 A) + (10)(877 A) = 8.72mW mW = 17.4mW ECE Electronics - Dr. S. Kozaitis- 102
103 ac Power Measurements of Load average power instantaneous power Vout Vin ECE Electronics - Dr. S. Kozaitis- 103
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