1 BJT Bipolar Junction Transistor

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1 BJT Bipolar  Junction  Transistor • Widely used in amplifier circuits • Formed by1 BJT Bipolar Junction Transistor • Widely used in amplifier circuits • Formed by junction of 3 materials • npn or pnp structure

2 pnp transistor 2 pnp transistor

3 Large current. Operation of npn transistor 3 Large current. Operation of npn transistor

4 Modes of operation of a BJT transistor Mode BE junction BC junction cutoff reverse biased4 Modes of operation of a BJT transistor Mode BE junction BC junction cutoff reverse biased linear(active) forward biased reverse biased saturation forward biased

5 Summary of npn transistor behavior npn collector emitterbase I B I EI C small current5 Summary of npn transistor behavior npn collector emitterbase I B I EI C small current large current + V BE —

6 Summary of pnp transistor behavior pnp collector emitterbase I B I EI C small current6 Summary of pnp transistor behavior pnp collector emitterbase I B I EI C small current large current + V BE —

7 Summary of equations for a BJT I E I C =  I B 7 Summary of equations for a BJT I E I C = I B is the current gain of the transistor 100 V BE = 0. 7 V(npn) V BE = -0. 7 V(pnp)

8 Graphical representation of transistor characteristics I B I C I E Output circuit Input circuit8 Graphical representation of transistor characteristics I B I C I E Output circuit Input circuit

9 Input characteristics • Acts as a diode • V BE  0. 7 VI B9 Input characteristics • Acts as a diode • V BE 0. 7 VI B V BE 0. 7 V

10 Output characteristics I C V CEI B = 10 AI B = 20 AI B10 Output characteristics I C V CEI B = 10 AI B = 20 AI B = 30 AI B = 40 A Cutoff region • At a fixed I B , I C is not dependent on V CE • Slope of output characteristics in linear region is near 0 (scale exaggerated) Early voltage

11 Biasing a transistor • We must operate the transistor in the linear region.  •11 Biasing a transistor • We must operate the transistor in the linear region. • A transistor’s operating point (Q-point) is defined by I C , V CE , and I B.

12 A small ac signal v be is superimposed on the DC voltage V BE. 12 A small ac signal v be is superimposed on the DC voltage V BE. It gives rise to a collector signal current i c , superimposed on the dc current I C. Transconductance ac input signal(DC input signal 0. 7 V) ac output signal DC output signal. I B The slope of the i c — v BE curve at the bias point Q is the transconductance g m : the amount of ac current produced by an ac voltage.

13 Analysis of transistor circuits at DC For all circuits:  assume transistor operates in linear13 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 -1 B-E junction acts like a diode V E = V B — V BE = 4 V — 0. 7 V = 3. 3 V I EI C I E = (V E — 0)/R E = 3. 3/3. 3 K = 1 m. A I C I E = 1 m. A V C = 10 — I C R C = 10 — 1(4. 7) = 5. 3 V

14 Example-2 B-E Voltage loop 5 = I B R B + V BE,  solve14 Example-2 B-E Voltage loop 5 = I B R B + V BE, solve for I B = (5 — V BE )/R B = (5 -. 7)/100 k = 0. 043 m. A I C = I B = (100)0. 043 m. A = 4. 3 m. A V C = 10 — I C R C = 10 — 4. 3(2) = 1. 4 VI EI C I B =

15 Exercise-3 V E = 0 -. 7 = -0. 7 V I EI C I15 Exercise-3 V E = 0 -. 7 = -0. 7 V I EI C I B = 50 I E = (V E — -10)/R E = (-. 7 +10)/10 K = 0. 93 m. A I C I E = 0. 93 m. A I B = I B / m V C = 10 — I C R C = 10 -. 93(5) = 5. 35 V

16 Prob.  • Use a voltage divider, R B 1 and R B 2 to16 Prob. • Use a voltage divider, R B 1 and R B 2 to bias V B to avoid two power supplies. • Make the current in the voltage divider about 10 times I B to simplify the analysis. Use V B = 3 V and I = 0. 2 m. A. I BI (a) R B 1 and R B 2 form a voltage divider. Assume I >> I B I = V CC /(R B 1 + R B 2 ) . 2 m. A = 9 /(R B 1 + R B 2 ) AND V B = V CC [R B 2 /(R B 1 + R B 2 )] 3 = 9 [R B 2 /(R B 1 + R B 2 )], Solve for R B 1 and R B 2. R B 1 = 30 K , and R B 2 = 15 K .

17 Prob. Find the operating point • Use the Thevenin equivalent circuit for the base •17 Prob. Find the operating point • Use the Thevenin equivalent circuit for the base • Makes the circuit simpler • V BB = V B = 3 V • R BB is measured with voltage sources grounded • R BB = R B 1 || R B 2 = 30 K 15 K . 10 K

18 Prob. Write B-E loop and C-E loop B-E loop V BB = I B R18 Prob. Write B-E loop and C-E loop B-E loop V BB = I B R BB + V BE +I E R E I E =2. 09 m. A C-E loop V CC = I C R C + V CE +I E R E V CE =4. 8 V This is how all DC circuits are analyzed and designed!

19 Exercise-4 (a) Find V C , V B , and V E , given: =19 Exercise-4 (a) Find V C , V B , and V E , given: = 100, V A = 100 V I E = 1 m. A I B I E / = 0. 01 m. A V B = 0 — I B 10 K = -0. 1 V V E = V B — V BE = -0. 1 — 0. 7 = -0. 8 V V C = 10 V — I C 8 K = 10 — 1(8) = 2 VV

20 Example-5 • 2 -stage amplifier, 1 st stage has an npn transistor; 2 nd stage20 Example-5 • 2 -stage amplifier, 1 st stage has an npn transistor; 2 nd stage has an pnp transistor. I C = I B I C I E V BE = 0. 7 (npn) = -0. 7 (pnp) = 100 Find I C 1 , I C 2 , V CE 1 , V CE 2 • Use Thevenin circuits.

21 Example -5 • R BB 1 = R B 1 ||R B 2 = 3321 Example -5 • R BB 1 = R B 1 ||R B 2 = 33 K • V BB 1 = V CC [R B 2 /(R B 1 +R B 2 )] V BB 1 = 15[50 K/150 K] = 5 V Stage 1 • B-E loop V BB 1 = I B 1 R BB 1 + V BE +I E 1 R E 1 Use I B 1 I E 1 / 5 = I E 1 33 K / 100 +. 7 + I E 1 3 K I E 1 = 1. 3 m. AI B 1 I

22 Example -5 C-E loop neglect IB 2 because it is IB 2  IC 122 Example -5 C-E loop neglect IB 2 because it is IB 2 << IC 1 I E 1 I C 1 V CC = I C 1 R C 1 + V CE 1 +I E 1 R E 1 15 = 1. 3(5) + V CE 1 +1. 3(3) V CE 1 = 4. 87 V

23 Example-5 Stage 2 • B-E loop I B 2 I E 2 V CC =23 Example-5 Stage 2 • B-E loop I B 2 I E 2 V CC = I E 2 R E 2 + V EB +I B 2 R BB 2 + V BB 2 15 = I E 2 (2 K) +. 7 +I B 2 (5 K) + 4. 87 + 1. 3(3) Use I B 2 I E 2 / solve for I E 2 = 2. 8 m.

24 Example-5 Stage 2 • C-E loop I E 2 I C 2 V CC =24 Example-5 Stage 2 • C-E loop I E 2 I C 2 V CC = I E 2 R E 2 + V EC 2 +I C 2 R C 2 15 = 2. 8(2) + V EC 2 + 2. 8 (2. 7) solve for V EC 2 V CE 2 = 1. 84 V

25 Summary of DC problem • Bias transistors so that they operate in the linear region25 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 V BE = 0. 7 (npn), I C I E , I C = I B • Represent base portion of circuit by the Thevenin circuit • Write B-E, and C-E voltage loops. • For analysis, solve for I C , and V CE. • For design, solve for resistor values (I C and V CE specified).

26 Summary of npn transistor behavior npn collector emitterbase I B I EI C small current26 Summary of npn transistor behavior npn collector emitterbase I B I EI C small current large current + V BE —

27 Transistor as an amplifier • Transistor circuits are analyzed and designed in terms of DC27 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 I C and V CE ) of the transistor affects the ac operation of the circuit. • There at least two ac parameters determined from DC quantities.

28 A small ac signal v be is superimposed on the DC voltage V BE. 28 A small ac signal v be is superimposed on the DC voltage V BE. It gives rise to a collector signal current i c , superimposed on the dc current I C. Transconductance ac input signal(DC input signal 0. 7 V) ac output signal DC output signal. I B The slope of the i c — v BE curve at the bias point Q is the transconductance g m : the amount of ac current produced by an ac voltage.

29 Transconductance = slope at Q point g m = d i c /d v BE29 Transconductance = slope at Q point g m = d i c /d v BE | i c = I CQ where I C = I S [exp(-V BE /V T )-1]; the equation for a diode. Transconductance ac input signal. DC input signal (0. 7 V) ac output signal DC output signal g m = I S exp(-V BE /V T ) (1/V T ) g m I C /V T (A/V)

30 ac input resistance  1/slope at Q point r  = d v BE /d30 ac input resistance 1/slope at Q point r = d v BE /d i b | i c = I CQ r V T /I B r e V T /I E ac input resistance of transistor ac input signal. DC input signal (0. 7 V) ac output signal DC output signal. I

31 Small-signal equivalent circuit models • ac model • Hybrid-  model • They are equivalent31 Small-signal equivalent circuit models • ac model • Hybrid- model • They are equivalent • Works in linear region only

32 Steps to analyze a transistor circuit 1 DC problem Set ac sources to zero, solve32 Steps to analyze a transistor circuit 1 DC problem Set ac sources to zero, solve for DC quantities, I C and V CE. 2 Determine ac quantities from DC parameters Find g m , r and r e. 3 ac problem Set DC sources to zero, replace transistor by hybrid- model, find ac quantites, Rin, Rout, Av, and Ai.

33 Example-6 Find v out /v in , (  = 100) DC problem Short v33 Example-6 Find v out /v in , ( = 100) DC problem Short v i , determine I C and V CE B-E voltage loop 3 = I B R B + V BE I B = (3 -. 7)/R B = 0. 023 m. A C-E voltage loop V CE = 10 — I C R C V CE = 10 — (2. 3)(3) V CE = 3. 1 V Q point: V CE = 3. 1 V, I C = 2. 3 m.

34 Example -6 ac problem Short DC sources, input and output circuits are separate, only coupled34 Example -6 ac problem Short DC sources, input and output circuits are separate, only coupled mathematically g m = I C /V T = 2. 3 m. A/25 m. V = 92 m. A/V r = V T / I B = 25 m. V/. 023 m. A = 1. 1 K v be = v i [ r / (100 K + r 011 v i v out = — g m v be R C v out = — 92 ( 011 v i ) 3 K v out /v i = -3. 04 + v out — eb c + v be —

35 Exercise-7  Find g m ,  r , and r , given: = 100,35 Exercise-7 Find g m , r , and r , given: = 100, V A = 100 V, I C =1 m. A g m = I C /V T = 1 m. A/25 m. V = 40 m. A/V r = V T / I B = 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 = | V A |/I C = 100 K

36 Summary of transistor analysis • Transistor circuits are analyzed and designed in terms of DC36 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 I C and V CE ) of the transistor affects the ac operation of the circuit. • There at least two ac parameters determined from DC quantities.

37 Steps to analyze a transistor circuit 1 DC Analysis Set ac sources to zero, solve37 Steps to analyze a transistor circuit 1 DC Analysis Set ac sources to zero, solve for DC quantities, I C and V CE. 2 Determine ac quantities from DC parameters Find g m , r and r o. 3 AC Analysis Set DC sources to zero, replace transistor by hybrid- model, find ac quantities, Rin, Rout, Av, and Ai. r o

38 Circuit I E = 1 m. A V C = 10 V - I C38 Circuit I E = 1 m. A V C = 10 V — I C 8 K = 10 — 1(8) = 2 V I B I E / = 0. 01 m. A g m = I C /V T = 1 m. A/25 m. V = 40 m. A/V V B = 0 — I B 10 K = -0. 1 V r = V T / I B = 25 m. V/. 01 m. A = 2. 5 K V E = V B — V BE = -0. 1 — 0. 7 = -0. 8 V + V out —

39 ac equivalent circuit b e c v be = (R b ||R pi )/ [(R39 ac equivalent circuit b e c v be = (R b ||R pi )/ [(R b ||R pi ) +R s ]v i v be = 0. 5 v i v out = -(g m v be )||(R o ||R c ||R L ) v out = -154 v be A v = v out /v i = — 77 + v out — Neglecting R o v out = -(g m v be )||(R c ||R L ) A v = v out /v i = —

40 Prob.  + V out - =100 40 Prob. + V out — =

41 Prob.  + V out - (a) Find R in = R pi = V41 Prob. + V out — (a) Find R in = R pi = V T /I B = (25 m. V)100/. 1 = 2. 5 K (c) Find R out = R c = 47 K R in R out (b) Find A v = v out /v in v out = — i b R c v in = i b (R + R pi) A v = v out /v in = — i b R c / i b (R + R pi = — R c /(R + R pi) = — (47 K)/(100 K + 2. 5 K) = — = ibb e c i b

42 Graphical analysis Input circuit B-E voltage loop V BB = I B R B +V42 Graphical analysis Input circuit B-E voltage loop V BB = I B R B +V BE I B = (V BB — V BE )/R

43 Graphical construction of I B and V BE I B = (V BB - V43 Graphical construction of I B and V BE I B = (V BB — V BE )/R B If V BE = 0 , I B = V BB /R B If I B = 0 , V BE = V BB /R

44 Load line Output circuit C-E voltage loop V CC = I C R C +V44 Load line Output circuit C-E voltage loop V CC = I C R C +V CE I C = (V CC — V CE )/R

45 Graphical construction of I C and V CE V CC /R C I C =45 Graphical construction of I C and V CE V CC /R C I C = (V CC — V CE )/R C If V CE = 0 , I C = V CC /R C If I C = 0 , V CE = V

46 Graphical analysis Input signal Output signal 46 Graphical analysis Input signal Output signal

47 • Load-line A results in bias point Q A which is too close to V47 • 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. Bias point location effects

48 Basic single-stage BJT amplifier configurations   We will study 3 types of BJT amplifiers48 Basic single-stage BJT amplifier configurations We will study 3 types of BJT amplifiers • CE — common emitter, used for A V , A i , and general purpose • CE with R E — common emitter with R E , same as CE but more stable • CC common collector, used for A i , low output resistance, used as an output stage CB common base (not covered)

49 Common emitter amplifier ac equivalent circuit 49 Common emitter amplifier ac equivalent circuit

50 Common emitter amplifier R in R out+ V out -R in (Does not include source)50 Common emitter amplifier R in R out+ V out -R in (Does not include source) R in = R pi R out (Does not include load) R out = R C A V = V out /V in V out = — i b R C V in = i b (R s + R pi ) A V = — R C / (R s + R pi ) A i = i out /i in i out = — i b i in = i b A i = — i b i out

51 Common emitter with R E amplifier ac equivalent circuit 51 Common emitter with R E amplifier ac equivalent circuit

52 Common emitter with R E amplifier R in R out + V out -R in52 Common emitter with R E amplifier R in R out + V out -R in = V/i b V = i b R pi + (i b + i b )R E R in = R pi + (1 + )R E (usually large) R out = R C A V = V out /V in V out = — i b R C V in = i b R s + i b R pi + (i b + i b )R E A V = — R C / (R s + R pi + (1 + )R E ) (less than CE, but less sensitive to variations) A i = i out /i in i out = — i b i in = i b A i = — i b i out i b + i b+ V —

53 Common collector (emitter follower) amplifier b c e + v out - (v out at53 Common collector (emitter follower) amplifier b c e + v out — (v out at emitter) ac equivalent circuit

54 Common collector amplifier R in + v out -R in =  V/i b V54 Common collector amplifier R in + v out -R in = V/i b V = i b R pi + (i b + i b )R L R in = R pi + (1 + )R L A V = v out /v s v out = (i b + i b )R L v s = i b R s + i b R pi + (i b + i b )R L A V = (1+ R L / (R s + R pi + (1 + )R L ) (always < 1)i b + i b+ V —

55 Common collector amplifier R out+ v out - R out (don’t include RL, set Vs55 Common collector amplifier R out+ v out — R out (don’t include RL, set Vs = 0) R out = v out /- ( i b + i b ) v out = -i b R pi + -i b R s R out = (R pi + R s ) / (1+ (usually low)A i = i out /i in i out = i b + i b i in = i b A i = i b + i b

56 Prob + v out -  = 50 ac circuit CE with R E amp,56 Prob + v out — = 50 ac circuit CE with R E amp, because R E is in ac circuit Given R pi =V T /I B = 25 m. V(50)/. 2 m. A = 6. 25 K

57 Prob.  (a) Find Rin R in =  V/i b V  = i57 Prob. (a) Find Rin R in = V/i b V = i b R pi + (i b + i b )R E R in = R pi + (1 + )R E R in = 6. 25 K + (1 + )125 R in 12. 62 KR in+ V — i b + i b

58 Prob.  (b) Find A V  = v out /v s v out =58 Prob. (b) Find A V = v out /v s v out = — i b (R C ||R L ) v s = i b R s + i b R pi + (i b + i b )R E A V = — (R C ||R L ) / (R s + R pi + (1 + )R E ) A V = — (10 K||10 K) /(10 K + 6. 25 K i + (1 + )125) A V — i b + v out — i b

59 Prob.  (c) If v be is limited to 5 m. V, what is the59 Prob. (c) If v be is limited to 5 m. V, what is the largest signal at input and output? v be = i b R pi = 5 m. V i b = v be / R pi = 5 m. V/6. 25 K = 0. 8 A (ac value) v s = i b R s + i b R pi + (i b + i b )R E v s = (0. 8 A)10 K + (0. 8 A) 6. 25 K + (0. 8 A + ( 0. 8 A )125 v s 18 m. V i b + v out -+ v be —

60 Prob.  (c) If v be is limited to 5 m. V, what is the60 Prob. (c) If v be is limited to 5 m. V, what is the largest signal at input and output? v out = v s A V v out = 17. 4 m. V(-11) v out -191 m. V (ac value) i b + v out -+ v be —

61 Prob. Using this circuit, design an amp with: I E = 2 m. A A61 Prob. Using this circuit, design an amp with: I E = 2 m. A A V = -8 current in voltage divider I = 0. 2 m. A (CE amp because RE is not in ac circuit) = 100 Voltage divider Vcc/I = 9/0. 2 m. A = 45 K = R 1 + R 2 Choose V B 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

62 Prob. = 100 Find RE (input circuit) Use Thevenin equivalent B-E loop V BB =I62 Prob. = 100 Find RE (input circuit) Use Thevenin equivalent B-E loop V BB =I B R BB +V BE +I E R E using I B I E / R E = [V BB — V BE — (I E / )R BB ]/I E R E = [3 -. 7 — (2 m. A/ )10 K]/2 m. A R E = 1. 05 K + V BE — I EI

63 Prob. Find Rc (ac circuit) Rpi = V T /I B = 25 m. V(100)/263 Prob. Find Rc (ac circuit) Rpi = V T /I B = 25 m. V(100)/2 m. A = 1. 25 K Ro = V A /I C = 100/2 m. A = 50 K Av = v out /v in v out = -g m v be (Ro||Rc||RL) v be = 10 K||1. 2 K / [10 K+ 10 K||1. 2 K]v i Av = -g m (Ro||Rc||RL)(10 K||1. 2 K) / [10 K||1. 2 K +Rs] Set Av = -8, and solve for Rc, Rc 2 K + v out —

64 CE amplifier 64 CE amplifier

65 CE amplifier Av  -12. 2 65 CE amplifier Av -12.

66 FOURIER COMPONENTS OF TRANSIENT RESPONSE V($N_0009)  DC COMPONENT = -1. 226074 E-01  HARMONIC66 FOURIER COMPONENTS OF TRANSIENT RESPONSE V($N_0009) DC COMPONENT = -1. 226074 E-01 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONENT (DEG) PHASE (DEG) 1 1. 000 E+03 1. 581 E+00 1. 000 E+00 -1. 795 E+02 0. 000 E+00 2 2. 000 E+03 1. 992 E-01 1. 260 E-01 9. 111 E+01 2. 706 E+02 3 3. 000 E+03 2. 171 E-02 1. 374 E-02 -1. 778 E+02 1. 668 E+00 4 4. 000 E+03 3. 376 E-03 2. 136 E-03 -1. 441 E+02 3. 533 E+01 TOTAL HARMONIC DISTORTION = 1. 267478 E+01 PERCENT CE amplifier

67 CE amplifier with R E 67 CE amplifier with R

68 CE amplifier with R E Av  - 7. 5 68 CE amplifier with R E Av — 7.

69 FOURIER COMPONENTS OF TRANSIENT RESPONSE V($N_0009)  DC COMPONENT = -1. 353568 E-02  HARMONIC69 FOURIER COMPONENTS OF TRANSIENT RESPONSE V($N_0009) DC COMPONENT = -1. 353568 E-02 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONENT (DEG) PHASE (DEG) 1 1. 000 E+03 7. 879 E-01 1. 000 E+00 -1. 794 E+02 0. 000 E+00 2 2. 000 E+03 1. 604 E-02 2. 036 E-02 9. 400 E+01 2. 734 E+02 3 3. 000 E+03 5. 210 E-03 6. 612 E-03 -1. 389 E+02 4. 056 E+01 4 4. 000 E+03 3. 824 E-03 4. 854 E-03 -1. 171 E+02 6. 231 E+01 TOTAL HARMONIC DISTORTION = 2. 194882 E+00 PERCENT

70 Summary    Av THD CE -12. 2 12. 7 CE w/R E (R70 Summary Av THD CE -12. 2 12. 7% CE w/R E (R E = 100) -7. 5 2. 19%

71 Prob. + v out - • 2 stage amplifier (a) Find I C and V71 Prob. + v out — • 2 stage amplifier (a) Find I C and V C of each transistor • Both stages are the same (same for each stage) • Capacitively coupled = 100 RL=2 KRc=6. 8 K

72 Prob.  (a) Find I C and V C of each transistor (same for each72 Prob. (a) Find I C and V C of each transistor (same for each stage) B-E voltage loop V BB = I B R BB + V BE + I E R E where R BB = R 1 ||R 2 = 32 K V BB = V CC R 2 /(R 1 +R 2 ) = 4. 8 V, and I B I E / I E = [V BB — V BE ]/[R BB / + R E ] I E = 0. 97 m. A V C = V CC — I C R C V C = 15 -. 97(6. 8) V C = 8. 39 V + V C —

73 Prob.  b c e + v out -(b) find ac circuit b c e73 Prob. b c e + v out -(b) find ac circuit b c e R BB = R 1 ||R 2 = 100 K||47 K = 32 K Rpi = V T /I B = 25 m. V(100)/. 97 m. A 2. 6 K g m = I C /V T =. 97 m. A/25 m. V 39 m. A/V RL=2 K

74 Prob.  b c e + v out - (c) find Rin 1 = R74 Prob. b c e + v out — (c) find Rin 1 = R BB ||Rpi = 32 K||2. 6 K = 2. 4 K b c e Rin 1 (d) find Rin 2 = R BB ||Rpi = 2. 4 K Rin 2 find v b 1 /v i = Rin 1 / [Rin 1 + R S ] = 2. 4 K/[2. 4 K + 5 K] = 0. 32+ v b 1 — find v b 2 /v b 1 v b 2 = -g m v be 1 [R C ||R BB ||Rpi] v b 2 /v be 1 = -g m [R C ||R BB ||Rpi] v b 2 /v b 1 = -(39 m. A/V)[6. 8||32 K||2. 6 K] = -69. 1+ v b 2 — RL=2 K

75 Prob.  b c e + v out - (e) find v out /v b75 Prob. b c e + v out — (e) find v out /v b 2 v out = -g m v be 2 [R C ||R L ] v out /v be 2 = -g m [R C ||R L ] v b 2 /v b 1 = -(39 m. A/V)[6. 8 K||2 K] = -60. 3 b c e (f) find overall voltage gain v out /v i = (v b 1 /v i ) (v b 2 /v b 1 ) (v out /v b 2 ) v out /v i = (0. 32) (-69. 1) (-60. 3) v out /v i = 1332+ v b 1 — + v b 2 — RL=2 K

76 Prob.  Find I E 1 , I E 2 , V B 1 ,76 Prob. Find I E 1 , I E 2 , V B 1 , and V B 2 I E 2 = 2 m. A I E 1 = I 20 A + I B 2 I E 1 = I 20 A + I E 2 / 2 I E 1 = 20 A + 10 A I E 1 = 30 A IE 1 IE 2 Q 1 has = 20 Q 2 has = 200 Q 1 Q 2 I

77 Prob.  Find V B 1 , and V B 2 Use Thevenin equivalent V77 Prob. Find V B 1 , and V B 2 Use Thevenin equivalent V B 1 = V BB 1 — I B 1 (R BB 2 ) = 4. 5 — (30 A/20)500 K = 3. 8 V V B 2 = V B 1 — V BE = 3. 8 V — 0. 7 = 3. 1 V Q 1 has = 20 Q 2 has = 200 IB 2 + v B 1 — + v B 2 -I

78 Prob.  (b) find v out /v b 2 v out = (i b 278 Prob. (b) find v out /v b 2 v out = (i b 2 + i b 2 )R L v b 2 = (i b 2 + i b 2 )R L + i b 2 Rpi 2 v out /v b 2 = (1 + 2 )R L /[(1 + 2 )R L + Rpi 2 ] = (1 + )1 K/[(1 + )1 K + 2. 5 K] 0. 988 b 1 e 1 c 1 b 2 e 2 c 2 + v out -+ v B 2 — Rpi 2 = V T /I B 2 = V T 2 /I E 2 = 25 m. V(200)/2 m. A = 2. 5 K

79 Prob.  (b) find R in 2 = v b 2 /i b 2 v79 Prob. (b) find R in 2 = v b 2 /i b 2 v b 2 = (i b 2 + i b 2 )R L + i b 2 Rpi 2 R in 2 = v b 2 /i b 2 = (1 + )R L + Rp = (1 + )1 K + 2. 5 K 204 Kb 1 e 1 c 1 b 2 e 2 c 2+ v B 2 — R in

80 Prob. (c) find R in 1 = R BB 1 ||(v b 1 /i b80 Prob. (c) find R in 1 = R BB 1 ||(v b 1 /i b 1 ) = R BB 1 || [i b 1 Rpi 1 + (i b 1 + i b 1 )R in 2 ]/i b 1 = R BB 1 || [Rpi 1 + (1+ )R in 2 ], where Rpi 1 = V T 1 /I E 1 = 25 m. V(20)/30 A = 16. 7 K = 500 K||[16. 7 K + (1+ )204 K] 500 K b 1 e 1 c 1 b 2 e 2 c 2+ v B 1 -R in 1 i

81 Prob.  (c) find v e 1 /v b 1 v e 1 = (i81 Prob. (c) find v e 1 /v b 1 v e 1 = (i b 1 + i b 1 )R in 2 v b 1 = (i b 1 + i b 1 )R in 2 + i b 1 Rpi 1 v e 1 /v b 1 = (1 + 1 ) R in 2 /[(1 + 1 ) R in 2 + Rpi 1 ] = (1 + )204 K/[(1 + )204 K + 16. 7 K] 0. 996 b 1 e 1 c 1 b 2 e 2 c 2+ v e 1 -i B 1 + v B 1 —

82 Prob.  (d) find v b 1 /v i = R in 1 /[R S82 Prob. (d) find v b 1 /v i = R in 1 /[R S + R in 1 ] = 0. 82 b 1 e 1 c 1 b 2 e 2 c 2+ v b 1 — (e) find overall voltage gain v out /v i = (v b 1 /v i ) (v e 1 /v b 1 ) (v out /v e 1 ) v out /v i = (0. 82) (0. 99) v out /v i = 0.

83 Voltage outputs at each stage Output of  stage 2 Output of  stage 183 Voltage outputs at each stage Output of stage 2 Output of stage 1 Input

84 Current Input current. Input to stage 2 (ib 2) 84 Current Input current. Input to stage 2 (ib 2)

85 Current output current. Input to stage 2 (ib 2) 85 Current output current. Input to stage 2 (ib 2)

86 Current output current. Input to stage 2 (ib 2) Input current 86 Current output current. Input to stage 2 (ib 2) Input current

87 Power and current gain Input current = (Vi)/R in = 1/500 K = 2. 087 Power and current gain Input current = (Vi)/R in = 1/500 K = 2. 0 A output current= (Vout)/R L = (0. 81 V)/1 K= 0. 81 m. A current gain = 0. 81 m. A/ 2. 0 A = 405 Input power = (Vi)/R in = 1 x 1/500 K = 2. 0 W output power = (Vout)/R L = (0. 81 V)/1 K= 656 W power gain = 656 W/2 W =

88 BJT Output Characteristics • Plot Ic vs. Vce for multiple values of Vce and Ib88 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

89 Probe: BJT Output Characteristics 1 Result of probe 2 Add plot (plot menu) - Add89 Probe: BJT Output Characteristics 1 Result of probe 2 Add plot (plot menu) -> Add trace (trace menu) -> IC(Q 1) 3 Delete plot (plot menu)

90 BJT Output Characteristics: current gain Ib = 5 A (Each plot 10 A difference) 90 BJT Output Characteristics: current gain Ib = 5 A (Each plot 10 A difference) at Vce = 4 V, and Ib = 45 A = 8 m. A/45 A =

91 BJT Output Characteristics: transistor output resistance Ib = 5 A (Each plot 10 A difference)Ro91 BJT Output Characteristics: transistor output resistance Ib = 5 A (Each plot 10 A difference)Ro = 1/slope At Ib = 45 A, 1/slope = (8. 0 — 1. 6)V/(8. 5 — 7. 8)m. A Rout = 9. 1 K

92 CE Amplifier: Measurements with Spice Rin Rout 92 CE Amplifier: Measurements with Spice Rin Rout

93 Input Resistance Measurement Using SPICE • Replace source, Vs and Rs with Vin, measure 93 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 (1 K)(100 E-6) 1. 6 • Vin should have a small value so operating point does not change Vin 1 m. V

94 Rin Measurement • Transient  analysis 94 Rin Measurement • Transient analysis

95 Probe results I(C 2) Rin = 1 m. V/204 n. A   = 4.95 Probe results I(C 2) Rin = 1 m. V/204 n. A = 4. 9 K

96 Output Resistance Measurement Using SPICE • Replace load, RL with Vin, measure Rin = Vin/Iin96 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 (1 K)(100 E-6) 1. 6 • Vin should have a small value so operating point does not change Vin 1 m. V

97 Rout Measurement • Transient  analysis 97 Rout Measurement • Transient analysis

98 Probe results -I(C 1) Rout = 1 m. V/111 n. A   = 998 Probe results -I(C 1) Rout = 1 m. V/111 n. A = 9 K -I(C 1) is current in Vin flowing out of + terminal

99 DC Power measurements Power delivered by + 10 sources:  (10)(872 A) + (10)(877 A)99 DC Power measurements Power delivered by + 10 sources: (10)(872 A) + (10)(877 A) = 8. 72 m. W + 8. 77 m. W = 17. 4 m. W

100 ac Power Measurements of Load instantaneous power   average power Vout Vin 100 ac Power Measurements of Load instantaneous power average power Vout Vin