0 Chap. 4 BJT transistors Widely used in amplifier circuits Formed by junction of 3 materials npn or pnp structure

1 pnp transistor

2 Large current Operation of npn transistor

3 Modes of operation of a BJT transistor ModeBE junctionBC junction cutoffreverse biased reverse biased linear(active)forward biased reverse biased saturation forward biased forward biased

4 Summary of npn transistor behavior npn collector emitter base IBIB IEIE ICIC small current large current + V BE -

5 Summary of pnp transistor behavior pnp collector emitter base IBIB IEIE ICIC small current large current + V BE -

6 Summary of equations for a BJT I E  I C I C =  I B  is the current gain of the transistor  100 V BE = 0.7V(npn) V BE = -0.7V(pnp)

7 4.5 Graphical representation of transistor characteristics IBIB ICIC IEIE Output circuit Input circuit

8 Input characteristics Acts as a diode V BE  0.7V IBIB IBIB V BE 0.7V

9 Output characteristics ICIC ICIC V CE I B = 10  A I B = 20  A I B = 30  A I 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

10 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.

11 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 0.7V) ac output signal DC output signal IBIB 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.

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 V E = V B - V BE = 4V - 0.7V = 3.3V IEIE ICIC I E = (V E - 0)/R E = 3.3/3.3K = 1mA I C  I E = 1mA V C = 10 - I C R C = (4.7) = 5.3V

13 Example 4.6 B-E Voltage loop 5 = I B R B + V BE, solve for I B I B = (5 - V BE )/R B = (5-.7)/100k = 0.043mA I C =  I B = (100)0.043mA = 4.3mA V C = 10 - I C R C = (2) = 1.4V IEIE ICIC IBIB  = 100

14 Exercise 4.8 V E = = -0.7V IEIE ICIC IBIB  = 50 I E = (V E - -10)/R E = ( )/10K = 0.93mA I C  I E = 0.93mA I B = I B /  m  V C = 10 - I C R C = (5) = 5.35V

15 Prob Use a voltage divider, R B1 and R B2 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 = 3V and I = 0.2mA. IBIB I (a) R B1 and R B2 form a voltage divider. Assume I >> I B I = V CC /(R B1 + R B2 ).2mA = 9 /(R B1 + R B2 ) AND V B = V CC [R B2 /(R B1 + R B2 )] 3 = 9 [R B2 /(R B1 + R B2 )], Solve for R B1 and R B2. R B1 = 30K , and R B2 = 15K .

16 Prob Find the operating point Use the Thevenin equivalent circuit for the base Makes the circuit simpler V BB = V B = 3V R BB is measured with voltage sources grounded R BB = R B1 || R B2 = 30K  15K . 10K 

17 Prob Write B-E loop and C-E loop B-E loop C-E loop B-E loop V BB = I B R BB + V BE +I E R E I E =2.09 mA 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!

18 Exercise 4.24 (a) Find V C, V B, and V E, given:  = 100, V A = 100V I E = 1 mA I B  I E /  = 0.01mA V B = 0 - I B 10K = -0.1V V E = V B - V BE = = -0.8V V C = 10V - I C 8K = (8) = 2V VBVB

19 Example stage amplifier, 1st stage has an npn transistor; 2nd stage has an pnp transistor. I C =  I B I C  I E V BE = 0.7 (npn) = -0.7 (pnp)  = 100 Find I C1, I C2, V CE1, V CE2 Use Thevenin circuits.

20 Example 4.8 R BB1 = R B1 ||R B2 = 33K V BB1 = V CC [R B2 /(R B1 +R B2 )] V BB1 = 15[50K/150K] = 5V Stage 1 B-E loop V BB1 = I B1 R BB1 + V BE +I E1 R E1 Use I B1  I E1 /  5 = I E1 33K / I E1 3K I E1 = 1.3mA I B1 I E1

21 Example 4.8 C-E loop neglect IB2 because it is IB2 << IC1 I E1 I C1 V CC = I C1 R C1 + V CE1 +I E1 R E1 15 = 1.3(5) + V CE1 +1.3(3) V CE1 = 4.87V

22 Example 4.8 Stage 2 B-E loop I B2 I E2 V CC = I E2 R E2 + V EB +I B2 R BB2 + V BB2 15 = I E2 (2K) +.7 +I B2 (5K) (3) Use I B2  I E2 /  solve for I E2 I E2 = 2.8mA

23 Example 4.8 Stage 2 C-E loop I E2 I C2 V CC = I E2 R E2 + V EC2 +I C2 R C2 15 = 2.8(2) + V EC (2.7) solve for V EC2 V CE2 = 1.84V

24 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).

25 Summary of npn transistor behavior npn collector emitter base IBIB IEIE ICIC small current large current + V BE -

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 are at least two ac parameters determined from DC quantities.

27 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 0.7V) ac output signal DC output signal IBIB 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.

28 Transconductance = slope at Q point g m = di c /dv 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 (0.7V) 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)

29 ac input resistance  1/slope at Q point r  = dv BE /di 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 (0.7V) ac output signal DC output signal IBIB

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

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

32 Example 4.9 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 = 0.023mA C-E voltage loop V CE = 10 - I C R C V CE = 10 - (2.3)(3) V CE = 3.1V Q point: V CE = 3.1V, I C = 2.3mA

33 Example 4.9 ac problem Short DC sources, input and output circuits are separate, only coupled mathematically g m = I C /V T = 2.3mA/25mV = 92mA/V r  = V T / I B = 25mV/.023mA = 1.1K v be = v i [r  / (100K + r   011v i v out = - g m v be R C v out = - 92 (  011v i )3K v out /v i = v out - e bc + v be -

34 Exercise 4.24 Find g m, r , and r , given:  = 100, V A = 100V,I C =1 mA g m = I C /V T = 1 mA/25mV = 40 mA/V r  = V T / I B = 25mV/.01mA = 2.5K r 0 = output resistance of transistor r 0 = 1/slope of transistor output characteristics r 0 = | V A |/I C = 100K

35 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 are at least two ac parameters determined from DC quantities.

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

37 Circuit from Exercise 4.24 I E = 1 mA V C = 10V - I C 8K = (8) = 2V I B  I E /  = 0.01mA g m = I C /V T = 1 mA/25mV = 40 mA/V V B = 0 - I B 10K = -0.1V r  = V T / I B = 25mV/.01mA = 2.5K V E = V B - V BE = = -0.8V + V out -

38 ac equivalent circuit b e c v be = (R b ||R pi )/ [(R b ||R pi ) +R s ]v i v be = 0.5v i v out = -(g m v be )||(R o ||R c ||R L ) v out = -154v be A v = v out /v i = v out - Neglecting R o v out = -(g m v be )||(R c ||R L ) A v = v out /v i = - 80

39 Prob V out -  =100

40 Prob V out - (a) Find R in R in = R pi = V T /I B = (25mV)100/.1 = 2.5K  (c) Find R out R out = R c = 47K  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) = -  (47K)/(100K + 2.5K) = -  =  ib b e c ibib

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

42 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 V BB /R B

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

44 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 CC

45 Graphical analysis Input signal Output signal

46 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

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)

48 Common emitter amplifier ac equivalent circuit

49 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 AVAV = 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 ) AiAi = i out /i in i out = -  i b i in = i b A i = -  ibib i out

50 Common emitter with R E amplifier ac equivalent circuit

51 Common emitter with R E amplifier R in R out + V out - R in 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 out = R C AVAV = 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) AiAi = i out /i in i out = -  i b i in = i b A i = -  ibib i out i b +  i b +V-+V-

52 Common collector (emitter follower) amplifier bc e + v out - (v out at emitter)ac equivalent circuit

53 Common collector amplifier R in + v out - R in R in = V/i b V = i b R pi + (i b +  i b )R L R in = R pi + (1 +  )R L AVAV = 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) ibib i b +  i b +V-+V-

54 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) AiAi = i out /i in i out = i b +  i b i in = i b A i =  ibib i b +  i b

55 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 = 25mV(50)/.2mA = 6.25K

56 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 = 6.25K + (1 +  )125 R in  12.62K R in +V-+V- ibib i b +  i b

57 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 = -  (10K||10K) /(10K K i + (1 +  )125) A V  -  ibib i b +  i b + v out -  i b

58 Prob (c) If v be is limited to 5mV, what is the largest signal at input and output? v be = i b R pi = 5mV i b = v be /R pi = 5mV/6.25K = 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)10K + (0.8  A) 6.25K + (0.8  A + (  0.8  A )125 v s  18mV ibib i b +  i b + v out - + v be -

59 Prob (c) If v be is limited to 5mV, what is the largest signal at input and output? v out = v s A V v out = 17.4mV(-11) v out  -191mV (ac value) ibib i b +  i b + v out - + v be -

60 Prob Using this circuit, design an amp with: I E = 2mA A V = -8 current in voltage divider I = 0.2mA (CE amp because RE is not in ac circuit)  = 100 Voltage divider Vcc/I = 9/0.2mA = 45K 45K = 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 ) = 3V Combining gives, R 1 = 30K, R 2 = 15K

61 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 = [ (2mA/  )10K]/2mA R E = 1.05K  + V BE - IEIE IBIB

62 Prob Find Rc (ac circuit) Rpi = V T /I B = 25mV(100)/2mA = 1.25K Ro = V A /I C = 100/2mA = 50K Av = v out /v in v out = -g m v be (Ro||Rc||RL) v be = 10K||1.2K / [10K+ 10K||1.2K]v i Av = -g m (Ro||Rc||RL)(10K||1.2K) / [10K||1.2K +Rs] Set Av = -8, and solve for Rc, Rc  2K + v out -

63 CE amplifier

64 CE amplifier Av  -12.2

65 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 CE amplifier

66 CE amplifier with R E

67 CE amplifier with R E Av  - 7.5

68 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

69 Summary AvTHD CE % CE w/R E (R E = 100) %

70 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=2K Rc=6.8K

71 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 = 32K V BB = V CC R 2 /(R 1 +R 2 ) = 4.8V, and I B  I E /  I E = [V BB - V BE ]/[R BB /  + R E ] I E = 0.97mA V C = V CC - I C R C V C = (6.8) V C = 8.39V +VC-+VC-

72 Prob bc e + v out - (b) find ac circuit b c e R BB = R 1 ||R 2 = 100K||47K = 32K  Rpi = V T /I B = 25mV(100)/.97mA  2.6K  g m = I C /V T =.97mA/25mV  39mA/V RL=2K

73 Prob bc e + v out - (c) find Rin 1 Rin 1 = R BB ||Rpi = 32K||2.6K = 2.4K  b c e Rin 1 (d) find Rin 2 Rin 2 = R BB ||Rpi = 2.4K  Rin 2 find v b1 /v i = Rin 1 / [Rin 1 + R S ] = 2.4K/[2.4K + 5K] = v b1 - find v b2 /v b1 v b2 = -g m v be1 [R C ||R BB ||Rpi] v b2 /v be1 = -g m [R C ||R BB ||Rpi] v b2 /v b1 = -(39mA/V)[6.8||32K||2.6K] = v b2 - RL=2K

74 Prob bc e + v out - (e) find v out /v b2 v out = -g m v be2 [R C ||R L ] v out /v be2 = -g m [R C ||R L ] v b2 /v b1 = -(39mA/V)[6.8K||2K] = b c e (f) find overall voltage gain v out /v i = (v b1 /v i ) (v b2 /v b1 ) (v out /v b2 ) v out /v i = (0.32) (-69.1) (-60.3) v out /v i = v b1 - + v b2 - RL=2K

75 Prob Find I E1, I E2, V B1, and V B2 I E2 = 2mA I E1 = I 20  A + I B2 I E1 = I 20  A + I E2 /  2 I E1 = 20  A + 10  A I E1 = 30  A IE1 IE2 Q1 has   = 20 Q2 has   = 200 Q1 Q2 IB2

76 Prob Find V B1, and V B2 Use Thevenin equivalent V B1 = V BB1 - I B1 (R BB2 ) = (30  A/20)500K = 3.8V V B2 = V B1 - V BE = 3.8V = 3.1V Q1 has   = 20 Q2 has   = 200 IB2 + v B1 - + v B2 - IB1

77 Prob (b) find v out /v b2 v out = (i b2 +   i b2 )R L v b2 = (i b2 +   i b2 )R L + i b2 Rpi 2 v out /v b2 = (1 +  2 )R L /[(1 +  2 )R L + Rpi 2 ] = (1 +  )1K/[(1 +  )1K + 2.5K]  b1b1 e1e1 c1c1 b2b2 e2e2 c2c2 + v out - + v B2 - Rpi 2 = V T /I B2 = V T  2 /I E2 = 25mV(200)/2mA = 2.5K 

78 Prob (b) find R in2 = v b2 /i b2 v b2 = (i b2 +   i b2 )R L + i b2 Rpi 2 R in2 = v b2 /i b2 = (1 +   )R L + Rp = (1 +  )1K + 2.5K  204K b1b1 e1e1 c1c1 b2b2 e2e2 c2c2 + v B2 - R in2

79 Prob (c) find R in1 = R BB1 ||(v b1 /i b1 ) = R BB1 || [i b1 Rpi 1 + (i b1 +   i b1 )R in2 ]/i b1 = R BB1 || [Rpi 1 + (1+   )R in2 ], where Rpi 1 = V T  1 /I E1 = 25mV(20)/30  A = 16.7K = 500K||[16.7K + (1+  )204K]  500K  b1b1 e1e1 c1c1 b2b2 e2e2 c2c2 + v B1 - R in1 i B1

80 Prob (c) find v e1 /v b1 v e1 = (i b1 +   i b1 )R in2 v b1 = (i b1 +   i b1 )R in2 + i b1 Rpi 1 v e1 /v b1 = (1 +  1 ) R in2 /[(1 +  1 ) R in2 + Rpi 1 ] = (1 +  )204K/[(1 +  )204K K]  b1b1 e1e1 c1c1 b2b2 e2e2 c2c2 + v e1 - i B1 + v B1 -

81 Prob (d) find v b1 /v i v b1 /v i = R in1 /[R S + R in1 ] = 0.82 b1b1 e1e1 c1c1 b2b2 e2e2 c2c2 + v b1 - (e) find overall voltage gain v out /v i = (v b1 /v i ) (v e1 /v b1 ) (v out /v e1 ) v out /v i = (0.82) (0.99) (0.99) v out /v i = 0.81

82 (Prob. 4.96) Voltage outputs at each stage Output of stage 2 Output of stage 1 Input

83 (Prob. 4.96) Current Input current Input to stage 2 (ib2)

84 (Prob. 4.96) Current output current Input to stage 2 (ib2)

85 (Prob. 4.96) Current output current Input to stage 2 (ib2) Input current

86 (Prob. 4.96) Power and current gain Input current = (Vi)/R in = 1/500K = 2.0  A output current= (Vout)/R L = (0.81V)/1K= 0.81mA current gain = 0.81mA/ 2.0  A = 405 Input power = (Vi) (Vi)/R in = 1 x 1/500K = 2.0  W output power = (Vout) (Vout)/R L = (0.81V) (0.81V)/1K= 656  W power gain = 656  W/2  W = 329

87 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

88 Probe: BJT Output Characteristics 1 Result of probe 2 Add plot (plot menu) -> Add trace (trace menu) -> IC(Q1) 3 Delete plot (plot menu)

89 BJT Output Characteristics: current gain Ib = 5  A (Each plot 10  A difference)  at Vce = 4V, and Ib = 45  A  = 8mA/45  A = 178

90 BJT Output Characteristics: transistor output resistance Ib = 5  A (Each plot 10  A difference) Ro = 1/slope At Ib = 45  A, 1/slope = ( )V/( )mA Rout = 9.1K 

91 CE Amplifier: Measurements with Spice Rin Rout

92 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

93 Rin Measurement Transient analysis

94 Probe results I(C2) Rin = 1mV/204nA = 4.9K 

95 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

96 Rout Measurement Transient analysis

97 Probe results -I(C1) Rout = 1mV/111nA = 9K  -I(C1) is current in Vin flowing out of + terminal

98 DC Power measurements Power delivered by + 10 sources: (10)(872  A) + (10)(877  A) = 8.72mW mW = 17.4mW

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