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Bjt transistors converted

This document contains lecture slides about BJT transistors and amplifier circuits. It discusses BJT biasing, modes of operation, input and output characteristics, and analysis of transistor circuits using voltage loops. Examples are provided to demonstrate analyzing DC operating points and determining AC quantities like transconductance and input resistance. The document also introduces small-signal models for analyzing transistor amplifiers.

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• BJT biasing and Amplifier • Unit-1 and 2 Prepared by S.Vijayakumar, AP/ECE Ramco Institute of Technology Academic year (2018-2019 odd sem)
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 1 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 2
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 3 pnp transistor
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 4
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 5
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 6 Large current Operation of npn transistor
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 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 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 8 Summary of npn transistor behavior npn collector emitter base IB IE IC small current large current + VBE -
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 9 Summary of pnp transistor behavior pnp collector emitter base IB IE IC small current large current + VBE -
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 10 Summary of equations for a BJT IE  IC IC = bIB b is the current gain of the transistor  100 VBE = 0.7V(npn) VBE = -0.7V(pnp)
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 11 4.5 Graphical representation of transistor characteristics IB IC IE Output circuit Input circuit
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 12 Input characteristics •Acts as a diode •VBE 0.7V IB IB VBE 0.7V
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 13 Output characteristics IC IC VCE IB = 10mA IB = 20mA IB = 30mA IB = 40mA Cutoff region •At a fixed IB, IC is not dependent on VCE •Slope of output characteristics in linear region is near 0 (scale exaggerated) Early voltage
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 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 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 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 IE IC IE = (VE - 0)/RE = 3.3/3.3K = 1mA IC  IE = 1mA VC = 10 - ICRC = 10 - 1(4.7) = 5.3V
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 16 Example 4.6 B-E Voltage loop 5 = IBRB + VBE, solve forIB IB = (5 - VBE)/RB = (5-.7)/100k = 0.043mA IC = bIB = (100)0.043mA = 4.3mA VC = 10 - ICRC = 10 - 4.3(2) = 1.4V IE IC IB b = 100
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 17 Exercise 4.8 VE = 0 - .7 = - 0.7V IE IC IB b = 50 IE = (VE - -10)/RE = (-.7 +10)/10K = 0.93mA IC  IE = 0.93mA IB = IC/b = .93mA/50 = 18.6mA VC = 10 - ICRC = 10 - .93(5) = 5.35V VCE = 5.35 - -0.7 = 6.05V
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 18 Summary of npn transistor behavior npn collector emitter base IB IE IC small current large current + VBE -
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 19 Summary of equations for a BJT IE  IC IC = bIB b is the current gain of the transistor  100 VBE = 0.7V(npn) VBE = -0.7V(pnp)
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 20 Prob. 4.32 •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. IB I (a) RB1 and RB2 form a voltage divider. Assume I >> IB I = VCC/(RB1 + RB2) .2mA = 9 /(RB1 + RB2) AND VB = VCC[RB2/(RB1 + RB2)] 3 = 9 [RB2/(RB1 + RB2)], Solve for RB1 and RB2. RB1 = 30KW, and RB2 = 15KW.
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 21 Prob. 4.32 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 = 30KW || 15KW = . 10KW
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 22 Prob. 4.32 Write B-E loop and C-E loop B-E loop C-E loop B-E loop VBB = IBRBB + VBE +IERE 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 23 Example 4.8 •2-stage amplifier, 1st stage has an npn transistor; 2nd stage has an pnp transistor. IC = bIB IC  IE VBE = 0.7(npn) = -0.7(pnp) b = 100 Find IC1, IC2, VCE1, VCE2 •Use Thevenin circuits.
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 24 Example 4.8 •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/ b 5 = IE133K/ 100 + .7 + IE13K IE1 = 1.3mA IB1 IE1
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 25 Example 4.8 C-E loop neglect IB2 because it is IB2 << IC1 IE1 IC1 VCC = IC1RC1 + VCE1 +IE1RE1 15 = 1.3(5) + VCE1 +1.3(3) VCE1= 4.87V
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 26 Example 4.8 Stage 2 •B-E loop IB2 IE2 VCC = IE2RE2 + VEB +IB2RBB2 + VBB2 15 = IE2(2K) + .7 +IB2 (5K)+ 4.87 + 1.3(3) Use IB2  IE2/ b, solve for IE2 IE2 = 2.8mA
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 27 Example 4.8 Stage 2 •C-E loop IE2 IC2 VCC = IE2RE2 + VEC2 +IC2RC2 15 = 2.8(2) + VEC2 + 2.8(2.7) solve for VEC2 VCE2 = 1.84V
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 28 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 = bIB •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 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 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 30 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. Transconductance ac input signal (DC input signal 0.7V) ac output signal DC output signal IB 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 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 31 Transconductance = slope at Q point gm = dic/dvBE|ic = ICQ where IC = IS[exp(-VBE/VT)-1]; the equation for a diode. Transconductance ac input signal DC input signal (0.7V) ac output signal DC output signal gm = ISexp(-VBE/VT) (1/VT) gm  IC/VT (A/V)
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 32 ac input resistance  1/slope at Q point rp = dvBE/dib|ic = ICQ rp  VT /IB re  VT /IE ac input resistance of transistor ac input signal DC input signal (0.7V) ac output signal DC output signal IB
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 33 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 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, 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 35 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 = 0.023mA C-E voltage loop VCE = 10 - ICRC VCE = 10 - (2.3)(3) VCE = 3.1V Q point: VCE = 3.1V, IC = 2.3mA
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 36 Example 4.9 ac problem Short DC sources, input and output circuits are separate, only coupled mathematically gm = IC/VT = 2.3mA/25mV = 92mA/V rp = VT/ IB = 25mV/.023mA = 1.1K vbe= vi [rp / (100K + rp)] = 0.011vi vout = - gm vbeRC vout = - 92(0.011vi)3K vout/vi = -3.04 + vout - e b c + vbe -
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 37 Exercise 4.24 (a) Find VC, VB, and VE, given: b = 100, VA = 100V IE = 1 mA IB  IE/b= 0.01mA VB = 0 - IB10K = -0.1V VE = VB - VBE = -0.1 - 0.7 = -0.8V VC = 10V - IC8K = 10 - 1(8) = 2V VB
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 38 Exercise 4.24 (b) Find gm, rp, and r0, given: b = 100, VA = 100V gm = IC/VT = 1 mA/25mV = 40 mA/V rp = VT/ IB = 25mV/.01mA = 2.5K r0 = output resistance of transistor r0 = 1/slope of transistor output characteristics r0 = | VA|/IC = 100K
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 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 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 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, 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 41 Circuit from Exercise 4.24 IE = 1 mA VC = 10V - IC8K = 10 - 1(8) = 2V IB  IE/b= 0.01mA gm = IC/VT = 1 mA/25mV = 40 mA/V VB = 0 - IB10K = -0.1V rp = VT/ IB = 25mV/.01mA = 2.5K VE = VB - VBE = -0.1 - 0.7 = -0.8V + Vout -
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 42 ac equivalent circuit b e c vbe = (Rb||Rpi)/ [(Rb||Rpi) +Rs]vi vbe = 0.5vi vout = -(gmvbe)(Ro||Rc ||RL) vout = -154vbe Av =vout/vi = - 77 + vout - Neglecting Ro vout = -(gmvbe)(Rc ||RL) Av =vout/vi = - 80
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 43 Prob. 4.76 + Vout - b=100
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 44 Prob. 4.76 + Vout - (a) Find Rin Rin = Rpi = VT/IB = (25mV)100/.1 = 2.5KW (c) Find Rout Rout = Rc = 47KW Rin Rout (b) Find Av = vout/vin vout = - bib Rc vin = ib (R +Rpi) Av = vout/vin = - bib Rc/ ib (R +Rpi ) = - bRc/(R +Rpi) = - 100(47K)/(100K + 2.5K) = - 37.6 = bib b e c ib
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 45 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 46 Graphical construction of IB and VBE IB = (VBB - VBE)/RB If VBE = 0, IB = VBB/RB If IB = 0, VBE = VBB VBB/RB
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 47 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 48 Graphical construction of IC and VCE VCC/RC IC = (VCC - VCE)/RC If VCE = 0, IC = VCC/RC If IC = 0, VCE = VCC
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 49 Graphical analysis Input signal Output signal
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 50 •Load-line A results in bias point QA which is too close to VCC and thus limits the positive swing of vCE. •Load-line B results in an operating point too close to the saturation region, thus limiting the negative swing of vCE. Bias point location effects
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 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 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 52 Common emitter amplifier ac equivalent circuit
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 53 Common emitter amplifier Rin Rout + Vout - Rin (Does not include source) Rin =Rpi Rout (Does not include load) Rout =RC AV = Vout/Vin Vout = - bibRC Vin = ib(Rs + Rpi) AV = - bRC/ (Rs + Rpi) Ai = iout/iin iout = - bib iin = ib Ai = - b ib iout
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 54 Common emitter with RE amplifier ac equivalent circuit
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 55 Common emitter with RE amplifier Rin Rout + Vout - Rin Rin =V/ib V = ib Rpi + (ib + bib)RE Rin = Rpi + (1 + b)RE (usually large) Rout Rout =RC AV = Vout/Vin Vout = - bibRC Vin = ib Rs + ib Rpi + (ib + bib)RE AV = - bRC/ (Rs + Rpi + (1 + b)RE) (less than CE, but less sensitive to b variations) Ai = iout/iin iout = - bib iin = ib Ai = - b ib iout ib + bib + V -
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 56 Common collector (emitter follower) amplifier b c e + vout - (vout at emitter) ac equivalent circuit
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 57 Common collector amplifier Rin + vout - Rin Rin =V/ib V = ib Rpi + (ib + bib)RL Rin = Rpi + (1 + b)RL AV = vout/vs vout = (ib + bib)RL vs = ib Rs + ib Rpi + (ib + bib)RL AV = (1+ b)RL/ (Rs + Rpi + (1 + b)RL) (always < 1) ib ib + bib + V -
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 58 Common collector amplifier Rout + vout - Rout (don’t include RL, set Vs = 0) Rout = vout /- (ib + bib) vout = -ib Rpi + -ibRs Rout = (Rpi + Rs)/ (1+ b) (usually low) Ai = iout/iin iout = ib + bib iin = ib Ai = b + 1 ib ib + bib
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 59 Prob. 4.84 + vout - b = 50 ac circuit CE with RE amp, because RE is in ac circuit Given Rpi =VT/IB = 25mV(50)/.2mA = 6.6K
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 60 Prob. 4.84 (a) Find Rin Rin =V/ib V = ib Rpi + (ib + bib)RE Rin = Rpi + (1 + b)RE Rin = 6.6K+ (1 + 50)125 Rin  13K Rin + V - ib ib + bib
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 61 Prob. 4.84 (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 (10K||10K)/(10K + 6.6Ki + (1 + 50)125) AV  - 11 ib ib + bib + vout - -bib
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 62 Prob. 4.84 (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 = 0.76mA (ac value) vs = ib Rs + ib Rpi + (ib + bib)RE vs = (0.76mA)10K + (0.76mA) 6.6K + (0.76mA + (50)0.76mA )125 vs  17.4mV ib ib + bib + vout - + vbe -
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 63 Prob. 4.84 (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) ib ib + bib + vout - + vbe -
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 64 Prob. 4.79 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) b = 100 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) = 3V Combining gives, R1 = 30K, R2 = 15K
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 65 Prob. 4.79 b = 100Find RE (input circuit) Use Thevenin equivalent B-E loop VBB=IBRBB+VBE+IERE using IB  IE/b RE = [VBB - VBE - (IE/b)RBB]/IE RE = [3 - .7 - (2mA/100)10K]/2mA RE = 1.05KW + VBE - IE IB
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 66 Prob. 4.79 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 + vout -
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 67 CE amplifier
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 68 CE amplifier Av  -12.2
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 69 FOURIER COMPONENTS OF TRANSIENT RESPONSE V($N_0009) DC COMPONENT = -1.226074E-01 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG) 1 1.000E+03 1.581E+00 1.000E+00 -1.795E+02 0.000E+00 2 2.000E+03 1.992E-01 1.260E-01 9.111E+01 2.706E+02 3 3.000E+03 2.171E-02 1.374E-02 -1.778E+02 1.668E+00 4 4.000E+03 3.376E-03 2.136E-03 -1.441E+02 3.533E+01 TOTAL HARMONIC DISTORTION = 1.267478E+01 PERCENT CE amplifier
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 70 CE amplifier with RE
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 71 CE amplifier with RE Av  - 7.5
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 72 FOURIER COMPONENTS OF TRANSIENT RESPONSE V($N_0009) DC COMPONENT = -1.353568E-02 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG) 1 1.000E+03 7.879E-01 1.000E+00 -1.794E+02 0.000E+00 2 2.000E+03 1.604E-02 2.036E-02 9.400E+01 2.734E+02 3 3.000E+03 5.210E-03 6.612E-03 -1.389E+02 4.056E+01 4 4.000E+03 3.824E-03 4.854E-03 -1.171E+02 6.231E+01 TOTAL HARMONIC DISTORTION = 2.194882E+00 PERCENT
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 73 Summary Av THD CE -12.2 12.7% CE w/RE (RE = 100) -7.5 2.19%
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 74 Prob. 4.83 + vout - •2 stage amplifier (a) Find IC and VC of each transistor •Both stages are the same (same for each stage) •Capacitively coupled b = 100 RL=2K Rc=6.8K
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 75 Prob. 4.83 (a) Find IC and VC of each transistor (same for each stage) B-E voltage loop VBB = IBRBB + VBE + IERE where RBB = R1||R2 = 32K VBB = VCCR2/(R1+R2) = 4.8V, and IB  IE/b IE = [VBB - VBE ]/[RBB/b + RE] IE = 0.97mA VC = VCC - ICRC VC = 15 - .97(6.8) VC = 8.39V + VC -
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 76 Prob. 4.83 b c e + vout - (b) find ac circuit b c e RBB = R1||R2 = 100K||47K = 32KW Rpi = VT/IB = 25mV(100)/.97mA  2.6KW gm = IC/VT = .97mA/25mV  39mA/V RL=2K
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 77 Prob. 4.83 b c e + vout - (c) find Rin1 Rin1 = RBB||Rpi = 32K||2.6K = 2.4KW b c e Rin1 (d) find Rin2 Rin2 = RBB||Rpi = 2.4KW Rin2 find vb1/vi = Rin1/[Rin1 + RS] = 2.4K/[2.4K + 5K] = 0.32 + vb1 - find vb2/vb1 vb2 = -gmvbe1[RC||RBB||Rpi] vb2/vbe1 = -gm[RC||RBB||Rpi] vb2/vb1 = -(39mA/V)[6.8||32K||2.6K] = -69.1 + vb2 - RL=2K
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 78 Prob. 4.83 b c e + vout - (e) find vout/vb2 vout = -gmvbe2[RC||RL] vout/vbe2 = -gm[RC||RL] vb2/vb1 = -(39mA/V)[6.8K||2K] = -60.3 b c e (f) find overall voltage gain vout/vi = (vb1/vi) (vb2/vb1) (vout/vb2) vout/vi = (0.32) (-69.1) (-60.3) vout/vi = 1332 + vb1 - + vb2 - RL=2K
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 79 Prob. 4.96 Find IE1, IE2, VB1, and VB2 IE2 = 2mA IE1 = I20mA + IB2 IE1 = I20mA + IE2/b2 IE1 = 20mA + 10mA IE1 = 30mA IE1 IE2 Q1 has b1 = 20 Q2 has b2 = 200 Q1 Q2 IB2
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 80 Prob. 4.96 Find VB1, and VB2 Use Thevenin equivalent VB1 = VBB1 - IB1(RBB1) = 4.5 - (30mA/20)500K = 3.8V VB2 = VB1 - VBE = 3.8V - 0.7 = 3.1V Q1 has b1 = 20 Q2 has b2 = 200 IB2+ vB1 - + vB2 - IB1
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 81 Prob. 4.96 (b) find vout/vb2 vout = (ib2 + b2ib2)RL vb2 = (ib2 + b2ib2)RL + ib2Rpi2 vout/vb2 = (1 + b2)RL/[(1 + b2)RL + Rpi2] = (1 + 200)1K/[(1 + 200)1K+ 2.5K]  0.988 b1 e1 c1 b2 e2 c2 + vout - + vB2 - Rpi2 = VT/IB2 = VT b2/IE2 = 25mV(200)/2mA = 2.5KW
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 82 Prob. 4.96 (b) find Rin2 = vb2/ib2 vb2 = (ib2 + b2ib2)RL + ib2Rpi2 Rin2 = vb2/ib2 = (1 + b2)RL + Rpi2 = (1 + 200)1K + 2.5K  204K b1 e1 c1 b2 e2 c2 + vB2 -Rin2
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 83 Prob. 4.96 (c) find Rin1 = RBB1||(vb1/ib1) = RBB1|| [ib1Rpi1 +(ib1 + b1ib1)Rin2]/ib1 = RBB1|| [Rpi1 +(1+ b1)Rin2], where Rpi1 = VT b1/IE1 = 25mV(20)/30mA = 16.7K = 500K||[16.7K + (1+ 20)204K]  500KW b1 e1 c1 b2 e2 c2 + vB1 - Rin1 iB1
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 84 Prob. 4.96 (c) find ve1/vb1 ve1 = (ib1 + b1ib1)Rin2 vb1 = (ib1 + b1ib1)Rin2 + ib1Rpi1 ve1/vb1 = (1 + b1) Rin2 /[(1 + b1) Rin2 + Rpi1] = (1 + 20)204K/[(1 + 20)204K+ 16.7K]  0.996 b1 e1 c1 b2 e2 c2 + ve1 - iB1+ vB1 -
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 85 Prob. 4.96 (d) find vb1/vi vb1/vi = Rin1/[RS + Rin1] = 0.82 b1 e1 c1 b2 e2 c2 + vb1 - (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 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 86 (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 87 (Prob. 4.96) Current Input current Input to stage 2 (ib2)
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 88 (Prob. 4.96) Current output current Input to stage 2 (ib2)
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 89 (Prob. 4.96) Current output current Input to stage 2 (ib2) Input current
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 90 (Prob. 4.96) Power and current gain Input current = (Vi)/Rin = 1/500K = 2.0mA output current= (Vout)/RL = (0.81V)/1K= 0.81mA current gain = 0.81mA/ 2.0mA = 405 Input power = (Vi) (Vi)/Rin = 1 x 1/500K = 2.0mW output power = (Vout) (Vout)/RL = (0.81V) (0.81V)/1K= 656mW power gain = 656mW/2mW = 329
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 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 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 92 Probe: BJT Output Characteristics 1 Result of probe 2 Add plot (plot menu) -> Add trace (trace menu) -> IC(Q1) 3 Delete plot (plot menu)
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 93 BJT Output Characteristics: current gain Ib = 5mA (Each plot 10mA difference) b at Vce = 4V, and Ib = 45mA b = 8mA/45mA = 178
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 94 BJT Output Characteristics: transistor output resistance Ib = 5mA (Each plot 10mA difference) Ro = 1/slope At Ib = 45mA, 1/slope = (8.0 - 1.6)V/(8.5 - 7.8)mA Rout = 9.1KW
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 95 CE Amplifier: Measurements with Spice Rin Rout
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 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/wC|. For C = 100mF, w = 1KHz, Rcap = 1/2p(1K)(100E-6)  1.6W •Vin should have a small value so operating point does not change Vin  1mV
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 97 Rin Measurement •Transient analysis
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 98 Probe results I(C2) Rin = 1mV/204nA = 4.9KW
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 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/wC|. For C = 100mF, w = 1KHz, Rcap = 1/2p(1K)(100E-6)  1.6W •Vin should have a small value so operating point does not change Vin  1mV
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 100 Rout Measurement •Transient analysis
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 101 Probe results -I(C1) Rout = 1mV/111nA = 9KW -I(C1) is current in Vin flowing out of + terminal
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 102 DC Power measurements Power delivered by + 10 sources: (10)(872mA) + (10)(877mA) = 8.72mW + 8.77mW = 17.4mW
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 103 ac Power Measurements of Load instantaneous power average power Vout Vin
Reference • www.electronics.teipir.gr/personalpages/.../1/.. ./ECE3111BJTTransistors. ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 104

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Bjt transistors converted

  • 1.
    • BJT biasingand Amplifier • Unit-1 and 2 Prepared by S.Vijayakumar, AP/ECE Ramco Institute of Technology Academic year (2018-2019 odd sem)
  • 2.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 1 Chap. 4 BJT transistors •Widely used in amplifier circuits •Formed by junction of 3 materials •npn or pnp structure
  • 3.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 2
  • 4.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 3 pnp transistor
  • 5.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 4
  • 6.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 5
  • 7.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 6 Large current Operation of npn transistor
  • 8.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 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
  • 9.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 8 Summary of npn transistor behavior npn collector emitter base IB IE IC small current large current + VBE -
  • 10.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 9 Summary of pnp transistor behavior pnp collector emitter base IB IE IC small current large current + VBE -
  • 11.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 10 Summary of equations for a BJT IE  IC IC = bIB b is the current gain of the transistor  100 VBE = 0.7V(npn) VBE = -0.7V(pnp)
  • 12.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 11 4.5 Graphical representation of transistor characteristics IB IC IE Output circuit Input circuit
  • 13.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 12 Input characteristics •Acts as a diode •VBE 0.7V IB IB VBE 0.7V
  • 14.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 13 Output characteristics IC IC VCE IB = 10mA IB = 20mA IB = 30mA IB = 40mA Cutoff region •At a fixed IB, IC is not dependent on VCE •Slope of output characteristics in linear region is near 0 (scale exaggerated) Early voltage
  • 15.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 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.
  • 16.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 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 IE IC IE = (VE - 0)/RE = 3.3/3.3K = 1mA IC  IE = 1mA VC = 10 - ICRC = 10 - 1(4.7) = 5.3V
  • 17.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 16 Example 4.6 B-E Voltage loop 5 = IBRB + VBE, solve forIB IB = (5 - VBE)/RB = (5-.7)/100k = 0.043mA IC = bIB = (100)0.043mA = 4.3mA VC = 10 - ICRC = 10 - 4.3(2) = 1.4V IE IC IB b = 100
  • 18.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 17 Exercise 4.8 VE = 0 - .7 = - 0.7V IE IC IB b = 50 IE = (VE - -10)/RE = (-.7 +10)/10K = 0.93mA IC  IE = 0.93mA IB = IC/b = .93mA/50 = 18.6mA VC = 10 - ICRC = 10 - .93(5) = 5.35V VCE = 5.35 - -0.7 = 6.05V
  • 19.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 18 Summary of npn transistor behavior npn collector emitter base IB IE IC small current large current + VBE -
  • 20.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 19 Summary of equations for a BJT IE  IC IC = bIB b is the current gain of the transistor  100 VBE = 0.7V(npn) VBE = -0.7V(pnp)
  • 21.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 20 Prob. 4.32 •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. IB I (a) RB1 and RB2 form a voltage divider. Assume I >> IB I = VCC/(RB1 + RB2) .2mA = 9 /(RB1 + RB2) AND VB = VCC[RB2/(RB1 + RB2)] 3 = 9 [RB2/(RB1 + RB2)], Solve for RB1 and RB2. RB1 = 30KW, and RB2 = 15KW.
  • 22.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 21 Prob. 4.32 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 = 30KW || 15KW = . 10KW
  • 23.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 22 Prob. 4.32 Write B-E loop and C-E loop B-E loop C-E loop B-E loop VBB = IBRBB + VBE +IERE C-E loop VCC = ICRC + VCE +IERE Solve for,IC, VCE, and IB. This is how all DC circuits are analyzed and designed!
  • 24.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 23 Example 4.8 •2-stage amplifier, 1st stage has an npn transistor; 2nd stage has an pnp transistor. IC = bIB IC  IE VBE = 0.7(npn) = -0.7(pnp) b = 100 Find IC1, IC2, VCE1, VCE2 •Use Thevenin circuits.
  • 25.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 24 Example 4.8 •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/ b 5 = IE133K/ 100 + .7 + IE13K IE1 = 1.3mA IB1 IE1
  • 26.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 25 Example 4.8 C-E loop neglect IB2 because it is IB2 << IC1 IE1 IC1 VCC = IC1RC1 + VCE1 +IE1RE1 15 = 1.3(5) + VCE1 +1.3(3) VCE1= 4.87V
  • 27.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 26 Example 4.8 Stage 2 •B-E loop IB2 IE2 VCC = IE2RE2 + VEB +IB2RBB2 + VBB2 15 = IE2(2K) + .7 +IB2 (5K)+ 4.87 + 1.3(3) Use IB2  IE2/ b, solve for IE2 IE2 = 2.8mA
  • 28.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 27 Example 4.8 Stage 2 •C-E loop IE2 IC2 VCC = IE2RE2 + VEC2 +IC2RC2 15 = 2.8(2) + VEC2 + 2.8(2.7) solve for VEC2 VCE2 = 1.84V
  • 29.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 28 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 = bIB •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).
  • 30.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 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.
  • 31.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 30 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. Transconductance ac input signal (DC input signal 0.7V) ac output signal DC output signal IB 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.
  • 32.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 31 Transconductance = slope at Q point gm = dic/dvBE|ic = ICQ where IC = IS[exp(-VBE/VT)-1]; the equation for a diode. Transconductance ac input signal DC input signal (0.7V) ac output signal DC output signal gm = ISexp(-VBE/VT) (1/VT) gm  IC/VT (A/V)
  • 33.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 32 ac input resistance  1/slope at Q point rp = dvBE/dib|ic = ICQ rp  VT /IB re  VT /IE ac input resistance of transistor ac input signal DC input signal (0.7V) ac output signal DC output signal IB
  • 34.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 33 4.8 Small-signal equivalent circuit models •ac model •Hybrid-p model •They are equivalent •Works in linear region only
  • 35.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 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, 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.
  • 36.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 35 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 = 0.023mA C-E voltage loop VCE = 10 - ICRC VCE = 10 - (2.3)(3) VCE = 3.1V Q point: VCE = 3.1V, IC = 2.3mA
  • 37.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 36 Example 4.9 ac problem Short DC sources, input and output circuits are separate, only coupled mathematically gm = IC/VT = 2.3mA/25mV = 92mA/V rp = VT/ IB = 25mV/.023mA = 1.1K vbe= vi [rp / (100K + rp)] = 0.011vi vout = - gm vbeRC vout = - 92(0.011vi)3K vout/vi = -3.04 + vout - e b c + vbe -
  • 38.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 37 Exercise 4.24 (a) Find VC, VB, and VE, given: b = 100, VA = 100V IE = 1 mA IB  IE/b= 0.01mA VB = 0 - IB10K = -0.1V VE = VB - VBE = -0.1 - 0.7 = -0.8V VC = 10V - IC8K = 10 - 1(8) = 2V VB
  • 39.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 38 Exercise 4.24 (b) Find gm, rp, and r0, given: b = 100, VA = 100V gm = IC/VT = 1 mA/25mV = 40 mA/V rp = VT/ IB = 25mV/.01mA = 2.5K r0 = output resistance of transistor r0 = 1/slope of transistor output characteristics r0 = | VA|/IC = 100K
  • 40.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 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.
  • 41.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 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, 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
  • 42.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 41 Circuit from Exercise 4.24 IE = 1 mA VC = 10V - IC8K = 10 - 1(8) = 2V IB  IE/b= 0.01mA gm = IC/VT = 1 mA/25mV = 40 mA/V VB = 0 - IB10K = -0.1V rp = VT/ IB = 25mV/.01mA = 2.5K VE = VB - VBE = -0.1 - 0.7 = -0.8V + Vout -
  • 43.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 42 ac equivalent circuit b e c vbe = (Rb||Rpi)/ [(Rb||Rpi) +Rs]vi vbe = 0.5vi vout = -(gmvbe)(Ro||Rc ||RL) vout = -154vbe Av =vout/vi = - 77 + vout - Neglecting Ro vout = -(gmvbe)(Rc ||RL) Av =vout/vi = - 80
  • 44.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 43 Prob. 4.76 + Vout - b=100
  • 45.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 44 Prob. 4.76 + Vout - (a) Find Rin Rin = Rpi = VT/IB = (25mV)100/.1 = 2.5KW (c) Find Rout Rout = Rc = 47KW Rin Rout (b) Find Av = vout/vin vout = - bib Rc vin = ib (R +Rpi) Av = vout/vin = - bib Rc/ ib (R +Rpi ) = - bRc/(R +Rpi) = - 100(47K)/(100K + 2.5K) = - 37.6 = bib b e c ib
  • 46.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 45 4.9 Graphical analysis Input circuit B-E voltage loop VBB = IBRB +VBE IB = (VBB - VBE)/RB
  • 47.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 46 Graphical construction of IB and VBE IB = (VBB - VBE)/RB If VBE = 0, IB = VBB/RB If IB = 0, VBE = VBB VBB/RB
  • 48.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 47 Load line Output circuit C-E voltage loop VCC = ICRC +VCE IC = (VCC - VCE)/RC
  • 49.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 48 Graphical construction of IC and VCE VCC/RC IC = (VCC - VCE)/RC If VCE = 0, IC = VCC/RC If IC = 0, VCE = VCC
  • 50.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 49 Graphical analysis Input signal Output signal
  • 51.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 50 •Load-line A results in bias point QA which is too close to VCC and thus limits the positive swing of vCE. •Load-line B results in an operating point too close to the saturation region, thus limiting the negative swing of vCE. Bias point location effects
  • 52.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 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)
  • 53.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 52 Common emitter amplifier ac equivalent circuit
  • 54.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 53 Common emitter amplifier Rin Rout + Vout - Rin (Does not include source) Rin =Rpi Rout (Does not include load) Rout =RC AV = Vout/Vin Vout = - bibRC Vin = ib(Rs + Rpi) AV = - bRC/ (Rs + Rpi) Ai = iout/iin iout = - bib iin = ib Ai = - b ib iout
  • 55.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 54 Common emitter with RE amplifier ac equivalent circuit
  • 56.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 55 Common emitter with RE amplifier Rin Rout + Vout - Rin Rin =V/ib V = ib Rpi + (ib + bib)RE Rin = Rpi + (1 + b)RE (usually large) Rout Rout =RC AV = Vout/Vin Vout = - bibRC Vin = ib Rs + ib Rpi + (ib + bib)RE AV = - bRC/ (Rs + Rpi + (1 + b)RE) (less than CE, but less sensitive to b variations) Ai = iout/iin iout = - bib iin = ib Ai = - b ib iout ib + bib + V -
  • 57.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 56 Common collector (emitter follower) amplifier b c e + vout - (vout at emitter) ac equivalent circuit
  • 58.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 57 Common collector amplifier Rin + vout - Rin Rin =V/ib V = ib Rpi + (ib + bib)RL Rin = Rpi + (1 + b)RL AV = vout/vs vout = (ib + bib)RL vs = ib Rs + ib Rpi + (ib + bib)RL AV = (1+ b)RL/ (Rs + Rpi + (1 + b)RL) (always < 1) ib ib + bib + V -
  • 59.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 58 Common collector amplifier Rout + vout - Rout (don’t include RL, set Vs = 0) Rout = vout /- (ib + bib) vout = -ib Rpi + -ibRs Rout = (Rpi + Rs)/ (1+ b) (usually low) Ai = iout/iin iout = ib + bib iin = ib Ai = b + 1 ib ib + bib
  • 60.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 59 Prob. 4.84 + vout - b = 50 ac circuit CE with RE amp, because RE is in ac circuit Given Rpi =VT/IB = 25mV(50)/.2mA = 6.6K
  • 61.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 60 Prob. 4.84 (a) Find Rin Rin =V/ib V = ib Rpi + (ib + bib)RE Rin = Rpi + (1 + b)RE Rin = 6.6K+ (1 + 50)125 Rin  13K Rin + V - ib ib + bib
  • 62.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 61 Prob. 4.84 (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 (10K||10K)/(10K + 6.6Ki + (1 + 50)125) AV  - 11 ib ib + bib + vout - -bib
  • 63.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 62 Prob. 4.84 (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 = 0.76mA (ac value) vs = ib Rs + ib Rpi + (ib + bib)RE vs = (0.76mA)10K + (0.76mA) 6.6K + (0.76mA + (50)0.76mA )125 vs  17.4mV ib ib + bib + vout - + vbe -
  • 64.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 63 Prob. 4.84 (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) ib ib + bib + vout - + vbe -
  • 65.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 64 Prob. 4.79 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) b = 100 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) = 3V Combining gives, R1 = 30K, R2 = 15K
  • 66.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 65 Prob. 4.79 b = 100Find RE (input circuit) Use Thevenin equivalent B-E loop VBB=IBRBB+VBE+IERE using IB  IE/b RE = [VBB - VBE - (IE/b)RBB]/IE RE = [3 - .7 - (2mA/100)10K]/2mA RE = 1.05KW + VBE - IE IB
  • 67.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 66 Prob. 4.79 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 + vout -
  • 68.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 67 CE amplifier
  • 69.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 68 CE amplifier Av  -12.2
  • 70.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 69 FOURIER COMPONENTS OF TRANSIENT RESPONSE V($N_0009) DC COMPONENT = -1.226074E-01 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG) 1 1.000E+03 1.581E+00 1.000E+00 -1.795E+02 0.000E+00 2 2.000E+03 1.992E-01 1.260E-01 9.111E+01 2.706E+02 3 3.000E+03 2.171E-02 1.374E-02 -1.778E+02 1.668E+00 4 4.000E+03 3.376E-03 2.136E-03 -1.441E+02 3.533E+01 TOTAL HARMONIC DISTORTION = 1.267478E+01 PERCENT CE amplifier
  • 71.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 70 CE amplifier with RE
  • 72.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 71 CE amplifier with RE Av  - 7.5
  • 73.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 72 FOURIER COMPONENTS OF TRANSIENT RESPONSE V($N_0009) DC COMPONENT = -1.353568E-02 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG) 1 1.000E+03 7.879E-01 1.000E+00 -1.794E+02 0.000E+00 2 2.000E+03 1.604E-02 2.036E-02 9.400E+01 2.734E+02 3 3.000E+03 5.210E-03 6.612E-03 -1.389E+02 4.056E+01 4 4.000E+03 3.824E-03 4.854E-03 -1.171E+02 6.231E+01 TOTAL HARMONIC DISTORTION = 2.194882E+00 PERCENT
  • 74.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 73 Summary Av THD CE -12.2 12.7% CE w/RE (RE = 100) -7.5 2.19%
  • 75.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 74 Prob. 4.83 + vout - •2 stage amplifier (a) Find IC and VC of each transistor •Both stages are the same (same for each stage) •Capacitively coupled b = 100 RL=2K Rc=6.8K
  • 76.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 75 Prob. 4.83 (a) Find IC and VC of each transistor (same for each stage) B-E voltage loop VBB = IBRBB + VBE + IERE where RBB = R1||R2 = 32K VBB = VCCR2/(R1+R2) = 4.8V, and IB  IE/b IE = [VBB - VBE ]/[RBB/b + RE] IE = 0.97mA VC = VCC - ICRC VC = 15 - .97(6.8) VC = 8.39V + VC -
  • 77.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 76 Prob. 4.83 b c e + vout - (b) find ac circuit b c e RBB = R1||R2 = 100K||47K = 32KW Rpi = VT/IB = 25mV(100)/.97mA  2.6KW gm = IC/VT = .97mA/25mV  39mA/V RL=2K
  • 78.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 77 Prob. 4.83 b c e + vout - (c) find Rin1 Rin1 = RBB||Rpi = 32K||2.6K = 2.4KW b c e Rin1 (d) find Rin2 Rin2 = RBB||Rpi = 2.4KW Rin2 find vb1/vi = Rin1/[Rin1 + RS] = 2.4K/[2.4K + 5K] = 0.32 + vb1 - find vb2/vb1 vb2 = -gmvbe1[RC||RBB||Rpi] vb2/vbe1 = -gm[RC||RBB||Rpi] vb2/vb1 = -(39mA/V)[6.8||32K||2.6K] = -69.1 + vb2 - RL=2K
  • 79.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 78 Prob. 4.83 b c e + vout - (e) find vout/vb2 vout = -gmvbe2[RC||RL] vout/vbe2 = -gm[RC||RL] vb2/vb1 = -(39mA/V)[6.8K||2K] = -60.3 b c e (f) find overall voltage gain vout/vi = (vb1/vi) (vb2/vb1) (vout/vb2) vout/vi = (0.32) (-69.1) (-60.3) vout/vi = 1332 + vb1 - + vb2 - RL=2K
  • 80.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 79 Prob. 4.96 Find IE1, IE2, VB1, and VB2 IE2 = 2mA IE1 = I20mA + IB2 IE1 = I20mA + IE2/b2 IE1 = 20mA + 10mA IE1 = 30mA IE1 IE2 Q1 has b1 = 20 Q2 has b2 = 200 Q1 Q2 IB2
  • 81.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 80 Prob. 4.96 Find VB1, and VB2 Use Thevenin equivalent VB1 = VBB1 - IB1(RBB1) = 4.5 - (30mA/20)500K = 3.8V VB2 = VB1 - VBE = 3.8V - 0.7 = 3.1V Q1 has b1 = 20 Q2 has b2 = 200 IB2+ vB1 - + vB2 - IB1
  • 82.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 81 Prob. 4.96 (b) find vout/vb2 vout = (ib2 + b2ib2)RL vb2 = (ib2 + b2ib2)RL + ib2Rpi2 vout/vb2 = (1 + b2)RL/[(1 + b2)RL + Rpi2] = (1 + 200)1K/[(1 + 200)1K+ 2.5K]  0.988 b1 e1 c1 b2 e2 c2 + vout - + vB2 - Rpi2 = VT/IB2 = VT b2/IE2 = 25mV(200)/2mA = 2.5KW
  • 83.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 82 Prob. 4.96 (b) find Rin2 = vb2/ib2 vb2 = (ib2 + b2ib2)RL + ib2Rpi2 Rin2 = vb2/ib2 = (1 + b2)RL + Rpi2 = (1 + 200)1K + 2.5K  204K b1 e1 c1 b2 e2 c2 + vB2 -Rin2
  • 84.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 83 Prob. 4.96 (c) find Rin1 = RBB1||(vb1/ib1) = RBB1|| [ib1Rpi1 +(ib1 + b1ib1)Rin2]/ib1 = RBB1|| [Rpi1 +(1+ b1)Rin2], where Rpi1 = VT b1/IE1 = 25mV(20)/30mA = 16.7K = 500K||[16.7K + (1+ 20)204K]  500KW b1 e1 c1 b2 e2 c2 + vB1 - Rin1 iB1
  • 85.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 84 Prob. 4.96 (c) find ve1/vb1 ve1 = (ib1 + b1ib1)Rin2 vb1 = (ib1 + b1ib1)Rin2 + ib1Rpi1 ve1/vb1 = (1 + b1) Rin2 /[(1 + b1) Rin2 + Rpi1] = (1 + 20)204K/[(1 + 20)204K+ 16.7K]  0.996 b1 e1 c1 b2 e2 c2 + ve1 - iB1+ vB1 -
  • 86.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 85 Prob. 4.96 (d) find vb1/vi vb1/vi = Rin1/[RS + Rin1] = 0.82 b1 e1 c1 b2 e2 c2 + vb1 - (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
  • 87.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 86 (Prob. 4.96) Voltage outputs at each stage Output of stage 2 Output of stage 1 Input
  • 88.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 87 (Prob. 4.96) Current Input current Input to stage 2 (ib2)
  • 89.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 88 (Prob. 4.96) Current output current Input to stage 2 (ib2)
  • 90.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 89 (Prob. 4.96) Current output current Input to stage 2 (ib2) Input current
  • 91.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 90 (Prob. 4.96) Power and current gain Input current = (Vi)/Rin = 1/500K = 2.0mA output current= (Vout)/RL = (0.81V)/1K= 0.81mA current gain = 0.81mA/ 2.0mA = 405 Input power = (Vi) (Vi)/Rin = 1 x 1/500K = 2.0mW output power = (Vout) (Vout)/RL = (0.81V) (0.81V)/1K= 656mW power gain = 656mW/2mW = 329
  • 92.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 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
  • 93.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 92 Probe: BJT Output Characteristics 1 Result of probe 2 Add plot (plot menu) -> Add trace (trace menu) -> IC(Q1) 3 Delete plot (plot menu)
  • 94.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 93 BJT Output Characteristics: current gain Ib = 5mA (Each plot 10mA difference) b at Vce = 4V, and Ib = 45mA b = 8mA/45mA = 178
  • 95.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 94 BJT Output Characteristics: transistor output resistance Ib = 5mA (Each plot 10mA difference) Ro = 1/slope At Ib = 45mA, 1/slope = (8.0 - 1.6)V/(8.5 - 7.8)mA Rout = 9.1KW
  • 96.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 95 CE Amplifier: Measurements with Spice Rin Rout
  • 97.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 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/wC|. For C = 100mF, w = 1KHz, Rcap = 1/2p(1K)(100E-6)  1.6W •Vin should have a small value so operating point does not change Vin  1mV
  • 98.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 97 Rin Measurement •Transient analysis
  • 99.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 98 Probe results I(C2) Rin = 1mV/204nA = 4.9KW
  • 100.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 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/wC|. For C = 100mF, w = 1KHz, Rcap = 1/2p(1K)(100E-6)  1.6W •Vin should have a small value so operating point does not change Vin  1mV
  • 101.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 100 Rout Measurement •Transient analysis
  • 102.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 101 Probe results -I(C1) Rout = 1mV/111nA = 9KW -I(C1) is current in Vin flowing out of + terminal
  • 103.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 102 DC Power measurements Power delivered by + 10 sources: (10)(872mA) + (10)(877mA) = 8.72mW + 8.77mW = 17.4mW
  • 104.
    ECE 3111 -Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 103 ac Power Measurements of Load instantaneous power average power Vout Vin
  • 105.
    Reference • www.electronics.teipir.gr/personalpages/.../1/.. ./ECE3111BJTTransistors. ECE 3111- Electronics - Dr. S. Kozaitis- Florida Institute of Technology – Fall 2002 104