Lecture 20 Bipolar Junction Transistors (BJT): Part 4 Small Signal BJT Model Reading: Jaeger 13.5-13.6, Notes
Further Model Simplifications (useful for circuit analysis) Ebers-Moll Forward Active Mode Neglect Small Terms I EB B EB T T T I 0 e 1 I 0 e 1 I I 0 e 1 I 0 F F R F F R I I S e EB T
Modeling the Early Effect (non-zero slopes in I curves) I i B1 < i B2 <i B3 Base width changes due to changes in the base- collector depletion width with changes in B. i B3 (theory) i B2 (theory) i B1 (theory) This changes T T, which changes I, D and B F A E Major BJT ircuit Relationships i EB EB T T E I e i I e 1 1 S S v A F FO v E A i B i F I S FO e EB T
Small Signal Model of a BJT Just as we did with a p-n diode, we can break the BJT up into a large signal analysis and a small signal analysis and linearize the non-linear behavior of the Ebers-Moll model. Small signal lmodels are only useful lfor Forward active mode and thus, are derived under this condition. (Saturation and cutoff are used for switches which involve very large voltage/current swings from the on to off states.) Small signal models are used to determine amplifier characteristics (Example: Gain = Increase in the magnitude of a signal at the output of a circuit relative to it s magnitude at the input of the circuit). Warning: Just like when a diode voltage exceeds a certain value, the non-linear behavior of the diode leads to distortion of the current/voltage curves (see previous lecture), if the inputs/outputs t t exceed certain limits, it the full llebers-moll model must be used.
onsider the BJT as a two-port Network + 1 - i 1 i 2 Two Port + Network 2 - General y-parameter Network i 1 =y 11 v 1 + y 12 v 2 i 2 =y 21 v 1 + y 22 v 2 BJT y-parameter Network i b =y 11 v be + y 12 v ce i c =y 21 v be + y 22 v ce
onsider the BJT as a two-port Network i b =y 11 v be + y 12 v ce i c =y 21 v be + y 22 v ce
onsider the BJT as a two-port Network o is most often taken as a constant, F
Alternative Representations Transconductance Input Resistance Output Resistance g r r o m y 21 I T 40I 1 o T o y I g 11 1 y 22 A I E m Y-parameter Model Hybrid-pi Model v 1
Alternative Representations g m v be g m r i b i o b oltage ontrolled urrent source version of Hybrid-pi Model urrent ontrolled urrent source version of Hybrid-pi Model
Single Transistor Amplifier Analysis: Summary of Procedure Steps to Analyze a Transistor Amplifier 1.) Determine D operating point and calculate small signal parameters (see next page) 2.) onvert to the A only model. D oltage sources are shorts to ground D urrent sources are open circuits Large capacitors are short circuits Large inductors are open circuits Step 1 Step 2 Step 3.) Use a Thevenin circuit (sometimes a 3 Norton) where necessary. Ideally the base should be a single resistor + a single source. Do not confuse this with the D Step Thevenin you did in step 1. 4 4.) Replace transistor with small signal model 5.) Simplify the circuit as much as Step necessary. 5 6.) alculate the small signal parameters (r, g m, r o etc ) and then gains etc Important!
Step 1 detail Single Transistor Amplifier Analysis D Bias Point Important! Thevenin I b R TH be I e 3=I E R E + be +I B R TH 3=I (( o +1)/ o )R e +0.7+I B R TH 3=I B o (( o +1)/ o ) R e +0.7+I B R TH 3= I B (100+1)1300+0.7+ I B 7500 B I B =16.6 ua, I = I B o =1.66 ma, I E =( o +1) c / o =1.67 ma B
Step 6 detail Single Transistor Amplifier Analysis Transconductance Input Resistance Output Resistance alculate small signal parameters g r r o m y 1 y 21 11 1 y 22 I T o T I A I 40 I E g o m 0.0664 S 1506 I A 45.2 K Important! 880 out R R TH r L = R R 3 r o TH =0.88 S be g m be For Extra Examples see: Jaeger section 13.6, and pages 627-630 (top of 630) A v r v out g mv ber L and v be v Th and v Th 0. 88v R r vout vout vbe vth r oltage Gain m L v v v v R r S be th S Th 1506 Av 0.066445,200 4300 100,000 0.88 880 1506 A 139 / v Th S g R 0.88
Important! Supplement Added to describe more details of the Solution of this Problem Bipolar Junction Transistors (BJT): Part 5 Details of Amplifier Analysis Reading: Jaeger 13.5-13.6, Notes
Detailed Example: Single Transistor Amplifier Analysis Important! Notes on slides 14-25 were prepared by a previous student.
Step 1: Determine D Operating Point Remove the apacitors Important! Because the impedance of a capacitor is Z = 1/(jω), capacitors have infinite impedance or are open circuits in D (ω = 0). Inductors (not present in this circuit) have an impedance Z = jωl, and are shorts in D.
Step 1: Determine D Operating Point Determine the D Thevenin Equivalent Important! Replace all connections to the transistor with their Thevenin equivalents.
Step 1: Determine D Operating Point alculate Small Signal Parameters Important! Identify the type of transistor (npn in this example) and draw the base, collector, and emitter currents in their + proper direction and their corresponding voltage polarities. Applying KL to the controlling loop - (loop 1): THB I B R THB BE I E R E = 0 Applying KL to the transistor: I E = I B + I + - I B + BE - + I Because I = βi B, I E = I B + I = I B + βi B = I B (1+β) 1 - I E Substituting for I E in the loop equation: THB I B R THB BE I B (1+β)R E = 0
Step 1: Determine D Operating Point Plug in the Numbers Important! THB I B R THB BE I B (1+β)R E = 0 THB BE I B (R THB +(1+β)R E ) = 0 THB = 12R 1 /(R 1 +R 2 ) = 3 R THB = R 1 R 2 = 7.5 kω Assume BE = 0.7 Assume β for this particular transistor is given to be 100. + - I B + BE + - I 07 B 3 0.7 I B (7500 + (1+100)*1300) = 0 I B = 16.6 μa I = βi B = 1.66 ma I E =I B +I = 1.676 ma BE - + 1 - I E
Step 1: Determine D Operating Point heck Assumptions: Forward Active? Important! = 12 I R = 12 - (1.66 ma)(4300) = 4.86 E =I E R E = (1.67 ma)(1300) = 2.18 B = THB I B R THB = 3 (16.6μA)(7500) = 2.88 heck: For an npn transistor in forward active: > B + - B + + - I 4.86 > 2.88 I B BE - E + B E = BE = 0.7 2.88 2.18 = 0.7 1 - I E
Step 2: onvert to A-Only Model Short the apacitors and D urrent Sources Important! D voltage sources are shorts (no voltage drop/gain through a short circuit). D current sources are open (no current flow through an open circuit). Large capacitors are shorts (if is large, 1/jω is small). Large inductors are open (if L is large, jωl is large).
Step 2: onvert to A-Only Model (Optional) Simplify Before Thevenizing Important! r c r 2 43kΩ 4.3 30 kω r s r r c r L s r L 100 2 kω 4.12 kω 2kΩ r kω r 1 r 2 1 v s 10 kω v s 7.5 kω
Step 3: Thevenize the A-Only Model Important! v s r r r s r thb = r s r 1 r 2 2 kω c L 4.12 kω 1.58 kω r 1 r 2 7.5 kω v thb = v s * (r 1 r 2 )/(r s +[r 1 r 2 ]) r th = r c r L 412kΩ 4.12 v th = 0 r the = 0 Ω v the = 0
Step 4: Replace Transistor With Small Signal Model Important! r th = r c r L r thb = r s r 1 r 2 4.12 kω v th = 0 1.58 kω v thb = v s * [(r 1 r 2 )/(r s +r 1 r 2 )] r the =0Ω Ω v the = 0 After replacing the transistor, apply Ohm s Law: = IR to find v out. r o and r th are in parallel, so that Ohm s Law becomes: v out = -IR = -(g m v BE )(r o r th ) v thb Because r th = r c r L v out = -(g m v BE )(r o r c r L ) r thb B + + v BE r π g m v BE - r o r th - v out /v BE = -g m (r o r c r L ) is the gain from transitor input (v BE ) to transistor/circuit output v v out ) out E E TRANSISTOR EXTERNAL
Step 5: alculate Gain and Small Signal Parameters Important! r thb B + + As previously determined: v thb /v s = (r 1 r 2 )/([r 1 r 2 ] + r s ) v thb TRANSISTOR EXTERNAL - E v BE r π g m v BE r o r th E Gain = v out /v s = (v thb /v s )(v BE /v thb )(v out /v BE ) v out - Applying a voltage divider: v BE /v thb = r π /(r π +r thb ) Gi Gain factor: v out /v BE = -g m (r o r c r L ) Because calculating the D operating point was done first, we have equations for g m, r π, and r o in terms of previously calculated D currents and voltages. Plugging in the numbers: Gain = v out /v s = -139 /
Interpretation/Analysis of Results Important! Gain = v out /v s = (v thb /v s )(v BE /v thb )(v out /v BE ) = -139 / v thb /v s = (r 1 r 2 )/([r 1 r 2 ] + r s ) v BE /v thb =r/(r π π +r thb ) Both terms are loss factors, i.e. they can never be greater than 1 in magnitude and thus cause the gain to decrease. This term is the gain factor and is responsible for amplifying the signal. v out /v BE = -g m (r o r c r L ) The A input signal has been amplified 139 times in magnitude. The negative sign indicates there has been a phase shift of 180.
ompleting the Small Signal Model of the BJT Base harging apacitance (Diffusion apacitance) In active mode when the emitter-base is forward biased, the capacitance of the emitter-base junction is dominated by the diffusion capacitance (not depletion capacitance). Recall for a diode we started out by saying: Sum up all the minority carrier charges on either side of the junction Q D qa 0 p no e v ' D T Diffusion dq dt D Neglect charge dqd injected from the ' dv D base into the dt emitter due to p+ ' dv emitter in pnp D ' x vd x Lp T Ln 1 e dx qa n po e 1 0 e dx Excess charge stored is due almost entirely to the charge injected from the emitter.
ompleting the Small Signal Model of the BJT Base harging apacitance (Diffusion apacitance) The BJT acts like a very efficient i siphon : As majority carriers from the emitter are injected into the base and become excess minority carriers, the ollector siphons them out of the base. We can view the collector current as the amount of excess charge in the base collected by the collector per unit time. Thus, we can express ess the charge due to the excess hole concentration in the base as: Q B i or the excess charge in the base depends on the magnitude of current flowing and the forward base transport time, F, the average time the carriers spend in the base. It can be shown (see Pierret section 12.2.2) 2 2) that: 2 W F where 2D, D B Minority carrier Base Quasi neutral WB F region diffusion width coefficient
ompleting the Small Signal Model of the BJT Base harging apacitance (Diffusion apacitance) Thus, the diffusion capacitance is, B Q v W 2 B i Q po BE D int 2 B vbe B I F F g T m Q po int The upper operational frequency of the transistor is limited by the forward base transport time: 1 f 2 F Note the similarity to the Diode Diffusion capacitance we found previously: pnol p n poln qa Diffusion g d t where t is the transit time I S
ompleting the Small Signal Model of the BJT Base harging apacitance (Total apacitance) In active mode for small forward biases the depletion capacitance of the base-emitter junction can contribute to the total capacitance je where, jeo 1 jeo EB bi for emitterbase zero bias depletion capaci tan ce bi for emitterbase Thus, the total emitter-base capacitance is: built in voltage for the E B junction B je
ompleting the Small Signal Model of the BJT Base harging apacitance (Depletion apacitance) In active mode when the collector-base is reverse biased, the capacitance of the collector-base junction is dominated by the depletion capacitance (not diffusion capacitance). 1 bi o B for collectorbase where, o zero bias depletion capaci tan ce built in voltage for the bi for collectorbase B junction
ompleting the Small Signal Model of the BJT ollector to Substrate apacitance (Depletion apacitance) S In some integrated circuit BJTs (lateral BJTs in particular) the device has a capacitance to the substrate wafer it is fabricated in. This results from a buried reverse biased junction. Thus, the collector-substrate junction is reverse biased and the capacitance of the collector-substrate junction is dominated by the depletion capacitance (not diffusion capacitance). Emitter where, S S n-base S 1 p-collector bi for collectorsubstrate zero bias depletion capaci tan ce bi for collectorsubstrate p n-substrate built in voltage for the substrate junction
ompleting the Small Signal Model of the BJT Parasitic Resistances r b = base resistance between metal interconnect and B- E junction r c = parasitic collector resistance r ex = emitter resistance due to polysilicon contact These resistance's can be included in SPIE simulations, but are usually ignored in hand calculations.
ompleting the Small Signal Model of the BJT omplete Small Signal Model