ch1_BJT_01-14

ch1_BJT_01-14 - ELEC202 – Electronic Circuits II BJT...

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Unformatted text preview: ELEC202 – Electronic Circuits II BJT Bipolar Transistor (BJT) Symbol (npn) Ideal Structure (npn) Actual Structure in IC (npn) Collector C C n B Base IC IB p E IE E C n+ p n− B n B NEmitter > NBase > Ncollector Emitter E Modes of Operation VBC VCE =VCB +VBE Reverse E/B reverse C/B forward VBC(on) Cutoff Both reverse Note: VCE(sat) = −VBC(on) + VBE(on) ≈ −0.5V + 0.7V ≈ 0.2V Saturation Both forward 0 VBE(on) VBE Active Forward E/B forward C/B reverse Typical values VBE(on) = 0.6V - 0.7V VCE(sat) = 0.2V- 0.3V Amplifier - BJT operates at Active Forward Region Switch - BJT operates between Cutoff and Saturation Regions BJT in Active Mode Operations: C VCB >0 B VBE >VBE(on) Injected holes Collected n n− electrons p p diffusing electrons n n+ W C/B junction Reverse bias E/B junction Forward bias Good BJT transistor requires: (1) doping concentration in Emitter (ND) >> in Base (NA) (2) very narrow base width W (3) large Emitter-Base junction AE E Last updated: 8-Sep-03 1 ELEC202 – Electronic Circuits II BJT Equations: + VD − V I D = I S exp D nV T Recall ID Collector Current: V I C = I S exp BE , n = 1 V T A qD n IS = E n i N AW A ∝E N AW IC is a function of VBE and independent of collector voltage as long as C/B junction is reverse biased. 2 AE = E/B junction area, Dn = diffusion constant NA = Base doping concentration, W = base width ni = intrinsic carrier concentration = 1.5×1010 carrier/cm3 Base Current: IB is a factor of IC. I IB = C where β- current gain (common-emitter current gain) - function of W and ND/ NA ratio β β ↑ with W ↓ and/or ND/ NA ↑ typical values of β is between 50 - 200 Emitter Current: I E = IC + I B I β +1 = IC + C = IC β or iC = α i E β where α = β β +1 or β= α 1−α α – common-base current gain 100 for large β, say 100, α = ≈ 0.99 ≈ 1 101 20 β = 20, α = ≈ 0.95 21 Last updated: 8-Sep-03 2 ELEC202 – Electronic Circuits II BJT Large Signal Model → IB B Common Emitter Configuration ( π-model ) ← IC + VBE C V I S exp BE V T − ← IC (IS /β) B → IB βIB Size of this diode = (IS /β) E β = common-emitter current gain E C α = common-base current gain C V I S exp BE VT Common Base Configuration ( T-model ) C αIE B B (IS /α) IE (IS /α) E E Voltage-Controlled Current Source Current-Controlled Current Source pnp transistor Symbol (pnp) Ideal Structure (pnp) Actual Structure in IC (pnp) Collector E E E p B Base n IE IB B p IC C p+ C B p+ n p-substrate In general, βpnp < βnpn Emitter C Last updated: 8-Sep-03 3 ELEC202 – Electronic Circuits II BJT Voltage Polarities and Current Flow npn BJT pnp BJT C VCB >0 IB IC B VBE > VBE(on) E IE E VBE > |VBE(on)| IB IE B IC VBC >0 C V I C = I S exp BE V T IC = β I B V I C = I S exp EB V T IC = β I B IC = α I E IC = α I E Replace all VBE by VEB DC Analysis: Example 1: Find VC , VB , VE , IC , IB and IE. Assume β = 100, VBE = 0.7V at IC = 1mA. 10V Assume the BJT operates in active region. VB = 0 VE = VB – VBE = 0 − 0.7 = −0.7V V − (−10) − 0.7 + 10 IE = E = = 0.93mA RE 10k 100 β IC = α I E = IE = 0.93m = 0.9208mA β +1 101 VC = 10 − ICRC = 10 − (0.7208m)(5k) = 5.396V 0.9208m I IB = C = = 9.208µA 100 β RC = 5kΩ IC VC IB VE RE = 10kΩ IE −10V Check: VC >VB , ∴BJT operates in active region More Accurate Calculation: Given: VBE = 0.7V at IC = 1mA Now IC = 0.9208mA ⇒ VBE = ? V V Since I C1 = I S exp BE1 and I C 2 = I S exp BE 2 V V T T Last updated: 8-Sep-03 4 ELEC202 – Electronic Circuits II BJT V I S exp BE 2 V T = exp(V − V ) / V = BE 2 BE1 T VBE1 I S exp VT Therefore IC 2 I C1 or I V BE 2 − VBE1 = VT ln C 2 I C1 where VT = kT = 25.9mV @ 300k q 0.9208m So, IC = 0.9208mA ⇒ V BE = 0.7 + 25.9m ln = 0.6979V 1m V + 10 0 − 0.6979 + 10 IE = E = = 0.9302mA 10k RE % error in VBE = 0.7 − 0.6979 = 0.3% 0.6979 0.9302 − 0.93 negligible = 0.02% 0.9302 Now, assume the calculated IC = 10mA (10 times larger than the rated current) 0.7596 − 0.7 = 7.9% VBE = 0.7 + 25.9m ln(10) = 0.7596V ⇒ % error = 0.7596 0.9240 − 0.93 10 − 0.7596 IE = = 0.9240mA ⇒ % error = = 0.6% Still very small 10k 0.9240 % error in IE = Conclusion: We always assume VBE is a constant in DC analysis, VBE(on) = 0.7V Example 2: Find VC , VB , VE , IC , IB and IE. Assume β = 100, |VBE(on) | = 0.7V. Again, you need to assume the BJT operates in active region. VB = 0 VE = VB – VEB = 0 + 0.7= 0.7V 10 − V E 10 − 0.7 IE = = = 0.93mA RE 10k 100 β IC = α I E = IE = 0.93m = 0.921mA 101 β +1 VC = −10 + ICRC = −10 + (0.921m)(5k) = −5.40V I 0.921m IB = C = = 9.21µA β 100 Check: Note: VC < VB, ∴BJT operates in active region. Usually keep 3 significant figures in the calculation. Last updated: 8-Sep-03 10V RE = 10kΩ IB IE VE VC RC = 5kΩ IC −10V 5 ELEC202 – Electronic Circuits II BJT Example 3: Find VC , VB , VE , IC , IB and IE. Assume β = 100, VBE(on) = 0.7V. VCC = 10V IB ≠ 0 ⇒ VB ≠ 0 KVL: VEE + IE RE +VBE(on)+ IB RB = 0 But I E = ( β + 1) I B − V EE − V BE ( on ) 10 − 0.7 IB = = = 8.38µA (β + 1)RE + RB (101)(10k ) + 100k ∴ I C = β I B = 0.838mA I E = ( β + 1) I B = 101(8.38µ ) = 0.846mA VB = 0 − IBRB = −(8.38µ)(100k) = −0.838V ≠ 0 VE = VB – VBE(on) = −0.838 − 0.7 = −1.508V VC = VCC – ICRC = 10 − (0.838m)(5k) = 5.81V RC = 5kΩ RB = 100kΩ RE = 10kΩ VEE = −10V Again, need to check if VC > VB. Example 4: VCC = 15V Find VB and IE. Assume β = 100, VBE(on) = 0.7V. RB 2 50k × 15 = 5V VB = VCC = 50k + 100k R B1 + R B 2 VE = VB – VBE(on) = 5 − 0.7 = 4.3V V − 0 4 .3 IE = E = = 1.433mA RE 3k I 1.433m IB = E = = 14.19 µA β +1 101 Since IB ≠ 0, more current will be flowing through RB1 and VB should be lower than the above calculated value. Therefore, the above calculation is wrong. RC = 5kΩ RB1 = 100kΩ RB2 =50kΩ RE = 3kΩ Correct calculation: VCC = 15V RB1 = 100kΩ VCC = 15V RC = 5kΩ RBB RC = 5kΩ where V BB = RB 2 VCC R B1 + RB 2 = RB2 =50kΩ Last updated: 8-Sep-03 50k × 15 = 5V 50k + 100k RBB = RB1//RB2 RE = 3kΩ RE = 3kΩ = 100k//50k = 33.3kΩ 6 ELEC202 – Electronic Circuits II BJT Now use KVL: 0 + IE RE + VBE(on) + IB RBB = VBB VBB − VBE ( on ) IB = ........(*) (β + 1)RE + RBB 5 − 0.7 = 12.78µA (101)(3k ) + 33.3k I E = ( β + 1) I B = 101(12.78µA) = 1.291mA VB = VBB – IB RBB = 5 − (12.78µ)(33.3k) = 4.57V, not 5V = % error in the initial calculation: 1.433m − 1.291m error in IE = ≈ 11% 1.291m 5 − 4.57 error in VB = ≈ 9% 4.57 In fact, from (*), the initial calculation is correct if (β+1) RE >> RB or (β+1) RE >> RB1, RB2. However, RB1 & RB2 cannot be too small; otherwise, lots of energy will be lost in the “biasing resistors” RB1 & RB2. Note: Example 5: Find the voltages in all nodes & current in all branches. Assume β = 100, |VBE(on) | = 0.7V. Similar to Example 4, we can find IB1 = 12.78µA, IE1 = 1.291mA, and IC1 = βIB1 = 1.278mA VCC = 15V Now there is a problem. We do not know the voltage of VC1 as the current IRC1 ≠ IC1. IRC1 RB1 = 100kΩ In fact, IC1 = IRC1 + IB2. So, how are we going to find out the voltage of VC1? Solution 1: Use Algebra 1 VCC − VE 2 I ⋅ I B2 = E 2 = β +1 β +1 RE 2 = VCC − (VC1 + VEB ( on ) ) (β + 1)RE 2 = Q1 RE2 = 2kΩ RC1 = 5kΩ IE2 IB2 Q2 IC1 RC2 = 2.7kΩ RB2 =50kΩ RE1 = 3kΩ 15 − Vc1 − 0.7 .......(1) 101(2k ) But VC1 = VCC – IRC1RC1= VCC – (IC1 − IB2 ) RC1 ……………..(2) Sub (1) into (2), VC1 = 15 − (1.278m − 14.3 − VC1 ) ⋅ 5k 202k 5k 5k 1 + VC1 = 15 − (1.278m)(5k ) + 14.3 ⋅ 202k 202k ⇒ VC1 = 8.75V Once you fix VC1, you can find the rest. Last updated: 8-Sep-03 7 ELEC202 – Electronic Circuits II BJT Solution 2: Use approximation & iteration Assume IB2 is very small compare with IC1. VC1 = VCC – IRC1RC1 ≈ VCC – IC1RC1 = 15 − (1.278m)(5k) = 8.61V VE2 = VC1 + VEB(on) = 8.61 + 0.7 = 9.31V V − VE 2 15 − 9.31 I 2.85m = CC = = 2.85mA ⇒ I B 2 = E 2 = = 0.0282mA 2k β + 1 101 RE 2 Then, and I E2 ∴ I B 2 0.0282m = ≈ 2.2% error I C1 1.278m and % error in I E 2 ≈ ⇒ VC1 we calculated has ≈ 2% error 2.85 − 2.775 × 100% ≈ 2.7% . 2.775 To get a more accurate answer, we can use iteration method. Use the IB2 value in the above calculation. IRC1 ≈ IC1 – IB2 = 1.278m − 0.0282m = 1.250mA VC1 = 15 − (1.250m)(5k) = 8.75V , ∴ VE2 = 8.75 + 0.7 = 9.45V I E2 = 15 − 9.45 = 2.78mA 2k ⇒ I B2 = 2.78m = 0.0275mA 101 The new IRC1 = IC1 – IB2 = 1.278m − 0.0275m = 1.251mA, which is very close to the one we used in the 1st iteration. So, we conclude that VC1 ≈ 8.75V. Note that the value is the same as the one calculated by 1st method. Conclusion: In DC Analysis, you can assume IB is small and calculate the voltage of each node. Then estimate the values of IB and see if it is really small compare with other currents. The assumption is acceptable if IB is less then 1-2% of the other currents. Also, if β > 100, we can assume α = 1 or IC = IE in analyzing a big circuit. Graphical Representation IC IC-VBE characteristic IC IC-VCE characteristic (ideal) V I C = I S exp BE V T VBE3 >VBE2 for VCB > 0 VBE2 >VBE1 Diode like I-V char. VBE Last updated: 8-Sep-03 VBE1 >VBE(on) 0 VCE 0.2-0.3V 8 ELEC202 – Electronic Circuits II BJT IC Saturation Practical IC-VCE characteristic Active VBE4 >VBE3 VBE3 >VBE2 VBE2 >VBE1 IC + VCE + VBE VBE1 >VBE(on) − − 0 −VA VA = Early Voltage - due to the decrease in effective base width Weff - typical value 50-100V - depends on the process VCE(sat) VCE n n p VBE VCE1 Weff1 p n VBE VCE2 Weff2 n VCE2 > VCE1 ⇒ Weff1 >Weff2 Modified Equation V I C = I S exp BE V T VCE 1 + VA Define ro ≡ output resistance looking into the collector ∂i ≡ C ∂VCE VBE = consta nt VBE I S exp V T = VA −1 ro −1 = VA = ro IC Note: ro is fixed with a given VBE (or a given IC). Last updated: 8-Sep-03 9 ELEC202 – Electronic Circuits II BJT Transistor as an Amplifier VCC iC ic RC vC iC + vbe − + VBE − Note: t IC VBE 0 vBE vBE = total signal VBE = lare signal, DC signal vbe = small signal, AC signal t Now vBE = VBE – vbe v V + v be i C = I S exp BE = I S exp BE V V T T v 1 v = I C 1 + be + be VT 2! VT I = C V T iC 2 1 v + be 3! VT v be = g m v be where V = I S exp BE exp V T 3 + .... gm = IC VT ≈ I c (1 + v be V T v = I C exp be V T I v be ) = I C + C v be V VT T for Vbe << VT = 25.9mV = transconducta nce gm ∝ IC , i.e. the small signal output current ic depends on the large signal DC biasing current IC. iC ic Short segment of curve such that it is almost a straight line. t VBE 0 vBE t ∂i gm = C ∂VBE Note: In general, we consider vbe < 10mV to be a small signal. v v ~5% error in exp be ≈ (1 + be ) for vbe < 10mV V VT T = slope of the iC-vBE characteristic curve at iC = IC. ic = I C gm increases with the increase of IC. Last updated: 8-Sep-03 10 ELEC202 – Electronic Circuits II BJT Base Current & Input Resistance at the Base iC iB = = β I C + ic = I B + ib β rπ ≡ Rin| looking into the base ≡ Define β rπ = β = g m v be β = v be β ⇒ gm vbe β = ib gm vbe β = ib gm I 1mA gm = C = ≈ 40mA / V VT 25.9mV If IC =1mA, If β = 100, rπ = Since rπ ∝ Note: ic = Base resistance (small signal) gm Eg. ib = where β gm = rπ 100 = 2.5kΩ 40m 1 1 ∝ , if the BJT operates at high current, it has a lower Rin = rπ. gm IC Emitter Current & Input Resistance at the Emitter IE = ie = Define re = Note: IC α ic α = g m vbe α ⇒ vbe α = ie gm re ≡ Rin| looking into the emitter ≡ α gm rπ = α gm ≈ re 1 gm = Emitter resistance (small signal) β gm , re = α gm ⇒ β rπ = α re ∴ rπ = (β + 1) re i.e . rπ >> re for large β > 100. Voltage Gain vC = VCC – iC RC = VCC – (IC + ic) RC = (VCC – IC RC) – ic RC = VC – ic RC ⇒ vc – ic RC = −gm vbe RC = (−gm RC ) vbe I output voltage vc = = − g m RC ⇒ AV = − g m RC = − C RC Voltage gain ≡ input voltage vbe VT Note that the voltage gain ∝ I C only, and independent of β. i.e. independent to the characteristic of the BJT. Last updated: 8-Sep-03 11 ELEC202 – Electronic Circuits II BJT Small-signal Equivalent Models VCC VCC RC RC iC iB vC + vbe − + VBE − RC IC VC IB iE ic vc ib + vbe − IE + VBE − ie DC Analysis Large signal model AC Analysis Small signal model Note: All DC voltage sources become AC signal ground. Hybrid-π Small-signal Model b b c + vbe rπ gmvbe rπ ro e rπ = ro = T-model (small signal) c − re = ro = e Last updated: 8-Sep-03 gm = ro re β gmvbe = gm ib rπ = βib e gm ro VA IC c gmvbe + vbe βib I gm = c VT − b c ib Ic VT α gm VA IC αie ro b ie re e gmvbe = gm ie re = (β/rπ )iere = αie 12 ELEC202 – Electronic Circuits II BJT Example: VCC =10V RC = 3Ω Find the voltage gain of the amplifier. Assume β = 100, VT = 25mV, VBE(on) = 0.7V. Vo RBB = 100kΩ ± Vi VBB =3V VCC =10V VBB =3V Step 1: Find the DC biasing current IC. V BB − V BE ( on ) 3 − 0.7 IB = = = 23µA R BB 100k ∴ I C = β I B = 100(23µ ) = 2.3mA Step 2: IC RB = 100kΩ Find out the small-signal model parameters. I 2.3m gm = C = = 92mA / V VT 25m β rπ = re = Step 3: RC gm = 100 = 1.087 kΩ 92m rπ 1.087 k = = 10.76Ω 101 β +1 Find out the AC voltage gain using one of the small-signal model. RBB = 100kΩ vi + vbe − vo rπ gmvbe vbe = RC = 3kΩ rπ vi rπ + RBB vo = − g mvbe RC = ⇒ − g m rπ vi RC rπ + RBB vo − g m rπ = RC vi rπ + RBB 1.087k (92m)(3k ) 1.087k + 100k = −2.97V / V =− Last updated: 8-Sep-03 13 ELEC202 – Electronic Circuits II BJT Signal Waveforms VCC = 10V VCC = 10V RC = 3kΩ vO RBB = 100kΩ VO = VC RBB = 100kΩ with no ac signal i.e. DC condition vi vI RC = 3kΩ + VBE − VBB = 3V VBB = 3V From previous example, IB = 23µA, IC = 2.3mA ⇒ Vo = VC = VC – ICRC = 10 − (2.3m)(3k) = 3.1V and VBE = 0.7V Include the ac small signal, vI 4V ˆ vi = 1V ˆ Assume vi = 1V VBB = 3V 2V t vBE 0.711V ˆ vbe VBE =0.7V ˆ vbe = 0.689V t 0 vO 6.07V ˆ vo VO =3.1V 0.3V 0 rπ ˆ ⋅ vi rπ + RBB 1.087k ⋅ 1V 1.087k + 100k = 10.75mV = ˆ ˆ vo = AV vi = −2.97 ×1V = −2.97V 0.13V clipping t Two limitations: 1. Linearity: If vbe is too large (> 10mV), the output waveform will be distorted. 2. Saturation: If vO is too small, the BJT will go into saturation. Note: VCE > VCE(sat) ≈ 0.3V when operates in active region. The waveform will also be clipped when vO > VCC, i.e. IC ≈ 0 (BJT operates in cutoff region) Last updated: 8-Sep-03 14 ...
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This note was uploaded on 01/05/2010 for the course EENG 202 taught by Professor P.mok during the Spring '03 term at Arizona.

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