ch1_BJT_15-27

# ch1_BJT_15-27 - ELEC202 – Electronic Circuits II BJT DC...

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Unformatted text preview: ELEC202 – Electronic Circuits II BJT DC Biasing Technique Since the gain of the BJT amplifier depends on the gm and gm depends on IC, it is very important to bias the BJT such that a constant dc current flows into the collector/emitter of the transistor. This current has to be insensitive to variations of VBE (on) and β with temperature and process variation. Example: VCC VCC RC R1 VBB RC RBB VBE(on) + IBRBB = VBB V − VBE ( on ) I B = BB RBB I E = ( β + 1) I B = ( β + 1) VBB − VBE ( on ) RBB ⇒ IE highly depends on β. R2 If β changes 10%, IE changes 10% No Good! A. Single Power Supply VCC R1 VCC RC VBB RC VRC = 1/3 VCC RBB VCE = 1/3 VCC R2 RE RE VRE = 1/3 VCC IERE + VBE(on) + IBRBB = VBB IE = VBB − VBE ( on ) R RE + BB β +1 Now (i) IE is insensitive to β if RE >> RBB . However, RBB cannot be too small; otherwise β +1 lots of power will be lost in the biasing resistors R1 & R2. (ii) IE is insensitive to VBE(on) if VBB >> VBE(on). However, VBB cannot be too large; otherwise the BJT will go into saturation region. Last updated: 8-Sep-03 15 ELEC202 – Electronic Circuits II BJT Conditions for RE & RC 1. Voltage across RE must be large to ensure stable bias current. 2. Voltage across RC must be large to ensure large signal swing before BJT cutoff 3. Voltage across collector & emitter must be large to ensure large signal swing before BJT saturation. Allow 1/3 VCC to each section. Rule of Thumb: Conditions for R1 & R2 1. RBB = R1 // R2 must be small to reduce the effect of β variation to IE. i.e. RBB << (β+1)RE. 2. RBB must be large to reduce the power dissipation in the biasing network and to reduce and the loading of the input signal. i.e. RBB >> input resistance of the amplifier >> Rin|looking into base >> rπ =(β+1)re Rule of Thumb: Take the average 1 R +r R R1 // R2 = [( β + 1) RE + ( β + 1)re ] ≈ β ( E e ) ≈ β ( E ) 2 2 2 (i) arithmetic mean: Not Good! Qβ( RE )not << ( β + 1) RE 2 R1 // R2 = ( β + 1) RE × ( β + 1)re ≈ β RE re (ii) geometric mean: (iii) 10% power overhead: I R1 ≈ I R 2 ≈ 0.1I E E.g. VCC = 15V, I C ≈ I E = 1mA, β = 100, VT = 25mV. Set 1/3 VCC to each: ⇒ VRE = 5V, VRC = 5V & VCE =5V ∴ RE = RC = 5V = 5kΩ 1mA VB = VE + VBE = 5 + 0.7 = 5.7V R2 V R 5.7 15 ⇒ = R= ⇒ 1+ 1 = R1 + R2 VCC 15 R2 5.7 Now re = or R1 = 1.6KK(1) R2 VT 25m = = 25Ω 1m IE Use (i) R1 // R2 = β RE re = 100 5k × 25 = 35.4kΩKK(2) Solve (1) & (2) ⇒ R1 R 2 1 1 1 2 1 = 35.4k ⇒ + = ⇒ = R1 + R 2 1.6 R 2 R 2 35.4k 1.6 R 2 35.4k Give R2 = 57.5kΩ & R1 = 92.0kΩ Last updated: 8-Sep-03 16 ELEC202 – Electronic Circuits II Use (iii) R1 + R2 = BJT VCC 15 = = 150kΩKK(3) 0.1I E 0.1×1m 150k = 57.7 kΩ 2 .6 R1 = 92.3kΩ 1.6 R 2 + R 2 = 150kΩ ⇒ R 2 = Close to (ii) VCC B. Two Power Supplies RC −VEE + IERE + VBE(on) + IBRB = 0 VEE − VBE ( on ) IE = similar to single power supply. R RE + B β +1 RB RE Note: (1) For ground-base amplifier, set RB = 0. (2) Otherwise, use previous “ Rule of Thumb” for RB. −VE VCC C. Biasing Using Feedback VBE(on) + IBRB + (IB+IC)RC = VCC VBE(on) + IBRB + IERC = VCC IB = VCC − VBE ( on ) same as single power supply. RB RC + β +1 Advantage: RB vi IB RC IRC IC VCB = IBRB >0 ⇒ VC always >VB. i.e. BJT always in active region (never operates in saturation region) Disadvantage: small signal swing Q to stabilize IE, we need RB<< (β+1)RC IE But VCE = VCB + VBE = IBRB + VBE = R + VBE β +1 B = I RC R + VBE β +1 B << IRCRC + VBE << VRC ⇒ Not a symmetric output swing. Note: This biasing technique can only be used in common-emitter configuration. Last updated: 8-Sep-03 17 ELEC202 – Electronic Circuits II BJT D. Using Current Source VCC VCC Q1 and Q2 have the same size. (i.e. IS1 = IS2) Since VBE1 = VBE2, IC1 = IC2 = IC & I IB1 = IB2 = I B = C Iref RC R I Q1 RB + VBE1 − I Q2 + VBE2 − β −VEE I = I C 2 = I C ⇒ I ref = I C1 + I B1 + I B 2 = I C + Now ∴ I I ref = IC 1 = ⇒ I ≈ I ref 2IC 2 IC + 1+ β I ref = where 2I C β for large β β VCC + VEE − VBE ( on ) R Single-Transistor Amplifiers • • BJT has 3 terminals ⇒ 3 basic configurations (CB, CE, CC) How to characterize an amplifier? Transconductance Amplifier Voltage Amplifier R ii + vi − Rin ± AVvi High Low AV = + vo − + vi − Rin Gmvi Ro + vo − High Rin High Ro vo vi High Also High Ai = Last updated: 8-Sep-03 io ii Rin Ro High Want io Gm = io = Transconduc tan ce ii io ii 18 ELEC202 – Electronic Circuits II BJT A. Common-Emitter Configuration (CE) DC Transfer Characteristic (a) Cut-off region: VCC RC VI < VBE(on) ≈ 0.7V IC = 0 VO = VCC – IC RC = VCC (b) Forward-Active region: VBE(on) < VI < VBE(sat) VO VI V I C = βI B = I S exp I V T VO (a) V VO = VCC − I C RC = VCC − RC I S exp I V T VCC slope = voltage gain (exponential function) (b) (c) Saturation region: (c) VCE(sat) VI ≥ VBE(sat) VO = VCE(sat) ≈ 0.3 IC ≠ β IB VI VBE(on) VBE(sat) Small-signal characteristic (Use hybrid-π model including ro) Rin = rπ Ro = ro // RC v AV = o = − g m (ro // RC ) vi For RC << ro , AV = −gmRC Ro vi Rin + vπ − vo io rπ gmvπ ro RC RC >> ro , AV = − AV(max) IV V = − g m ro = − C ⋅ A = − A VT I C VT Typical VA = 100V, VT ≈ 25mV ⇒ Av(max) ≈ 4,000 V/V. Note: The gain also has other limitation. Av = g m ( RC || ro ) < g m RC = I C RC VCC < VT VT For VCC = 10V, VT ≈ 25mV ⇒ |Av| < 10/25mV = 400 V/V. Ai = Last updated: 8-Sep-03 io = ii ro ro ro ro ro + RC = −G m Rin = − g m rπ = −β vi ro + RC ro + RC ro + RC Rin − Gm vi 19 ELEC202 – Electronic Circuits II BJT (Use T-model including ro ) Rin – not easy to calculate Ro = ro // RC v vo = −α iC (ro // RC ) = −α i (ro // RC ) re AV = =− α ( β + 1) rπ β rπ io = ii vo io αie ro RC vi vo α = − (ro // RC ) vi re =− Ai = Ro Rin ie re (ro // RC ) = − g m (ro // RC ) (ro // RC ) Same as before ro α ro ro ro + RC =− = −β (ie − α ie ) 1 − α ro + RC ro + RC − α ie Same as before Actual Amplifier Circuit VCC RC Rin RS vS RC ∞ CC vi ∞ I AC DC vo vo VB RL + RS CE RC RL RS vS vi I −VEE −VEE CC and CE – bypass capacitors, also known as coupling capacitors – block dc voltage and pass ac signal – capacitance = ∞, i.e. open circuit for DC, short circuit for ac v vv Rin gr = − m π (rO // RC // RL ) Gain of the amplifier O = O i = − g m (rO // RC // RL ) vS vi vS Rin + RS rπ + RS E.g.: I = 1mA, β = 100, VA = 100V, RC = 5kΩ, RS = 100Ω and RL = 5kΩ I 1mA 1mA ≈ = 40mA / V ∴ gm = C = VT 26mV 25mV rπ = ∴ β gm = 100 = 2.5kΩ, 40m rO = VA 100 = = 100kΩ I C 1m vO r 2 .5 k = − g m (rO // RC // RL ) π = −40m(100k // 5k // 5k ) = −(40m)(2.44k )(0.962) = −93.9V / V vS rπ + RS 2.5k + 100 Note: In order to have a high voltage gain, RS << rπ = Rin, RL >> RC // ro . Last updated: 8-Sep-03 20 ELEC202 – Electronic Circuits II BJT Common Base Configuration (CB) VCC Ro RC RO v o ∞ io αie ∞ vi , ii ii = −ie + ro RC Rin ie vI re Rin ii vi − vO vi vi − vO =+ rO re rO but vi rO = −VEE Rin = vo vA VV & re = T ≈ T re I E IC Since VA (typically = 100V) >> VT (26mV), rO >> re vi v 1 α ⇒ Rin = i = re = ≈ re ii gm gm ∴ ii ≈ vO = −(α ie + RO = RC // rO vO (1 + RC α1 α ) = vi ( + ) RC ≈ vi RC rO re rO re ∴ AV = AV = vo − vi αv R v − v ) RC = i C − O i RC rO re rO α vo α rO RC = = ( RC // rO ) = g m ( RC // rO ) vi re rO + RC re For ro >> RC, Av = gmRC α r io vo Rin α = ⋅ = ( RC // rO ) ⋅ e = ( RC // rO ) ≈ α RC RC ii RC vi re for large ro Common-Collector Configuration (CC) Or Emitter Follower (EF) VCC vi Rin ∞ vi RO vo RL −VEE Last updated: 8-Sep-03 Rin αie ro ie re Ro io vo RL 21 ELEC202 – Electronic Circuits II AV = BJT vo RL ( RL // rO ) = ≈ vi ( RL // rO ) + re RL + re for ro >> RL ≈ 1 for RL >> re RO = re // rO = re = α (QVA >> VT & rO >> re ) gm ii vi + vi = ii rπ + ( β + 1)ii RL vπ v Rin = i = rπ + ( β + 1) RL ii βii rπ ro − vo Resistance–reflection rule: RL R(looking into the base) = (β+1) R(total resistance in emitter) iO = rO rO ( β + 1)ii ⇒ ie = rO + RL rO + RL Ai = io rO = ( β + 1) ≈ ( β + 1) for ro >> RL ii rO + RL Common-Emitter with Emitter Degeneration VCC vo RC vI RC ∞ αie Rin RE ro re ∞ RE -VEE Input Impedance – same as CC conf. Rin = rπ + (β+1)RE Output Impedance RO = RO′// RC ii vi iO′ RO′ RO vO + vπ rπ gmvπ − ro RC ve RE Last updated: 8-Sep-03 22 ELEC202 – Electronic Circuits II iO ' = ve ( BJT ve v −v v −v = g mvπ + O e = − g m ve + O e rπ // RE rO rO v 1 1 + gm + ) = O rπ // RE rO rO RO ' = (Q g m ≈ 1 1 >> ) re rO vO r // R E 1 = v e rO ( + gm ) ⋅ π = [ 1 + g m (rπ // R E ) ]rO iO ' rπ // R E ve ∴ RO = RC // [ 1 + g m (rπ // R E )] rO RO ≈ RC //(1 + g m rπ )rO ≈ RC // β rO For R E >> rπ , Voltage-gain vO = −iO ' RO = − β ii = − β AV = v i Ro rπ + ( β + 1) R E vO − βRO gR = ≈− m O vi rπ + ( β + 1) R E 1 + g m RE RE >> For 1 ≈ re , gm AV ≈ − vO io′ Ro′ RL RO RE OR use T-model: For large ro, vo = − α iC RC = − α AV = vi RC re + R E α RC α RO α {RC //[1 + g m (rπ // RE )]rO } vO =− =− =− vi re + RE re + RE re + RE For simplicity, AV ≈ − Note: AV = − α RO re + RE for more accurate result total resistance in collector total resistance in emitter =− β RO gR ≈− m O rπ + ( β + 1) RE 1 + g m RE Same as before Conclusion: with RE, Last updated: 8-Sep-03 - Rin ↑ by (β+1)RE - Av ↓ by (1 + gm RE) times - Ro ↑ by [1 + gm (rπ//RE) times, only for RC is large - ∆vi is not limited by 10mV - prevent thermal runaway 23 ELEC202 – Electronic Circuits II BJT Resistance-Reflection Rule: 1. 2. Since current in emitter = (β+1)·current in base R(looking into the base) = (β+1)R(total resistance in emitter) ⇒ Req = (β+1)( re + RE ) = rπ + (β+1)RE Since current in base = R(looking into the emitter) ⇒ Req = Req RE 1 · current in emitter β +1 RB 1 = R(total resistance in base) β +1 rπ + R B R = re + B β +1 β +1 Req Proof: Assume ro is very large. ib RB it = −ib − βib = −(β + 1) ib vt = −ib (RB + rπ ) v − i ( R + rπ ) rπ + R B R Req = t = b B = = re + B it − ( β + 1)ib β +1 β +1 + βib rπ vπ − it vt + − ro Req E.g. Rin = ( β + 1)[re + ( RE // RL )] = rπ + ( β + 1)( RE // RL ) rπ + ( RB1 // RB 2 // RS ) β +1 R // RB 2 // RS = RE //( re + B1 ) β +1 v vv AV = o = o ⋅ X vi v X vi VCC RO = RE // Rin // RB1 // RB 2 RE // RL = ⋅ RE // RL + re Rin // RB1 // RB 2 + RS Last updated: 8-Sep-03 RC RB1 RS ∞ vx Rin vi ∞ RB2 RE Ro vo RL 24 ELEC202 – Electronic Circuits II BJT BJT as a switch (ON and OFF) VCC RC In Cutoff mode (OFF) At VI < VBE(on) Vo VI IC =0 ⇒ VO = VCC In Saturation mode (ON) VCC In active region, VC = VCC − IC RC and IB increase, VC decrease until it reached the edge of active region at VBC = 0 OR VB = VC RC IC IB VC ˆ ˆ V − VC = VCC − VB ≈ VCC − VBE ( on ) & I = I C ˆ Define I C = CC B β RC RC RC ˆ I C is the maximum current that can take before leaving the active region. ˆ Increase IB above I B , IC increase slightly & VC decrease slightly. The BJT is now in “saturation” and ∆iC = incremental β is slightly smaller than before. ∆i B Further increase IB, VC decrease until C-B junction turns on with VBE(on) ≈0.4-0.6V. BJT is now in saturation with VCE(sat) = VBE(on) − VBC(on) ≈0.2V. VCC − VCE ( sat ) = collector current in saturation. Define I C ( sat ) = RC I C ( sat ) To derive the BJT into saturation, min I B = I B ( EOS ) = (where EOS = Edge of Saturation) β To ensure the BJT is driven in Saturation, IB > IB(EOS). IB typical 2 to 10. Define overdrive factor = I B ( EOS ) Also define forced β , β forced = Mode for Saturation BJT B VBE ≈ 0.7V E npn Last updated: 8-Sep-03 I C ( sat ) IB <β and C β β forced = overdrive factor E VCE(sat) ≈ 0.2V VEB ≈ 0.7V B VEC(sat) ≈ 0.2V C pnp 25 ELEC202 – Electronic Circuits II BJT E.g. Find the voltage of each node. β = 100, VBE(ON) = 0.7V, VCE(sat) = 0.2V. VCC Assume the transistor in active region. 6V VB = 6V, VE = VB – VBE(on) = 6 − 0.7 = 5.3V V 5 .3 IE = E = = 1.61mA RE 3.3k RC =4.7kΩ I C = α I E = 1.59mA VC = VCC − I C RC = 10 − (1.59m)(4.7 k ) = 2.52V Check: VC < VB ⇒ assumption is wrong. i.e. BJT is in saturation. RE =3.3kΩ Now VC = VE + VCE ( sat ) = 5.3 + 0.2 = 5.5V 10 − 5.5 = 0.957mA 4.7 k I C ( sat ) 0.957 = = 9.57 µA and I B ( EOS ) = β 100 I B = I E − I C = 1.61m − 0.957m = 0.649mA I 0.957m ⇒ β forced = C ( sat ) = = 1.475 IB 0.649m ⇒ I C ( sat ) = ⇒ overdrive factor = IB I B ( EOS ) or overdrive factor = Note: = IB I B ( EOS ) IB ⋅ β I C ( sat ) = = β β forced = 100 = 67.8 1.475 0.649m = 67.8 9.57 µ BJT is not a perfect switch as in ON stage (BJT in saturation), VCE(sat) ≈0.2V ≠ 0 ∴ Power loss in the switch, Ploss ≈ ICVCE(sat) ≈ 0.2 IC E.g. = IC =1mA ⇒ Ploss = 0.2 mW Also, you need power to drive the switch. Pdriving = IBVBE ≈ 0.7 IB E.g. = IC =1mA ⇒ overdrive factor = 5, β = 100 I 1mA I B = 5 I B ( EOS ) = 5 C ( sat ) = 5 ⋅ = 0.05mA β 100 Pdriving ≈ 0.7 (0.05m) = 0.035 mW Last updated: 8-Sep-03 26 ELEC202 – Electronic Circuits II BJT BJT as a Logic Inverter Basic requirement for a logic inverter 1. low power dissipation in both ON & OFF states in BJT, OFF state is good ON state is not very good, Ploss = ICVCE(sat) VCC RC IC VO VI RB 2. well defined output state levels 1 = VCC (cutoff) 0 = VCE(sat) (saturation) IB Voltage Transfer Characteristic VO Example: RC = 1kΩ, RB = 10kΩ, β=50, VCC =5V (i) At VI = VIH I V − VCE ( sat ) 5 − 0.2 = = 96µA I B = I B ( EOS ) = C ( sat ) = CC β β RC 50(1k ) VOH = 5V ∴VIH = I B ( EOS ) RB + VBE ( on ) = 96µ (10k ) + 0.7 = 1.66V Cutoff (0.7, 5) Active VOL = 0.2V 0 Saturation (1.66, 0.2) (ii) Between VIL < VI < VIH, the circuit acts as an amplifier. VIL VIH = 0.7V β RC rπ 1 ⇒ AV = (− g m RC ) = − ∝ ⇒ Av is not constant rπ + RB rπ + RB rπ Assume rπ << RB , AV ≈ − β VI RC 1k = −50 = −5V / V RB 10k Note: AV can also be derived from the graph. AV ≈ 0.2 − 5 4.8 =− = −5V / V 1.66 − 0.7 0.96 Check if the inverter is stable. When VI = VOL = 0.2V < VIL BJT is in cutoff, VO = VOH = 5V When VI = VOH = 5V > VIH BJT is in saturation, VO = VCE(sat) ≈ 0.2V VCC − VCE ( sat ) (5 − 0.2) β forced = IC RC 4.8 1k = = = = 11.2 I B VOH − V BE ( on ) (5 − 0.7) 0.43 10k RB Noise Margin NMH =VOH – VIH = 5−1.66 = 3.34V Not symmetric NML= VIL − VOL = 0.7−0.2 = 0.5V Note: NMH can be adjusted by the RB/RC ratio. Since NML is fixed and small, this simple BJT circuit is not a very good logic inverter. Last updated: 8-Sep-03 27 ...
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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 University of Arizona- Tucson.

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