SmallSignal BJT Operation Part II Chris Allen calleneecs

Small-Signal BJT Operation Part II Chris Allen (callen@eecs. ku. edu) Course website URL people. eecs. ku. edu/~callen/412/EECS 412. htm 1

Outline Overview of small-signal models for the BJT Graphical analysis of BJT amplifiers Biasing the BJT amplifier configurations • • • Common-emitter configuration Common-base configuration Common-collector configuration (emitter follower) Higher-order effects • • • Breakdown Power dissipation Datasheet example High frequency characteristics of bipolar transistors • Miller effect 2

Small-signal BJT model overview Hybrid- Model 3

Small-signal BJT model overview T Model 4

Small-signal BJT model overview 5

Graphical analysis of BJT amplifiers Graphical techniques are useful in illustrating what is going on in the transistor circuit. Clearly this is simply a learning tool, not a practical analysis tool. For example, consider the circuit shown We want to find IB, the DC base bias current Previously, for VBB > 0. 7 V, we assume VBE = 0. 7 V so that IB = (VBB – 0. 7) / RB But how valid is this assumption that VBE = 0. 7 V? 6

Graphical analysis of BJT amplifiers But how valid is this assumption that VBE = 0. 7 V? If we consider the pn junction between the base & emitter, we know the I-V characteristic follows IB = (IS / ) exp(VBE /VT ) or graphically 7

Graphical analysis of BJT amplifiers But how valid is this assumption that VBE = 0. 7 V? From analysis of the base circuit we know v. BE = VBB – RBIB or graphically “load line” 8

Graphical analysis of BJT amplifiers But how valid is this assumption that VBE = 0. 7 V? We can solve these two equations analytically, however we gain some insight by looking at the solution graphically, and visualize the bias point Q 9

Graphical analysis of BJT amplifiers In a similar fashion we can look at the collector current, i. C, and collector voltage, v. CE In our discussion on the Early voltage, we saw a family of I-V curves that are dependent on i. B 10

Graphical analysis of BJT amplifiers And we can again graph the load line for the collector part of the circuit. From analysis of the collector circuit we know v. CE = VCC – RC i. C or graphically “load line” 11

Graphical analysis of BJT amplifiers Again, we can solve these two equations analytically, however we gain some insight by looking at the solution graphically, and visualize the bias point Q 12

Biasing the BJT The BJT must be biased to operate in the proper mode There a number of techniques for biasing a BJT In all cases it is desireable for the bias current to be stable (constant) for variations in: • • • Supply voltage (VCC, VEE) Transfer (unit to unit variations) Temperature 13

Biasing the BJT Why is it important for IC or IE to be stable? Recall in our small-signal models and analysis that several key parameters are dependent on IC gm = IC / VT r = / g m = V T / I B Therefore variations in IC , IE , re = / g m ro = V A / I C IB will result variations in AC operation To get a sense of this, consider this circuit. Find an expression for IE We know IE = IB ( + 1) And we can find IE using KVL as VCC – IER 3 /( + 1) - VBE - IE RE = 0 So (VCC - VBE) /[R 3 / ( + 1) + RE] = IE 14

Biasing the BJT Sensitivity to ((VCC - VBE) /[R 3 / ( + 1) + RE] = IE or IE = (VCC - VBE) ( + 1) /[RE ( + 1) + R 3] So what? Now perform a sensitivity analysis of this expression Consider variations in , i. e. , To make IE relatively insensitive to we need RE >> R 3 /( + 1) such that IE (VCC - VBE) / RE if we make R 3 0. 1 RE 15

Biasing the BJT Sensitivity to VCC or VBE ? VCC : due to supply voltage variations VBE : due to temperture variations IE / VCC = 1 / RE So larger values of RE smaller IE less sensitivity However we cannot make IE arbitrarily small 16

Biasing the BJT Now, consider this circuit. We know RB and = R 1 // R 2 VBB = VCC R 2 / (R 1 + R 2) Find an expression for IE As before, we get IE = (VBB - VBE) /[RE + RB / ( + 1)] IE = [VCC { R 2 / (R 1 + R 2)} - VBE] ( + 1) /[RE ( + 1) + RB] To make IE relatively insensitive to need RE >> R 3 ( + 1) or, as before RB 0. 1 RE Also, the voltage divider reduces the sensitivity to VCC 17

Biasing the BJT Sensitivity to VCC or VBE ? VCC : due to supply voltage variations VBE : due to temperature variations IE / VCC = {R 2 / (R 1 + R 2)} / RE Compared to the previous biasing technique, the voltage divider reduces the sensitivity to VCC

Biasing the BJT Other considerations regarding biasing BJTs for amplifier circuits. Rules of Thumb Where to bias VC , VE , VBB To maximize the usable range of operation in the active region, design the circuit such that: • VBB ~ VCC / 3 • VCE ~ VCC / 3 • IC RC ~ VCC / 3 Furthermore design the bias network such that IB << current through the voltage divider Select R 1 and R 2 such that I 1 or IB < 0. 1 I 1 >> IB 19

BJT amplifier configurations So far we have covered • Analysis techniques for BJT amplifier circuits • Use of small-signal models for BJT amplifier analysis • Various biasing techniques, and how to compare their goodness, for BJT amplifier circuits Now we look at various BJT amplifier configurations There are 3 basic configurations for BJT amplifier circuits The BJT is a 3 terminal device, and for any amplifier configuration: • We need 2 terminals to comprise an input port • We need 2 terminals to comprise an output port Therefore one of the BJTs terminals must be common to both input and output ports 20

BJT amplifier configurations There are 3 basic configurations for BJT amplifier circuits The BJT is a 3 terminal device and one of the BJT’s terminals must be common to both input and output ports These three configurations are as follows: 21

BJT amplifier configurations We will be examining the characteristics of BJT amplifiers in these 3 configurations For comparison purposes the parameters of interest are: • • • Ri input resistance Ro output resistance Gm short-circuit transconductance Av voltage gain Ai current gain 22

BJT amplifier configurations By finding the equivalent values for Ri (input resistance), Ro (output resistance), and Gm (short-circuit transconductance), for a given amplifier, we can derive Av (voltage gain) and Ai (current gain) to complete our characterization of the amplifier 23

BJT amplifier configurations Our circuit for evaluating these configurations is the Basic BJT Amplifier Structure By selectively grounding and applying input and output connections we can configure this structure as a CE, CB, CC amplifier Note that the capacitors are large enough such that their impedance, Z 0 for AC signals DC analysis yields IC , IE , IB , V C , V E , V B These values are independent of the configuration: CE, CB, CC Likewise we can find gm, r , re , ro for the small-signal models 24

BJT amplifier configurations – CE Common-emitter amplifier configuration (CE) X: input Y: ground Z: output 25

BJT amplifier configurations – CE Common-emitter amplifier configuration (CE) X: input Y: ground Z: output 26

BJT amplifier configurations – CE Common-emitter amplifier configuration (CE) X: input Y: ground Z: output 27

BJT amplifier configurations – CE Common-emitter amplifier configuration (CE) X: input Y: ground Z: output To find Ri & Ro follow the same procedure we used with the Op Amp analysis, i. e. , Ri = vin / iin with RL on output Ro = -vo(oc) / io(sc) with input applied with source resistance 28

BJT amplifier configurations – CE Common-emitter amplifier configuration (CE) Ri = vin / iin with RL on output Ro = -vo(oc) / io(sc) with input applied with source resistance, RS 29

BJT amplifier configurations – CE Common-emitter amplifier configuration (CE) Ri = ? Inspection reveals that Ri = r // RB And for the case where RB >> r then Ri r = / gm 30

BJT amplifier configurations – CE Common-emitter amplifier configuration (CE) Ro = ? Next evaluate the output resistance, Ro Inspection reveals that And io(sc) -vo(oc) / io(sc) vo(oc) = -gm v (ro // RC) [note: RL for o. c. ] = -gm v [note: RL 0 for s. c. ] Therefore Ro = - (gm v / -gm v ) (ro // RC) = ro // RC = Ro 31

BJT amplifier configurations – CE Common-emitter amplifier configuration (CE) Gm = ? Next evaluate the short-circuit transconductance, Gm Inspection reveals that Therefore Gm io(sc) / vin = v and io(sc) = -gm v = -gm 32

BJT amplifier configurations – CE Common-emitter amplifier configuration (CE) Av = ? Next evaluate the voltage gain, Av vo / vs vo = -gm v (ro // RC // RL ) or vo = -gm v (Ro // RL ) Inspection reveals that or vin = vs (RB // r ) / [(RB // r ) + RS ] vin = v. S Ri / (Ri + RS ) 33

BJT amplifier configurations – CE Common-emitter amplifier configuration (CE) Av = ? So vo = -gm (ro // RC // RL ) (RB // r ) / [(RB // r ) + RS ] v. S or Av = -gm (RB // r ) (ro // RC // RL ) / [(RB // r ) + RS ] Av = -gm Ri (Ro // RL ) / (Ri + RS ) And if RB >> r then Ri r then Av = -gm r (ro // RC // RL ) / (r + RS ) Therefore 34

BJT amplifier configurations – CE Common-emitter amplifier configuration (CE) Ai = ? Next evaluate the voltage gain, Ai io / iin io = -gm v (ro // RC // RL ) / RL or vo = vo / RL iin = vs / [(RB // r ) + RS ] = v / (RB // r ) so Ai = -gm (RB // r ) (ro // RC // RL ) / RL 35

Common-Emitter Example Given the Common-Emitter circuit shown find Ri , Ro , Gm , Av , Ai if RS = RL = 10 k Step 1: DC analysis find VC , VB , VE , IC , IB , IE IE = (VCC - VBE) / [RE + RB ( + 1)] IE = 846 A , VE = -1. 54 V IB = IE / ( + 1) IB = 8. 38 A , VB = -0. 838 V IC = IE / ( + 1) IC = 838 A , VC = 1. 62 V VCC = 10 V, VEE = 10 V RC = 10 k , RE = 10 k RB = 100 k , = 100 VA = 100 V, VBE = 0. 7 V 36

Common-Emitter Example Given the Common-Emitter circuit shown find Ri , Ro , Gm , Av , Ai if RS = RL = 10 k Step 2: small-signal parameters find gm , ro , re gm = IC / VT gm = 33. 5 m. A/V r = / gm r = 2. 98 k ro = V A / I C ro = 119 k re = / gm re = 29. 6 gm = 33. 5 m. S r = 2. 98 k ro = 119 k re = 29. 6 VCC = 10 V, VEE = 10 V RC = 10 k , RE = 10 k RB = 100 k , = 100 VA = 100 V, VBE = 0. 7 V VC = 1. 62 V, IC = 838 A VB = -0. 838 V, IB = 8. 38 A VE = -1. 54 V, IE = 846 A 37

Common-Emitter Example Given the Common-Emitter circuit shown find Ri , Ro , Gm , Av , Ai if RS = RL = 10 k Step 3: amplifier parameters find Ri , Ro , Gm Ri = RB // r Ri = 2. 89 k r = 2. 98 k Ro = RC // ro Ro = 9. 22 k RC = 10 k Gm = -gm Gm = -33. 5 m. S Ri = 2. 98 k Ro = 9. 22 k gm = 33. 5 m. S r = 2. 98 k ro = 119 k re = 29. 6 VCC = 10 V, VEE = 10 V RC = 10 k , RE = 10 k RB = 100 k , = 100 VA = 100 V, VBE = 0. 7 V VC = 1. 62 V, IC = 838 A VB = -0. 838 V, IB = 8. 38 A VE = -1. 54 V, IE = 846 A 38

Common-Emitter Example Given the Common-Emitter circuit shown find Ri , Ro , Gm , Av , Ai if RS = RL = 10 k Step 4: gain parameters find Av , Ai Av = Ri / (Ri + RS ) Gm (Ro // RL ) Av = 36. 9 V/V Ai = Gm Ri (ro // RC // RL ) / RL Ai = -46. 5 A/A Gm = -33. 5 m. S Ri = 2. 98 k Ro = 9. 22 k gm = 33. 5 m. S r = 2. 98 k ro = 119 k re = 29. 6 VCC = 10 V, VEE = 10 V RC = 10 k , RE = 10 k RB = 100 k , = 100 VA = 100 V, VBE = 0. 7 V VC = 1. 62 V, IC = 838 A VB = -0. 838 V, IB = 8. 38 A VE = -1. 54 V, IE = 846 A 39

BJT amplifier configurations – CE Common-emitter amplifier configuration (CE) Ri = 2. 89 k Ro = 9. 22 k Av = -36. 9 V/V Ai = -46. 5 A/A Good voltage / current gain Medium input / output resistance Good voltage amplifier gm = 33. 5 m. S r = 2. 98 k ro = 119 k re = 29. 6 VCC = 10 V, VEE = 10 V RC = 10 k , RE = 10 k RB = 100 k , = 100 VA = 100 V, VBE = 0. 7 V VC = 1. 62 V, IC = 838 A VB = -0. 838 V, IB = 8. 38 A VE = -1. 54 V, IE = 846 A 40

BJT amplifier configurations – CB Common-base amplifier configuration (CB) X: ground Y: input Z: output 41

BJT amplifier configurations – CB Common-base amplifier configuration (CB) X: ground Y: input Z: output 42

BJT amplifier configurations – CB Common-base amplifier configuration (CB) X: ground Y: input Z: output T model without Early effect 43

BJT amplifier configurations – CB Common-base amplifier configuration (CB) Ri = vin / iin with RL on output Ro = -vo(oc) / io(sc) with input applied with source resistance, RS T model without Early effect 44

BJT amplifier configurations – CB Common-base amplifier configuration (CB) Ri = ? Inspection reveals that Ri = RE // re Ro = ? Turn off vs (replace with short circuit) therefore Ro vi = 0, ie = 0 = RC 45

BJT amplifier configurations – CB Common-base amplifier configuration (CB) Gm = ? Recall that Gm io(sc) / vin with RL = 0 io(sc) = ie ie = - vin / re vin = - re ie Therefore Gm = - ie / (- re ie) = / re or Gm = gm 46

BJT amplifier configurations – CB Common-base amplifier configuration (CB) Av = ? vo / vs vo = - ie (RC // RL ) ie = - vin / re vin = vs (RE // re ) / [(RE // re ) + RS ] vo = / re vs (RC // RL ) (RE // re ) / [(RE // re ) + RS ] Therefore Av = - / re ( RC // RL ) (RE // re ) / [(RE // re ) + RS ] Recall that Av 47

BJT amplifier configurations – CB Common-base amplifier configuration (CB) Av = ? = / re ( RC // RL ) (RE // re ) / [(RE // re ) + RS ] or Av = gm ( RC // RL ) (RE // re ) / [(RE // re ) + RS ] and if re << RE, then Av = gm (re ) / (re + RS ) ( RC // RL ) Therefore Av 48

BJT amplifier configurations – CB Common-base amplifier configuration (CB) Ai = ? io / iin io = vo / RL = ie ( RC // RL ) / RL iin = - vin / (RE // re ) ii = -ie re / (re // RE) ie = - vin / re vin = - ie re if re << RE Therefore Ai = ( RC // RL ) (RE // re ) / ( re RL ) (RC // RL ) / RL Recall that Ai } 49

Common-Base Example Given the Common-Base circuit shown find Ri , Ro , Gm , Av , Ai if RS = RL = 10 k Step 3: amplifier parameters find Ri , Ro , Gm Ri = RE // re Ri = 29. 5 Ro = RC Ro = 10 k Gm = gm Gm = 33. 5 m. S gm = 33. 5 m. S r = 2. 98 k ro = 119 k re = 29. 6 VCC = 10 V, VEE = 10 V RC = 10 k , RE = 10 k RB = 100 k , = 100 VA = 100 V, VBE = 0. 7 V VC = 1. 62 V, IC = 838 A VB = -0. 838 V, IB = 8. 38 A VE = -1. 54 V, IE = 846 A 50

Common-Base Example Given the Common-Base circuit shown find Ri , Ro , Gm , Av , Ai if RS = RL = 10 k Step 4: gain parameters find Av , Ai Av = gm (re ) / (re + RS ) ( RC // RL ) Av = 0. 5 V/V Ai = (RC // RL ) / RL Ai = 0. 5 A/A Gm = 33. 5 m. S Ri = 29. 5 k Ro = 10 k gm = 33. 5 m. S r = 2. 98 k ro = 119 k re = 29. 6 VCC = 10 V, VEE = 10 V RC = 10 k , RE = 10 k RB = 100 k , = 100 VA = 100 V, VBE = 0. 7 V VC = 1. 62 V, IC = 838 A VB = -0. 838 V, IB = 8. 38 A VE = -1. 54 V, IE = 846 A 51

BJT amplifier configurations – CB Common-base amplifier configuration (CB) Ri = 29. 5 Ro = 10 k Av = 0. 5 V/V Ai = 0. 5 A/A Low input resistance Small current gain Good current buffer gm = 33. 5 m. S r = 2. 98 k ro = 119 k re = 29. 6 VCC = 10 V, VEE = 10 V RC = 10 k , RE = 10 k RB = 100 k , = 100 VA = 100 V, VBE = 0. 7 V VC = 1. 62 V, IC = 838 A VB = -0. 838 V, IB = 8. 38 A VE = -1. 54 V, IE = 846 A 52

BJT amplifier configurations – CC Common-collector amplifier configuration (CC) X: input Y: output Z: ground 53

BJT amplifier configurations – CC Common-collector amplifier configuration (CC) X: input Y: output Z: ground 54

BJT amplifier configurations – CC Common-collector amplifier configuration (CC) X: input Y: output Z: ground 55

BJT amplifier configurations – CC Common-collector amplifier configuration (CC) Ri = vin / iin with RL on output Ro = -vo(oc) / io(sc) with input applied with source resistance, RS 56

BJT amplifier configurations – CC Common-collector amplifier configuration (CC) Ri = ? Ri = vin / iin First break problem into pieces, as shown R = ro // RE // RL Vo = ( + 1) ib R ib = (Vi – Vo) / r r ib = Vi – ( + 1) R ib 57

BJT amplifier configurations – CC Common-collector amplifier configuration (CC) Ri = vin / iin r ib = vin – ( + 1) R ib [r + ( + 1) R ] ib = vin / ib = r + ( + 1) R R ib vin / ib = r + ( + 1) (ro // RE // RL ) R i = RB // Rib Note that Rib will be very large due to the ( +1) multiplier 58

BJT amplifier configurations – CC Common-collector amplifier configuration (CC) Ro = -vo(oc) / io(sc) Turn off vs (replace with short circuit) and find the equivalent resistance looking back into the output 59

BJT amplifier configurations – CC Common-collector amplifier configuration (CC) Ro = v x / ix from KCL we know - ix = vx ( + 1) ib – vx / ro – vx / RE or - ix = -vx [( + 1) / (r + RS // RB ) + 1 / ro + 1 / RE ] so Ro = ro // RE // [(r + RS // RB ) / ( + 1) ] Note that RL appears in Ri expression & RS appears in Ro expression 60

BJT amplifier configurations – CC Common-collector amplifier configuration (CC) Gm = ? Gm is not meaningful because Ri depends on RL and Ro and is not unilateral 61

BJT amplifier configurations – CC Common-collector amplifier configuration (CC) Av = ? It can be shown that Av = RL / ( RL // Ro ) < 1 Therefore is useful as buffer at output of amplifier chain 62

BJT amplifier configurations – CC Common-collector amplifier configuration (CC) Ai = ? It can be shown that Ai ( + 1) ro / (ro + RL) and for RL << ro this simplifies to Ai ( + 1) 63

Common-Collector Example Given the Common-Collector circuit shown find Ri , Ro , Av , Ai if RS = RL = 10 k Step 3: amplifier parameters find Ri , Ro , Av , Ai Ri = RB // Rib Ri = 83 k Ro = ro // RE // [(r + RS // RB ) / ( + 1)] Ro = 118 Av = RL / ( RL // Ro ) Av = 0. 89 V/V Ai ( + 1) ro / (ro + RL) Ai = 8. 3 A/A gm = 33. 5 m. S r = 2. 98 k ro = 119 k re = 29. 6 VCC = 10 V, VEE = 10 V RC = 10 k , RE = 10 k RB = 100 k , = 100 VA = 100 V, VBE = 0. 7 V VC = 1. 62 V, IC = 838 A VB = -0. 838 V, IB = 8. 38 A VE = -1. 54 V, IE = 846 A 64

BJT amplifier configurations – CC Common-collector amplifier configuration (CC) Ri = 83 k Ro = 118 Av = 0. 89 V/V Ai = 8. 3 A/A High input resistance Low output resistance Small voltage gain Good current gain Good current amplifier Good voltage buffer gm = 33. 5 m. S r = 2. 98 k ro = 119 k re = 29. 6 VCC = 10 V, VEE = 10 V RC = 10 k , RE = 10 k RB = 100 k , = 100 VA = 100 V, VBE = 0. 7 V VC = 1. 62 V, IC = 838 A VB = -0. 838 V, IB = 8. 38 A VE = -1. 54 V, IE = 846 A 65

BJT amplifier configurations – summary Common-Emitter Common Base Common Collector Ri = 2. 89 k Ro = 9. 22 k Av = -36. 9 V/V Ai = -46. 5 A/A Ri = 29. 5 Ro = 10 k Av = 0. 5 V/V Ai = 0. 5 A/A Ri = 83 k Ro = 118 Av = 0. 89 V/V Ai = 8. 3 A/A Medium input resistance Medium output resistance Good voltage gain Good current gain Low input resistance High output resistance Low voltage gain Low current gain High input resistance Low output resistance Voltage gain 1 Good current gain voltage amp current buffer amp unity-gain voltage amp Note that the CE Ri can be increased by including an unbypassed RE component 66

Higher-order effects p-n junction breakdown Recall the junction status for the modes of BJT operation Mode EBJ Cutoff reverse biased Active (forward) forward biasedreverse biased Active (reverse) reverse biased forward biased Saturation forward biased Anytime we have a reverse-biased p-n junction breakdown is a concern. In Active Mode (forward) (Amplifier configuration) The CBJ is reverse biased and breakdown is an issue In Cutoff Mode (Switching configuration) Both CBJ and EBJ are reverse biased In Active Mode (reverse) The EBJ is reverse biased and breakdown is an issue 67

Higher-order effects How to determine the breakdown level ? Consider the circuit Note: VB = 0 for IE > 0 VE < 0 therefore CBJ is reverse biased for VCB >0 So the BJT is in Active Mode 68

Higher-order effects How to determine the breakdown level ? Consider the circuit Note: VB = 0 for IE > 0 VE < 0 therefore CBJ is reverse biased for VCB >0 So the BJT is in Active Mode Notice: When i. E = 0 (open ckt) breakdown occurs at BVCBO (Collector-Base Breakdown Voltage emitter Open circuit) when VCB > ~ BVCBO 69

Higher-order effects How to determine the breakdown level ? Consider the circuit Note: VB = 0 for IE > 0 VE < 0 therefore CBJ is reverse biased for VCB >0 So the BJT is in Active Mode Notice: For cases where i. E > 0 the breakdown voltage (BV) is reduced BV < BVCBO 70

Higher-order effects How to determine the breakdown level ? Consider the circuit Note: VB = 0 for IE > 0 VE < 0 therefore CBJ is reverse biased for VCB >0 So the BJT is in Active Mode Notice: In active mode i. C is somewhat dependent on VCB. Ideally the i. C vs. VCB curves would be flat, i. e. , a slope = 0, instead the slope is > 0 and it increases with VCB 71

Higher-order effects How does IC increase when IE is constant ? • If IC increases (as the plot indicates) while IE remains constant, then the base current (IB) must be changing. • The Early effect cannot explain this phenomenon as ro connects Emitter and Collector and cannot change. • Therefore our model of the BJT is incomplete. New resistance term, r is added joining B to C, with r > ro, i. e. , very large. 72

Higher-order effects Breakdown can also occur between the Collector-Emitter junction as well as at the Base-Emitter junction. These are typically characterized for cases where third terminal is open circuited. (o. c. ) Example: BVCEO (base is o. c. ) is typically ½ of BVCBO BVEBO (collector is o. c. ) is typically small (~ 6 to 8 V) • • This will not occur in active mode as EBJ is forward biased, not reverse biased. Also, breakdown of the emitter-base junction is destructive, i. e. , it results in permanent change in device characteristics ( ). 73

Higher-order effects Power dissipation in the BJT As we’ve seen previously, power dissipation for any multi-terminal device is found using PD = In. Vn, for n = 1, 2, …, N where N is the number of terminals and all currents flows are into the device Applying this to the BJT we can find the power dissipated Consider the example where: VC = 5 V, IC = 10 m. A VB = 2 V, IB = 0. 1 m. A VE = 1. 3 V, IE = 10. 1 m. A PD = (5 V)(10 m. A) + (2 V)(0. 1 m. A) + (1. 3 V)(-10. 1 m. A) PD = 50 m. W + 0. 2 m. W – 13. 1 m. W = 37. 1 m. W 74

Device Higher-order effects We’ve treated as a device constant throughout our analysis. In fact is dependent on temperature (T) and collector current (IC) Furthermore DC IC / IB AC = ic / ib = hfe (when v. CE = constant) 75

High Frequency Characteristics of BJTs Internal capacitances Capacitance within the BJT is due to the p-n junctions. The depletion region and charge distribution act as a capacitance at high frequencies. High-frequency models of the BJT include capacitance between the base -emitter junction, CBE, (C ) and between the collector-base junction, CBC (C ). There are in fact other capacitances in the BJT, but C and C are the dominant ones. Adding these new capacitances to our existing low-frequency BJT hybrid- model we get Typically C few 10 s of p. F C ~1 p. F 76

High Frequency Characteristics of BJTs Internal capacitances At moderate and high frequencies, the admittance of r is much less than the admittance of C and therefore r can be neglected without compromising the model’s fidelity. The resulting simplified high-frequency small-signal model for the active BJT is Computer simulation models (e. g. , Spice) does not neglect these terms, in fact it included many terms we don’t even bother with here. 77

High Frequency Characteristics of BJTs BJT as a Current Amplifier In active mode the BJT is a current amplifier, ic = ib In evaluating its frequency response, current gain is a parameter of interest Ai = ic / ib = hfe where h : hybrid f : forward e : common emitter 78

High Frequency Characteristics of BJTs BJT’s unity gain frequency, f. T We can define the BJT’s unity gain frequency, f. T The frequency at which |Ai| = 1 for this circuit model To determine f. T we derive expressions for ib and ic to find Ai Admittance ic = gmv + (0 - v ) j C ic = v ( gm - j C ) v = ib [ r // (1/ j C ) ] v = ib / [ (1/r ) + j C ] Ai = ic / ib = ( gm - j C ) / [ (1/r ) + j (C + C )] 79

High Frequency Characteristics of BJTs BJT’s unity gain frequency, f. T The frequency at which |Ai| = 1 for this circuit model Ai = ( gm - j C ) / [ (1/r ) + j (C + C )] Ai = ( gm - j C ) r / [ 1 + j r (C + C )] and if gm >> C then Ai = gm r / [ 1 + j r (C + C )] and since r = DC/gm then gm Ai = DC r = DC / [ 1 + j r (C + C )] Therefore the break frequency, b, is b = 1/ [r (C + C )] A ( ) = /(1 +j) The frequency where Ai is -3 d. B i b of the mid-band value 80

High Frequency Characteristics of BJTs BJT’s unity gain frequency, f. T If gm >> C and DC >> 1 then Ai = DC / [ 1 + j r (C + C )] when = DC / [r (C + C )] = gm / (C + C ) = T Such Ai( T) Therefore the break frequency, b, is b = 1/ [r (C + C )] The frequency where Ai is -3 d. B of the mid-band value 81

High Frequency Characteristics of BJTs BJT’s unity gain frequency, f. T If gm >> C and DC >> 1 then Ai = DC / [ 1 + j r (C + C )] when = DC / [r (C + C )] = or gm / (C + C ) = T since Ai( T) DC / DC = 1 Example Consider a BJT with DC = 100, IC = 1 m. A, C = 0. 3 p. F, and C = 12 p. F Sketch |Ai| vs. freq At DC, Ai = DC = 100 The -3 d. B frequency, fb, is fb = b/ 2 = 1/ 2 [r (C + C )] gm = IC/VT = 1 m. A / 25 m. V = 40 m. S r = DC / gm = 100 / 0. 04 = 2. 5 k fb = 5. 18 MHz 82

High Frequency Characteristics of BJTs BJT’s unity gain frequency, f. T Example f. T = DC fb, = (100) (5. 18 MHz) f. T = 518 MHz 83

High Frequency Characteristics of BJTs Miller effect Analyzing the performance of a BJT amplifier at high frequencies is complicated by the fact that the base-collector capacitance provides a current path between the base and the collector. Miller’s theorem provides an effective technique for simplifying this analysis. For an impedance, Z, connecting two nodes (1) and (2) where the voltages at these two nodes are related as V 1 = K V 2, where K is a gain factor (can be positive or negative) it can be shown that the series impedance can be replaced by two shunt impedances as shown. 84

High Frequency Characteristics of BJTs Miller effect Consider the Common-Emitter BJT amplifier circuit The small-signal equivalent circuit is shown with a signal source connected to the input and a load resistance connected to the output. note at high frequencies that the board-level capacitors (C 1, C 2, C 3) are replaced by short circuits while BJT’s internal capacitances remain. 85

High Frequency Characteristics of BJTs Miller effect Ignoring the effects of C we find the voltage gain, G G = vo / vi = -gm R Applying Miller’s theorem (K = -gm R) We find values for Cm and Cm’ in the new small-signal equivalent circuit as Cm = C [1 + gm. R] Cm = C [1 + 1 / (gm. R)] When gm. R >> 1 then Cm becomes large and Cm is becomes small 86

High Frequency Characteristics of BJTs Miller effect Example Continuing our previous example DC = 100, IC = 1 m. A, C = 0. 3 p. F, C = 12 p. F gm = 40 m. S, r = 2. 5 k , R = 2 k , RB = 2 k Gain = -80 V/V = gm R = (40 m. S) (2 k ) Cm = 0. 3 [1+80] = 24. 3 p. F C m = 0. 304 p. F Now the break frequency, fb = (1/2 ) [(RB//r )(C + Cm)]-1 = 1. 8 MHz and f. T = (Gain)(f. B ) = (80)(1. 8 MHz) f. T = 144 MHz 87