In this section we will consider a number of BJT circuits and perform the DC circuit analysis. For those circuits with an active mode BJT, we’ll assume that VBE = 0.7V (npn) or V EB = 0.7V (pnp). Example N12.1 (text example 5.4). Compute the node voltages and currents in the circuit below assuming ß = 100.

 If the BJT is in the active mode, Vbe = 0.7V then

 With Ic Iae = then

 Consequently, using KVL

 Vc=10-IeRc= 10-0.99*10-3>-4.7*103=5.3V Finally, using KCL IB+IC=IE,or IB=IE-IC=1-0.99=0.01mANow we’ll check to see if these values mean the BJT is in the active mode (as assumed). • VCB = 5.3-4 =1.3 V. This is greater than zero, which means the CBJ is reversed biased. • VBE =0.7 V. This is greater than zero, which means the EBJ is forward biased. Because the CBJ is reversed biased and the EBJ is forward biased, the BJT is operating in the active mode. Note that in the text, they show a technique for analyzing such circuits right on the circuit diagram in . 

 Example N12.2 (text example 5.5). Repeat the previous example but with VB =6 V. Assuming the BJT is operating in the active mode: 

 From the last calculation 2.57 C Vc = 2.57VÞ VCB = 3.43 V. Consequently, the BJT is not in the active mode because the CBJ is forward biased. A better assumption is the transistor is operating in the saturation mode. We’ll talk more about this later. For now, suffice it to say that in the saturation mode VCE|

sat »0.2 V (see Section 5.3.4). Assuming this and reanalyzing the circuit: 

 

Notice that

 This ratio is often called forced ß . Observe that it’s not equal to 100, as this ratio would be if the transistor were operating in the active mode (see Section 5.3.4). Example N12.3 (text example 5.7). Compute the node voltages and currents in the circuit below assuming ß = 100. To begin, we’ll assume the pnp transistor is operating in the active mode. 

 Now check if the BJT is in the active mode: • EBJ? Forward biased. • CBJ? Reversed biased. So the BJT is in the active mode, as originally assumed. Example N12.4 (text exercise D5.25). Determine the largest RC that can be used in the circuit below so that the BJT remains in the active mode. (This circuit is very similar to the one in the previous example.) 

 We’ll begin by assuming the BJT is operating in the active mode. In the active mode, the CBJ needs to be reversed biased. The lowest voltage across this junction for operation in the active mode is VCB=0 VC=VB=0VTherefore, by KVL -10+R CIc=0or

 This value of RC and smaller is required for the BJT to operate in the active mode. Example N12.5 (text example 5.10). Determine the node voltages and currents in the circuit shown below. Assume the BJT is operating in the active mode with ß=100 . 

 First, we’ll use Thévenin’s theorem to simplify the

base circuit

 The Thévenin equivalent resistance and voltage are then

 Using this Thévenin equivalent for the base circuit, the overall circuit is then 

 To find the emitter current, we’ll apply KVL over the loop shown giving 5=33.3*103.IB+0.7+3,000.IEThe quantity of interest is IB. With CB IC= Iß B=and IC =IaE for a BJT in the active mode, we find

 Using this in the KVL equation 5-0.7=[33.3*103+3,000(ß+1)]Ib With ß=100 then solving this equation we find 

 Next, by KCL

 The node voltages are then

 Lastly, let’s check if the BJT is operating in the active mode. • BE B E VBE= VB- VE = 4.57 -3.87 =0.7 V This is 0.7 V originally assumed for a forward biased EBJ. • VBC= VB -V C=4.57 -8.6=-4.03 V. This is less than zero, which means the CBJ is reversed biased. Therefore the BJT is operating in the active mode, as originally assumed. 

One very useful application of the transistor is an amplifier of time varying signals. Consider the “conceptual BJT amplifier” circuit shown below: 

 The DC voltages provide the biasing. The input signal is vbe and the output signal is vc. We will assume the transistor is biased so that VC is greater than VB by an amount that llows for sufficient “signal swing” at the collector, but the transistor remains in the active

mode at all times. That is, the transistor does not become saturated or cutoff during the cycle. From the circuit above, the total base-to-emitter voltage is 

 Correspondingly, the collector current is

 or using (5.53) ic=Ice

For small vbe such that Vbe<<2VVr (i.e., the small-signal approximation), then (3) can be approximated by

 This is a familiar result: We saw something very similar with small signals and diodes back in Lecture 4.The time varying current in (4) 

 Can be written as ic=gwVbe

 is defined as the transistor small-signal transconductance. Its units are Siemens. Note that gmIc. Significance of the BJT Small-Signal Transconductance What is the physical significance of gm? First, gm is the slope of the iC-vBE characteristic curve at the Q point: 

 Consider the plot shown in Fig. 

 with ic= Ise from (2), the right-hand side of (8) becomes

 

 therefore as we defined in (6). Observe that: • The small-signal vbe assumption restricts the operation of the BJT to nearly linear portions of the iC-vBE characteristic curve. • From (6), the BJT behaves as a voltage controlled current source for small signals: The small-signal be v controls the small-signal icSignal Voltage GainSecond, gm has an important relationship to the signal voltage gain in this circuit. Using KVL in Fig. 5.48a, the total collector voltage is

 where VC is the DC voltage at the collector. So from (11), the AC signal at the collector is Vc=-IcRc This result is negative, which means this circuit operates as an inverting amplifier for small, time varying

signals. From (6), c mbe ic= gm vbe = . Using this result in (12) gives Vc=-gmVbeRc=-(gmRc)vbe Consequently, the small-signal AC voltage gain Av is

 In a broad sense, we can see that this transistor circuit can act an amplifier of the time varying input signal, provided this input voltage remains small enough. 

 gm is a very important amplifier parameter since the voltage gain in (14) is directly proportional to gm. BJTs have a relatively large gm compared to field effect transistors, which we will consider in the next chapter. Consequently, BJTs have better voltage gain in such circuits. 

Our next objective is to develop small-signal circuit models for the BJT. We’ll focus on the npn variant in this lecture. Recall that we did this for the diode back in 

 In order to develop theseBJT_Small-signal models, there are two small-signal resistances that we must first determine. These are: 

 1. r p: the small-signal, active mode input resistance betweenthe base and emitter, as “seen looking into the base.” 2. re: the small-signal, active mode output resistance between the base and emitter, as “seen looking into the emitter.” These resistances are NOT the same. Why? Because the transistor is not a reciprocal device. Like a diode, the behavior of the BJT in the circuit changes if we interchange the terminals. Determine r pAssuming the transistor in this circuit

 is operating in the active mode, then

 The AC small-signal equivalent circuit from

 Notice the “AC ground” in the circuit. This is an extremely important concept. Since the voltage at this terminal is held constant at VCC, there is no time variation of the voltage. Consequently, we can set this terminal to be an “AC ground” in the small-signal circuit. For AC grounds, we “kill” the DC sources at that terminal: short circuit voltage sources and open circuit current sources. So, from the small-signal equivalent circuit above: 

 we see that

 Hence, using (2) in (3) 

 This r p is the BJT active mode small-signal input resistance ofthe BJT between the base and the emitter as seen looking into the base terminal. (Similar to a Thévenin resistance, this statement means we are fictitiously separating the source from the base of the BJT and observing the input resistance, as indicated by the

dashed line in Fig. 2.) Determine reWe’ll determine re following a similar procedure as for r p, but beginning with

 The AC component of iE in (5) is

 or with IE =I c|c , 

 As indicated in Fig. 1 above, re is theBJT_Small-signal resistance between the emitter and base seen looking into the emitter: 

 Mathematically, this is stated as

 Assuming an ideal signal voltage source, then e be ve=-vbe and

 Using (7) in this equation we find

 But from (5.87) 

 

Therefore, using this last result in (10) gives

 This is the BJT active mode small-signal resistance between the base and emitter seen looking into the emitter. It can be shown that Rz=(b+1)re[ed] It is quite apparent

from this equation that e r r p . . This result is not unexpected because the active mode BJT

is a non-reciprocal device, as mentioned on page 1 of these notes. 

BJT Small-Signal Equivalent Circuits

We are now in a position to construct the equivalent active mode, small-signal circuit

models for the BJT. There are two families of such circuits: 

1. Hybrid- p model

2. T model. 

Both are equally valid models, but choosing one over the other sometimes leads to simpler

analysis of certain circuits. 

Hybrid- p Model

Version A. 

 

Let’s verify that this circuit incorporates all of the necessarysmall-signal characteristics of

the BJT: ib= vbe|rp as required by (3). ic= gm vbe as required by (5.86), which we saw in

the last lecture. ib+ ic= ie as required by KCL. 

We can also show from these relationships that ie= vbe| re.

Version B. We can construct a second equivalent circuit by using

 

Hence, using this result the second hybrid- p model is

 

T Model

The hybrid- p model is definitely the most popular small-signal model for the BJT. The

alternative is the T model, which is useful in certain situations. The T model also has two

versions: 

Version A. 

 

Version B. 

 

The small-signal models for pnp BJTs are identically the same as those shown here for the

npn transistors. It is important to note that there is no change in any polarities (voltage or

current) for the pnp models relative to the npn models. Again, these small-signal models are

identically the same. 

We will now consider three examples in this lecture of BJTs used as linear amplifiers. Here are the steps to follow when solving small-signal transistor amplifier problems: 1. Determine the Q point of the BJT using DC analysis. Compute IC.2. Calculate the small-signal model parameters for the BJT: 

 3. Rewrite the small-signal circuit: short out DC sources and open DC current sources. Use the small-signal model for the BJT. 4. Analyze the small-signal circuit for the desired quantities such as voltage, small-signal voltage gain, etc. Example N15.1 (text example 5.14). Determine the small-signal AC voltage gain for the circuit below, assuming ß = 100 and the output voltage taken at the collector terminal. 

 The first step in the solution is to determine the Q point through DC analysis. By superposition, we’ll force 0 i v = for this analysis. Assuming the BJT is in the active mode, the results of the DC analysis are: 

 We see that the CBJ is reversed biased so this npn BJT is in the active mode because of this and the EBJ is forward biased. Next, we determine the BJT small-signal model parameters for the hybrid- p model: 

 Now, we insert a small-signal equivalent model of the BJT into the circuit of Fig. 5.53(a) after shorting the DC voltage sources (VBB and VCC). This gives the small-signal equivalent circuit: 

 Notice the AC ground at RC. This is an “AC ground” because the voltage at this node does not vary with time. For the purposes of the AC signal analysis, we can set this node to an AC ground. (As a side note, in the lab power supplies have a finite internal resistance. This Thévenin equivalent resistance must be included in the AC circuit for analysis purposes.) Next, we perform the small-signal analysis referring to Fig. 5.53c. At the input

 while at the outputvo=-Rcie=-RcgmVbe

Substituting for vbe from (4) gives

 Therefore, the small-signal AC voltage gain, Av, is

 For this particular problem

 The negative sign indicates this is an inverting

amplifier: the AC output signal is inverted with respect to the input AC signal. Example N15.2 (text example 5.15). Repeat the analysis of the previous example but with a triangular input waveform of small amplitude. In the text, vip =0.8 V, is the peak amplitude of the triangularinput voltage (= Vi in the text). Then from (4) above (and the fact that there are only resistors in the circuit) 

 which is less that 10 mV. This is fairly small with respect to 2 VT = 50mV so we’ll go ahead and use the small-signal analysis and models. Sketches of the total voltages and currents from this circuit are shown 

 

 A few things to take special notice: • In Fig. 5.54c, vBE has a DC part and an AC part (see Fig.5.54a) that is “riding” on the former. Notice the enlarged vertical scale in Fig. 5.54c

• In Fig. 5.54d, ic is in-phase with the input voltage. • In Fig. 5.54e, vc= Vc -icRc is 180° out-of-phase with the input. As vi ,Ic  . Þ vc. We can see how the AC ground works here. Example N15.3 (text example 5.16). Determine the small-signal AC voltage gain for the BJT amplifier circuit shown in 

 The two capacitors in this circuit serve as DC blocks. They have a large enough C so that Xc=0 . at the operating frequency. With these capacitors, the DC bias is unchanged by the source or load attachments. We call this “capacitively coupled” input and output. As always, we first determine the DC bias. We’ll assume the BJT is in the active mode and that ß = 100: From this result

 From this resultIc=0.92mAÞvc=-10+Icrc=-5.4VNext, we construct the small-signal equivalent circuit and analyze the circuit to determine the voltage gain. We’ll use the T model, though the hybrid- p model would work as well. 

 Notice the two AC grounds in this circuit: one at RE and the other at RC. Also notice this is the first small-signal model of the pnp transistor we have used. The small-signal model of the pnp transistor is exactly the same as that for the npn with no change in the polarities of the currents or voltages. This can be a little confusing. Here, for example, ie is a negative quantity. Using (3) for the small-signal equivalent model of the BJT

 From the small-signal AC circuit: • v0=-a ieRc

• Because the base is grounded, Ie=-Vi|reTherefore, 

 

Notice that this small-signal voltage gain is a positive quantity. The reason for this is the input is tied to the emitter. (Note that this positive gain did not occur just because this is a pnp BJT.) Now, with a = 0.99 then from (10) 

 Lastly, for linear operation of this amplifier, veb mV. With Veb=-Vithen Vi@10 mV for linear operation of the amplifier, which implies that v0@1.8V A sketch of the output from this small-signal amplifier is shown in Fig. 5.57 for a sinusoidal input voltage: 

 We’re assuming the output remains linear and the BJT in the active mode at all times for the entire voltage swing in vC. If this input voltage were set to a larger value, this would no longer be the case and the BJT would first encounter nonlinear behavior and eventually saturate. Both of these effects would distort the output voltage and it would no longer be an amplified copy of the input voltage. 

We can use graphical analysis to approximately analyze the response of simple transistor amplifier circuits. This technique is primarily useful to develop physical insight. Consider once again the “conceptual BJT amplifier” circuit: 

 Similar to the analytical solution, there are two primary steps to the graphical solution of such small-signal amplifiers: 1. DC basis analysis2. AC small-signal analysis. DC BiasThe first step in the bias calculations is to determine IB. This is done with the iB-vBE characteristic curve and the load line: 

 Once IB has been determined we can compute IC knowing that Ic=b IB for a BJT in the active mode. With this IC value and the iC-vCE characteristic curve of the transistor, we can determine V We haven’t yet seen the iC-vCE characteristic curve of the BJT. This can be measured using the circuit in Fig. 5.19(a) below. vBEis fixed at some value, then vCE is swept while measuring iC. The results are shown below for different values of vBE.

 When vCE is very small, iC is nearly zero. This is the cutoff mode of the BJT. As vCE increases, the CBJ is forward biased and the BJT is in the saturation mode. When vCE becomes large enough, the CBJ becomes reversed biased and the BJT enters the active mode. The slopes of the lines in Fig. 5.19 in the active mode are quite exaggerated in this figure. So, back to the graphical solution. With the IC=bIB value from Fig. 5.28 and the iC-vCE characteristic curve of the transistor from Fig. 5.19, we can determine VCE: 

 Curve tracers are pieces of equipment that will measure and display families of iC-vCE characteristic curves for transistors. AC Small-Signal AnalysisThe first step in the AC small-signal analysis is to determine ib. This is performed using a slightly complicated interaction of the input waveform vi, the subsequent time variation of the load line, and the

iB-vBE characteristic curve of the BJT: 

 From this comes the small-signal quantities vbe and ib. With ib known and ic=biB , then we use these values on the vic characteristic curve to determine vce: 

 Cutoff and SaturationNotice that there are limits on vCE in which the BJT remains in the active mode: • Too large ( CC V = ) and the BJT cuts off• Too small (few tenths of a volt) and the transistor saturates. These limits are readily apparent if we reexamine the previous figure of the small-signal variation: 

 Because of these limits on vCE, it is important to choose the Q point properly to all for the desired swing in the signal voltage (vce). 

It is important for the biasing of a transistor amplifier that it remains largely invariant to fairly large changes in â and temperature. Proper biasing doesn’t happen by chance. For example, the npn and pnp inverter circuits in Laboratory #3 are highly sensitive to variations in â. That is usually a poor design (but was done on purpose for the lab, of course). In this lecture, we will study four BJT biasing methods: 1. Single power supply2. Dual power supply

3. Alternate method for common emitter amplifiers4. Current source. Single Power Supply Biasing MethodPerhaps the most common method for biasing BJT amplifier circuits with a single power supply is shown in 

 RE is part of this biasing method as well. When used as an amplifier, the input signal would be capacitively coupled to the base of the BJT while the output would be taken (through capacitive coupling) at the collector or emitter of the transistor, depending on the specific requirements for the amplifier. We analyzed a specific example of this type of circuit in Lecture 12 employing Thévenin’s theorem to simplify the analysis: 

 where VBB and RB are given in (5.68) and (5.69) in the text. Using KVL in the loop shown above V BB=IBRB +VBE +IE RE With IB= IE|(b+1) then (1) becomes

 Consequently, 

 We can use (3) to design the biasing circuit so that it is largely insensitive to variations in b. The question is then how do we make IE (and hence IC) largely insensitive to b variations? Examining (3), we deduce that the answer is to choose

 Furthermore, we can design this biasing circuit so that it is largely insensitive to variations in temperature. The effects of temperature enter this circuit because VBE is a relatively strong function of temperature having a temperature coefficient of -2 mV/°C. (We saw this same behavior with diodes.) From (3) we can see that if we chooseVBB >>VBE then we’ll have a biasing circuit design that is largely insensitive to variations in temperature. So physically how do these conditions (4) and (5) make a good biasing circuit? • Eqn. (4) makes the base voltage largely independent of â and determined almost solely by R1 and R2. How? Because the current in the divider is much greater than the base current. The rule of thumb for “much greater” is that the divider current should be on the order of IE to IE/10. • Eqn. (5) ensures that small variations in VBE (from its nominal 0.7 V) due to temperature changes are much smaller than VBB. Additionally, there is an upper limit to VBB because a higher VBB lowers VCB and affects the small values of the positive signal swing. The rule of thumb here is

that VBB» VCC|3 and VCB (or VCE) » VCC|3 and .IC RC »VCC|3 Example N17.1. Design the bias circuit below for VCC = 9 V to provide VCC/3 V across RE and RC, IE = 0.5 mA, and the voltage divider current of 0.2IE, as shown. Design the circuit for a large b, then find the actual value obtained for IE with a BJT having b=100. 

 For the resistors Reand Rc,Ic RE =VCC /3=3V For I E =0.5 mA , then RE =6Kb For the voltage divider , if this BJT is in the active mode then VBE »0.7V Hence VB =VBE+VE =0.7+3=3.7V Such that 

 A large b for a BJT in the active mode implies Ib»0. By Ohm’s law

 Hence, RI =90kb-R2 =53K b For the design with b = 100 it can be shown that IE =0.48 mA. (This is only a -4% change from 0.5 mA with b= ¥ ;) Dual Power Supply Biasing MethodWhen two power supplies are available, a possible biasing method is

 Using KVL around the loop L gives

 This is the same result as (3), but with VBB replaced by VEE. Consequently, the b- and temperature-invariant design equations for this circuit are the same as those given earlier in (4) and (5) with VBB replaced by VEE. Alternative Biasing for Common Emitter AmplifiersThis biasing method has a resistor tied from the collector to the base as 

 As shown in the text, for IE to be insensitive to b variations, choose

 and for VBE to be insensitive to temperature variations, choose VCC VBE This latter requirement is most often very easy to meet! Biasing with a Current Source

The last BJT amplifier biasing method we’ll consider is one using a current source. 

 In this circuit, E I I = . If we are using a “good” current source, then IE will not depend on â. Very nice. However, what we’ve done in this approach is to push the technical problem to the design of a good current source. Current MirrorSimple biasing methods often fail to provide constant collector currents if the supply voltage or ambient temperature change.This is a problem with mobile telephones, for example, where the battery voltage changes with use and the device operates in a range of temperatures. There are sophisticated circuits consisting of tens of evices that can produce “golden currents” that are supply voltage and temperature independent. These golden currents are replicated throughout a device using a current mirror: 

 There are better and more sophisticated approaches than this, of course. This is just a simple example. In this current mirror, Q1 is called a diode-connected BJT because the collector and base terminals are connected together. For proper operation of this

circuit, it is very important that the BJTs be “matched,” meaning they having the same b characteristic curves, etc. Usually this means that the BJTs must be fabricated at the same time on the same substrate. For the analysis of this circuit, we assume that â is very large and that Q1 and Q2 operate in the active mode. Because of this, we ignore the base currents in Q1 and Q2. Therefore, the collector (and emitter) current through Q1 is approximately equal to IREF. By KVL, 

 Now, since Q1 and Q2 are matched and they have the same VBE, then the collector currents must be the same. This implies that 

 This current mirror circuit will supply this current I as long as Q2 operates in the active region: V > VBE- VEE Notice that the diode-connected Q1 cannot saturate since thebase and collector terminals are shorted together. Hence, Q1operates in the active mode or is simply cutoff. 

We will now begin the analysis of the three basic types of linear BJT small-signal amplifiers: 1. Common emitter (CE) 2. Common base (CB) 3. Common collector (CC), which is oftentimes called the emitter follower amplifier. We’ll study the CE amplifier in this lecture and the next, followed by the CB and CC amplifiers. The CE amplifier is excited at the base of the BJT with

the output taken at the emitter: 

 The capacitor CE is called a bypass capacitor. At the operating frequency, its purpose is to shunt out the effects of the DC current source from the time varying signal. In other words, CE sets an AC ground at this node at the frequency of operation. There are a number of ways to bias this amplifier, other than that shown above. What we’re primarily interested in here is the small-signal characteristics. Common Emitter Small-Signal Amplifier AnalysisThe small-signal equivalent circuit for the CE amplifier above is shown below. Because the emitter is located at an AC ground is the reason this type of amplifier is called a “common emitter” amplifier. 

 Notice that we’ve included ro in this small-signal model. This is the finite output resistance of the BJT. This accounts for thefinite slope of the characteristic curves of iC versus vCE mentioned briefly in 

 where VA is called the Early voltage. Usually ro is fairly large, on the order of many tens of kb Our quest in the small-signal analysis of this amplifier is to determine these quantities: input resistance Rin, the “overall” small-signal voltage gain GV = vo/ vsig , the “partial” small-signal voltage gain v o i Av= vo v i , the overall small-signal current gain Gi= i0/ I i , the short circuit small-signal current gain is os i Ais= ias ii ,and the output resistance Rout. • Input resistance, Rin. Directly from the small-signal equivalent circuit, we see that Rin= RB || r p Oftentimes we select RB rp so that i Rin rp Oftentimes we select RB rp so that r p will often be a few kb, which means this CE amplifier presents a moderately large value of input impedance. • Overall small-signal voltage gain, Gv. By “overall” voltage gain we mean

 which is the actual small-signal voltage gain that would be realized in the circuit above. At the output of this circuitVo=-gmVp(ro||Rc ||Rl)while at the input

 Substituting (4) into (3) gives an expression for the overall (i.e., realized) gain of this CE amplifier

 In the usual case that B RB>> rp , then

 Recall that rp = ß/gm If it also turned out Rsig>> rp , then we see from (6) that Gv would be directly dependent on b. This is not a favorable condition since, as we learned when discussing biasing of such BJT circuits, bita can vary considerably between transistors. • Partial small-signal voltage gain, Av. This is only a partial voltage gain since we are calculating

 At the input, vi = Vp while at the output, V0=-gmVp(ro||Rc ||RL)Therefore, the partial small-signal voltage gain is

Av=-gm(ro||Rc||RL   • Overall small-signal current gain, Gi. By definition

Referring to the small-signal equivalent circuit shown above, we see that

 Forming the ratio of these two currents, we find that the current gain is

 or, using (9) 

 • Short circuit small-signal current gain, Ais. This is the smallsignal current gain of the amplifier but with a short circuitedload ( R L = 0): 

 Equivalently, A is=G i| R l=0 Using (11) in (13) with R l=0 . givesA is=-g m(r p||R B)

In the usual case that R B r p then A is »-bita This result is not unexpected because bita is by definition the short circuit current gain for the BJT when operating in the active mode. • Output resistance Rout. Using the small-signal

equivalent circuit above, we short out the source vsig =0 which means that vp =0 as well. Therefore, gm vp = 0, which is an open circuit for a current source. Consequently, Rout= Rc|| r o which is generally fairly large. Summary of CE Amplifier CharacteristicsSummary for theCommon_Emitter_Amplifier: Big voltage and current gains are possible. Input resistance is moderately large. Output resistance is fairly large. This last characteristic is often not desirable. Why? Consider this simple Thévenin equivalent for the output of a small-signal amplifier: 

 The output signal voltage provided to this resistive load is

 Now, if Rout<< RL then

 This is not a favorable result if this Thévenin equivalent circuit is for an amplifier because the output voltage is beingattenuated. Con versely, if there were a small output resistance such that Rout<< RL then then (17) becomesvout vo which is much more favorable for an amplifier. 

We will cover the second of the three families of BJT amplifiers in this lecture by discussing

theCommon_Base_Amplifier shown in Fig. 5.62a: 

 The small-signal equivalent circuit for this amplifier is shown in Fig. 5.62b (ignoring ro): 

 As before, let’s determine the small-signal AC characteristics of this amplifier by solving or Rin, Gv, Gi, Ais, and Rout. • Input resistance, Rin. From direct inspection of the smallsignal equivalent circuit, we see that Rin=re

Since re is often small (on the order of 20 to 30bita), then Rin of the CB amplifier is very small. Generally this is not desirable, though in the case of certain high frequency amplifiers input impedances near 50 bita is very useful (to reduce so-called “mismatch reflections” at the input). • Small-signal voltage gain, Gv. We’ll first calculate the partial voltage gain

 

At the output, vo=-aie(Rc||RL

The small-signal emitter current is

 Substituting (3) and (4) into (2) gives the partial voltage gain to be

 This is the same gain as for the CE amplifier (without ro), except the gain here for the CB amplifier is positive. The overall (from the input to the output) small-signal voltage gain Gv is defined as

 We can equivalently write this voltage gain as

 with Av given in (5). By simple voltage division at the input to the small-signal equivalent circuit

 Substituting this result and (5) into (7) yields the final expression for the overall small-signal voltage gain

 Since from (1) Rin= re then Gv simplifies to

 If 1 we can interpret this small-signal overall voltage expression in (10) as the ratio of the total resistance in the collector lead to the total resistance in the emitter lead. This gain can be fairly large, though if Rsig is nearly the same size as the total emitter

resistance the gain will be small. In other words, if this amplifier is connected to a high output impedance stage, it will be difficult to realize high gain. • Overall small-signal current gain, Gi. By definition

 Using current division at the output of the small-signal equivalent circuit above

 Because ii = -ie this expression gives

 • Short circuit current gain, Ais. In the case of a short circuit load (RL = 0), Gi in (13) reduces to the short circuit current gain: 

 • Output resistance, Rout. Referring to the small-signal equivalent circuit above and shorting out the input vsig = 0 Rout= Rc which is the same as the CE amplifier (when ignoring ro).SummarySummary of the CB small-signal amplifier: 1. Low input resistance. 2. Gv can be very large, though critically dependent on Rsig. 3. Ais= á4. Potentially large output resistance (dependent on RC). One very important use of the CB amplifier is as a unity-gain current amplifier, which is also called a current buffer amplifier. This type of amplifier accepts an input signal current at a low impedance level and outputs nearly the same current amplitude,

but at a high output impedance level. Even though this is a buffer amplifier, there is still power gain. 

The third, and final, small-signal BJT amplifier we will consider is the common collector amplifier shown below: 

 The small-signal equivalent circuit is shown in 

 We’ve included ro in this model since it can have an appreciable effect on the operation of this amplifier. Notice that ro is connected from the emitter to an AC ground. We can simplify the AC small-signal analysis of this circuit by moving the collector-side lead of ro to the DC ground, as shown in 

 Similar to the previous BJT amplifiers, we’ll determine the characteristics of this one by solving for Rin, Gv, Gi, Ais, and Rout. 

• Input resistance, Rin. Looking into the base of the BJT, 

 From the circuit above, we see thatVb=ie(re+r0||RL

Substituting this and 1b=i e/( ß +1) into (1) yieldsRib=( ß +1)(r e+r 0||R L)This expression for Rib follows the so-called resistance reflection rule: the input resistance is ( ß+1) times the total resistance in the emitter lead of the amplifier. (We saw a similar result in Lecture 19 for the CE amplifier with emitter degeneration.) In the special case when r e<< RL<< r 0 thenR ab  (ß +1)RL

which can potentially be a large value. Referring to circuit above, the input resistance to the amplifier is

 • Small-signal voltage gain, Gv. We’ll first calculate the partial voltage gain

 Beginning at the output, 

 from which we can directly determine that

 The overall (from the input to the output) small-signal voltage gain Gv is defined as

 We can equivalently write this voltage gain as

 with Av given in (8). By simple voltage division at the input to the small-signal equivalent circuit

 Substituting this result into (10) yields an expression for the overall small-signal voltage gain

 We can observe directly that each of the two factors in this expression are less than one, so this overall small-signal voltage gain is less than unity. In the special instance that r0<< RL then (12) simplifies to

 and if RB >>( ß+1)( re+RL) then this further simplifies to

 We see from this expression that under the above two assumptions and a third RL>>re+ Rsig (ß + 1) , the smallsignal voltage gain is less than but approximately equal to one. This

means that

 Because of this result, the common collector amplifier is also called an emitter follower amplifier. • Overall small-signal current gain, Gi. By definition

 Using current division at the output of the small-signal equivalent circuit above

 while using current division at the input

 Substituting this into (17) gives

 from which we find that

 • Short circuit current gain, Ais. In the case of a short circuit load (RL = 0), Gi in (21) reduces to the short circuit current gain: 

 In the case that RB >>( ß + 1)(re+RL)=( ß + 1)re, as was used earlier, then Ais ß + 1 which can be very large. So even though the amplifier has a voltage

gain less than one (and approaching one in certain circumstances), it has a very large small-signal current gain. Overall, the amplifier does provide power gain to the AC signal. • Output resistance, Rout. With vsig = 0 in the small-signal equivalent circuit, we’re left with

 It is a bit difficult to determine Rout directly from this circuit because of the dependent current source. The trick here is to apply a signal source vx and then determine ix. The output resistance is computed from the ratio of these quantities as

 Applying KVL from the output through the input of this circuit gives

 Using KCL at the output

 Substituting (26) into (25) 

 Forming the ratio of vx and ix in (27) gives

 

 SummarySummary of the CC (emitter follower) small-signal amplifier: 1. High input resistance. 2. Gv less than one, and can be close to one. 3. Ais can be large. 4. Low output resistance.These characteristics mean that the emitter follower amplifier is highly suited as a voltage buffer amplifier. 

The BJT amplifiers we have examined so far are all low frequency amplifiers. For large valued DC blocking capacitors and for frequencies of tens to hundreds of kHz, the simple smallsignal models we used will work well. As the frequency increases, though, there are multiple sources of effects that will limit the performance of these amplifiers including: 1. Internal capacitances of th BJT. These are due to charge storage effects at and near the two pn

junctions. 2. Parasitic effects. These are due to packaging and transistor construction that create additional capacitances, lead inductances, and resistances. Additionally, the performance of many BJT amplifiers we’ve already examined will be sharply curtailed by DC blocking capacitors that have finite value (i.e., less than infinity). For these reasons, all real transistor amplifiers operate effectively only over a limited (but hopefully large) range of signal frequencies. Referring to Fig. 5.71(b), our analysis of small-signal BJT amplifiers up to this point has focused on the “Midband” frequency region. This frequency band is bounded by the frequencies fL and fH, which are the -3-dB gain frequencies, or half-power frequencies. 

 The roll off in gain near fL and lower is due to effects of the DC blocking capacitors CC1 and CC2, and the bypass capacitor CE. It’s not possible to eliminate this effect, though fL can be moved about by choosing different values for these capacitors. Butlarge capacitors take up lots of space and can be expensive. The primary focus of this lecture,

however, is the origin of the roll off in gain xperienced at higher frequencies near fH and higher. Capacitance of pn JunctionsThere are basically two types of capacitances associated with pn junctions: 1. Junction capacitance. This is related to the space charge that exists in the depletion region of the pn junction. 2. Diffusion capacitance, or charge storage capacitance. This is a new phenomenon we haven’t yet considered in this course. The junction capacitance effect was briefly mentioned earlier in this course in Lecture 4. The width of the depletion region will change depending on the applied voltage and whether the junction is reversed or forward biased: 

 The time-varying E due to the space charge in the depletion region is a so-called displacement current that can be modeled by a junction capacitance. The second basic type of capacitance, diffusion capacitance, is associated with pn junctions that are forward biased. 

 In this state, current will flow across the junction, of

course. Because of the current source in Fig. 3.49 and the voltage dropV, holes are injected a cross the junction into the n region while electrons are injected across the junction into the p region. 

 The concentrations of these electrons and holes decrease in value away from the junction, as shown in Fig. 3.50, due to recombination effects. The important point here is that these concentrations of charges create an electric field across the pn junction that will vary with time when a signal source is connected to this device. This electric field is directed from the n to p region, and the overall effect can be modeled by what is called the charge storage capacitance, or diffusion capacitance. To summarize, the capacitive effects of a reversed biased pn junction are described by the junction capacitance while those of a forward biased pn junction are described by both a junction and a diffusion capacitance. In the latter case, though, the diffusion capacitance usually dominates. BJT High Frequency Small-Signal ModelThe active mode BJT has one forward biased pn junction (the EBJ) and one reversed biased pn junction (the CBJ). In the case of an npn BJT the capacitances associated with the pn junctions in the device are labe ed as: 

 As we just discussed, there is a junction capacitance associated with the reversed biased CBJ, which is labeled C ì as shown above. There will be a junction capacitance, Cje, associated with the forward biased EBJ as well as a diffusion capacitance abeled Cde. These latter tw o capacitances appear in parallel and so can be combined as Typically C ì ranges from a fraction of pF to a few pF while C ð ranges from a few pF to tens of pF, which is dominated by Cde. With these capacitances, the high frequency small-signal model of the BJT becomes

 Note the use of the V ð notation in this small-signal model. Your textbook has switched to sinusoidal steady state notation for this high frequency discussion. The high frequency small-signal model in Fig. 5.67 also includes the resistance rx, which is mostly important at high frequencies. It’s there to approximately model the resistance of the base region from the terminal to a point somewhere directly below the emitter: 

 C ì is sometimes referred to as Cob (or Cobo) in

datasheets. This designation reflects the fact that C ì can be the output resistance when the BJT is used as a common base amplifier. The values of these small-signal circuit model elements may or may not be available in a datasheet for your transistor. For example, from the Motorola P2N2222A datasheet: 

 Actually, we would expect these capacitances to vary with the voltage across the respective pn junction. In the following figure from the Motorola P2N2222A datasheet, we see the dependence of “Ceb” (= C ð ?) and “Ccb” (= C ì) for a range of junction voltages. (Perhaps t he labeled voltage for Ceb should be “forward voltage”?) 

 Unity-Gain BandwidthAn important high frequency characteristic of transistors that is usually specified is the unity-gain bandwidth, fT. This is defined as the frequency at which the short-circuit current gain

 has decreased to a value of one. A test circuit for this measurement would look something like: 

 The small-signal high frequency model of this test circuit is: 

 Applying KCL at the collector terminal provides an equation for the short-circuit collector current

 At the input terminal B’

 Substituting (4) into (3) gives

 Using the definition of hfe from (2) we find from this last equation that

 It turns out that C ì is typically quite small and for the purposes of determining the unity-gain bandwidth, gm is | | j Cì . for the frequencies of interest here. In other words, the frequency at which ùC ì is important relative to gm is much higher than what is of interest here. Consequently, from (5) 

 We can recognize this frequency response of hfe in (6) as that for a single pole low pass circuit: 

 0=gmr in this plot is the low frequency value of | hfe|, as we’ve used in the past [see eqn. (5.93)], while the 3-dB frequency of | hfe| is given by 

 The frequency at which hfe in (6) declines to a value of 1 is denoted by ùT, which we can determine from (6) to be

 such that

 Therefore

 

 This unity-gain frequency fT (or bandwidth) is often specified on transistor datasheets. On page 8, for

example, f r =300 MHz for the Motorola P2N2222A. Using (9), this fT can be used to determine C + C  for a particular DC bias current. Lastly, the high frequency, hybrid-ð, small-signal model of Fig.5.67 is fairly accurate up to frequencies of about 0.2 fT. Furthermore, at frequencies above 5f  to 10 f , the effects of r ðare small compared to the impedance effects of C ð. Above that, rx becomes the only resistive part of the input impedance at high frequencies. Consequently, rx is a very important element of the small-signal model at these high frequencies, but much less so atlow frequencies.