4 bi polar transistor notes

8/14/2019 4 Bi Polar Transistor Notes

1/21

Bipolar Junction Transistors

Murray Thompson

Sept. 5, 1999

Contents1 Introduction 1

2 A Reassurance 1

3 Summary of Bipolar Transistor theory so far 2

4 Transistor Model 5

5 Various Simple Ampliers 6

6 A Useful form of the Common Emitter Amplier 207 A More Useful Form of the Common Emitter Amplier 21

-

1 IntroductionThese notes follow those on pn junctions, in which the basic components of npn transistors were discussed.

2 A ReassuranceAt this point, some students, who prefer to have precise calculations andexact theory, may start to become concerned about these sloppy methods,

1


2/21

bad approximations and imprecise calculations. We can reassure them. In

spite of these rough methods, the general concept of negative feedback (tobe introduced later) will allow us to use these imperfect circuits yet gainextremely linearcircuits, extremely precise amplications and precision measurements of ourinput signals.

First we must consider these imperfect ampliers. The later feedback andprecision circuits would be useless without them!

3 Summary of Bipolar Transistor theory sofar

1. The fraction of the electrons leaving the emitter and going to the col-lector instead of the base is normally called I collectorI emitter = .

More strictly, in case of non-linear behaviour, we dene = dI collectordI emitter

2. The ratio of the collector current to the base current is normally called . Thus, = I collectorI base From this we can get = 1 .

More strictly, in case of non-linear behaviour, we dene = dI collectordI emitter

3. Hence (1 ) =

= 1+

4. By arranging the geometry so that the base region is very thin, most of the electrons entering from the emitter into the base will be fall over theedge down the steep electric potential fall of the base-collector junctionbefore they meet and recombine with holes in the base.

Thus, by making the base be very thin so that the emitter-base junctionand the collector-base junction are very close, and doping the collector-base junction so that it is very thin (with a high electric eld E = dV dx ,the fraction of the electrons leaving the emitter and going to the

2


3/21

collector instead of the base can be made close to 1 .0 and = 1 can

be made large. This can make the transistor more useful.While a transistor with a large , such as = 100, can be used tomake ampliers with a high gain, the thin base and collector-base junc-tion have a general defect. This defect is that the thin collector-base junction will have high electric elds E and avalanche breakdown mayoccur.

5. So long as the collector-base voltage is greater than about 2 or 3 volts,very few of the electrons which fall over the collector-base junction canbounce back.

Thus, So long as the collector-base voltage is greater than about 2 or3 volts, the fraction of electrons passing through both junctions isonly slightly dependent upon the collector-base voltage.

A typical set of I collector and V Collector Emitter curves for equispaced basecurrents are shown below.

Although the following curves are not usually shown in specications,the following shows I Collector and I Base curves for equispaced collectorvoltages.

3


4/21

6. Thus and particularly are useful parameters because they are nearlyconstant for any given bipolar junction transistor over a range of work-ing voltages and currents.

The

may vary from one transistor to another due to manufacturingvariations and due to changing temperature. Note that I C in the fol-lowing graph has a logarithmic horizontal scale and covers a very widerage.

Even though does not change much with voltage and current, inany given transistor, the variation in due to temperature and man-ufacturing is signicant. The variation due to manufacturing, can beminimized by selecting transistors with particular values of .

7. In most practical circuits, we try to make the action be independent of .This can be done in a variety of ways but will require that be large

so that >> 1 and we can approximate functions like + 1 as .8. The transistor is said to be in saturation if the collector voltage is too

low.

4


5/21

9. Any transistor has maximum ratings for

(a) The maximum power giving a maximum product of I C max andV CE max .

(b) A maximum current into the collector I C max through its connect-ing wire and weld.

(c) A maximum voltage on the collector-base junction V CE max abovewhich the junction may break down.

4 Transistor ModelBy inspecting the transistor I Collector , V CE and curves above, we can makea model of the actions within a transistor. Like some other models, the

reproduction is imperfect but will allow us to make systems where the actionis nearly independent of .We want a 3-terminal model which we can use to replace the transistor

when want to make calculations. In this model we assume

5


6/21

1. The base-emitter junction is kept forward biased (turned on) by a

small but sufficient base-emitter current to keep the base emitter volt-age near 0.7 volt.

2. The Base-Collector voltage is sufficient for the transistor to have a > 10

A common convention is to use lower case r for each of the internal resis-tances to distinguish them from the normal external resistors we attach tothe emitter, base and collector.

We will call the join of the 3 lines as the point join.We replace the base to emitter junction with a small 0.7 volt battery with

a small (perhaps negligible resistor r b.The current through the base is called I b.From the previous discussion of p,n junctions, we know that the emitter

has the dynamic junction impedance (which may also be negligible) of aboutr e = 2 ohm + 0.026 voltI base .

We insert a Current Generator in the collector to cause a current I C = I b.

5 Various Simple AmpliersWe will now consider four general circuits corresponding to the 3 permuta-

tions of 1 terminal being shared between the circuitry of the input andoutput attached circuitry the permutation of the common emitter is usedtwice. Each amplier amplies an input DC voltage which may change.By using an additional coupling capacitor, each can amplify only the ACcomponent of the input voltage.

6


7/21

1. Grounded Emitter AmplierThe simple transistor model suggests how an amplier might be made.We call this a grounded emitter amplier.

Variations V

input in the base input voltage will cause variations inthe emitter current through r b. The transistor can be replaced forcalculation purposes by the transistor model as in the next diagram.If, as usual, r e is small, then

I b = V inputR B + r band these will cause variations in the collector current

I c = I b = V input

R B + r b

The changing collector current ows through the collector resistor R cfrom a constant voltage usually called V cc and so the changing col-lector current causes the collector voltage to change by

V C = R C V inputR B + r b V C = ( R C R B + r b ). V inputThus a small change in the input to the base can cause a larger changein the voltage of the collector.

(a) The voltage gain or amplication V cV b = (R C

R B + r b ) is negativemeaning that the output signal is inverted relative to the inputsignal. This is not an disadvantage but is actually an advantagefor negative feedback and for digital circuitry.

(b) This circuit is very easy to understand but is seldom used for thefollowing reasons.

(c) It is hard to make transistors with a particular precise value of .

(d) In any transistor, depends upon temperature

7


8/21

(e) also depends, slightly, upon the collector voltage and collec-

tor current I c . (In other words, the transistor model is not veryprecise.)

(f) r b depends, a little, upon temperature, I b and I c.

(g) r e depends, a little, upon temperature, I b and I c.

(h) The voltage difference across the junction (near 0.7 volt) is tem-perature dependent.

In general, the grounded emitter gives a high amplication but thevoltage gain V cV b is too unpredictable for this circuit to be used exceptin digital circuits.

The following three amplifying circuits include some type of negativefeedback to overcome these problems. Other circuits use transistorsin identical pairs and use them to amplify pairs of voltages (one thesignal and the other a constant voltage or a negative signal) so thatsome of the unwanted effects can cancel.

8


9/21

2. Common Emitter Amplier

This has 3 resistors as shown R B , R E and R C attached to the base,emitter and collector. The top rail is attached to a positive powersupply (perhaps +10 Volt). The lower rail is the ground.

The input is on the left (via the resistor R B . The output is on the rightfrom the collector pin attached to the resistor R C .

Apply a small incremental voltage V in to the resistor R B leading tothe base.

The transistor can be replaced for calculation purposes by the transistormodel as in the next diagram.

We concern ourselves only with the current and voltage increments socan ignore the +0.7 volt battery. Dene the voltage on the point below

the current generator at the join of the 3 lines as V join We can obtain2 equations.

(a) The current increment I Base into the base due to the voltageincrement V in is

9


10/21

I Base = V in V join

R B + r b

(b) The current increment I Collector into the Collector due to thevoltage increment V in is I Collector = I BaseThe voltage increment across the actual resistor R E and the inter-nal resistance r e totalling R E + r e , due to the two currents causedby the voltage increment V in is V join = (1 + ) I Base (R E + r e )

From these two equations

I Base = V in (1+ ) I Base (R E + r e )

R B + r b

(R B + r b) I Base = V in (1 + ) I Base (R E + r e)

(R B + r b) I Base + (1 + ) I Base (R E + r e ) = V in(R B + r b + (1 + )(R E + r e)) I Base = V in I Base = V in(R B + r b +(1+ )( R E + r e ))From this, we can calculate 3 important parameters for this circuit.

(a) The Voltage Gain of the common emitter circuitThe collector voltage will decrement due to the increased currentthrough the resistor R C from the positive +10 Volt supply. Thus

the voltage increment on the collector will be negative; V C = I C R C V C = I Base R C V C = V in(R B + r b +(1+ )( R E + r e )) R C Thus the voltage gain of this common emitter circuit isAV = V C V in =

R C(R B + r b +(1+ )( R E + r e ))

If we neglect both r b and r e , thenAV

(R C )(R C +(1+ )R E )

again, if we assume that >> 1, the voltage gain, AV , becomes

AV R CR ENotes.

The gain is negative meaning that the output voltage signalis inverted.

10


11/21

Using the concepts of feedback, we say that the resistor R E

is providing negative feedback so that in the approximationsof being high and r b and r e being low,the voltage gain AV is minus(the ratio of the collector andemitter resistances)

(b) The Input Impedance of the common emitter circuitThe Input Impedance is the ratioZ in = V in I BaseUse I Base from aboveZ in = ( R B + r b + (1 + )(R E + r e ))Again, if we neglect r b and r e and assume that >> R BR

E

thenZ in = R E

(c) The Output Impedance of the common emitter circuitRemembering that the impedance of a current generator is in-nite, the output impedance of the circuit looking back into thecircuit is only that of the resistor R C to the power supply whichis a virtual ground. Thus we have

Z out = R C

11


12/21

3. Common Collector Amplier (The Emit-ter Follower)

This has 2 resistors as shown R B and R E attached to the base andemitter. The top rail is attached to a positive power supply (perhaps+10 Volt). The lower rail is the ground.

The Collector is attached directly to the top rail at +10 Volt.

The input is on the left (via the resistor R B . The output is on the rightfrom the emitter pin attached to the resistor R E .

Apply a small incremental voltage vin to the resistor R B leading to thebase.

The transistor can be replaced for calculation purposes by the transistor

model.Apply a small incremental voltage V in to the resistor R B leading tothe base.

12


13/21

We concern ourselves only with the current and voltage increments so

can ignore the +0.7 volt battery. Dene the voltage on the point belowthe current generator at the join of the 3 lines as V joinAs with the Common Emitter before, we can obtain 2 equations.

(a) The current increment I Base into the base due to the voltageincrement V in is I Base = V in

V joinR B + r b

(b) The current increment I Collector into the Collector due to thevoltage increment V in is I Collector = I BaseThe voltage increment across the actual resistor R E and the inter-nal resistance r e totalling R E + r e , due to the two currents causedby the voltage increment V in is V join = (1 + ) I Base (R E + r e )

From these two equations

I Base = V in (1+ ) I Base R E

R B + r b

(R B + r b) I Base = V in (1 + ) I Base (R E + r e)

(R B + r b) I Base + (1 + ) I Base (R E + r e ) = V in

(R B + r b + (1 + )(R E + r e) I Base = V in I Base = V in(R B + r b +(1+ )( R E + r e ))From this, we can calculate 3 important parameters for this circuit.

(a) The Voltage Gain of the common collector (emitter follower) cir-cuitThe current through the emitter is I Emitter = (1 + ) I BaseThe voltage increment across R E is the output signal V out

V out = I Emitter R E V out = (1 + ) V in(R B + r b +(1+ )( R E + r e )) R E

V out = (1+ )R E(R B + r b +(1+ )( R E + r e )) V inIf we neglect r e and r b

13


14/21

V out (1+ )R E(R B +(1+ )R E ) V in

If >> 1, then V out (1+ )R E(R B +(1+ )R E ) V inThe voltage GainAV = V out V in

(1+ )R E(R B +(1+ )( R E ))

If (R B > R BR E thenZ in = R E

(c) The Output Impedance of the common collector (emitter follower)circuitTo calculate this, we assume no change to V in (ie V in = 0) butdraw a small current I out from the output (emitter pin) and

calculate or measure the resulting V out .We concern ourselves only with the current and voltage incrementsso can ignore the +0.7 volt battery. Again, dene the voltage onthe point below the current generator at the join of the 3 lines as V joinAs with the Common Emitter before, we can obtain 2 equations.

i. V join = I base (R B + r bii. V join = ( + 1) I base r e + ((1 + ) I base I out )R E

I base (R B + r b = ( + 1) I base r e + ((1 + ) I base I out )R E Grouping the I base terms on the right I out R E = I base [R B + r b + (1 + )( r e + R E )] I base = I out R E[R B + r b +(1+ )( r e + R E )]The change in the output voltage V out is due to the two changesin current through R E

14


15/21

V out = (( increase in emitter current ) (current drawn from output ))R E

V out = ((1 + ) I base I out )R E V out = ((1 + ) I out R E[R B + r b +(1+ )( r e + R E )] I out )R E

The Output impedance isZ out = V out I out = (1 + )

R 2E[R B + r b +(1+ )( r e + R E )] + R E

Z out = (1 + )R 2E

[R B + r b +(1+ )( r e + R E )] + R E

Z out = (1+ )R 2E

[R B + r b +(1+ )( r e + R E )] + R E

Z out = (1+ )R 2E +[ R B + r b +(1+ )( r e + R E )]R E

[R B + r b +(1+ )( r e + R E )]

Z out

= +[ R B + r b +(1+ )( r e )]R E[R B + r b +(1+ )( r e + R E )]

If r b and r e are very smallZ out = R B R E[R B +(1+ )R E ]If is largeZ out = R B1+ The output impedance has been lowered to a fraction 11+ of theR B . In this way, an emitter follower (common collector circuit)can be used to lower the effective output impedance of any devicewhich might have had an output impedance of R E .

The emitter follower (common collector) circuit is often used togetherwith another circuit which might already provide sufficient voltage gain.

By adding an emitter follower before the another circuit, the inputimpedance of the combination may be raised by a factor of about .

By adding an emitter follower after the another circuit, the outputimpedance of the combination may be lowered by a factor of about .

15


16/21


17/21

The current through the resistor R E and internal resistance r e is

I C + I b = ( + 1) I bThe voltage at point join isV join = ( I C + I b)(r e + R E )

V join = ( + 1) I b(r e + R E ) (2)

V C = 10 volt I C R C = 10 volt ( + 1) I bR C (3)

Use equations 1 and 2 to eliminate V join .2.7 volt r bI b = ( + 1) I b(r e + R E )

I b =2.7 volt

r b +( +1)( r e + R E )Substitute this I b into equation 3.

V C = 10 volt ( + 1)2.7 volt

r b + ( + 1)( r e + R E )R C (4)

(b) Now calculate these again but with the addition of an input cur-

rent I in due to the signal introduced via a capacitor. Consider thecircuit with an injected input current signal, I in = 0.The Transistor Model gives us the equivalent circuit for incremen-

tal voltages and currents. Calculate the new values for V join , I C ,I b, and V C .If V b = +3 .0 volt , then

V join = +3 .0 volt 0.7 volt r bI b (5)

The current through internal resistance r e is I C + I b = ( + 1) I b.However the current through the resistor R E will differ due to theinput signal.The voltage at point join isV join = ( I C + I b)r e + ( I C + I b + I in )R E

V join = (( + 1) I b)r e + (( + 1) I b + I in )R E (6)

V C = 10 volt I C R C = 10 volt ( + 1) I bR C (7)

17


18/21

Use equations 5 and 6 to eliminate V join .

2.7 volt r bI b = (( + 1) I b)r e + (( + 1) I b + I in )R E This can be simplied2.7 volt I in R E = ( r b + ( + 1)( r e + R E ))I bI b = 2.7 volt

I in R Er b +( +1)( r e + R E )

Substitute this I b into equation 7.

V C = 10 volt ( + 1)2.7 volt I in R E

r b + ( + 1)( r e + R E )R C (8)

Compare this equation 8 with the previous equation 4. The injected input current I in has changed the output voltage by

V out = New V C Old V C = ( + 1) I in R E

r b + ( + 1)( r e + R E )R C

(9)If r e and r b are small,

V out ( + 1) I in R E

( + 1)( R E )R C (10)

V out I in R C (11)

Note a positive current injection through the input coupling capac-

itor causes a positive voltage excursion on the collector NOINVERSION.

The output V out is the same as if the injected current I inwere subtracted by the transistor from its current through theresistor R C to keep the current through R E constant. Thecurrent through the resistor R C changes by I in causing anoutput voltage excursion V out + I in R C which has the

same polarity as the input current signal.From this, as before we can calculate 3 important parameters for thiscircuit.

(a) The Voltage Gain of the common base circuit

18


19/21

(b) The Input Impedance of the common base circuit

(c) The Output Impedance of the common base circuit

19


20/21

6 A Useful form of the Common Emitter Am-plier

The common emitter amplier needs the input voltage to be a few volts aboveground so that, some current ows through the equivalent junction batteryand some current ows through the collector to the emitter. We say we needto bias the base input to ensure the transistor stays in this working regionwhere is fairly constant. This is particularly important in the amplicationof AC since the coupling capacitor blocks any DC input.

A simple way of biassing the AC common emitter is to add two resistanceswith resistances higher than the emitter resistor so that the drain currentdraining down through both bias resistors satises two requirements.

1. The drain current is much greater than the current into the transistorbase so that the DC voltage at the base is determined by the biasresistors and the supply voltage rather than the transistor which isunpredictable and

2. the two resistors, when seen from the input, do not present a lowimpedance. Since the amplier may have been preceeded by a pream-plier, this requirement is usually that the input impedance of the twobias resistors should be large compared with the collector resistance of the preamplier.

Often, these two conditions imply that the bias resistors must be kept be-tween two limits which are a factor of about apart. For example, if isabout = 65, then choose bias resistors giving the following.

1. A total bias drain current from supply to ground of about 8 times thetransistors base current and 865 times the emitter current.

20


21/21

2. The parallel impedance of the two bias resistors about 8 times the

output impedance of the previous amplier.

7 A More Useful Form of the Common Emit-ter Amplier

The circuit is similar to the previous common emitter amplier but the bi-asing structure has been changed to include some negative feedback fromoutput signal on the collector. As a result, this circuit gives an amplicationwhich is more predictable in spite of variations in the of the transistor.

21

4 bi polar transistor notes

Documents