BJT Device Equations and Small-Signal Models

8/14/2019 BJT Device Equations and Small-Signal Models

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It follows that =

iC iE

=

1 + (6)

Thus the currents are related by the equations

iC = i B = i E (7)

Transfer and Output Characteristics

The transfer characteristics are a plot of the collector current iC as a function of the base-to-emittervoltage vBE with the collector-to-emitter voltage vCE held constant. From Eqs. 1 and 3, we canwrite

iC = I S 0 1 + vCE V A exp vBE V T (8)It follows that iC varies exponentially with vBE . A plot of this variation is given in Fig. 2. Itcan be seen from the plot that the collector current is essentially zero until the base-to-emittervoltage reaches a threshold value. Above this value, the collector current increases rapidly. Thethreshold value is typically in the range of 0.5 to 0.6 V. For high current transistors, it is usuallysmaller. The plot shows a single curve. If vCE is increased, the current for a given vBE is larger.

However, the displacement between the curves is so small that it can be di ffi cult to distinguishbetween them. The small-signal transconductance gm de ned below is the slope of the transfercharacteristics curve evaluated at the quiescent or dc operating point for a device.

Figure 2: BJT transfer characteristics.

The output characteristics are a plot of the collector current iC as a function of the collector-to-emitter voltage vCE with the base current iB held constant. From Eqs. 1 and 4, we can write

iC = 0 1 + vCE V A iB (9)It follows that iC varies linearly with vCE . A plot of this variation is given in Fig. 3. For smallvCE such that 0 vCE < v BE , Eq. (9) does not hold. In the region in Fig. 3 where this holds, thebjt is saturated. The small-signal collector-to-emitter resistance r0 de ned below is the reciprocalof the slope of the transfer characteristics curve evaluated at the quiescent or dc operating pointfor a device.

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Figure 3: BJT output characteristics.

Bias Equation

Figure 4(a) shows the BJT with the external circuits represented by Thvenin dc circuits. If theBJT is biased in the active region, we can write

V BB V EE = I B RBB + V BE + I E REE

=I C

RBB + V BE +I C

REE (10)

This equation can be solved for I C to obtain

I C =V BB V EE V BE RBB / + REE /

(11)

It can be seen from Fig. 2 that large changes in I C are associated with small changes in V BE . Thismakes it possible to calculate I C by assuming typical values of V BE . Values in the range from 0.6 Vto 0.7 V are commonly used. Two sets of values for and are convenient for hand calculations.One is = 0 .99 and = 99 . The other is = 0 .995 and = 199 .

Figure 4: (a) BJT dc bias circuit. (b) Circuit for Example 1.

Example 1 Figure 4(b) shows a BJT dc bias circuit. It is given that V + = 15V , R1 = 20k ,R2 = 10 k , R3 = R4 = 3 k , R5 = R6 = 2 k . Solve for I C 1 and I C 2. Assume V BE = 0 .7 V and = 100 for each transistor.

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Solution. For Q1, we have V BB 1 = V + R2/ (R1 + R2), RBB 1 = R1kR2, V EE 1 = I B 2R4 = I C 2R4/ , V EE 1 = 0 , and REE 1 = R4. For Q2, we have V BB 2 = I E 1R4 = I C 1R4/ , RBB 2 = R4,V EE 2 = 0 , REE 2 = R6. Thus the bias equations are

V +R2

R1 + R2+

I C 2

R4 = V BE +I C 1

R1kR2 +I C 1

R4

I C 1

R4 = V BE +I C 2

R4 +I C 2

R6

These equations can be solved simultaneously to obtain I C 1 = 1 .41mA and I C 2 = 1 .74mA.

Hybrid- Model

Let each current and voltage be written as the sum of a dc component and a small-signal accomponent as follows:

iC = I C + ic iB = I B = ib (12)

vBE = V BE + vbe vCE = V CE + vce (13)

If the ac components are su ffi ciently small, we can write

ic = I C

V BE vbe + I

C

V CE vce ib = I

B

V BE vbe (14)

where the derivatives are evaluated at the dc bias values. Let us de ne the transconductance gm ,the collector-to-emitter resistance r0, and the base-to-emitter resistance r as follows:

gm =I C

V BE =

I S V T

exp V BE V T = I C V T (15)r0 = I C V CE

1= I S 0V A exp V BE V T

1=

V A + V CE I C

(16)

r = I BV BE 1

= I S 0 0V T exp V BE V T 1

= V T I B (17)

The collector and base currents can thus be written

ic = i0

c +vcer0

ib =vr

(18)

wherei 0c = gm v v = vbe (19)

The small-signal circuit which models these equations is given in Fig. 5(a). This is called thehybrid- model. The resistor r x , which does not appear in the above equations, is called the basespreading resistance. It represents the resistance of the connection to the base region inside thedevice. Because the base region is very narrow, the connection exhibits a resistance which oftencannot be neglected.

The small-signal base-to-collector ac current gain is de ned as the ratio i0c/i b. It is given by

=i0cib

=gm v

ib= gm r =

I C V T

V T I B

=I C I B

(20)

Note that ic diff ers from i0

c by the current through r 0. Therefore, ic/i b 6= unless r 0 = .

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Figure 5: (a) Hybrid- model. (b) T model.

T Model

The T model replaces the resistor r in series with the base with a resistor r e in series with theemitter. This resistor is called the emitter intrinsic resistance. The current i0e can be written

i0e = ib + i0

c = 1 + 1i0c = 1 + i0c = i0

c

(21)

where is the small-signal emitter-to-collector ac current gain given by

=

1 + (22)

Thus the current i0c can be writteni0c = i

0

e (23)

The voltage v can be related to i0

e as follows:

v = ibr =i 0c

r =i 0e

r = i0

er

= i0er

1 + = i 0er e (24)

It follows that the intrinsic emitter resistance r e is given by

r e =vi0e

=r

1 + =

V T (1 + ) I B

=V T I E

(25)

The T model of the BJT is shown in Fig. 5(b). The currents in both models are related by theequations

i 0c = gm v = i b = i0

e (26)

Simpli ed T Model

Figure 6 shows the T model with a Thvenin source in series with the base. We wish to solve foran equivalent circuit in which the source i

0

c connects from the collector node to ground rather thanfrom the collector node to the B node.

The rst step is to replace the source i 0e with two identical series sources with the commonnode grounded. The circuit is shown in Fig. 7(a). The object is to absorb the left i 0e source intothe base-emitter circuit. For the circuit, we can write

ve = vtb i0e

1 + (R tb + r x ) i

0

er e = vtb i0

e R tb + r x1 + + r e (27)5


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Figure 6: T model with Thvenin source connected to the base.

Let us de ne the resistance r 0e by

r 0e =R tb + r x

1 + + r e

or=R tb + r x + r

1 + (28)

With this de nition, ve is given byve = vtb i

0

er0

e (29)

Figure 7: (a) Circuit with the i0

c source replaced by identical series sources. (b) Simpli ed T model.

The circuit which models Eq. (29) is shown in Fig. 7(b). We will call this the simpli ed Tmodel. It predicts the same emitter and collector currents as the circuit in Fig. 6. Note that theresistors R tb and r x do not appear in this circuit because they are contained in r

0

e .

Norton Collector Circuit

The Norton equivalent circuit seen looking into the collector can be used to solve for the responseof the common-emitter and common-base stages. It consists of a parallel current source ic(sc ) anda resistor r ic from the collector to signal ground. Fig. 8(a) shows the BJT with Thvenin sources

connected to its base and emitter. To solve for the Norton equivalent circuit seen looking into thecollector, we use the simpli ed T model in Fig. 8(b).

By superposition of vc, i0

e , vtb , and vte , the following equations for ic and i0

e can be written

ic =vc

r0 + r 0ekR te+ i 0e

vtbr 0e + R te kr0

R teR te + r0

vte

R te + r 0ekr 0r 0e

r 0e + r0(30)

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Figure 9: (a) Circuit for calculating r ic . (b) Norton collector circuit.

Figure 10: (a) BJT with Thvenin source connected to the base. (b) T model circuit for calculatingve(oc) .

whereGm =

r 0e + R te

(39)

The value of ic(sc ) calculated with this approximation is simply the value of i0

e , where i0

e is cal-culated with r0 considered to be an open circuit. The term r 0 approximations is used in thefollowing when r 0 is neglected in calculating ic(sc ) but not neglected in calculating r ic .

Thvenin Emitter CircuitThe Thvenin equivalent circuit seen looking into the emitter is useful in calculating the responseof common-collector stages. It consists of a voltage source ve(oc) in series with a resistor r ie from theemitter node to signal ground. Fig. 10(a) shows the BJT symbol with a Thvenin source connectedto the base. The resistor R tc represents the external load resistance in series with the collector.To solve for the Thvenin equivalent circuit seen looking into the emitter, we use the simpli ed Tmodel in Fig. 10(b).

By superposition of vtb , ie , and i0

e , the following equations for ve and i0

e can be written

ve = vtbr0 + R tc

r 0e + r0 + R tc ie

r 0ek (r0 + R tc )

i 0e R tc r0e

r 0e + r0 + R tc(40)

i 0e =vtb ve

r 0e(41)

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Figure 11: (a) Circuit for calculating ie(sc ) . (b) Thvenin emitter circuit.

These can be solved to obtain

ve 1 R tcr 0e + r 0 + R tc =vtb

r 0e + r0 + R tc[r0 + (1 ) R tc ]

ier0

ek (r0 + R tc ) (42)

which simpli es to

ve = vtbr0 + (1 ) R tc

r 0e + r0 + (1 ) R tc ie

r 0e (r0 + R tc )r 0e + r 0 + (1 ) R tc

(43)

This equation is of the formve = ve(oc) ier ie (44)

where ve(oc) and r ie are given by

ve(oc) = vtbr0 + (1 ) R tc

r 0e + r0 + (1 ) R tcor= vtb

r0 + R tc / (1 + )r 0e + r0 + R tc / (1 + )

(45)

r ie = r0

er0 + R tc

r 0e + r0 + (1 ) R tcor= r 0e

r 0 + R tcr 0e + r 0 + R tc / (1 + )

(46)

The Thvenin equivalent circuit seen looking into the emitter is shown in Fig. ?? .

Thvenin Base Circuit

Although the base is not an output terminal, the Thvenin equivalent circuit seen looking into thebase is useful in calculating the base current. It consists of a voltage source vb(oc) in series with aresistor r ib from the base node to signal ground. Fig. 12(a) shows the BJT symbol with a Thveninsource connected to its emitter. Fig. 12(b) shows the Pi model for calculating the base voltage.

By superposition of vte , ib, and i b, treating each branch of i b seperately in the superposition,we can write the following equation for vb

vb = vter 0 + R tc

R te + r0 + R tc + ib [r x + r + R te k (r0 + R tc )]

+ i br0R te

R te + r0 + R tc

= vter 0 + R tc

R te + r0 + R tc+ ib r x + r + R te k (r0 + R tc )

+ r0R te

R te + r0 + R tc (47)9


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Figure 12: (a) BJT with Thevenin source connected to the emitter. (b) T model for calculatingvb(oc) .

Figure 13: (a) Circuit for calculating vb. (b) Thvenin base circuit.

This equation is of the formvb = vb(oc) + ibr ib (48)

where vb(oc) and r ib are given by

vb(oc) = vter 0 + R tc

R te + r 0 + R tc(49)

r ib = r x + r + R te k (r0 + R tc )

+ r0R te

R te + r0 + R tcor= r x + (1 + ) r e + R te k (r 0 + R tc )

+ r0R te

R te + r0 + R tc(50)

The equivalent circuit which models these equations is shown in Fig. 13.

The r 0 Approximations

The r0 approximations approximate r0 as an open circuit in all equations except the one for r ic .In this case, the simpli ed T model reduces to the one shown in Fig. 14. Because r0 no longerconnects to the emitter, there is only one emitter current and i 0e = ie . This circuit is very useful in

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Figure 14: Emitter equivalent circuit with the r0 approximations.

making rapid hand calculations and gives little error in most applications. If r0 = , then r ic isan open circuit.

The equations for the Norton collector circuit and the Thvenin emitter and base circuits aresummarized as follows:

Norton Collector Circuit

ic(sc ) = i e = Gm (vtb vte ) Gm =

r 0e + R te(51)

r ic = r0 + r0

ekR te1 R te / (r 0e + R te )(52)

Thvenin Emitter Circuitve(oc) = vtb r ie = r

0

e (53)

Thvenin Base Circuit

vb(oc) = vte r ib = r x + r + (1 + ) R teor= r x + (1 + ) ( r e + R te ) (54)

Summary of Models

Figure 15: Summary of the small-signal equivalent circuits.

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Example Ampli er Circuits

This section describes several examples which illustrate the use of the small-signal equivalent circuitsderived above to write by inspection the voltage gain, the input resistance, and the output resistanceof both single-stage and two-stage ampli ers.

The Common-Emitter Ampli er

Figure 16(a) shows the ac signal circuit of a common-emitter ampli er. We assume that the biassolution and the small-signal resistances r 0e and r0 are known. The output voltage and outputresistance can be calculated by replacing the circuit seen looking into the collector by the Nortonequivalent circuit of Fig. 9(b). With the aid of this circuit, we can write

vo = ic(sc ) (r ic kR tc ) = Gmb (r ic kR tc ) vtb (55)

r out = r ic kR tc (56)where Gmb and r ic , respectively, are given by Eqs. ( ?? ) and ( ?? ). The input resistance is given by

r in = R tb + r ib (57)

where r ib is given by Eq. ( ?? ).

Figure 16: (a) Common-emitter ampli er. (b) Common-collector ampli er. (c) Common-baseampli er.

The Common-Collector Ampli er

Figure 16(b) shows the ac signal circuit of a common-collector ampli er. We assume that the biassolution and the small-signal resistances r 0e and r0 are known. The output voltage and outputresistance can be calculated by replacing the circuit seen looking into the emitter by the Thveninequivalent circuit of Fig. 10(b). With the aid of this circuit, we can write

vo = ve(oc) R ter ie + R te= r 0 + R tc / (1 + )r 0e + r0 + R tc / (1 + )

R ter ie + R tevtb (58)

r out = r ie kR te (59)where r ie is given by Eq. ( ?? ). The input resistance is given by

r in = R tb + r ib (60)

where r ib is given by Eq. ( ?? ).

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The Common-Base Ampli er

Figure 16(c) shows the ac signal circuit of a common-base ampli er. We assume that the biassolution and the small-signal parameters r 0e and r0 are known. The output voltage and outputresistance can be calculated by replacing the circuit seen looking into the collector by the Nortonequivalent circuit of Fig. 9(b). The input resistance can be calculated by replacing the circuit seenlooking into the emitter by the Thvenin equivalent circuit of Fig. 10(b) with ve(oc) = 0 . With theaid of these circuits, we can write

vo = ic(sc ) (r ic kR tc ) = Gme (r ic kR tc ) vte (61)

r out = r ic kR tc (62)

r in = R te + r ie (63)

where Gme , r ic , and r ie , respectively, are given by Eqs. ( ?? ), (?? ), and ( ?? ).

The CE/CC Ampli er

Figure 17(a) shows the ac signal circuit of a two-stage ampli er consisting of a CE stage followedby a CC stage. Such a circuit is used to obtain a high voltage gain and a low output resistance.The voltage gain can be written

vovtb1

=ic1(sc )vtb1

vtb2

ic1(sc )

ve2(oc)vtb2

vo

ve2(oc)

= Gmb 1 [ (r ic 1kRC 1)] r 0

r 0e2 + r 0

R te 2r ie 2 + R te 2

(64)

where r 0e2 is calculated with R tb2 = r ic 1kRC 1. The input and output resistances are given by

r in = R tb1 + r ib1 (65)

r out = r ie 2kR te 2 (66)

Although not a part of the solution, the resistance seen looking out of the collector of Q1 isR tc 1 = RC 1kr ib2.

Figure 17: (a) CE-CC ampli er. (b) Cascode ampli er.

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The Cascode Ampli er

Figure 17(b) shows the ac signal circuit of a cascode ampli er. The voltage gain can be written

vovtb1

=ic1(sc )vtb1

vte 2

ic1(sc )

ic2(sc )vte 2

vo

ic2(sc )= Gm 1 ( r ic 1) ( Gme 2) ( r ic 2kR tc 2)

where Gme 2 and r ic 2 are calculated with R te 2 = r ic 1. The input and output resistances are givenby

r in = R tb1 + r ib1

r out = R tc 2kr ic 2

The resistance seen looking out of the collector of Q1 is R tc 1 = r ie 2.A second cascode ampli er is shown in Fig. 18(a) where a pnp transistor is used for the second

stage. The voltage gain is given by

vovtb1

=ic1(sc )vtb1

vte 2

ic1(sc )

ic2(sc )vte 2

vo

ic2(sc )= Gm 1 ( r ic 1kRC 1) ( Gme 2) ( r ic 2kR tc 2)

The expressions for r in and r out are the same as for the cascode ampli er in Fig. 17(b). Theresistance seen looking out of the collector of Q1 is R tc 1 = RC 1kr ie 2.

Figure 18: (a) Second cascode ampli er. (b) Di ff erential ampli er.

The Di ff erential Ampli er

Figure 18(b) shows the ac signal circuit of a di ff erential ampli er. For the case of an active tail biassupply, the resistor RQ represents its small-signal ac resistance. We assume that the transistorsare identical, biased at the same currents and voltages, and have identical small-signal parameters.Looking out of the emitter of Q1, the Thvenin voltage and resistance are given by

vte 1 = ve2(oc)RQ

RQ + RE + r ie

= vtb2r0 + R tc / (1 + )

r 0e + r0 + R tc / (1 + )RQ

RQ + RE + r ie(67)

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R te 1 = RE + RQ k (RE + r ie ) (68)

The small-signal collector voltage of Q1 is given by

vo1 = ic1(sc ) (r ic kR tc ) = (Gmb vtb1 Gme vte 1) ( r ic kR tc )= Gmb (r ic kR tc ) vtb1

+ Gmer0 + R tc / (1 + )

r 0e + r0 + R tc / (1 + )RQ

r ie + RE + RQvtb2 (69)

By symmetry, vo2 is obtained by interchanging the subscripts 1 and 2 in this equation. The small-signal resistance seen looking into either output is

r out = R tc kr ic (70)

where r ic calculated from Eq. ( ?? ) with R te = RE + RQ k (RE + r ie ). Although not labeled on thecircuit, the input resistance seen by both vtb1 and vtb2 is r in = r ib .

A second solution of the di ff amp can be obtained by replacing vtb1 and vtb2 with di ff erentialand common-mode components as follows:

vtb1 = vi(cm ) +vi(d)

2(71)

vtb2 = vi(cm ) vi(d)

2(72)

where vi(d) = vtb1 vtb2 and vi(cm ) = ( vtb1 + vtb2) / 2. Superposition of vi(d) and vi(cm ) can be used tosolve for vo1 and vo2. With vi(cm ) = 0 , the eff ects of vtb1 = vi(d) / 2 and vtb2 = vi(d) / 2 are to causevq = 0 . Thus the vq node can be grounded and the circuit can be divided into two common-emitterstages in which R te (d) = RE for each transistor. In this case, vo1(d) can be written

vo1(d) =ic1(sc )vtb1(d)

vo1(d)ic1(sc )

vtb1(d) = Gm (d) ( r ic kR tc )vi (d)

2

= Gm (d) ( r ic kR tc )vtb1 vtb2

2(73)

By symmetry vo2(d) = vo1(d) .With vi(d) = 0 , the eff ects of vtb1 = vtb2 = vi(cm ) are to cause the emitter currents in Q1 and

Q2 to change by the same amounts. If RQ is replaced by two parallel resistors of value 2RQ , itfollows by symmetry that the circuit can be separated into two common-emitter stages each withR te (cm ) = RE + 2 RQ . In this case, vo1(cm ) can be written

vo1(cm ) =ic1(sc )

vtb1(cm )

vo1(cm )ic1(sc )

vtb1(cm ) = Gm (cm ) ( r ic kR tc ) vi (cm )

= Gm (cm ) ( r ic kR tc )vtb1 + vtb2

2(74)

By symmetry vo2(cm )

= vo1(cm )

.Because R te is diff erent for the di ff erential and common-mode circuits, Gm and r ib are diff erent.

However, the total solution vo1 = vo1(d) + vo1(cm ) is the same as that given by Eq. (69), andsimilarly for vo2. Note that r ic is the same for both solutions and is calculated with R te = RE +RQ k (RE + r ie ). The small-signal base currents can be written ib1 = vi(cm ) /r ib(cm ) + vi(d) /r ib(d) andib2 = vi (cm ) /r ib(cm ) vi(d) /r ib(d) . If RQ , the common-mode gain is very small, approaching 0 asr0 . In this case, the di ff erential solutions can be used for the total solutions. If RQ RE + r ie ,the common-mode solutions are often approximated by zero.

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Small-Signal High-Frequency Models

Figure 19 shows the hybrid- and T models for the BJT with the base-emitter capacitance c andthe base-collector capacitance c added. The capacitor ccs is the collector-substrate capacitancewhich in present in monolithic integrated-circuit devices but is omitted in discrete devices. Thesecapacitors model charge storage in the device which a ff ects its high-frequency performance. Thecapacitors are given by

c = c je + F I C

V T (75)

c =c jc

[1 + V CB / C ]m c (76)

ccs =c jcs

[1 + V CS / C ]m c (77)

where I C is the dc collector current, V CB is the dc collector-base voltage, V CS is the dc collector-substrate voltage, c je is the zero-bias junction capacitance of the base-emitter junction, F isthe forward transit time of the base-emitter junction, c jc is the zero-bias junction capacitance of the base-collector junction, c jcs is the zero-bias collector-substrate capacitance, C is the built-inpotential, and mc is the junction exponential factor. For integrated circuit lateral pnp transistors,ccs is replaced with a capacitor cbs from base to substrate, i.e. from the B node to ground.

Figure 19: High-frequency small-signal models of the BJT. (a) Hybrid- model. (b) T model.

In these models, the currents are related by

i 0c = gm v = i0

b = i0

e (78)

These relations are the same as those in Eq. (26) with ib replaced with i0

b.

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BJT Device Equations and Small-Signal Models