EE101: BJT circuits (Part 1) - ee.iitb.ac.insequel/ee101/ee101_bjt_2.pdf · EE101: BJT circuits (Part 1) M. B. Patil ... BJT ampli er: basic operation E C C ... BJT ampli er biasing:

EE101: BJT circuits (Part 1)

M. B. [email protected]

www.ee.iitb.ac.in/~sequel

Department of Electrical EngineeringIndian Institute of Technology Bombay

M. B. Patil, IIT Bombay

BJT amplifier: basic operation

E

C

0.2 0.4 0.6 0.8 0

IC

α IE

IE

IB

RC

VCC

VB

Vo

RC

VCC

VB

t

t

VBE

IC

* In the active mode, IC changes exponentially with VBE .

* Vo = VCC − IC RC ⇒ the amplitude of Vo is simply IC RC which can be mademuch larger than the input voltage amplitude.

* Note that both the input (VBE ) and output (Vo) voltages have DC (“bias”)components.



E

C

0.2 0.4 0.6 0.8 0

IC

α IE

IE

IB

RC

VCC

VB

Vo

RC

VCC

VB

t

t

VBE

IC






E

C

0.2 0.4 0.6 0.8 0

IC

α IE

IE

IB

RC

VCC

VB

Vo

RC

VCC

VB

t

t

VBE

IC






E

C

0.2 0.4 0.6 0.8 0

IC

α IE

IE

IB

RC

VCC

VB

Vo

RC

VCC

VB

t

t

VBE

IC





BJT amplifier biasing: why?

E

B

C

0 1 2 3 4 5

1

2

3

4

5

0

Vi

RC

VCC

Vo

RB Vo(Volts)

Vi (Volts)

Consider a more realistic BJT amplifier circuit, with RB added to limit the basecurrent (and thus protect the transistor).

* The gain of the amplifier is given bydVo

dVi.

* Since Vo is nearly constant for Vi < 0.7 V (due to cut-off) and Vi > 1.3 V (dueto saturation), the circuit will not work an an amplifier in this range.

* Further, to get a large swing in Vo without distortion, the DC bias of Vi shouldbe at the centre of the amplifying region, i.e., Vi ≈ 1 V .



E

B

C

0 1 2 3 4 5

1

2

3

4

5

0

Vi

RC

VCC

Vo

RB Vo(Volts)

Vi (Volts)



dVi.





E

B

C

0 1 2 3 4 5

1

2

3

4

5

0

Vi

RC

VCC

Vo

RB Vo(Volts)

Vi (Volts)



dVi.





E

B

C

0 1 2 3 4 5

1

2

3

4

5

0

Vi

RC

VCC

Vo

RB Vo(Volts)

Vi (Volts)



dVi.





t (msec) t (msec) t (msec)

E

B

C

A B C

A B C

4.50

4.60

4.70

4.80

4.90

5.00

0 0.2 0.4 0.6 0.8 1

0.70

0.72

0.74

0.76

0.78

0.80

0.95

0.97

0.99

1.01

1.03

1.05

1.25

1.27

1.29

1.31

1.33

1.35

0.15

0.25

0.35

0.45

0.55

0.65

0 0.2 0.4 0.6 0.8 12.40

2.60

2.80

3.00

3.20

3.40

0 0.2 0.4 0.6 0.8 1

0 1 2 3 4 5

1

2

3

4

5

0

Vi

Vo

Vi

RB

RC

VCC

Vo Vo

Vi

(SEQUEL file: ee101 bjt amp1.sqproj)M. B. Patil, IIT Bombay

BJT amplifier

E

B

C

0 1 2 3 4 5

1

2

3

4

5

0

Vi

RC

VCC

Vo

RB Vo(Volts)

Vi (Volts)

* The key challenges in realizing this amplifier in practice are

- adjusting the input DC bias to ensure that the BJT remains in the linear(active) region with a certain bias value of IC .

- mixing the input DC bias with the signal voltage.

* The first issue is addressed by using a suitable biasing scheme, and the secondby using “coupling” capacitors.


BJT amplifier

E

B

C

0 1 2 3 4 5

1

2

3

4

5

0

Vi

RC

VCC

Vo

RB Vo(Volts)

Vi (Volts)






BJT amplifier

E

B

C

0 1 2 3 4 5

1

2

3

4

5

0

Vi

RC

VCC

Vo

RB Vo(Volts)

Vi (Volts)






BJT amplifier

E

B

C

0 1 2 3 4 5

1

2

3

4

5

0

Vi

RC

VCC

Vo

RB Vo(Volts)

Vi (Volts)






BJT amplifier: a simple biasing scheme

15 V

1 k

VCC

RBRC

“Biasing” an amplifier ⇒ selection of component values for a certain DC value of IC(i.e., when no signal is applied).

Equivalently, we may bias an amplifier for a certain DC value of VCE , since IC andVCE are related: VCE + ICRC = VCC .

As an example, for RC = 1 k, β = 100, let us calculate RB which will giveIC = 3.3 mA, assuming the BJT to be operating in the active mode.

IB =IC

β=

3.3 mA

100= 33µA =

VCC − VBE

RB

→ RB =14.3V

33µA= 430 kΩ .



15 V

1 k

VCC

RBRC




IB =IC

β=

3.3 mA

100= 33µA =

VCC − VBE

RB

→ RB =14.3V

33µA= 430 kΩ .



15 V

1 k

VCC

RBRC




IB =IC

β=

3.3 mA

100= 33µA =

VCC − VBE

RB

→ RB =14.3V

33µA= 430 kΩ .



15 V

1 k

VCC

RBRC




IB =IC

β=

3.3 mA

100= 33µA =

VCC − VBE

RB

→ RB =14.3V

33µA= 430 kΩ .



15 V

1 k

VCC

RBRC




IB =IC

β=

3.3 mA

100= 33µA =

VCC − VBE

RB

→ RB =14.3V

33µA= 430 kΩ .


BJT amplifier: a simple biasing scheme (continued)

15 V

1 k

VCC

RBRC

With RB = 430 k, we expect IC = 3.3 mA, assuming β = 100.

However, in practice, there is a substantial variation in the β value (even for the sametransistor type). The manufacturer may specify the nominal value of β as 100, but theactual value may be 150, for example.

With β = 150, the actual IC is,

IC = β ×VCC − VBE

RB= 150×

(15− 0.7)V

430 k= 5 mA ,

which is significantly different than the intended value, 3.3 mA.

⇒ need a biasing scheme which is not so sensitive to β.



15 V

1 k

VCC

RBRC





RB= 150×

(15− 0.7)V

430 k= 5 mA ,





15 V

1 k

VCC

RBRC





RB= 150×

(15− 0.7)V

430 k= 5 mA ,





15 V

1 k

VCC

RBRC





RB= 150×

(15− 0.7)V

430 k= 5 mA ,




BJT amplifier: improved biasing scheme

3.6 k

1 k

10 k

2.2 k

10 V

IB

VCC

RE

RC

R1IC

IER2

IB VCC

VCC R2

R1

IC

IE

RC

RE

IB

IE

IC

VCCRTh

VThRE

RC

VTh =R2

R1 + R2VCC =

2.2 k

10 k + 2.2 k× 10V = 1.8V , RTh = R1 ‖ R2 = 1.8 k

Assuming the BJT to be in the active mode,

KCL: VTh = RTh IB + VBE + RE IE = RTh IB + VBE + (β + 1) IB RE

→ IB =VTh − VBE

RTh + (β + 1)RE, IC = β IB =

β (VTh − VBE )

RTh + (β + 1)RE.

For β = 100, IC=1.07 mA.

For β = 200, IC=1.085 mA.



3.6 k

1 k

10 k

2.2 k

10 V

IB

VCC

RE

RC

R1IC

IER2

IB VCC

VCC R2

R1

IC

IE

RC

RE

IB

IE

IC

VCCRTh

VThRE

RC

VTh =R2

R1 + R2VCC =

2.2 k

10 k + 2.2 k× 10V = 1.8V , RTh = R1 ‖ R2 = 1.8 k



→ IB =VTh − VBE

RTh + (β + 1)RE, IC = β IB =

β (VTh − VBE )

RTh + (β + 1)RE.

For β = 100, IC=1.07 mA.

For β = 200, IC=1.085 mA.



3.6 k

1 k

10 k

2.2 k

10 V

IB

VCC

RE

RC

R1IC

IER2

IB VCC

VCC R2

R1

IC

IE

RC

RE

IB

IE

IC

VCCRTh

VThRE

RC

VTh =R2

R1 + R2VCC =

2.2 k

10 k + 2.2 k× 10V = 1.8V , RTh = R1 ‖ R2 = 1.8 k



→ IB =VTh − VBE

RTh + (β + 1)RE, IC = β IB =

β (VTh − VBE )

RTh + (β + 1)RE.

For β = 100, IC=1.07 mA.

For β = 200, IC=1.085 mA.



3.6 k

1 k

10 k

2.2 k

10 V

IB

VCC

RE

RC

R1IC

IER2

IB VCC

VCC R2

R1

IC

IE

RC

RE

IB

IE

IC

VCCRTh

VThRE

RC

VTh =R2

R1 + R2VCC =

2.2 k

10 k + 2.2 k× 10V = 1.8V , RTh = R1 ‖ R2 = 1.8 k



→ IB =VTh − VBE

RTh + (β + 1)RE, IC = β IB =

β (VTh − VBE )

RTh + (β + 1)RE.

For β = 100, IC=1.07 mA.

For β = 200, IC=1.085 mA.



3.6 k

1 k

10 k

2.2 k

10 V

IB

VCC

RE

RC

R1IC

IER2

IB VCC

VCC R2

R1

IC

IE

RC

RE

IB

IE

IC

VCCRTh

VThRE

RC

VTh =R2

R1 + R2VCC =

2.2 k

10 k + 2.2 k× 10V = 1.8V , RTh = R1 ‖ R2 = 1.8 k



→ IB =VTh − VBE

RTh + (β + 1)RE, IC = β IB =

β (VTh − VBE )

RTh + (β + 1)RE.

For β = 100, IC=1.07 mA.

For β = 200, IC=1.085 mA.



3.6 k

1 k

10 k

2.2 k

10 V

IB

VCC

RE

RC

R1IC

IER2

IB VCC

VCC R2

R1

IC

IE

RC

RE

IB

IE

IC

VCCRTh

VThRE

RC

VTh =R2

R1 + R2VCC =

2.2 k

10 k + 2.2 k× 10V = 1.8V , RTh = R1 ‖ R2 = 1.8 k



→ IB =VTh − VBE

RTh + (β + 1)RE, IC = β IB =

β (VTh − VBE )

RTh + (β + 1)RE.

For β = 100, IC=1.07 mA.

For β = 200, IC=1.085 mA.



3.6 k

1 k

10 k

2.2 k

10 V

IB

VCC

RE

RC

R1IC

IER2

IB VCC

VCC R2

R1

IC

IE

RC

RE

IB

IE

IC

VCCRTh

VThRE

RC

VTh =R2

R1 + R2VCC =

2.2 k

10 k + 2.2 k× 10V = 1.8V , RTh = R1 ‖ R2 = 1.8 k



→ IB =VTh − VBE

RTh + (β + 1)RE, IC = β IB =

β (VTh − VBE )

RTh + (β + 1)RE.

For β = 100, IC=1.07 mA.

For β = 200, IC=1.085 mA.



3.6 k

1 k

10 k

2.2 k

10 V

IB

VCC

RE

RC

R1IC

IER2

IB VCC

VCC R2

R1

IC

IE

RC

RE

IB

IE

IC

VCCRTh

VThRE

RC

VTh =R2

R1 + R2VCC =

2.2 k

10 k + 2.2 k× 10V = 1.8V , RTh = R1 ‖ R2 = 1.8 k



→ IB =VTh − VBE

RTh + (β + 1)RE, IC = β IB =

β (VTh − VBE )

RTh + (β + 1)RE.

For β = 100, IC=1.07 mA.

For β = 200, IC=1.085 mA.


BJT amplifier: improved biasing scheme (continued)

1.8 V

6 V

1.1 V

2.2 k 1 k

3.6 k10 k

10 V

IB

VCC

RE

IER2

RC

R1

IC

With IC = 1.1 mA, the various DC (“bias”) voltages are,

VE = IE RE ≈ 1.1 mA× 1 k = 1.1V ,

VB = VE + VBE ≈ 1.1V + 0.7V = 1.8V ,

VC = VCC − IC RC = 10V − 1.1 mA× 3.6 k ≈ 6V ,

VCE = VC − VE = 6− 1.1 = 4.9V .



1.8 V

6 V

1.1 V

2.2 k 1 k

3.6 k10 k

10 V

IB

VCC

RE

IER2

RC

R1

IC


VE = IE RE ≈ 1.1 mA× 1 k = 1.1V ,

VB = VE + VBE ≈ 1.1V + 0.7V = 1.8V ,

VC = VCC − IC RC = 10V − 1.1 mA× 3.6 k ≈ 6V ,

VCE = VC − VE = 6− 1.1 = 4.9V .



1.8 V

6 V

1.1 V

2.2 k 1 k

3.6 k10 k

10 V

IB

VCC

RE

IER2

RC

R1

IC


VE = IE RE ≈ 1.1 mA× 1 k = 1.1V ,

VB = VE + VBE ≈ 1.1V + 0.7V = 1.8V ,

VC = VCC − IC RC = 10V − 1.1 mA× 3.6 k ≈ 6V ,

VCE = VC − VE = 6− 1.1 = 4.9V .



1.8 V

6 V

1.1 V

2.2 k 1 k

3.6 k10 k

10 V

IB

VCC

RE

IER2

RC

R1

IC


VE = IE RE ≈ 1.1 mA× 1 k = 1.1V ,

VB = VE + VBE ≈ 1.1V + 0.7V = 1.8V ,

VC = VCC − IC RC = 10V − 1.1 mA× 3.6 k ≈ 6V ,

VCE = VC − VE = 6− 1.1 = 4.9V .



1.8 V

6 V

1.1 V

2.2 k 1 k

3.6 k10 k

10 V

IB

VCC

RE

IER2

RC

R1

IC


VE = IE RE ≈ 1.1 mA× 1 k = 1.1V ,

VB = VE + VBE ≈ 1.1V + 0.7V = 1.8V ,

VC = VCC − IC RC = 10V − 1.1 mA× 3.6 k ≈ 6V ,

VCE = VC − VE = 6− 1.1 = 4.9V .



2.2 k 1 k

3.6 k10 k

10 V

IB

VCC

RE

IER2

RC

R1

IC

A quick estimate of the bias values can be obtained by ignoring IB (which is fair if β islarge). In that case,

VB =R2

R1 + R2VCC =

2.2 k

10 k + 2.2 k× 10V = 1.8V .

VE = VB − VBE ≈ 1.8V − 0.7V = 1.1V .

IE =VE

RE=

1.1V

1 k= 1.1 mA.

IC = α IE ≈ IE = 1.1 mA.

VCE = VCC − IC RC − IE RE = 10V − (3.6 k× 1.1 mA)− (1 k× 1.1 mA) ≈ 5V .



2.2 k 1 k

3.6 k10 k

10 V

IB

VCC

RE

IER2

RC

R1

IC


VB =R2

R1 + R2VCC =

2.2 k

10 k + 2.2 k× 10V = 1.8V .

VE = VB − VBE ≈ 1.8V − 0.7V = 1.1V .

IE =VE

RE=

1.1V

1 k= 1.1 mA.

IC = α IE ≈ IE = 1.1 mA.




2.2 k 1 k

3.6 k10 k

10 V

IB

VCC

RE

IER2

RC

R1

IC


VB =R2

R1 + R2VCC =

2.2 k

10 k + 2.2 k× 10V = 1.8V .

VE = VB − VBE ≈ 1.8V − 0.7V = 1.1V .

IE =VE

RE=

1.1V

1 k= 1.1 mA.

IC = α IE ≈ IE = 1.1 mA.




2.2 k 1 k

3.6 k10 k

10 V

IB

VCC

RE

IER2

RC

R1

IC


VB =R2

R1 + R2VCC =

2.2 k

10 k + 2.2 k× 10V = 1.8V .

VE = VB − VBE ≈ 1.8V − 0.7V = 1.1V .

IE =VE

RE=

1.1V

1 k= 1.1 mA.

IC = α IE ≈ IE = 1.1 mA.




2.2 k 1 k

3.6 k10 k

10 V

IB

VCC

RE

IER2

RC

R1

IC


VB =R2

R1 + R2VCC =

2.2 k

10 k + 2.2 k× 10V = 1.8V .

VE = VB − VBE ≈ 1.8V − 0.7V = 1.1V .

IE =VE

RE=

1.1V

1 k= 1.1 mA.

IC = α IE ≈ IE = 1.1 mA.




2.2 k 1 k

3.6 k10 k

10 V

IB

VCC

RE

IER2

RC

R1

IC


VB =R2

R1 + R2VCC =

2.2 k

10 k + 2.2 k× 10V = 1.8V .

VE = VB − VBE ≈ 1.8V − 0.7V = 1.1V .

IE =VE

RE=

1.1V

1 k= 1.1 mA.

IC = α IE ≈ IE = 1.1 mA.



Adding signal to bias

VCC

RC

RE

R1

R2

vB

CB

vs

* As we have seen earlier, the input signal vs(t) = V sin ωt (for example) needs tobe mixed with the desired bias value VB so that the net voltage at the base isvB(t) = VB + V sin ωt.

* This can be achieved by using a coupling capacitor CB .

* Let us take a simple circuit to illustrate how a coupling capacitor works.



VCC

RC

RE

R1

R2

vB

CB

vs






VCC

RC

RE

R1

R2

vB

CB

vs






VCC

RC

RE

R1

R2

vB

CB

vs





Coupling capacitor: example

C

A

i3i1 i2

vCR2

R1vs(t) V0 (DC)

Let vA be the instantaneous node voltage at A. KCL gives,

−i1 + i3 + i2 = 0→ −CdvC

dt+

vA

R1+

vA − V0

R2= 0.

−Cd(vs − vA)

dt+

vA

R1+

vA − V0

R2= 0. (1)

Let vA = VA + va(t), where VA = constant (DC).

In sinusoidal steady state, Eq. (1) can be split into two equations, one with only DCterms and the other with only sinusoidal terms.

−Cd(−VA)

dt+

VA

R1+

VA − V0

R2= 0, and −C

d(vs − va)

dt+

va

R1+

va

R2= 0.



C

A

i3i1 i2

vCR2

R1vs(t) V0 (DC)


−i1 + i3 + i2 = 0→ −CdvC

dt+

vA

R1+

vA − V0

R2= 0.

−Cd(vs − vA)

dt+

vA

R1+

vA − V0

R2= 0. (1)



−Cd(−VA)

dt+

VA

R1+

VA − V0

R2= 0, and −C

d(vs − va)

dt+

va

R1+

va

R2= 0.



C

A

i3i1 i2

vCR2

R1vs(t) V0 (DC)


−i1 + i3 + i2 = 0→ −CdvC

dt+

vA

R1+

vA − V0

R2= 0.

−Cd(vs − vA)

dt+

vA

R1+

vA − V0

R2= 0. (1)



−Cd(−VA)

dt+

VA

R1+

VA − V0

R2= 0, and −C

d(vs − va)

dt+

va

R1+

va

R2= 0.



C

A

i3i1 i2

vCR2

R1vs(t) V0 (DC)


−i1 + i3 + i2 = 0→ −CdvC

dt+

vA

R1+

vA − V0

R2= 0.

−Cd(vs − vA)

dt+

vA

R1+

vA − V0

R2= 0. (1)



−Cd(−VA)

dt+

VA

R1+

VA − V0

R2= 0, and −C

d(vs − va)

dt+

va

R1+

va

R2= 0.



C

A

i3i1 i2

vCR2

R1vs(t) V0 (DC)


−i1 + i3 + i2 = 0→ −CdvC

dt+

vA

R1+

vA − V0

R2= 0.

−Cd(vs − vA)

dt+

vA

R1+

vA − V0

R2= 0. (1)



−Cd(−VA)

dt+

VA

R1+

VA − V0

R2= 0, and −C

d(vs − va)

dt+

va

R1+

va

R2= 0.


Coupling capacitor: example (continued)

C

A

i3i1 i2

vCR2

R1vs(t) V0 (DC)

SincedVA

dt= 0 (VA being constant), we get

VA

R1+

VA − V0

R2= 0, and

va

(1

R1+

1

R2

)= C

d(vs − va)

dt.

In other words, the original circuit can be thought of as

AND

C

AA

(DC circuit) (AC circuit)

C

A

i3i1 i2

vC

R1

R2 R2 R2

R1R1vs(t) vs(t)V0V0

We can get VA from the first circuit, va(t) from the second, and then combine them

to get the actual vA(t): vA(t) = VA + va(t)



C

A

i3i1 i2

vCR2

R1vs(t) V0 (DC)

SincedVA


VA

R1+

VA − V0

R2= 0, and

va

(1

R1+

1

R2

)= C

d(vs − va)

dt.


AND

C

AA


C

A

i3i1 i2

vC

R1

R2 R2 R2

R1R1vs(t) vs(t)V0V0





C

A

i3i1 i2

vCR2

R1vs(t) V0 (DC)

SincedVA


VA

R1+

VA − V0

R2= 0, and

va

(1

R1+

1

R2

)= C

d(vs − va)

dt.


AND

C

AA


C

A

i3i1 i2

vC

R1

R2 R2 R2

R1R1vs(t) vs(t)V0V0




Capacitors in an amplifier circuit

* Split the original circuit into two circuits:

- DC circuit: replace each capacitor with an open circuit, each AC voltage

source with a short circuit.

- AC circuit: replace each DC voltage source with a short circuit.

* Analyse the two circuits separately.

* Combine the results of the two circuits to obtain the net instantaneous voltagesand currents.

* The procedure described above also applies to a nonlinear circuit such as a BJTamplifier IF the “small-signal approximation” is satisfied. We will soon see whatthat means.





























Common-emitter amplifier

couplingcapacitor

couplingcapacitor

bypasscapacitor

loadresistor

vs

CB

CC

CE

VCC

RL

RE

R2

R1

RC

DC circuit

VCC

RER2

R1

RC

AND

AC circuit

vs

CB

CC

CE

RL

RE

R2

R1

RC

* The coupling capacitors ensure that the singal source and the load resistor donot affect the DC bias of the amplifier. (We will see the purpose of CE a littlelater.)

* This enables us to bias the amplifier without worrying about what load it isgoing to drive.



couplingcapacitor

couplingcapacitor

bypasscapacitor

loadresistor

vs

CB

CC

CE

VCC

RL

RE

R2

R1

RC

DC circuit

VCC

RER2

R1

RC

AND

AC circuit

vs

CB

CC

CE

RL

RE

R2

R1

RC





couplingcapacitor

couplingcapacitor

bypasscapacitor

loadresistor

vs

CB

CC

CE

VCC

RL

RE

R2

R1

RC

DC circuit

VCC

RER2

R1

RC

AND

AC circuit

vs

CB

CC

CE

RL

RE

R2

R1

RC





couplingcapacitor

couplingcapacitor

bypasscapacitor

loadresistor

vs

CB

CC

CE

VCC

RL

RE

R2

R1

RC

DC circuit

VCC

RER2

R1

RC

AND

AC circuit

vs

CB

CC

CE

RL

RE

R2

R1

RC





couplingcapacitor

couplingcapacitor

bypasscapacitor

loadresistor

vs

CB

CC

CE

VCC

RL

RE

R2

R1

RC

DC circuit

VCC

RER2

R1

RC

AND

AC circuit

vs

CB

CC

CE

RL

RE

R2

R1

RC




Common-emitter amplifier: AC circuit

CB

CC

CE

RL

RE

R2vs

R1

RC

RL

vs

R2

R1

RC

vsRL

R2R1

RC

* The coupling and bypass capacitors are “large” (typically, a few µF ), and atfrequencies of interest, their impedance is small.

For example, for C = 10µF , f = 1 kHz,

ZC =1

2π × 103 × 10× 10−6= 16 Ω,

which is much smaller than typical values of R1, R2, RC , RE (a few kΩ).

⇒ CB , CC , CE can be replaced by short circuits at the frequencies of interest.

* The circuit can be re-drawn in a more friendly format.

* We now need to figure out the AC description of a BJT.



CB

CC

CE

RL

RE

R2vs

R1

RC

RL

vs

R2

R1

RC

vsRL

R2R1

RC



ZC =1

2π × 103 × 10× 10−6= 16 Ω,







CB

CC

CE

RL

RE

R2vs

R1

RC

RL

vs

R2

R1

RC

vsRL

R2R1

RC



ZC =1

2π × 103 × 10× 10−6= 16 Ω,







CB

CC

CE

RL

RE

R2vs

R1

RC

RL

vs

R2

R1

RC

vsRL

R2R1

RC



ZC =1

2π × 103 × 10× 10−6= 16 Ω,







CB

CC

CE

RL

RE

R2vs

R1

RC

RL

vs

R2

R1

RC

vsRL

R2R1

RC



ZC =1

2π × 103 × 10× 10−6= 16 Ω,







CB

CC

CE

RL

RE

R2vs

R1

RC

RL

vs

R2

R1

RC

vsRL

R2R1

RC



ZC =1

2π × 103 × 10× 10−6= 16 Ω,







C

B

E

1.1

0 0.2 0.4 0.6 0.8 1

t (msec)

0.9

0.7

0.5

vBE

iCiB

iE

i C(m

A)

V0 = 0.65 V,

vBE(t) = V0 + Vm sinωt

f = 1 kHz

Vm = 10 mV

Vm = 5 mV

Vm = 2 mV

* As the vBE amplitude increases, the shape of iC (t) deviates from a sinusoid→ distortion.

* If vbe(t), i.e., the time-varying part of vBE , is kept small, iC varies linearlywith vBE . How small? Let us look at this in more detail.



C

B

E

1.1

0 0.2 0.4 0.6 0.8 1

t (msec)

0.9

0.7

0.5

vBE

iCiB

iE

i C(m

A)

V0 = 0.65 V,


f = 1 kHz

Vm = 10 mV

Vm = 5 mV

Vm = 2 mV





C

B

E

1.1

0 0.2 0.4 0.6 0.8 1

t (msec)

0.9

0.7

0.5

vBE

iCiB

iE

i C(m

A)

V0 = 0.65 V,


f = 1 kHz

Vm = 10 mV

Vm = 5 mV

Vm = 2 mV




BJT: small-signal model

C

B

E

vBE

iCiB

iE vBE

iC

iB

α iE

iE

Let vBE (t) = VBE + vbe(t) (bias+signal), and iC (t) = IC + ic (t).

Assuming active mode, iC (t) = α iE (t) = α IES

[exp

(vBE (t)

VT

)− 1

].

Since the B-E junction is forward-biased, exp

(vBE (t)

VT

) 1, and we get

iC (t) = α IES exp

(vBE (t)

VT

)= α IES exp

(VBE + vbe(t)

VT

)= α IES exp

(VBE

VT

)× exp

(vbe(t)

VT

)If vbe(t) = 0, iC (t) = IC (the bias value of iC ), i.e.,

IC = α IES exp

(VBE

VT

)⇒ iC (t) = IC exp

(vbe(t)

VT

).



C

B

E

vBE

iCiB

iE vBE

iC

iB

α iE

iE



[exp

(vBE (t)

VT

)− 1

].


(vBE (t)

VT

) 1, and we get

iC (t) = α IES exp

(vBE (t)

VT

)= α IES exp

(VBE + vbe(t)

VT

)= α IES exp

(VBE

VT

)× exp

(vbe(t)

VT


IC = α IES exp

(VBE

VT


(vbe(t)

VT

).



C

B

E

vBE

iCiB

iE vBE

iC

iB

α iE

iE



[exp

(vBE (t)

VT

)− 1

].


(vBE (t)

VT

) 1, and we get

iC (t) = α IES exp

(vBE (t)

VT

)= α IES exp

(VBE + vbe(t)

VT

)= α IES exp

(VBE

VT

)× exp

(vbe(t)

VT

)

If vbe(t) = 0, iC (t) = IC (the bias value of iC ), i.e.,

IC = α IES exp

(VBE

VT


(vbe(t)

VT

).



C

B

E

vBE

iCiB

iE vBE

iC

iB

α iE

iE



[exp

(vBE (t)

VT

)− 1

].


(vBE (t)

VT

) 1, and we get

iC (t) = α IES exp

(vBE (t)

VT

)= α IES exp

(VBE + vbe(t)

VT

)= α IES exp

(VBE

VT

)× exp

(vbe(t)

VT


IC = α IES exp

(VBE

VT


(vbe(t)

VT

).



C

B

E

vBE

iCiB

iE vBE

iC

iB

α iE

iE

iC (t) = IC exp

(vbe(t)

VT

)= IC

[1 + x + x2 + · · ·

], x = vbe(t)/VT .

If x is small, i.e., if the amplitude of vbe(t) is small compared to the thermal voltage VT , we get

iC (t) = IC

[1 +

vbe(t)

VT

].

We can now see that, for |vbe(t)| VT , the relationship between iC (t) and vbe(t) is linear, as wehave observed previously.

iC (t) = IC + ic (t) = IC

[1 +

vbe(t)

VT

]⇒ ic (t) =

IC

VT

vbe(t)



C

B

E

vBE

iCiB

iE vBE

iC

iB

α iE

iE

iC (t) = IC exp

(vbe(t)

VT

)= IC

[1 + x + x2 + · · ·

], x = vbe(t)/VT .


iC (t) = IC

[1 +

vbe(t)

VT

].



[1 +

vbe(t)

VT

]⇒ ic (t) =

IC

VT

vbe(t)



C

B

E

vBE

iCiB

iE vBE

iC

iB

α iE

iE

iC (t) = IC exp

(vbe(t)

VT

)= IC

[1 + x + x2 + · · ·

], x = vbe(t)/VT .


iC (t) = IC

[1 +

vbe(t)

VT

].



[1 +

vbe(t)

VT

]⇒ ic (t) =

IC

VT

vbe(t)



C

B

E

vBE

iCiB

iE vBE

iC

iB

α iE

iE

iC (t) = IC exp

(vbe(t)

VT

)= IC

[1 + x + x2 + · · ·

], x = vbe(t)/VT .


iC (t) = IC

[1 +

vbe(t)

VT

].



[1 +

vbe(t)

VT

]⇒ ic (t) =

IC

VT

vbe(t)



C

B

E

B C

E

vBE

iCiB

iE vBE

rπgm vbe

iC

iB

α iE

iE

The relationship, ic (t) =IC

VTvbe(t) can be represented by a VCCS, ic (t) = gm vbe(t),

where gm = IC/VT .

For the base current, we have,

iB(t) = IB + ib(t) =1

β[IC + ic (t)]

→ ib(t) =1

βic (t) =

1

βgm vbe(t)→ vbe(t) = (β/gm) ib(t).

The above relationship is represented by a resistance, rπ = β/gm, connected betweenB and E.

The resulting model is called the π-model for small-signal description of a BJT.



C

B

E

B C

E

vBE

iCiB

iE vBE

rπgm vbe

iC

iB

α iE

iE



where gm = IC/VT .


iB(t) = IB + ib(t) =1

β[IC + ic (t)]

→ ib(t) =1

βic (t) =

1






C

B

E

B C

E

vBE

iCiB

iE vBE

rπgm vbe

iC

iB

α iE

iE



where gm = IC/VT .


iB(t) = IB + ib(t) =1

β[IC + ic (t)]

→ ib(t) =1

βic (t) =

1






C

B

E

B C

E

vBE

iCiB

iE vBE

rπgm vbe

iC

iB

α iE

iE



where gm = IC/VT .


iB(t) = IB + ib(t) =1

β[IC + ic (t)]

→ ib(t) =1

βic (t) =

1






C

B

E

B C

E

vBE

iCiB

iE vBE

rπgm vbe

iC

iB

α iE

iE

* The transconductance gm depends on the biasing of the BJT, sincegm = IC/VT . For IC = 1 mA, VT ≈ 25 mV (room temperature),gm = 1 mA/25 mV = 40 mf.

* rπ also depends on IC , since rπ = β/gm = β VT /IC . For IC = 1 mA,VT ≈ 25 mV , β = 100, rπ = 2.5 kΩ.

* Note that the small-signal model is valid only for small vbe (compared to VT ), aswe have seen earlier.



C

B

E

B C

E

vBE

iCiB

iE vBE

rπgm vbe

iC

iB

α iE

iE






C

B

E

B C

E

vBE

iCiB

iE vBE

rπgm vbe

iC

iB

α iE

iE






C

B

E

B C

E

vBE

iCiB

iE

rπgm vbe

0

1

0 1 2 3 4 5

I C(m

A)

VCE (V)

CB

E

rπgm vbe

ro

* In the above model, note that ic is independent of vce .

* In practice, ic does depend on vce because of the Early effect, anddIC

dVCE≈ constant = 1/ro , where ro is called the output resistance.

* A more accurate model includes ro as well.



C

B

E

B C

E

vBE

iCiB

iE

rπgm vbe

0

1

0 1 2 3 4 5I C

(mA)

VCE (V)

CB

E

rπgm vbe

ro







C

B

E

B C

E

vBE

iCiB

iE

rπgm vbe

0

1

0 1 2 3 4 5I C

(mA)

VCE (V)

CB

E

rπgm vbe

ro







C

B

E

B C

E

vBE

iCiB

iE

rπgm vbe

0

1

0 1 2 3 4 5I C

(mA)

VCE (V)

CB

E

rπgm vbe

ro







C

B

E

CB

E

Cπ

iC

iE

iB

rb

rπgm vbe

ro

Cµ

* A few other components are required to make the small-signal model complete:rb: base spreading resistanceCπ : base charging capacitance + B-E junction capacitanceCµ: B-C junction capacitance

* The capacitances are typically in the pF range. At low frequencies, 1/ωC islarge, and the capacitances can be replaced by open circuits.

* Note that the small-signal models we have described are valid in the activeregion only.



C

B

E

CB

E

Cπ

iC

iE

iB

rb

rπgm vbe

ro

Cµ






C

B

E

CB

E

Cπ

iC

iE

iB

rb

rπgm vbe

ro

Cµ





Documents

EE101: BJT circuits (Part 1) - ee.iitb.ac.insequel/ee101/ee101_bjt_2.pdf · EE101: BJT circuits (Part 1) M. B. Patil ... BJT ampli er: basic operation E C C ... BJT ampli er biasing: