session 12 analysis of dc circuits with bjt...

Session 12Session 12

Analysis of DC circuits with BJT (polarization)

Electronic Components and Circuits

José A. Garcia Souto

www.uc3m.es/portal/page/portal/dpto_tecnologia_electronica/Personal/JoseAntonioGarcia

DC circuits with BJT and

polarization of transistors

OBJETIVESOBJETIVES• Summarizing the operating regions and introducing the

amplification (active)

Active, Cut-off, Saturation

• Analyzing basic BJT circuits in DC. Static characteristics and Load line.

• Understanding the need of polarization circuits• Understanding the need of polarization circuits

• Knowing and analyzing typical polarization circuits of transistors. Bias (quiescent) point.

2UC3M 2010 ECC - Session 12

Amplification concept with BJT

60

50

IC (mA) SATURATES

Vo ≈ 0 V

0 0,2 0,4 0,6 0,8

50

40

30

20

10

VBE (V)

CUT-OFF

Vo = Vcc

AMPLIFIES

Vo = G·Vi

Small changes of Vi result in greater variation of

Vo, thus providing gain Vo/Vi

It should be around a bias point or quiescent

point VBE-Q, VCE-Q


0 0,2 0,4 0,6 0,8 VBE (V)

Superposition of DC Bias and Signal

VBB is a continuous source that

provides along with RB a bias

point (polarization) : 0=vI point (polarization) : 0=gvCI

EI

BI

BEVCEV

60

50

40

30

20

10

IB (µA)

IB

A variable signal is introduced

and it is added to VBB

gBB vV +


EI

VBEVBB

10

VBE (V)

Amplification focuses on Active region

• Datasheet VBE(on) VCE(sat) hFE VBE(sat)

Search BC547 in http://www.fairchildsemi.com/

Region Base-Emitter Junction Base-Collector Junction

Cut-off Reverse(OFF) Reverse

Active Forward (ON) Reverse (Transistor Effect)

Saturation Forward (ON) Forward (Saturated)

Region Conditions NPN Operation NPN

5

Region Conditions NPN Operation NPN

Cut-off VBE < VBE-ON o IB = 0 IB=0, IC=IE=0

Active VBE-ON y VCE > VCE-SAT IC ≈≈≈≈ hFE·IB [VBE=VBE-ON]

Saturation VBE-ON y VCE-SAT VCE =VCE-SAT [VBE=VBE-SAT]

UC3M 2010 ECC - Session 12

Graphical DC Analysis

Cancel the signal (superposition)

0=vIAnalyze the input

Device Equation

(Input Characteristic)

Circuit Equation

(Load line)

Analyze the output

Device Equation

EXAMPLE

RC = 100 Ω

RB = 10 kΩ

VCC = 10 V

VBB = 10 V

VBE(on) = 0,7 V

VCE(sat) = 0,2 V

0=gvCI

EI

BI

BEVCEV

Device Equation

(Output Characteristic)

Circuit Equation

(Load line)

hFE = 100


E

Input Characteristic and

Load Line (I)

Graphically

Characteristic:60

IB (µA)

Characteristic:

Load line:

Analytically

VBB > VBE(on)

B

BEBBB

R

vVi

−=

Vv =

( )CEBEBB vvii ,=50

40

30

20

10

IB

7

)(onBEBE Vv =

BBBEBB RivV ⋅+=

)(onBEBE VV ≈B

onBEBB

BR

VVI

)(−=


VBEVBB VBE (V)

Output Characteristic and

Load Line (II)

Graphically

Characteristic:6

IC

(mA)

60 µA Characteristic:

Load line :

Analytically

VCE > VCE(sat)

( )BCECC ivii ,=

C

CECCC

R

vVi

−=

⋅=

5

4

3

2

1

IB

=40 µA

50 µA

30 µA

20 µA

10 µA

IC

8

BFEC ihi ⋅=

CCCECC RivV ⋅+=

CCCCCE RIVV ⋅−=BFEC IhI ⋅=


0µA

0 2 4 6 8 10 12 14 16

1 10 µA

VCE

(V)

VCE VCC

EXAMPLE

RC = 100 Ω

DC equivalent circuit: Analytically

Analyze the input RC = 100 Ω

RB = …

VCC = 10 V

VBB = …

VBE(on) = 0,7 V

VCE(sat) = 0,2 V

hFE = 100

Case 3

CI

I

BI

BEVCEV

Analyze the input

Device Equation

Circuit Equation

Analyze the output

Device Equation

Circuit Equation

9

Case 3

VBB = 10 V

RB = 10 kΩ


EI Circuit Equation

CI

Solution: DC Analysis Analyze the input

VBB > VBE(on)

Device Equation )(onBEBE Vv =CI

EI

BI

BEVCEV

Device Equation

Circuit Equation

Analyze the output

VCE > VCE(sat)

Device Equation

Circuit Equation

)(onBEBE Vv =

BBBEBB RivV ⋅+=

BFEC ihi ⋅=

RivV ⋅+=


Circuit Equation

)(onBEBE VV ≈B

onBEBB

BR

VVI

)(−=

CCCECC RivV ⋅+=

CCCCCE RIVV ⋅−=BFEC IhI ⋅=

Bias circuits

11

Set a stable operating point desensitized against the transistor

parameters.

Optimize the signal amplifier circuit.

Separate bias circuit if necessary (signal AC coupling).


Example

Reasoning operating region

Calculate the voltage VE

Calculate the current IE Calculate the current IE

Derive the relationship between IE, IB

and IC

Calculate the curents IC and IB

Calculate the voltage VC

DATA

VCC = 12 V

RC = RE = 10 kΩ


RC = RE = 10 kΩ

VBE(on) = 0,7 V

VCE(sat) = 0,2 V

hFE = 100

Types of polarization circuits

Biasing with REPractical circuit

13

Lower sensitivity to the transistor current gain

Stabilizes the bias point (quiescent point) against VBE, hFE, etc.


Case 1: Biasing with RB

DATA:

VBB = 5 V

VCC = 10 V

RE = 2 kΩ

VBE(on) = 0,7 VVCC = 10 V

RB = 50 kΩ

RC = 3 kΩ

Reasoning the operating region

Indicate the value of VBE

Calculate the currents IB, IC e IE

Calculate the voltage VCE

Check that is in active

Recalculate if you change hFE = 200

VCE(sat) = 0,2 V

hFE = 100


Case 2: Biasing with REDATA:

VBB = 5 V

VCC = 10 V

RB = 10 kΩ

RE = 2 kΩ

VBE(on) = 0,7 V

VCE(sat) = 0,2 V

Reasoning the operating region

Indicate the value of VBE RB = 10 kΩ

RC = 3 kΩ

VCE(sat) = 0,2 V

hFE = 100

Indicate the value of VBE

Calculate the current IE (you can

neglect the base current IB)

Calculate the currents IC, IB

Check that IB is negligible

Calculate the voltage VCE

Check that the BJT is in active




Generalize if not negligible IB (eg if

RB = 50 kΩ)

Biasing with current mirrors

16

refO II ≈1refrefO IR

RII ⋅== 10

3

22


Proposed exercise: Current sources

Zener ideal:

VZ = 3,3 V

IZmín = 10 mA

• Obtain the ratio IL/Iref

IL

IZmín = 10 mA

Transistor:

VCEsat = 0,2 V

VBEon = 0,7 V

β = 270

• Calculate IE, IC, IB and VB, VC, VE, VCE, IZ.

Iref

17

• Obtain the ratio IL/Iref

• Reason if IB is negligible

• Obtain Iref

• There is the scale factor IC2 =

2·IC1 between both transistor.

Calculate Vo.

• Calculate IE, IC, IB and VB, VC, VE, VCE, IZ.

• For RL = 1 kΩ, calculate IC y VC

• For RL = 22 kΩ, calculate IC y VC

• Plot the load lines (three cases).

• Maximum RL as a current source?.

• In the case RL = 22 kΩ, calculate IB, IZ


session 12 analysis of dc circuits with bjt...

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