Biasing Biasing: Application of dc voltages to establish a fixed level of current and voltage. BJT...

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BiasingBiasing: Application of dc voltages to establish a fixed

level of current and voltage.

BJT Biasing Circuits:• Fixed Bias Circuit

• Fixed Bias with Emitter Resistor Circuit

• Voltage-Divider Bias Circuit

• Feedback Bias Circuit

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Fixed Bias Circuit

• This is a Common Emitter (CE) configuration.

• Solve the circuit using KVL.• 1st step: Locate capacitors and replace them with an

open circuit• 2nd step: Locate 2 main loops

which areBE loopCE loop

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Fixed Biased Circuit• 1st step: Locate capacitors and replace them with an

open circuit

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Fixed Biased Circuit

• 2nd step: Locate 2 main loops.

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1

2

BE Loop CE Loop

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Fixed Biased Circuit• BE Loop Analysis:

1 From KVL:

IB

0VRIV BEBBCC

B

BECCB R

VVI

Solving for IB

(1)

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Fixed Biased Circuit• CE Loop Analysis: From KVL:

0VRIV CECCCC

B

BECCdcC R

VVI

As we know IC = dcIB

2

IC

CECCCC VRIV

Substituting IB from equation (1)

Also note thatVCE = VC - VE

But VE = 0

VCE = VC In addition, since

VBE = VB - VE

Since VE = 0

VBE = VB

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Fixed Biased Circuit

• DISADVANTAGE Unstable – because it is too dependent on β and

produce width change of Q-pointFor improved bias stability , add emitter resistor to

dc bias.

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Example 1: Determine the Followingfor the given Fixed Biased Circuit.(a) IBQ and ICQ (b) VCEQ

(c) VBC

Solution:

A08.471000240

7.012

R

VVI

B

BECCBQ

mAII BQdcCQ 35.21008.4750 6

V83.6)k240)(mA35.2(12RIVV CCCCCEQ

V83.6VV

V7.0VV

CEC

BEB

V13.683.67.0VVV CBBC

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Example 2: For the given fixed bias circuit, determine IBQ, ICQ, VCEQ, VC, VB and VE.

Solution:

A55.321000270

7.016

R

VVI

B

BECCBQ

mA93.21055.3290II 6BQdcCQ

V09.8)k7.2)(mA93.2(16

RIVV CCCCCEQ

V09.8VV

V7.0VV

CEC

BEB

0VE

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Example 3: Given the information appearing in the Fig. (a) , determine IC, RC, RB, and VCE.

Solution: IC = 3.2 mA, RC = 1.875k,

RB = 282.5 k, VCE = 6 V.

Example 4: Given the information appearing in Fig. (b), determine IC,

VCC, and RB.

Solution:IC = 3.98 mA, VCC = 15.96 V,

= 199, RB = 763 k.

Fig. (a)

Fig. (b)

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Load line Analysis with Fixed Bias Circuit

DC load line is drawn by using the following equations:

VCE = VCC

C

CCC R

VI

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Load line Analysis with Fixed Bias Circuit

Effect of Varying IB on the Q-Point:

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Load line Analysis with Fixed Bias Circuit

Effect of varying VCC on the Q-Point:

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Load line Analysis with Fixed Bias Circuit

Effect of varying RC on the Q-Point:

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Example: Given the load line and the defined Q-point , determine the required values of VCC, RC, and RB for a fixed bias configuration.

Solution:VCE = VCC = 20 V

k2mA10

20

I

VR

R

VI

C

CCC

C

CCC

B

BECCB R

VVI

k772A25

7.020

I

VVR

B

BECCB

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Emitter-Stabilized Bias Circuit

Resistor RE added

• An emitter resistor, RE is added to improve stability

• Solve the circuit using KVL.• 1st step: Locate capacitors

and replace them with an open circuit

• 2nd step: Locate 2 main loops which; BE loop CE loop

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Emitter-Stabilized Bias Circuit • 1st Step: Locate capacitors and replace them

with an open circuit.

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Emitter-Stabilized Bias Circuit • 2nd Step: Locate two main loops

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1

2

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Emitter-Stabilized Bias Circuit

BE Loop Analysis:

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Using KVL:

EEBEBBCC RIVRIV

Recall that IE = ( + 1)IB

EBBEBBCC RI)1(VRIV EBBBECC R)1(RIVV

EB

BECCB R1R

VVI

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Emitter-Stabilized Bias Circuit CE Loop Analysis:

2

From KVL:EECECCCC RIVRIV

Substituting IE IC we get

ECCECCCC RIVRIV

CEECCCC VRRIV

EC

CECCC RR

VVI

ECCE VVV

ECEC VVV

CCCCB RIVV

EBEB VVV

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Example: For the given emitter bias network, determine IB, IC, VCE, VC, VE, VB and VBC.

Solution:

A1.41)k1)(150(k430

7.020

R)1(R

VVI

EB

BECCB

mA01.2101.4050II 6BC

V97.13)k1k2)(mA01.2(20

RRIVV ECCCCCE

V98.1502.420)k2)(mA01.2(20RIVV CCCCC

V01.297.1398.15VVV CECE V71.201.27.0VVV EBEB

V27.1398.1571.2VVV CBBC

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Load line Analysis for Emitter Stabilized Bias Circuit

The collector-emitter loop equation that defines the load line is:

ECCCCCE RRIVV

Choosing IC = 0 gives

CCCE VV

And choosing VCE = 0 gives

EC

CCC RR

VI

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Example: For the given emitter stabilized bias circuit, determine IBQ, ICQ, VCEQ, VC, VB, VE.

Solution:

A18.29)k5.1)(1100(k510

7.020

R1R

VVI

EB

BECCBQ

mA92.2A18.29100II BQCQ

V61.8RRIVV ECCCCCEQ

V13RIVV CCCCC

V12.5RIVV BBCCB

V39.461.813VVV CECE

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Voltage Divider Bias• Provides good Q-point stability

with a single polarity supply voltage• Solve the circuit using KVL• 1st step: Locate capacitors and

replace them with an open circuit

• 2nd step: Simplify circuit using Thevenin Theorem

• 3rd step: Locate 2 main loops which; BE loop CE loop

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Voltage Divider Bias

• 1st step: Locate capacitors and replace them with an open circuit

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Voltage Divider Rule

2nd step: Simplify the circuit using Thevenin Theorem

From Thevenin’s Theorem

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2121TH RR

RRR//RR

CC21

2TH V

RR

RV

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Voltage Divider Bias

3rd step: Locate 2 main loops.

1

2

BE Loop

1

CE Loop

2

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Voltage Divider BiasBE Loop Analysis:

1

0RIVRIV EEBETHBTH

ERTH

BETHB

EBBETHBTH

R)1(R

VVI

0RI)1(VRIV

BE I)1(I

From KVL:

But

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Voltage Divider BiasCE Loop Analysis:

2

From KVL:0RIVRIV EECECCCC

AssumeIC IE

0RIVRIV ECCECCCC

)RR(IVV ECCCCCE

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IC

DC load line with Voltage Divider BiasThe dc load line can be drawn from the following

equation:)RR(IVV ECCCCCE

VCC

EC

CCC RR

VI

VCE

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Example: Determine the dc bias voltage VCE and the current IC for the given voltage divider configuration.

Solution:

K55.3K9.3k39

k9.3k39

RR

RRR

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21TH

V2k9.3k39

22k9.3

RR

VRV

21

CC2TH

A05.6R1R

VVI

ETH

BETHB

mA85.0A05.6140II BdcC

V22.12RRIVV ECCCCCE

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DC Bias with Voltage Feedback

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DC Bias with Voltage Feedback

Using KVL:

BE Loop

EEBEBBCCCC IRVRI'IRV In this circuit, IB is assumedto be very small andI’C IC = IC. The above Equation may therefore beRe-written as

CEBEBBCCCC IRVRIIRV

BEBEBBBCCC IRVRIIRV

BECBBECC RRRIVV

ECB

BECCB RRR

VVI

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DC Bias with Voltage Feedback

CE Loop

EECECCCC RIVR'IV Since I’C IC and IC IE, we have

ECCECCCC RIVRIV

ECCCCCE RRIVV

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Example: Determine the quiescent level of ICQ and VCEQ for the given network.

Solution:

ECB

BECCB RRR

VVI

A91.11k2.1k7.4)90(k250

7.010

mA07.11091.1190II 6BCQ

V69.3)k2.1k7.4)(mA07.1(10RRIVV ECCCCCEQ