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### Transcript of BJT Biasing Cont. - seas.upenn. ese319/Lecture_Notes/Lec_5_BJTBias2_12.pdfBJT Biasing Cont. Biasing...

• ESE319 Introduction to Microelectronics

12009 Kenneth R. Laker, update 21Sep12

BJT Biasing Cont. Biasing for DC Operating Point Stability BJT Bias Using Emitter Negative Feedback Single Supply BJT Bias Scheme Constant Current BJT Bias Scheme Rule of Thumb BJT Bias Design

• ESE319 Introduction to Microelectronics

22009 Kenneth R. Laker, update 21Sep12

Simple Base Biasing Schemes

What's wrong with these schemes?

• ESE319 Introduction to Microelectronics

32009 Kenneth R. Laker, update 21Sep12

Another Simple Biasing Scheme

RB

RC

I E

V CC

Is this scheme any better? If it is, why?

• ESE319 Introduction to Microelectronics

42009 Kenneth R. Laker, update 21Sep12

Biasing for Operating Point StabilityA practical biasing scheme must be insensitive to changes in transistor and operating temperature! Negative feedback is one solution.

IC

IB

IE R

E

RC

VCC

VB

• ESE319 Introduction to Microelectronics

52009 Kenneth R. Laker, update 21Sep12

I C I S eV BEV T

V BE=V BRE I E=V BREI C

Basic relationships:I E= I BI C=1 I B

I E=1

I C=

1

I C

Given the (DC bias) equations:

Assume: V BE=0.7 V

Let: VB, VCC and IC be specified;

Compute: RE to obtain IC:

RE=V BV BE

I C=

V B0.7I C

• ESE319 Introduction to Microelectronics

62009 Kenneth R. Laker, update 21Sep12

Negative feedback makes the collector current insensitive to VBE, IS, and . If IC increases due to an increase in IS then VBE will decrease; thus, limit-

ing the magnitude of the change in IC.

The equations that must be satisfied simultaneously are:

Negative Feedback via RE

I C=V BV BE1RE

=V BV BE

RE

V BE=V BREI C=V B

1

RE I CI C=I S eV BEV T and

if IS => IC when IC => VBE IC > VBE VBE insensitivityI C

V BRE

=>

• ESE319 Introduction to Microelectronics

72009 Kenneth R. Laker, update 21Sep12

Scilab Analysis of IC Insensitivity to ISSimultaneous equations:

I C= I S eV BEV T

I C=V BV BE

RE

or: V BV BE

RE=I S e

V BEV T

If we plot the exponential function and the straight line function,the solution values of IC and VBE for the circuit occur at theirintersection.

Let RE=4k V B=4.7Vand (want VB >> VBE)

• ESE319 Introduction to Microelectronics

82009 Kenneth R. Laker, update 21Sep12

Scilab Program//Calculate and plot npn BJT bias characteristicbeta=100;alpha=beta/(beta+1);VsubT=0.025;VTinv=1/VsubT;VBB=4.7;Re=4;vBE=0.0:0.01:1;iCline=alpha*(VBB-vBE)/Re;//mA.plot(vBE,iCline);iC=0.01:0.01:2; //mA.!IsubS =1E-16; //mA.for k= 1:1:8 IsubS=10*IsubS; vBE2=VsubT*log(iC/IsubS); plot(vBE2,iC); //Current in mA.end

I C=V BV BE

RE

V BE=V T lnI CI S

• ESE319 Introduction to Microelectronics

92009 Kenneth R. Laker, update 21Sep12

IC vs IS Results PlotI S=10

11 I S=1018

I C0.1mA Insensitive to I

S!

I C=V BV BE

RE

I C= I S eV BEV T

• ESE319 Introduction to Microelectronics

102009 Kenneth R. Laker, update 21Sep12

RE=V B0.7

I C

Example: =100V B=4.7VI C=1 mA

RE=0.994.70.7

103

=100101

0.99

I C=V B0.7

RE=

1

103 A

Writing IC as a function of :

Assume: 100 200100101

103=0.990 mA I C200201

103=0.995 mA

CONCLUSION: IC is insensitive to changes in !

RE4000

Insensitivity to Beta

• ESE319 Introduction to Microelectronics

112009 Kenneth R. Laker, update 21Sep12

Non-zero Base Resistance

I B

ICV B=I B RBV BE1 I B RE

I B=V BV BE

RB1RE

I E=1 I B

I C= I B=

RB1REV BV BE

I C= I B

1V BV BE

RE=

V BV BE RE

If RB1REI E

RC

RE

RB

VB

VCC

If RB1RE

I C= I BRB

V BV BE no feedback!

IE

• ESE319 Introduction to Microelectronics

122009 Kenneth R. Laker, update 21Sep12

Biasing for Operating Point Stability Quick Review

What is the purpose RE?

What does RB represent?

What is the condition on the value of V

B?

What is the condition on the value of R

E?

What is our rule of thumb for achieving x >> y?

• ESE319 Introduction to Microelectronics

132009 Kenneth R. Laker, update 21Sep12

Observations Emitter feedback stabilizes base voltage source bias. To reduce the sensitivity of IC to VBE, choose . RB 0, emitter feedback stabilization works well if RB is

not too large, i.e.RB1RE

Ideal rule of thumb (if possible):

RB1

101RE

V BV BE

RB=10010

RE=10 RELet 1=100

orRB1

101RE RB=

110

1RE

• ESE319 Introduction to Microelectronics

142009 Kenneth R. Laker, update 21Sep12

Emitter-Feedback Bias Design

Voltage bias circuit Single power supply version

R1

R2

VCC

RC

RE

IC

IE

IB

VB

+

-

• ESE319 Introduction to Microelectronics

152009 Kenneth R. Laker, update 21Sep12

RTh=RB=R1R2

RTh

VTh

V Th=V B=R2

R1R2V CC

Thevenin Equivalent

I 1

I B I 1 I C

• ESE319 Introduction to Microelectronics

162009 Kenneth R. Laker, update 21Sep12

Using Thevenin EquivalentUse Thevenin's theorem to simplify base circuit:

If VCC, VB and RB are inde-pendently specified then R1 and R2 are found by solving Eqs. (1) and (2):

Design Eqs.

I B0

V Th=V B=R2

R1R2V CC

I1

RTh=RB=R1R2=R1 R2

R1R2R1R2=

R1 R2RB

V B=R2 RBR1 R2

V CCR1=RBV CCV B

V B=R2

R1R2V CC R1 V BR2 V B=R2 V CC

R2=R1V B

V CCV B

R2=R1V B

V CCV B

R1=RBV CCV B

(1)

(2)

Can VCC, VB and RB be inde-pendently specified?

• ESE319 Introduction to Microelectronics

172009 Kenneth R. Laker, update 21Sep12

Using Thevenin Equivalent

If VCC, VB and RB are independently specified then R1 and R2 are found by solving Eqs. (1) and (2):

Design Eqs.

R2=R1V B

V CCV B

R1=RBV CCV B

Can VCC, VB and RB be independently specified?

RECALL: VB >> VBE and RB1RE

• ESE319 Introduction to Microelectronics

182009 Kenneth R. Laker, update 21Sep12

1st Rule of Thumb for Single Supply Biasing1. Choose RT = R1 + R2 so that I1

• ESE319 Introduction to Microelectronics

192009 Kenneth R. Laker, update 21Sep12

IC

R1

R2

RC

RE

V CCI

E

V CE+-

V RE+

-

+

-

Three design goals so farRB1REV BV BE

Constraint: V CC=V RCV CEV RETRADEOFF Increase VRC and/or VRE => Reduce VCE VRC too large => possible saturation & reduced op-erating range (i.e. V

CE -> 0.2 V or smaller).

VRC too small => possible cutoff & reduced operating range (i.e. I

C -> small or zero).

NEED A COMPROMISE!

I 1 I C also I 1 I B

(VCC

fixed)

V RC

V RE=V BV BE

B

• ESE319 Introduction to Microelectronics

202009 Kenneth R. Laker, update 21Sep12

2nd Useful Rule of Thumb 1/3, 1/3, 1/3 Rule

V RE=V CC

3V RC= I C RC=

V CC3

V CE=V CC

3

V CC=V RCV REV CE

RC=V RCI C

=V CC3 I C

RE=V REI E

=V CC3 I E

=V CC3 I C

V RE=V CCR2RT

V BE=V CC

3

Design Equations

where

IC

R1

R2

RC

RE

V CCI

E

V CE+-

+

-

+

-

V B=V BEV RE I 1 R2V BE=0.7V

VB >> VBE

V RE= I E REV B

I 1=I C10

V CCRT

R2=RTV BV CC

13

RT

R1=RT 1V BV CC

23

RT

RT = R1 + R2

I1 V RC

V RE

• ESE319 Introduction to Microelectronics

212009 Kenneth R. Laker, update 21Sep12

Biasing for Operating Point Stability Quick Review

What is the condition on the value of VB?

What is the condition on the value of RE?

What is the purpose for R1 & R

2?

What happened to RB?

What is the constraint on the value of I1?

What is the 1/3, 1/3, 1/3 Rule?

IC

R1

R2

RC

RE

V CCI

E

V CE

I1

IB

V RC

V CC=V RCV CEV RE

V RE

V CC=V RCV CEV RE

V B

V CCV RCV CEV B

• ESE319 Introduction to Microelectronics

222009 Kenneth R. Laker, update 21Sep12

Constant Emitter Current BiasThe current mirror is used to create a current source:

1. A BJT collector is the cur-rent source:

I C= I S eV BEV T

I REF

I

2. A diode-connected transistor sets the current.

3. Choose Rref for the desiredcurrent:

Rref =V CC0.7

I REF

V CC0.7I C1

V BE2=V BE1

I O

I REF=I C1I B1I B2 I C1 I E1

I REF=V CCV BE1

RrefRref

V CC IC1

I B1 I B2

V CE1=V BE1

4. If Q1 = Q2 then IC2 = IC1 => I O I refmatched

I O=I C2

and VCE2 = VCE1

• ESE319 Introduction to Microelectronics

232009 Kenneth R. Laker, update 21Sep12

Constant Emitter Current

If Q1 and Q

2 have the same saturation current:

Now: VBE1 = VBE2I S1= I S2

And the transistors are at the same temperature: T 1=T 2

The two collector currents set primarily by Rref are equal, as long as Q2 is not saturated.