1 13

We will study

Symbols

Operating principles

Operating regions

for npn and pnp BJT

Bipolar Junction Transistors (BJT)

2 13

Simplified structure of a npn BJT

3 13

Symbolsnpn pnp

An ohmmeterrsquos An ohmmeterrsquos

view of transistor view of transistor

terminalsterminals

There are interactions between the 2 junctions

4 13

Terminal characteristics ofnpn BJT

iC

=βiB

Transfer characteristicInput characteristic

T

BE

V

v

SB e

Ii

β=

T

BE

V

v

SC eIi =

5 13

Output characteristics of a npn BJT IS=210

-15A and β=100

Active region

iC=βiB

Saturation regioniC ltβiBVCEsatasymp02V

Off region

iC=iB=0

6 13

Operating regions of npn BJT

(off)

vBElt06V vBClt06V

(aF)

vBEgt06V vBClt06V

(exc)

vBEgt06V vBCgt=06V

(aR)

rarely used

7 13

The currents through BJT

iE

=iC+i

B

In the active region (aF)

iC=βiB )1

1(1

ββ+=+= CCCE iiii

iE=(β+1)iB asympβiB

This relations does not hold in the saturation region(exc) where iCltβiB

8 13

Limiting the control current for a BJT

difference BJT ndashMOSFET junction in the control circuit B-E

one have to use a series resistance in order to establish (limit) the base current

9 13

BJT Saturation

The values of the resistors and voltages should be chosen such that the BJT would operate in the desired region

BJT can be seen also as a current-controled current source iC=βiB for the active region (aF)

βCex

Bsat

ii =

10 13

Exemplification Operating regions RB=50K i) vCo=04V ii) vCo=17V iii) vCo=5V

vCo=27V βisin(25hellip200)

RB domain so that T is in i) (aF)

ii) (exc)

i) because vCo=04V lt VTh=06V T-(off)

ii) vCo gtVTh rArr T in (aF) or in (exc)

Consider vBE=07V for conduction and β=100Assume T is in (aF) so that iC =βmiddotiB

We should compare iB with iCex β

If iB gtiCex β T - (exc) if iB ltiCex β T - (aF)

11 13

mA952

2012=

minus=

minus=

C

CEsatPS

CexR

vVi

02050

7071

R

vvi

B

BECoB =

minus=

minus= mA

0590100

95

iCex ==β

mA

Because iB=20microA microA rArr T is in (aF)59=ltβCexi

V6037870 ltminus=minus=minus= VvvvCEBEBC

mA2020100 =sdot=sdot= BC ii β

V82212 =sdotminus=sdotminus= CCAlCE iRVvOP

12 13

mA086050

705=

minus=

minus=

B

BECo

BR

vvi

iii)

Because iB =86microAgtiCex β=59mA results that T is in (exc)

vBE asymp08V vBC=vBEsat -vCEsat asymp08V-02V=06V=VTh

Alternatively we can solve supposing T in (aF) rArr iC=βmiddotiB=86mA

vCE=VPS - RCmiddotiC =12-2middot86=-52V

Obviously an impossible value (vCE can be only positive) so our supposition is false Thus T is in (exc)

13 13

b) i) For T in (aF) we must be sure

that iBltiCex β regardless the β

value in the specified range the

worst case β=βmax= 200

max

CexB

ii

βlt

max

Cex

B

BECo i

R

vv

βlt

minusΩ=

minussdot=

minussdotgt K867

95

7072200max

Cex

BECo

Bi

vvR β

ii) For the saturation the following condition must be fullfield

min

CexB

ii

βgt

min

Cex

B

BEsatCo i

R

vv

βgt

minus

Ω=minus

sdot=minus

sdotgt K2495

807275min

Cex

BEsatCo

Bi

vvR β

2 13

Simplified structure of a npn BJT

3 13

Symbolsnpn pnp

An ohmmeterrsquos An ohmmeterrsquos

view of transistor view of transistor

terminalsterminals

There are interactions between the 2 junctions

4 13

Terminal characteristics ofnpn BJT

iC

=βiB

Transfer characteristicInput characteristic

T

BE

V

v

SB e

Ii

β=

T

BE

V

v

SC eIi =

5 13

Output characteristics of a npn BJT IS=210

-15A and β=100

Active region

iC=βiB

Saturation regioniC ltβiBVCEsatasymp02V

Off region

iC=iB=0

6 13

Operating regions of npn BJT

(off)

vBElt06V vBClt06V

(aF)

vBEgt06V vBClt06V

(exc)

vBEgt06V vBCgt=06V

(aR)

rarely used

7 13

The currents through BJT

iE

=iC+i

B

In the active region (aF)

iC=βiB )1

1(1

ββ+=+= CCCE iiii

iE=(β+1)iB asympβiB

This relations does not hold in the saturation region(exc) where iCltβiB

8 13

Limiting the control current for a BJT

difference BJT ndashMOSFET junction in the control circuit B-E

one have to use a series resistance in order to establish (limit) the base current

9 13

BJT Saturation

The values of the resistors and voltages should be chosen such that the BJT would operate in the desired region

BJT can be seen also as a current-controled current source iC=βiB for the active region (aF)

βCex

Bsat

ii =

10 13

Exemplification Operating regions RB=50K i) vCo=04V ii) vCo=17V iii) vCo=5V

vCo=27V βisin(25hellip200)

RB domain so that T is in i) (aF)

ii) (exc)

i) because vCo=04V lt VTh=06V T-(off)

ii) vCo gtVTh rArr T in (aF) or in (exc)

Consider vBE=07V for conduction and β=100Assume T is in (aF) so that iC =βmiddotiB

We should compare iB with iCex β

If iB gtiCex β T - (exc) if iB ltiCex β T - (aF)

11 13

mA952

2012=

minus=

minus=

C

CEsatPS

CexR

vVi

02050

7071

R

vvi

B

BECoB =

minus=

minus= mA

0590100

95

iCex ==β

mA

Because iB=20microA microA rArr T is in (aF)59=ltβCexi

V6037870 ltminus=minus=minus= VvvvCEBEBC

mA2020100 =sdot=sdot= BC ii β

V82212 =sdotminus=sdotminus= CCAlCE iRVvOP

12 13

mA086050

705=

minus=

minus=

B

BECo

BR

vvi

iii)

Because iB =86microAgtiCex β=59mA results that T is in (exc)

vBE asymp08V vBC=vBEsat -vCEsat asymp08V-02V=06V=VTh

Alternatively we can solve supposing T in (aF) rArr iC=βmiddotiB=86mA

vCE=VPS - RCmiddotiC =12-2middot86=-52V

Obviously an impossible value (vCE can be only positive) so our supposition is false Thus T is in (exc)

13 13

b) i) For T in (aF) we must be sure

that iBltiCex β regardless the β

value in the specified range the

worst case β=βmax= 200

max

CexB

ii

βlt

max

Cex

B

BECo i

R

vv

βlt

minusΩ=

minussdot=

minussdotgt K867

95

7072200max

Cex

BECo

Bi

vvR β

ii) For the saturation the following condition must be fullfield

min

CexB

ii

βgt

min

Cex

B

BEsatCo i

R

vv

βgt

minus

Ω=minus

sdot=minus

sdotgt K2495

807275min

Cex

BEsatCo

Bi

vvR β

3 13

Symbolsnpn pnp

An ohmmeterrsquos An ohmmeterrsquos

view of transistor view of transistor

terminalsterminals

There are interactions between the 2 junctions

4 13

Terminal characteristics ofnpn BJT

iC

=βiB

Transfer characteristicInput characteristic

T

BE

V

v

SB e

Ii

β=

T

BE

V

v

SC eIi =

5 13

Output characteristics of a npn BJT IS=210

-15A and β=100

Active region

iC=βiB

Saturation regioniC ltβiBVCEsatasymp02V

Off region

iC=iB=0

6 13

Operating regions of npn BJT

(off)

vBElt06V vBClt06V

(aF)

vBEgt06V vBClt06V

(exc)

vBEgt06V vBCgt=06V

(aR)

rarely used

7 13

The currents through BJT

iE

=iC+i

B

In the active region (aF)

iC=βiB )1

1(1

ββ+=+= CCCE iiii

iE=(β+1)iB asympβiB

This relations does not hold in the saturation region(exc) where iCltβiB

8 13

Limiting the control current for a BJT

difference BJT ndashMOSFET junction in the control circuit B-E

one have to use a series resistance in order to establish (limit) the base current

9 13

BJT Saturation

The values of the resistors and voltages should be chosen such that the BJT would operate in the desired region

BJT can be seen also as a current-controled current source iC=βiB for the active region (aF)

βCex

Bsat

ii =

10 13

Exemplification Operating regions RB=50K i) vCo=04V ii) vCo=17V iii) vCo=5V

vCo=27V βisin(25hellip200)

RB domain so that T is in i) (aF)

ii) (exc)

i) because vCo=04V lt VTh=06V T-(off)

ii) vCo gtVTh rArr T in (aF) or in (exc)

Consider vBE=07V for conduction and β=100Assume T is in (aF) so that iC =βmiddotiB

We should compare iB with iCex β

If iB gtiCex β T - (exc) if iB ltiCex β T - (aF)

11 13

mA952

2012=

minus=

minus=

C

CEsatPS

CexR

vVi

02050

7071

R

vvi

B

BECoB =

minus=

minus= mA

0590100

95

iCex ==β

mA

Because iB=20microA microA rArr T is in (aF)59=ltβCexi

V6037870 ltminus=minus=minus= VvvvCEBEBC

mA2020100 =sdot=sdot= BC ii β

V82212 =sdotminus=sdotminus= CCAlCE iRVvOP

12 13

mA086050

705=

minus=

minus=

B

BECo

BR

vvi

iii)

Because iB =86microAgtiCex β=59mA results that T is in (exc)

vBE asymp08V vBC=vBEsat -vCEsat asymp08V-02V=06V=VTh

Alternatively we can solve supposing T in (aF) rArr iC=βmiddotiB=86mA

vCE=VPS - RCmiddotiC =12-2middot86=-52V

Obviously an impossible value (vCE can be only positive) so our supposition is false Thus T is in (exc)

13 13

b) i) For T in (aF) we must be sure

that iBltiCex β regardless the β

value in the specified range the

worst case β=βmax= 200

max

CexB

ii

βlt

max

Cex

B

BECo i

R

vv

βlt

minusΩ=

minussdot=

minussdotgt K867

95

7072200max

Cex

BECo

Bi

vvR β

ii) For the saturation the following condition must be fullfield

min

CexB

ii

βgt

min

Cex

B

BEsatCo i

R

vv

βgt

minus

Ω=minus

sdot=minus

sdotgt K2495

807275min

Cex

BEsatCo

Bi

vvR β

4 13

Terminal characteristics ofnpn BJT

iC

=βiB

Transfer characteristicInput characteristic

T

BE

V

v

SB e

Ii

β=

T

BE

V

v

SC eIi =

5 13

Output characteristics of a npn BJT IS=210

-15A and β=100

Active region

iC=βiB

Saturation regioniC ltβiBVCEsatasymp02V

Off region

iC=iB=0

6 13

Operating regions of npn BJT

(off)

vBElt06V vBClt06V

(aF)

vBEgt06V vBClt06V

(exc)

vBEgt06V vBCgt=06V

(aR)

rarely used

7 13

The currents through BJT

iE

=iC+i

B

In the active region (aF)

iC=βiB )1

1(1

ββ+=+= CCCE iiii

iE=(β+1)iB asympβiB

This relations does not hold in the saturation region(exc) where iCltβiB

8 13

Limiting the control current for a BJT

difference BJT ndashMOSFET junction in the control circuit B-E

one have to use a series resistance in order to establish (limit) the base current

9 13

BJT Saturation

The values of the resistors and voltages should be chosen such that the BJT would operate in the desired region

BJT can be seen also as a current-controled current source iC=βiB for the active region (aF)

βCex

Bsat

ii =

10 13

Exemplification Operating regions RB=50K i) vCo=04V ii) vCo=17V iii) vCo=5V

vCo=27V βisin(25hellip200)

RB domain so that T is in i) (aF)

ii) (exc)

i) because vCo=04V lt VTh=06V T-(off)

ii) vCo gtVTh rArr T in (aF) or in (exc)

Consider vBE=07V for conduction and β=100Assume T is in (aF) so that iC =βmiddotiB

We should compare iB with iCex β

If iB gtiCex β T - (exc) if iB ltiCex β T - (aF)

11 13

mA952

2012=

minus=

minus=

C

CEsatPS

CexR

vVi

02050

7071

R

vvi

B

BECoB =

minus=

minus= mA

0590100

95

iCex ==β

mA

Because iB=20microA microA rArr T is in (aF)59=ltβCexi

V6037870 ltminus=minus=minus= VvvvCEBEBC

mA2020100 =sdot=sdot= BC ii β

V82212 =sdotminus=sdotminus= CCAlCE iRVvOP

12 13

mA086050

705=

minus=

minus=

B

BECo

BR

vvi

iii)

Because iB =86microAgtiCex β=59mA results that T is in (exc)

vBE asymp08V vBC=vBEsat -vCEsat asymp08V-02V=06V=VTh

Alternatively we can solve supposing T in (aF) rArr iC=βmiddotiB=86mA

vCE=VPS - RCmiddotiC =12-2middot86=-52V

Obviously an impossible value (vCE can be only positive) so our supposition is false Thus T is in (exc)

13 13

b) i) For T in (aF) we must be sure

that iBltiCex β regardless the β

value in the specified range the

worst case β=βmax= 200

max

CexB

ii

βlt

max

Cex

B

BECo i

R

vv

βlt

minusΩ=

minussdot=

minussdotgt K867

95

7072200max

Cex

BECo

Bi

vvR β

ii) For the saturation the following condition must be fullfield

min

CexB

ii

βgt

min

Cex

B

BEsatCo i

R

vv

βgt

minus

Ω=minus

sdot=minus

sdotgt K2495

807275min

Cex

BEsatCo

Bi

vvR β

5 13

Output characteristics of a npn BJT IS=210

-15A and β=100

Active region

iC=βiB

Saturation regioniC ltβiBVCEsatasymp02V

Off region

iC=iB=0

6 13

Operating regions of npn BJT

(off)

vBElt06V vBClt06V

(aF)

vBEgt06V vBClt06V

(exc)

vBEgt06V vBCgt=06V

(aR)

rarely used

7 13

The currents through BJT

iE

=iC+i

B

In the active region (aF)

iC=βiB )1

1(1

ββ+=+= CCCE iiii

iE=(β+1)iB asympβiB

This relations does not hold in the saturation region(exc) where iCltβiB

8 13

Limiting the control current for a BJT

difference BJT ndashMOSFET junction in the control circuit B-E

one have to use a series resistance in order to establish (limit) the base current

9 13

BJT Saturation

The values of the resistors and voltages should be chosen such that the BJT would operate in the desired region

BJT can be seen also as a current-controled current source iC=βiB for the active region (aF)

βCex

Bsat

ii =

10 13

Exemplification Operating regions RB=50K i) vCo=04V ii) vCo=17V iii) vCo=5V

vCo=27V βisin(25hellip200)

RB domain so that T is in i) (aF)

ii) (exc)

i) because vCo=04V lt VTh=06V T-(off)

ii) vCo gtVTh rArr T in (aF) or in (exc)

Consider vBE=07V for conduction and β=100Assume T is in (aF) so that iC =βmiddotiB

We should compare iB with iCex β

If iB gtiCex β T - (exc) if iB ltiCex β T - (aF)

11 13

mA952

2012=

minus=

minus=

C

CEsatPS

CexR

vVi

02050

7071

R

vvi

B

BECoB =

minus=

minus= mA

0590100

95

iCex ==β

mA

Because iB=20microA microA rArr T is in (aF)59=ltβCexi

V6037870 ltminus=minus=minus= VvvvCEBEBC

mA2020100 =sdot=sdot= BC ii β

V82212 =sdotminus=sdotminus= CCAlCE iRVvOP

12 13

mA086050

705=

minus=

minus=

B

BECo

BR

vvi

iii)

Because iB =86microAgtiCex β=59mA results that T is in (exc)

vBE asymp08V vBC=vBEsat -vCEsat asymp08V-02V=06V=VTh

Alternatively we can solve supposing T in (aF) rArr iC=βmiddotiB=86mA

vCE=VPS - RCmiddotiC =12-2middot86=-52V

Obviously an impossible value (vCE can be only positive) so our supposition is false Thus T is in (exc)

13 13

b) i) For T in (aF) we must be sure

that iBltiCex β regardless the β

value in the specified range the

worst case β=βmax= 200

max

CexB

ii

βlt

max

Cex

B

BECo i

R

vv

βlt

minusΩ=

minussdot=

minussdotgt K867

95

7072200max

Cex

BECo

Bi

vvR β

ii) For the saturation the following condition must be fullfield

min

CexB

ii

βgt

min

Cex

B

BEsatCo i

R

vv

βgt

minus

Ω=minus

sdot=minus

sdotgt K2495

807275min

Cex

BEsatCo

Bi

vvR β

6 13

Operating regions of npn BJT

(off)

vBElt06V vBClt06V

(aF)

vBEgt06V vBClt06V

(exc)

vBEgt06V vBCgt=06V

(aR)

rarely used

7 13

The currents through BJT

iE

=iC+i

B

In the active region (aF)

iC=βiB )1

1(1

ββ+=+= CCCE iiii

iE=(β+1)iB asympβiB

This relations does not hold in the saturation region(exc) where iCltβiB

8 13

Limiting the control current for a BJT

difference BJT ndashMOSFET junction in the control circuit B-E

one have to use a series resistance in order to establish (limit) the base current

9 13

BJT Saturation

The values of the resistors and voltages should be chosen such that the BJT would operate in the desired region

BJT can be seen also as a current-controled current source iC=βiB for the active region (aF)

βCex

Bsat

ii =

10 13

Exemplification Operating regions RB=50K i) vCo=04V ii) vCo=17V iii) vCo=5V

vCo=27V βisin(25hellip200)

RB domain so that T is in i) (aF)

ii) (exc)

i) because vCo=04V lt VTh=06V T-(off)

ii) vCo gtVTh rArr T in (aF) or in (exc)

Consider vBE=07V for conduction and β=100Assume T is in (aF) so that iC =βmiddotiB

We should compare iB with iCex β

If iB gtiCex β T - (exc) if iB ltiCex β T - (aF)

11 13

mA952

2012=

minus=

minus=

C

CEsatPS

CexR

vVi

02050

7071

R

vvi

B

BECoB =

minus=

minus= mA

0590100

95

iCex ==β

mA

Because iB=20microA microA rArr T is in (aF)59=ltβCexi

V6037870 ltminus=minus=minus= VvvvCEBEBC

mA2020100 =sdot=sdot= BC ii β

V82212 =sdotminus=sdotminus= CCAlCE iRVvOP

12 13

mA086050

705=

minus=

minus=

B

BECo

BR

vvi

iii)

Because iB =86microAgtiCex β=59mA results that T is in (exc)

vBE asymp08V vBC=vBEsat -vCEsat asymp08V-02V=06V=VTh

Alternatively we can solve supposing T in (aF) rArr iC=βmiddotiB=86mA

vCE=VPS - RCmiddotiC =12-2middot86=-52V

Obviously an impossible value (vCE can be only positive) so our supposition is false Thus T is in (exc)

13 13

b) i) For T in (aF) we must be sure

that iBltiCex β regardless the β

value in the specified range the

worst case β=βmax= 200

max

CexB

ii

βlt

max

Cex

B

BECo i

R

vv

βlt

minusΩ=

minussdot=

minussdotgt K867

95

7072200max

Cex

BECo

Bi

vvR β

ii) For the saturation the following condition must be fullfield

min

CexB

ii

βgt

min

Cex

B

BEsatCo i

R

vv

βgt

minus

Ω=minus

sdot=minus

sdotgt K2495

807275min

Cex

BEsatCo

Bi

vvR β

7 13

The currents through BJT

iE

=iC+i

B

In the active region (aF)

iC=βiB )1

1(1

ββ+=+= CCCE iiii

iE=(β+1)iB asympβiB

This relations does not hold in the saturation region(exc) where iCltβiB

8 13

Limiting the control current for a BJT

difference BJT ndashMOSFET junction in the control circuit B-E

one have to use a series resistance in order to establish (limit) the base current

9 13

BJT Saturation

The values of the resistors and voltages should be chosen such that the BJT would operate in the desired region

BJT can be seen also as a current-controled current source iC=βiB for the active region (aF)

βCex

Bsat

ii =

10 13

Exemplification Operating regions RB=50K i) vCo=04V ii) vCo=17V iii) vCo=5V

vCo=27V βisin(25hellip200)

RB domain so that T is in i) (aF)

ii) (exc)

i) because vCo=04V lt VTh=06V T-(off)

ii) vCo gtVTh rArr T in (aF) or in (exc)

Consider vBE=07V for conduction and β=100Assume T is in (aF) so that iC =βmiddotiB

We should compare iB with iCex β

If iB gtiCex β T - (exc) if iB ltiCex β T - (aF)

11 13

mA952

2012=

minus=

minus=

C

CEsatPS

CexR

vVi

02050

7071

R

vvi

B

BECoB =

minus=

minus= mA

0590100

95

iCex ==β

mA

Because iB=20microA microA rArr T is in (aF)59=ltβCexi

V6037870 ltminus=minus=minus= VvvvCEBEBC

mA2020100 =sdot=sdot= BC ii β

V82212 =sdotminus=sdotminus= CCAlCE iRVvOP

12 13

mA086050

705=

minus=

minus=

B

BECo

BR

vvi

iii)

Because iB =86microAgtiCex β=59mA results that T is in (exc)

vBE asymp08V vBC=vBEsat -vCEsat asymp08V-02V=06V=VTh

Alternatively we can solve supposing T in (aF) rArr iC=βmiddotiB=86mA

vCE=VPS - RCmiddotiC =12-2middot86=-52V

Obviously an impossible value (vCE can be only positive) so our supposition is false Thus T is in (exc)

13 13

b) i) For T in (aF) we must be sure

that iBltiCex β regardless the β

value in the specified range the

worst case β=βmax= 200

max

CexB

ii

βlt

max

Cex

B

BECo i

R

vv

βlt

minusΩ=

minussdot=

minussdotgt K867

95

7072200max

Cex

BECo

Bi

vvR β

ii) For the saturation the following condition must be fullfield

min

CexB

ii

βgt

min

Cex

B

BEsatCo i

R

vv

βgt

minus

Ω=minus

sdot=minus

sdotgt K2495

807275min

Cex

BEsatCo

Bi

vvR β

8 13

Limiting the control current for a BJT

difference BJT ndashMOSFET junction in the control circuit B-E

one have to use a series resistance in order to establish (limit) the base current

9 13

BJT Saturation

The values of the resistors and voltages should be chosen such that the BJT would operate in the desired region

BJT can be seen also as a current-controled current source iC=βiB for the active region (aF)

βCex

Bsat

ii =

10 13

Exemplification Operating regions RB=50K i) vCo=04V ii) vCo=17V iii) vCo=5V

vCo=27V βisin(25hellip200)

RB domain so that T is in i) (aF)

ii) (exc)

i) because vCo=04V lt VTh=06V T-(off)

ii) vCo gtVTh rArr T in (aF) or in (exc)

Consider vBE=07V for conduction and β=100Assume T is in (aF) so that iC =βmiddotiB

We should compare iB with iCex β

If iB gtiCex β T - (exc) if iB ltiCex β T - (aF)

11 13

mA952

2012=

minus=

minus=

C

CEsatPS

CexR

vVi

02050

7071

R

vvi

B

BECoB =

minus=

minus= mA

0590100

95

iCex ==β

mA

Because iB=20microA microA rArr T is in (aF)59=ltβCexi

V6037870 ltminus=minus=minus= VvvvCEBEBC

mA2020100 =sdot=sdot= BC ii β

V82212 =sdotminus=sdotminus= CCAlCE iRVvOP

12 13

mA086050

705=

minus=

minus=

B

BECo

BR

vvi

iii)

Because iB =86microAgtiCex β=59mA results that T is in (exc)

vBE asymp08V vBC=vBEsat -vCEsat asymp08V-02V=06V=VTh

Alternatively we can solve supposing T in (aF) rArr iC=βmiddotiB=86mA

vCE=VPS - RCmiddotiC =12-2middot86=-52V

Obviously an impossible value (vCE can be only positive) so our supposition is false Thus T is in (exc)

13 13

b) i) For T in (aF) we must be sure

that iBltiCex β regardless the β

value in the specified range the

worst case β=βmax= 200

max

CexB

ii

βlt

max

Cex

B

BECo i

R

vv

βlt

minusΩ=

minussdot=

minussdotgt K867

95

7072200max

Cex

BECo

Bi

vvR β

ii) For the saturation the following condition must be fullfield

min

CexB

ii

βgt

min

Cex

B

BEsatCo i

R

vv

βgt

minus

Ω=minus

sdot=minus

sdotgt K2495

807275min

Cex

BEsatCo

Bi

vvR β

9 13

BJT Saturation

The values of the resistors and voltages should be chosen such that the BJT would operate in the desired region

BJT can be seen also as a current-controled current source iC=βiB for the active region (aF)

βCex

Bsat

ii =

10 13

Exemplification Operating regions RB=50K i) vCo=04V ii) vCo=17V iii) vCo=5V

vCo=27V βisin(25hellip200)

RB domain so that T is in i) (aF)

ii) (exc)

i) because vCo=04V lt VTh=06V T-(off)

ii) vCo gtVTh rArr T in (aF) or in (exc)

Consider vBE=07V for conduction and β=100Assume T is in (aF) so that iC =βmiddotiB

We should compare iB with iCex β

If iB gtiCex β T - (exc) if iB ltiCex β T - (aF)

11 13

mA952

2012=

minus=

minus=

C

CEsatPS

CexR

vVi

02050

7071

R

vvi

B

BECoB =

minus=

minus= mA

0590100

95

iCex ==β

mA

Because iB=20microA microA rArr T is in (aF)59=ltβCexi

V6037870 ltminus=minus=minus= VvvvCEBEBC

mA2020100 =sdot=sdot= BC ii β

V82212 =sdotminus=sdotminus= CCAlCE iRVvOP

12 13

mA086050

705=

minus=

minus=

B

BECo

BR

vvi

iii)

Because iB =86microAgtiCex β=59mA results that T is in (exc)

vBE asymp08V vBC=vBEsat -vCEsat asymp08V-02V=06V=VTh

Alternatively we can solve supposing T in (aF) rArr iC=βmiddotiB=86mA

vCE=VPS - RCmiddotiC =12-2middot86=-52V

Obviously an impossible value (vCE can be only positive) so our supposition is false Thus T is in (exc)

13 13

b) i) For T in (aF) we must be sure

that iBltiCex β regardless the β

value in the specified range the

worst case β=βmax= 200

max

CexB

ii

βlt

max

Cex

B

BECo i

R

vv

βlt

minusΩ=

minussdot=

minussdotgt K867

95

7072200max

Cex

BECo

Bi

vvR β

ii) For the saturation the following condition must be fullfield

min

CexB

ii

βgt

min

Cex

B

BEsatCo i

R

vv

βgt

minus

Ω=minus

sdot=minus

sdotgt K2495

807275min

Cex

BEsatCo

Bi

vvR β

10 13

Exemplification Operating regions RB=50K i) vCo=04V ii) vCo=17V iii) vCo=5V

vCo=27V βisin(25hellip200)

RB domain so that T is in i) (aF)

ii) (exc)

i) because vCo=04V lt VTh=06V T-(off)

ii) vCo gtVTh rArr T in (aF) or in (exc)

Consider vBE=07V for conduction and β=100Assume T is in (aF) so that iC =βmiddotiB

We should compare iB with iCex β

If iB gtiCex β T - (exc) if iB ltiCex β T - (aF)

11 13

mA952

2012=

minus=

minus=

C

CEsatPS

CexR

vVi

02050

7071

R

vvi

B

BECoB =

minus=

minus= mA

0590100

95

iCex ==β

mA

Because iB=20microA microA rArr T is in (aF)59=ltβCexi

V6037870 ltminus=minus=minus= VvvvCEBEBC

mA2020100 =sdot=sdot= BC ii β

V82212 =sdotminus=sdotminus= CCAlCE iRVvOP

12 13

mA086050

705=

minus=

minus=

B

BECo

BR

vvi

iii)

Because iB =86microAgtiCex β=59mA results that T is in (exc)

vBE asymp08V vBC=vBEsat -vCEsat asymp08V-02V=06V=VTh

Alternatively we can solve supposing T in (aF) rArr iC=βmiddotiB=86mA

vCE=VPS - RCmiddotiC =12-2middot86=-52V

Obviously an impossible value (vCE can be only positive) so our supposition is false Thus T is in (exc)

13 13

b) i) For T in (aF) we must be sure

that iBltiCex β regardless the β

value in the specified range the

worst case β=βmax= 200

max

CexB

ii

βlt

max

Cex

B

BECo i

R

vv

βlt

minusΩ=

minussdot=

minussdotgt K867

95

7072200max

Cex

BECo

Bi

vvR β

ii) For the saturation the following condition must be fullfield

min

CexB

ii

βgt

min

Cex

B

BEsatCo i

R

vv

βgt

minus

Ω=minus

sdot=minus

sdotgt K2495

807275min

Cex

BEsatCo

Bi

vvR β

11 13

mA952

2012=

minus=

minus=

C

CEsatPS

CexR

vVi

02050

7071

R

vvi

B

BECoB =

minus=

minus= mA

0590100

95

iCex ==β

mA

Because iB=20microA microA rArr T is in (aF)59=ltβCexi

V6037870 ltminus=minus=minus= VvvvCEBEBC

mA2020100 =sdot=sdot= BC ii β

V82212 =sdotminus=sdotminus= CCAlCE iRVvOP

12 13

mA086050

705=

minus=

minus=

B

BECo

BR

vvi

iii)

Because iB =86microAgtiCex β=59mA results that T is in (exc)

vBE asymp08V vBC=vBEsat -vCEsat asymp08V-02V=06V=VTh

Alternatively we can solve supposing T in (aF) rArr iC=βmiddotiB=86mA

vCE=VPS - RCmiddotiC =12-2middot86=-52V

Obviously an impossible value (vCE can be only positive) so our supposition is false Thus T is in (exc)

13 13

b) i) For T in (aF) we must be sure

that iBltiCex β regardless the β

value in the specified range the

worst case β=βmax= 200

max

CexB

ii

βlt

max

Cex

B

BECo i

R

vv

βlt

minusΩ=

minussdot=

minussdotgt K867

95

7072200max

Cex

BECo

Bi

vvR β

ii) For the saturation the following condition must be fullfield

min

CexB

ii

βgt

min

Cex

B

BEsatCo i

R

vv

βgt

minus

Ω=minus

sdot=minus

sdotgt K2495

807275min

Cex

BEsatCo

Bi

vvR β

12 13

mA086050

705=

minus=

minus=

B

BECo

BR

vvi

iii)

Because iB =86microAgtiCex β=59mA results that T is in (exc)

vBE asymp08V vBC=vBEsat -vCEsat asymp08V-02V=06V=VTh

Alternatively we can solve supposing T in (aF) rArr iC=βmiddotiB=86mA

vCE=VPS - RCmiddotiC =12-2middot86=-52V

Obviously an impossible value (vCE can be only positive) so our supposition is false Thus T is in (exc)

13 13

b) i) For T in (aF) we must be sure

that iBltiCex β regardless the β

value in the specified range the

worst case β=βmax= 200

max

CexB

ii

βlt

max

Cex

B

BECo i

R

vv

βlt

minusΩ=

minussdot=

minussdotgt K867

95

7072200max

Cex

BECo

Bi

vvR β

ii) For the saturation the following condition must be fullfield

min

CexB

ii

βgt

min

Cex

B

BEsatCo i

R

vv

βgt

minus

Ω=minus

sdot=minus

sdotgt K2495

807275min

Cex

BEsatCo

Bi

vvR β

13 13

b) i) For T in (aF) we must be sure

that iBltiCex β regardless the β

value in the specified range the

worst case β=βmax= 200

max

CexB

ii

βlt

max

Cex

B

BECo i

R

vv

βlt

minusΩ=

minussdot=

minussdotgt K867

95

7072200max

Cex

BECo

Bi

vvR β

ii) For the saturation the following condition must be fullfield

min

CexB

ii

βgt

min

Cex

B

BEsatCo i

R

vv

βgt

minus

Ω=minus

sdot=minus

sdotgt K2495

807275min

Cex

BEsatCo

Bi

vvR β