1 / 13 We will study: Symbols Operating principles Operating regions for npn and pnp BJT Bipolar Junction Transistors (BJT)
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