Transcript Bipolar Junction Transistors (BJT)

Bipolar Junction Transistors (BJT)
NPN
PNP
BJT Cross-Sections
Emitter
Collector
NPN
PNP
Common-Emitter NPN Transistor
Reverse bias the CBJ
Forward bias the BEJ
Input Characteristics
• Plot IB as f(VBE, VCE)
• As VCE increases,
more VBE required to
turn the BE on so that
IB>0.
• Looks like a pn
junction volt-ampere
characteristic.
Output Characteristics
• Plot IC as f(VCE, IB)
• Cutoff region (off)
– both BE and BC
reverse biased
• Active region
– BE Forward biased
– BC Reverse biased
• Saturation region (on)
– both BE and BC
forward biased
Transfer Characteristics
Large-Signal Model of a BJT
KCL >> IE = IC + IB
βF = hFE = IC/IB
IC = βFIB + ICEO
IE = IB(1 + βF) + ICEO
IE = IB(1 + βF)
IE = IC(1 + 1/βF)
IE = IC(βF + 1)/βF
I E  I B  IC
 F  hFE
IC

IB
I C   F I B  I CEO
I E  I B (1   F )  I CEO  I B (  F  1)

1
I E  I C 1 
 F
IC   F I E
F

F 1
  IC
F

F
F 
 F 
F 1
1F
Transistor Operating Point
VB  VBE
IB 
RB
VCE VCC
IC  

RC
RC
VCE  VCC  I C RC
DC Load Line
VCC/RC
VCC
BJT Transistor Switch
VB  VBE
IB 
RB
VCE  VCC  I C RC
VCE  VCB  VBE
VCB  VCE  VBE
BJT Transistor Switch (continued)
I CM
VCC  VCE VCC  VBE


RC
RC
I BM 
I CM
F
BJT in Saturation
I CS 
I BS 
VCC  VCE ( sat )
RC
I CS
F
IB
ODF 
I BS
 forced
I CS

IB
Model with Current Gain
Miller Effect
vbe
iout
vce
Miller Effect (continued)
d
d
iout  Ccb (vbe  vce )  Ccb (vbe  Avbe )
dt
dt
d
d
iout  Ccb [1  A]vbe   Ccb [1  A]  vbe 
dt
dt
Ccb  Ccb [1  A]
Miller Effect (continued)
• Miller Capacitance, CMiller = Ccb(1 – A)
– since A is usually negative (phase inversion),
the Miller capacitance can be much greater
than the capacitance Ccb
• This capacitance must charge up to the
base-emitter forward bias voltage, causing
a delay time before any collector current
flows.
Saturating a BJT
• Normally apply more base current than
needed to saturate the transistor
• This results in charges being stored in the
base region
• To calculate the extra charge (saturating
charge), determine the emitter current
Ie  I B 
I cs

 ODF I BS  I BS  I BS  ODF  1
The Saturating Charge
• The saturating charge, Qs
Qs   s I e   s I BS (ODF  1)
storage time constant of the
transistor
Transistor Switching Times
Switching Times – turn on
• Input voltage rises from 0 to V1
• Base current rises to IB1
• Collector current begins to rise after the
delay time, td
• Collector current rises to steady-state
value ICS
• This “rise time”, tr allows the Miller
capacitance to charge to V1
• turn on time, ton = td + tr
Switching Times – turn off
• Input voltage changes from V1 to –V2
• Base current changes to –IB2
• Base current remains at –IB2 until the
Miller capacitance discharges to zero,
storage time, ts
• Base current falls to zero as Miller
capacitance charges to –V2, fall time, tf
• turn off time, toff = ts + tf
Charge Storage in Saturated BJTs
Charge storage in the Base
Charge Profile during turn-off
Example 4.2
Waveforms for the Transistor Switch
VCC = 250 V
VBE(sat) = 3 V
IB = 8 A
VCS(sat) = 2 V
ICS = 100 A
td = 0.5 µs
tr = 1 µs
ts = 5 µs
tf = 3 µs
fs = 10 kHz
duty cycle k = 50 %
ICEO = 3 mA
Power Loss due to IC for ton = td + tr
• During the delay time, 0 ≤t ≤td
• Instantaneous Power Loss
Pc (t )  vCE iC  VCC I CEO
Pc (t )  (250V )(3mA)  0.75W
• Average Power Loss
td
td
VCC I CEO
1
Pd   Pc (t )dt 
dt  VCC I CEO f std

T 0
T
0
Pd  (250V )(3mA)(10kHz )(0.5 s)  3.75mW
During the rise time, 0 ≤t ≤tr
Pc (t )  vCE ic

t  I CS
Pc (t )  VCC  (Vce ( sat )  VCC ) 
t
tr  tr


dPc (t ) Vce ( sat )  VCC I CS
t  I CS

t  VCC  (Vce ( sat )  VCC ) 
dt
tr
tr
tr  tr

Pc (t )  Pmax @ t  tm
trVCC
tm 
2[VCC  Vce ( sat ) ]
(1 s )(250V )
tm 
 0.504 s
2[250V  2V ]
Pmax
2
CC CS
V I

4[VCC  VCE ( sat ) ]
2
Pmax
(250V ) (100 A)

 6300W
4[250V  2V ]
Average Power during rise time
VCC VCE ( sat )  VCC 
1
Pr   Pc (t )dt  f s I CS tr 


T 0
3
 2

tr
 (250V ) (2V  250V ) 
Pr  (10kHz )(100 A)(1 s ) 


3
 2

Pr  42.33W
Total Power Loss during turn-on
Pon  Pd  Pr
Pon  0.00375  42.33  42.33375W
Pon  42.33W
Power Loss during
the Conduction Period
0  t  tn
ic (t )  I CS  100 A
vCE (t )  VCE ( sat )  2V
Pc (t )  ic vCE  (100 A)(2V )  200W
tn
tn
1
Pn   Pc (t )dt  VCE ( sat ) I CS f s  dt  VCE ( sat ) I CS f stn
T 0
0
Pn  (2V )(100 A)(10kHz )(48.5 s )  97W
Power Loss during turn off
Storage time
0  t  ts
ic (t )  I CS  100 A
vCE (t )  VCE ( sat )  2V
Pc (t )  vCE ic  VCE ( sat ) I CS  (2V )(100 A)
Pc (t )  200W
ts
ts
1
Ps   Pc (t )dt  VCE ( sat ) I CS f s  dt  VCE ( sat ) I CS f s t s
T 0
0
Ps  (2V )(100 A)(10kHz )(5 s )  10W
Power Loss during Fall time
0  t  tf

t 
ic (t )  I CS 1   , I CEO  0
 tf 


V
vCE (t )  CC t , I CEO  0
tf
Pc (t )  vCE ic  VCC I CS

t
1 
 t f
 t

 tf



 1  
t 
   t  1     0
 t f   t f  
t f 3 s
Pc (t )  Pm @ t  
 1.5 s
2
2
V I
(250V )(100 A)
Pm  CC CS 
 6250W
4
4
dPc (t ) VCC I CS

dt
tf
Power Loss during Fall time (continued)
tf
VCC I CS t f f s
1
Pf   Pc (t )dt 
T 0
6
(250V )(100 A)(3 s )(10kHz )
Pf 
 125W
6
VCC t f 

Poff  Ps  Pf  I CS f s  t sVCE ( sat ) 

6 

Poff  10  125  135W
Power Loss during the off time
0  t  to
vCE (t )  VCC
ic (t )  I CEO
Pc (t )  vCE iC  VCC I CEO  (250V )(3mA)  0.75W
to
1
Po   VCC I CEO dt  VCC I CEO f s to
T 0
Po  (250V )(3mA)(10kHz )((50  5  3)  s )
Po  0.315W
The total average power losses
PT  Pon  Pn  Poff  Po
PT  42.33  97  135  0.315
PT  274.65W
Instantaneous Power for Example 4.2
BJT Switch with an Inductive Load
Load Lines