Transistors.
Eugeniy E. Mikhailov
The College of William & Mary
Week 6
Eugeniy Mikhailov (W&M)
Electronics 1
Week 6
1 / 12
Transistors
invented in 1947
amplify current
lower power consumption
cheap for mass production
robust to vibration
long working time (decades) when properly used
replaced vacuum tube
building block of modern electronics
Some areas where vacuum tube are still good
ultra high voltage applications (more then 1000 V)
radiation prone locations
Eugeniy Mikhailov (W&M)
Electronics 1
Week 6
2 / 12
Bipolar junction Transistor (BJT)
NPN-transistor
PNP-transistor
Collector
C
N
Base
Collector
C
P
B
Base
P
N
N
P
E
Emitter
B
E
Emitter
C
C
B
B
E
Eugeniy Mikhailov (W&M)
E
Electronics 1
Week 6
3 / 12
Notation
Base-emitter current (Ibe )
C
Collector-emitter current (Ice )
Base-emitter voltage difference
(Vbe = Vb − Ve )
Collector-emitter voltage difference
(Vce = Vc − Ve )
Ice
B
Ibe
E
Eugeniy Mikhailov (W&M)
Electronics 1
Week 6
4 / 12
Simple NPN-transistor rules
To support shown currents direction
C
Ice
B
Ibe
E
C
B
E
Eugeniy Mikhailov (W&M)
Electronics 1
Week 6
5 / 12
Simple NPN-transistor rules
To support shown currents direction
Vce > 0
C
Ice
B
Ibe
E
C
B
E
Eugeniy Mikhailov (W&M)
Electronics 1
Week 6
5 / 12
Simple NPN-transistor rules
To support shown currents direction
Vce > 0
Vbe > 0
C
since, it is forward biased diode Vbe ≈ 0.6 V
Ice
B
Ibe
E
C
B
E
Eugeniy Mikhailov (W&M)
Electronics 1
Week 6
5 / 12
Simple NPN-transistor rules
To support shown currents direction
Vce > 0
Vbe > 0
C
since, it is forward biased diode Vbe ≈ 0.6 V
Vcb > 0
since, it is reversed biased diode, no current
goes from collector to base, all collector current
is directed to emitter
if Vcb < 0 transistor goes to saturation and
cannot be described by the following simple
rule.
Ice
B
Ibe
E
C
B
E
Eugeniy Mikhailov (W&M)
Electronics 1
Week 6
5 / 12
Simple NPN-transistor rules
To support shown currents direction
Vce > 0
Vbe > 0
C
since, it is forward biased diode Vbe ≈ 0.6 V
Vcb > 0
since, it is reversed biased diode, no current
goes from collector to base, all collector current
is directed to emitter
if Vcb < 0 transistor goes to saturation and
cannot be described by the following simple
rule.
Ice
B
Ibe
E
C
B
If above holds true then
E
Eugeniy Mikhailov (W&M)
Electronics 1
Week 6
5 / 12
Simple NPN-transistor rules
To support shown currents direction
Vce > 0
Vbe > 0
C
since, it is forward biased diode Vbe ≈ 0.6 V
Vcb > 0
since, it is reversed biased diode, no current
goes from collector to base, all collector current
is directed to emitter
if Vcb < 0 transistor goes to saturation and
cannot be described by the following simple
rule.
Ice
B
Ibe
E
C
B
If above holds true then
Ice = βIbe thus a BJT is a current amplifier
Eugeniy Mikhailov (W&M)
Electronics 1
E
Week 6
5 / 12
Simple NPN-transistor rules
To support shown currents direction
Vce > 0
Vbe > 0
C
since, it is forward biased diode Vbe ≈ 0.6 V
Vcb > 0
since, it is reversed biased diode, no current
goes from collector to base, all collector current
is directed to emitter
if Vcb < 0 transistor goes to saturation and
cannot be described by the following simple
rule.
Ice
B
Ibe
E
C
B
If above holds true then
Ice = βIbe thus a BJT is a current amplifier
E
the static forward current transfer ratio
β (or sometimes hfe ) ≈ 100 . . . 200
Eugeniy Mikhailov (W&M)
Electronics 1
Week 6
5 / 12
Simple NPN-transistor rules
To support shown currents direction
Vce > 0
Vbe > 0
C
since, it is forward biased diode Vbe ≈ 0.6 V
Vcb > 0
since, it is reversed biased diode, no current
goes from collector to base, all collector current
is directed to emitter
if Vcb < 0 transistor goes to saturation and
cannot be described by the following simple
rule.
Ice
B
Ibe
E
C
B
If above holds true then
Ice = βIbe thus a BJT is a current amplifier
E
the static forward current transfer ratio
β (or sometimes hfe ) ≈ 100 . . . 200
Ie = Ibe + Ice = (β + 1)Ibe ≈ βIbe
Eugeniy Mikhailov (W&M)
Electronics 1
Week 6
5 / 12
Simple PNP-transistor rules
Apply the same rules as before for NPN BJT
but multiply currents and voltages by -1.
Hints
the arrow indicates the direction in which
current is supposed to flow.
the arrow always connects the base and
emitter.
C
Ice
B
Ibe
E
C
B
E
Eugeniy Mikhailov (W&M)
Electronics 1
Week 6
6 / 12
Design considerations for β
Remember β is not a constant!
It depends on many parameters
temperature
collector current
varies from device to device even in the same batch
Good design should not depend on β value.
Eugeniy Mikhailov (W&M)
Electronics 1
Week 6
7 / 12
Constant current source
Current through the load resistor does not
depend on the load resistance.
Vcc
IL = Ic = βIbe = β
Vctrl − .6V
Rset
RL
A
Vctrl
Rset
B
C
E
Eugeniy Mikhailov (W&M)
Electronics 1
Week 6
8 / 12
Constant current source
Current through the load resistor does not
depend on the load resistance.
Vcc
RL
IL = Ic = βIbe = β
Vctrl − .6V
Rset
This is actually a sample of bad design since
the current through the load depends on β.
A
Vctrl
Rset
B
C
E
Eugeniy Mikhailov (W&M)
Electronics 1
Week 6
8 / 12
Constant current source
Current through the load resistor does not
depend on the load resistance.
Vcc
RL
IL = Ic = βIbe = β
This is actually a sample of bad design since
the current through the load depends on β.
A
Vctrl
Rset
B
Vctrl − .6V
Rset
Vc = Vcc − RL IL
C
E
Eugeniy Mikhailov (W&M)
Electronics 1
Week 6
8 / 12
Constant current source
Current through the load resistor does not
depend on the load resistance.
Vcc
RL
IL = Ic = βIbe = β
This is actually a sample of bad design since
the current through the load depends on β.
A
Vctrl
Rset
B
Vc = Vcc − RL IL
C
E
remember that Vc must be > Vb thus current
cannot be bigger then the saturation current
Isat = max(IL ) ≤
Eugeniy Mikhailov (W&M)
Vctrl − .6V
Rset
Electronics 1
Vcc − Vb
Vcc
≈
RL
RL
Week 6
8 / 12
Constant current source (continued)
From Vcc point of view, left schematic is
equivalent to the right one.
Vcc
Vcc
Rtrans
RL
A
Vctrl
Rset
B
Vcc − IL RL
Vc
=
=
IL
IL
C
RL
A
C
Rtrans
E
E
Transistor
Tran(sform)-(re)sistor
Eugeniy Mikhailov (W&M)
Electronics 1
Week 6
9 / 12
Constant current source. Power dissipation.
Transistor power dissipation
Ptrans = Pbe + Pce = Vbe Ibe + Vce Ice
Vcc
Since
Vbe ≤ Vce , Ibe = Ice /β Ice , and Ice = IL
Vcc
RL
RL
A
Vctrl
Rset
B
C
Ptrans ≈ Vce Ice =
Rtrans IL2
A
C
Rtrans
E
Eugeniy Mikhailov (W&M)
E
Electronics 1
Week 6
10 / 12
Constant current source. Power dissipation.
Transistor power dissipation
Ptrans = Pbe + Pce = Vbe Ibe + Vce Ice
Vcc
Since
Vbe ≤ Vce , Ibe = Ice /β Ice , and Ice = IL
Vcc
RL
RL
Ptrans ≈ Vce Ice =
A
Vctrl
Rset
B
C
E
Rtrans IL2
Maximum power dissipation in transistor
Eugeniy Mikhailov (W&M)
Electronics 1
A
C
Rtrans
E
Week 6
10 / 12
Constant current source. Power dissipation.
Transistor power dissipation
Ptrans = Pbe + Pce = Vbe Ibe + Vce Ice
Vcc
Since
Vbe ≤ Vce , Ibe = Ice /β Ice , and Ice = IL
Vcc
RL
RL
Ptrans ≈ Vce Ice =
A
Vctrl
Rset
B
C
E
Rtrans IL2
Maximum power dissipation in transistor
is when Rtrans = RL
Eugeniy Mikhailov (W&M)
Electronics 1
A
C
Rtrans
E
Week 6
10 / 12
Constant current source. Power dissipation.
Transistor power dissipation
Ptrans = Pbe + Pce = Vbe Ibe + Vce Ice
Vcc
Since
Vbe ≤ Vce , Ibe = Ice /β Ice , and Ice = IL
Vcc
RL
RL
Ptrans ≈ Vce Ice =
A
Vctrl
Rset
B
C
E
Rtrans IL2
Maximum power dissipation in transistor
is when Rtrans = RL
max(Ptrans ) =
Eugeniy Mikhailov (W&M)
A
C
Rtrans
E
2
Vcc
Vcc
, when IL =
4RL
2RL
Electronics 1
Week 6
10 / 12
Voltage controlled switch
When properly designed outcome does not depend on reasonable
variations of β
Recall that typically β = 100 . . . 200
We will assume the worst case scenario β = 10
Vcc
Notice that RL limits collector current
Bulb
RL
IL =
Vctrl Rb
Eugeniy Mikhailov (W&M)
Vcc
RL
Ibe =
Vctrl − .6V
IL
=
Rb
β
Rb ≤
Vctrl − .6V
βRL
Vcc
Electronics 1
Week 6
11 / 12
Emitter follower
Vcc
Vout = Vin − 0.6V
Vin
Vout
RL
Eugeniy Mikhailov (W&M)
Electronics 1
Week 6
12 / 12
Emitter follower
Vcc
Vout = Vin − 0.6V
Gain. What gain?
Vin
Vout
RL
Eugeniy Mikhailov (W&M)
Electronics 1
Week 6
12 / 12
Emitter follower
Vcc
Vout = Vin − 0.6V
Gain. What gain?
We achieved the input impedance increase.
Vin
Vout
RL
Eugeniy Mikhailov (W&M)
Rinput =
Vin
≈ RL (β + 1)
Ibe
As result our Vin source is not overloaded and
our load receive all required current (as long
as the collector power supply can support it).
Electronics 1
Week 6
12 / 12