Transistors. - physics.wm.eduphysics.wm.edu/~evmik/classes/Physics_252_Analog...Transistors invented...
Transcript of Transistors. - physics.wm.eduphysics.wm.edu/~evmik/classes/Physics_252_Analog...Transistors invented...
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Transistors.
Eugeniy E. Mikhailov
The College of William & Mary
Week 6
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Transistors
invented in 1947amplify currentlower power consumptioncheap for mass productionrobust to vibrationlong working time (decades) when properly usedreplaced vacuum tubebuilding block of modern electronics
Some areas where vacuum tube are still goodultra high voltage applications (more then 1000 V)radiation prone locations
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Bipolar junction Transistor (BJT)
NPN-transistor
N
N
PBase
Collector
Emitter
B
C
E
B
C
E
PNP-transistor
P
P
NBase
Collector
Emitter
B
C
E
B
C
E
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Notation
Base-emitter current (Ibe)Collector-emitter current (Ice)Base-emitter voltage difference(Vbe = Vb − Ve)Collector-emitter voltage difference(Vce = Vc − Ve)
B
C
E
Ice
Ibe
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Simple NPN-transistor rules
To support shown currents direction
Vce > 0Vbe > 0
since, it is forward biased diode Vbe ≈ 0.6 VVcb > 0
since, it is reversed biased diode, no currentgoes from collector to base, all collector currentis directed to emitterif Vcb < 0 transistor goes to saturation andcannot be described by the following simplerule.
If above holds true thenIce = βIbe thus a BJT is a current amplifierthe static forward current transfer ratioβ (or sometimes hfe) ≈ 100 . . . 200Ie = Ibe + Ice = (β + 1)Ibe ≈ βIbe
B
C
E
Ice
Ibe
B
C
E
Eugeniy Mikhailov (W&M) Electronics 1 Week 6 5 / 12
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Simple NPN-transistor rules
To support shown currents directionVce > 0
Vbe > 0since, it is forward biased diode Vbe ≈ 0.6 V
Vcb > 0since, it is reversed biased diode, no currentgoes from collector to base, all collector currentis directed to emitterif Vcb < 0 transistor goes to saturation andcannot be described by the following simplerule.
If above holds true thenIce = βIbe thus a BJT is a current amplifierthe static forward current transfer ratioβ (or sometimes hfe) ≈ 100 . . . 200Ie = Ibe + Ice = (β + 1)Ibe ≈ βIbe
B
C
E
Ice
Ibe
B
C
E
Eugeniy Mikhailov (W&M) Electronics 1 Week 6 5 / 12
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Simple NPN-transistor rules
To support shown currents directionVce > 0Vbe > 0
since, it is forward biased diode Vbe ≈ 0.6 V
Vcb > 0since, it is reversed biased diode, no currentgoes from collector to base, all collector currentis directed to emitterif Vcb < 0 transistor goes to saturation andcannot be described by the following simplerule.
If above holds true thenIce = βIbe thus a BJT is a current amplifierthe static forward current transfer ratioβ (or sometimes hfe) ≈ 100 . . . 200Ie = Ibe + Ice = (β + 1)Ibe ≈ βIbe
B
C
E
Ice
Ibe
B
C
E
Eugeniy Mikhailov (W&M) Electronics 1 Week 6 5 / 12
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Simple NPN-transistor rules
To support shown currents directionVce > 0Vbe > 0
since, it is forward biased diode Vbe ≈ 0.6 VVcb > 0
since, it is reversed biased diode, no currentgoes from collector to base, all collector currentis directed to emitterif Vcb < 0 transistor goes to saturation andcannot be described by the following simplerule.
If above holds true thenIce = βIbe thus a BJT is a current amplifierthe static forward current transfer ratioβ (or sometimes hfe) ≈ 100 . . . 200Ie = Ibe + Ice = (β + 1)Ibe ≈ βIbe
B
C
E
Ice
Ibe
B
C
E
Eugeniy Mikhailov (W&M) Electronics 1 Week 6 5 / 12
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Simple NPN-transistor rules
To support shown currents directionVce > 0Vbe > 0
since, it is forward biased diode Vbe ≈ 0.6 VVcb > 0
since, it is reversed biased diode, no currentgoes from collector to base, all collector currentis directed to emitterif Vcb < 0 transistor goes to saturation andcannot be described by the following simplerule.
If above holds true then
Ice = βIbe thus a BJT is a current amplifierthe static forward current transfer ratioβ (or sometimes hfe) ≈ 100 . . . 200Ie = Ibe + Ice = (β + 1)Ibe ≈ βIbe
B
C
E
Ice
Ibe
B
C
E
Eugeniy Mikhailov (W&M) Electronics 1 Week 6 5 / 12
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Simple NPN-transistor rules
To support shown currents directionVce > 0Vbe > 0
since, it is forward biased diode Vbe ≈ 0.6 VVcb > 0
since, it is reversed biased diode, no currentgoes from collector to base, all collector currentis directed to emitterif Vcb < 0 transistor goes to saturation andcannot be described by the following simplerule.
If above holds true thenIce = βIbe thus a BJT is a current amplifier
the static forward current transfer ratioβ (or sometimes hfe) ≈ 100 . . . 200Ie = Ibe + Ice = (β + 1)Ibe ≈ βIbe
B
C
E
Ice
Ibe
B
C
E
Eugeniy Mikhailov (W&M) Electronics 1 Week 6 5 / 12
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Simple NPN-transistor rules
To support shown currents directionVce > 0Vbe > 0
since, it is forward biased diode Vbe ≈ 0.6 VVcb > 0
since, it is reversed biased diode, no currentgoes from collector to base, all collector currentis directed to emitterif Vcb < 0 transistor goes to saturation andcannot be described by the following simplerule.
If above holds true thenIce = βIbe thus a BJT is a current amplifierthe static forward current transfer ratioβ (or sometimes hfe) ≈ 100 . . . 200
Ie = Ibe + Ice = (β + 1)Ibe ≈ βIbe
B
C
E
Ice
Ibe
B
C
E
Eugeniy Mikhailov (W&M) Electronics 1 Week 6 5 / 12
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Simple NPN-transistor rules
To support shown currents directionVce > 0Vbe > 0
since, it is forward biased diode Vbe ≈ 0.6 VVcb > 0
since, it is reversed biased diode, no currentgoes from collector to base, all collector currentis directed to emitterif Vcb < 0 transistor goes to saturation andcannot be described by the following simplerule.
If above holds true thenIce = βIbe thus a BJT is a current amplifierthe static forward current transfer ratioβ (or sometimes hfe) ≈ 100 . . . 200Ie = Ibe + Ice = (β + 1)Ibe ≈ βIbe
B
C
E
Ice
Ibe
B
C
E
Eugeniy Mikhailov (W&M) Electronics 1 Week 6 5 / 12
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Simple PNP-transistor rules
Apply the same rules as before for NPN BJTbut multiply currents and voltages by -1.Hints
the arrow indicates the direction in whichcurrent is supposed to flow.the arrow always connects the base andemitter.
B
C
E
Ice
Ibe
B
C
E
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Design considerations for β
Remember β is not a constant!It depends on many parameters
temperaturecollector currentvaries from device to device even in the same batch
Good design should not depend on β value.
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Constant current source
A
Vcc
Rset
RL
Vctrl
E
CB
Current through the load resistor does notdepend on the load resistance.
IL = Ic = βIbe = βVctrl − .6V
Rset
This is actually a sample of bad design sincethe current through the load depends on β.
Vc = Vcc − RLIL
remember that Vc must be > Vb thus currentcannot be bigger then the saturation current
Isat = max(IL) ≤Vcc − Vb
RL≈ Vcc
RL
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Constant current source
A
Vcc
Rset
RL
Vctrl
E
CB
Current through the load resistor does notdepend on the load resistance.
IL = Ic = βIbe = βVctrl − .6V
Rset
This is actually a sample of bad design sincethe current through the load depends on β.
Vc = Vcc − RLIL
remember that Vc must be > Vb thus currentcannot be bigger then the saturation current
Isat = max(IL) ≤Vcc − Vb
RL≈ Vcc
RL
Eugeniy Mikhailov (W&M) Electronics 1 Week 6 8 / 12
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Constant current source
A
Vcc
Rset
RL
Vctrl
E
CB
Current through the load resistor does notdepend on the load resistance.
IL = Ic = βIbe = βVctrl − .6V
Rset
This is actually a sample of bad design sincethe current through the load depends on β.
Vc = Vcc − RLIL
remember that Vc must be > Vb thus currentcannot be bigger then the saturation current
Isat = max(IL) ≤Vcc − Vb
RL≈ Vcc
RL
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Constant current source
A
Vcc
Rset
RL
Vctrl
E
CB
Current through the load resistor does notdepend on the load resistance.
IL = Ic = βIbe = βVctrl − .6V
Rset
This is actually a sample of bad design sincethe current through the load depends on β.
Vc = Vcc − RLIL
remember that Vc must be > Vb thus currentcannot be bigger then the saturation current
Isat = max(IL) ≤Vcc − Vb
RL≈ Vcc
RL
Eugeniy Mikhailov (W&M) Electronics 1 Week 6 8 / 12
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Constant current source (continued)
A
Vcc
Rset
RL
Vctrl
E
CB
From Vcc point of view, left schematic isequivalent to the right one.
Rtrans =Vc
IL=
Vcc − ILRL
IL
TransistorTran(sform)-(re)sistor
A
Vcc
RL
E
CRtrans
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Constant current source. Power dissipation.
Transistor power dissipation
A
Vcc
Rset
RL
Vctrl
E
CB
Ptrans = Pbe + Pce = VbeIbe + VceIce
SinceVbe ≤ Vce , Ibe = Ice/β � Ice, and Ice = IL
Ptrans ≈ VceIce = RtransI2L
Maximum power dissipation in transistoris when Rtrans = RL
max(Ptrans) =V 2
cc4RL
, when IL =Vcc
2RL
A
Vcc
RL
E
CRtrans
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Constant current source. Power dissipation.
Transistor power dissipation
A
Vcc
Rset
RL
Vctrl
E
CB
Ptrans = Pbe + Pce = VbeIbe + VceIce
SinceVbe ≤ Vce , Ibe = Ice/β � Ice, and Ice = IL
Ptrans ≈ VceIce = RtransI2L
Maximum power dissipation in transistor
is when Rtrans = RL
max(Ptrans) =V 2
cc4RL
, when IL =Vcc
2RL
A
Vcc
RL
E
CRtrans
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Constant current source. Power dissipation.
Transistor power dissipation
A
Vcc
Rset
RL
Vctrl
E
CB
Ptrans = Pbe + Pce = VbeIbe + VceIce
SinceVbe ≤ Vce , Ibe = Ice/β � Ice, and Ice = IL
Ptrans ≈ VceIce = RtransI2L
Maximum power dissipation in transistoris when Rtrans = RL
max(Ptrans) =V 2
cc4RL
, when IL =Vcc
2RL
A
Vcc
RL
E
CRtrans
Eugeniy Mikhailov (W&M) Electronics 1 Week 6 10 / 12
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Constant current source. Power dissipation.
Transistor power dissipation
A
Vcc
Rset
RL
Vctrl
E
CB
Ptrans = Pbe + Pce = VbeIbe + VceIce
SinceVbe ≤ Vce , Ibe = Ice/β � Ice, and Ice = IL
Ptrans ≈ VceIce = RtransI2L
Maximum power dissipation in transistoris when Rtrans = RL
max(Ptrans) =V 2
cc4RL
, when IL =Vcc
2RL
A
Vcc
RL
E
CRtrans
Eugeniy Mikhailov (W&M) Electronics 1 Week 6 10 / 12
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Voltage controlled switch
When properly designed outcome does not depend on reasonablevariations of β
RL
RbVctrl
Bulb
Vcc
Recall that typically β = 100 . . . 200We will assume the worst case scenario β = 10Notice that RL limits collector current
IL =Vcc
RL
Ibe =Vctrl − .6V
Rb=
ILβ
Rb ≤Vctrl − .6V
VccβRL
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Emitter follower
Vcc
RL
Vin
Vout
Vout = Vin − 0.6V
Gain. What gain?We achieved the input impedance increase.
Rinput =Vin
Ibe≈ RL(β + 1)
As result our Vin source is not overloaded andour load receive all required current (as longas the collector power supply can support it).
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Emitter follower
Vcc
RL
Vin
Vout
Vout = Vin − 0.6V
Gain. What gain?
We achieved the input impedance increase.
Rinput =Vin
Ibe≈ RL(β + 1)
As result our Vin source is not overloaded andour load receive all required current (as longas the collector power supply can support it).
Eugeniy Mikhailov (W&M) Electronics 1 Week 6 12 / 12
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Emitter follower
Vcc
RL
Vin
Vout
Vout = Vin − 0.6V
Gain. What gain?We achieved the input impedance increase.
Rinput =Vin
Ibe≈ RL(β + 1)
As result our Vin source is not overloaded andour load receive all required current (as longas the collector power supply can support it).
Eugeniy Mikhailov (W&M) Electronics 1 Week 6 12 / 12