AERO
ASTRO
16.682 - Prototyping Avionics
Spring 2006
LECTURE 6
February 27, 2006
DEPARTMENT OF AERONAUTICS AND ASTRONAUTICS
Alvar Saenz-Otero
Outline
•
Transistors
– Introduction
– NPN/PNP
– MOSFETs
•
Digital Logic Introduction
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16.682 - Prototyping Avionics
Transistor Introduction
• We want an electronic switch which allows us to open or close a
circuit…
– The diode could do that if we reversed voltage:
V+
V+
V+
“short”
“on”
D
V-
+
v >v
V-
“open”
“off”
V-
−
+
v < v −
– But can we do it by commanding a separate voltage, instead of the input
voltage?
• That is what a transistor is for:
C
C
B
iBE
vB B
vE
E
iCE
C
B
iBE
“short”
“on”
E
“open”
“off”
E
vB > vE → iCE >> iBE
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iCE
vB < vE → iCE = iBE = 0
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16.682 - Prototyping Avionics
Bipolar Transistors
• PNP and NPN Bipolar Transistors
– Current “amplification” components
•
iC = αiB
α ~ 40-200 E
PNP:
B
P
N
P
– Have three connections:
C
• Base - he “control” of the transistor
• Emitter - the connection that emits current
• Collector - the connect ion that collects current
– Have a “diode” between the Base and the
C
Emitter
NPN:
• Determines polarity: PNP or NPN • Turns “on” when current can fl ow t hrough the B
N
P
N
E
diode • The currentt is amplified bet ween the collector and the emitter Massachusetts Institute of Technology
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16.682 - Prototyping Avionics
Bipolar Transistors
•
Example with an NPN transistor: Vin
R1
If Vin>0.6V and
the current
source can
provide enough
iBE, then...
Q
NPN
Q
NPN
iB
Cutoff
vBE = vin − 0 = vin < 0.6V
GND
vBE = 0.6V
iBE
v − 0.6V
= in
R1
iCE = α ⋅iBE , vC ~ 0
•
R1
Vin
GND
But if the current
tries to flow the
wrong way, the
transistor is “off”
0.6V
iC
VBE
Saturation
Notes:
–
The diode dictates a maximum
–
The voltage drop between Cand E is usually 0.2V
–
The current amplification is usually determined by specific parts
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een B& Eof 0.6V
5
0.2V
VCE
16.682 - Prototyping Avionics
Transistor Use
Vpwr
• D
riving a high-current, separate voltage
Driving
voltage load
R2
100
– Vin does not have to be the same as Vpwr
– A large Vin saturates VBE
vE = 0
Vin
vin = 3.3V
R1
2k
Q
NPN
iB = (3.3V − 0.6V ) / 2k = 1.35ma
GND
iC ≈ 100ma
•
Negative input (inverse signal)
Vpwr
– Must be able to drive Vin@ Vpwr
Vin
– Vin= Vpwrmeans transistor is OFF
– Vin= 0V means transistor is ON
R1
2k
Q
PNP
R2
100
GND
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16.682 - Prototyping Avionics
Transistor Use
•
Vpwr
Current Source
–
⎛ R2 ⎞
⎟⎟
vB = Vin ⎜⎜
⎝ R1 + R2 ⎠
vE ≅ vB − 0.6V
iref =
iL
vB − 0.6V
Rref
R1
if iB ≈ 0 (α = 100) → iL =i ref
iB
– Example: Rref=10Ω, VB=1.6V
vB = 1.6V
iR
iref
Rref
vB − 0.6V
1
=
= 100mA
Rref
10
iB ≈ 1mA (Need R1 &R2 such that i B ≈ 0 WRT i R )
• Note limits: VB=10.6V then iB=100mA, too large!
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Q
NPN
R2
vE ≅ 1.0V
iref =
RL
Vin
7
GND
GND
16.682 - Prototyping Avionics
MOSFET Transistors
•
“Voltage” Transistors
– FET = Field-effect transistor =
“voltage” transistor
D
Q
MOSFET-N
G
– Gate/Source/Drain instead of BCE
S
– Voltage turns transistor on/off
• iSD=αvDS
• Need very small current (almost 0) at the gate
– VGS= 1 -> ON
S
– VGS= 0 -> OFF
G
Q
MOSFET-P
D
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16.682 - Prototyping Avionics
MOSFETS in use
•
Driving many “lines” from a single voltage
– Use small current
Q
MOSFET-N
Vin
Q
MOSFET-N
Q
MOSFET-N
Q
MOSFET
– Same ground, may change power
•
“CMOS” logic (Complementary MOS)
Vdd
– Example to create AND and OR circuits:
Out
In1
In1
0
V
0
V
In2
0
0
V
V
i
Off
Off
Off
On
In1
Q
MOSFET-N
Q
MOSFET-N
In2
Q
MOSFET-N
Out
In2
In1
0
V
0
V
Q
MOSFET-N
GND
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In2
0
0
V
V
i
Off
On
On
On
16.682 - Prototyping Avionics
Digital Logic
•
Basic Gates
– AND
– OR
A
B
Q
0
0
0
1
0
0
0
1
0
1
1
1
• Both inputs must be true
A
0
1
0
1
B
0
0
1
1
Q
0
1
1
1
• Any input is true
A
B
Q
– NOR
0
0
1
0
1
0
0
1
0
1
1
0
A
B
Q
1
0
0
1
1
0
0
1
1
1
1
0
• Neither input must be true
– NAND
• Neither input is true
– XOR
A
B
Q
0
0
0
1
0
1
0
1
1
1
1
0
• The inputs are different
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16.682 - Prototyping Avionics