• Lecture 6: Diodes
ECEN 1400 Introduction to Analog and Digital Electronics
Lecture 6
Diodes
•
•
•
•
•
•
What does it do (I/V curve)?
Hydraulic analogy
The actual physics
Anodes and cathodes
Use: rectification
Light emitting diodes (LEDs)
Robert R. McLeod, University of Colorado
70
• Lecture 6: Diodes
ECEN 1400 Introduction to Analog and Digital Electronics
I/V curve of the diode
I
+
V
Ideal
-
Threshold voltage
I
I
V>0
Short circuit
V>VT
Short circuit
V
V<0
Open circuit
Finite resistance
Actual nonlinearV>V
I
I
V>VT
Low resistance
V<VT
Open circuit
V
V<VT
Open circuit
V
T
Low resistance
V~VT
Nonlinear
V<VT
Leakage current
V
V<VBD
Breakdown
Robert R. McLeod, University of Colorado
http://en.wikipedia.org/wiki/Diode_modelling
71
• Lecture 6: Diodes
ECEN 1400 Introduction to Analog and Digital Electronics
Hydraulic analogy
I/V curve
I
Ball valve
4
3
2
V
I
1
+
V
1. Breakdown: Strong reverse voltage
collapses flap, reducing resistance to almost
zero.
2. Reverse bias: Moderate reverse voltage
closes valve, passing only very small leakage
current in reverse direction.
3. Turn-on: Forward bias overcomes
stiffness of valve to open it. Opening
changes with voltage, causing nonlinear
resistance.
4. Forward bias: A + voltage larger than
threshold fully opens valve, allowing large
current with only small resistance.
Robert R. McLeod, University of Colorado
http://en.wikipedia.org/wiki/Hydraulic_analogy
72
• Lecture 6: Diodes
ECEN 1400 Introduction to Analog and Digital Electronics
The physics
I
•  Semiconductors such as silicon can
be doped to have free electrons (Ntype) or free holes (P-type).
•  When N and P type regions are
placed in contact, the excess
electrons in the N-type material
diffuse over and fill the holes in the
P-type region, creating an
insulating depletion region.
•  This regions acts as the “valve”.
•  The charge in the depletion region
establishes a voltage which is about
0.7V for silicon.
+
V
2. Reverse bias
Hole
Electron
3. Turn-on
•  Reverse bias (V<0) pushes more
electrons and holes across the
junction to make the depletion
region larger. Little current flows.
•  At turn-on, a forward bias (V>0)
begins to force new holes from the
anode (P-type) side and electrons
from the cathode (N-type) side into
the depletion region, thinning it.
Some current begins to flow.
•  In forward bias (V>VT), the
depletion regions thins to zero and
a large current can flow.
4. Forward bias
Hole
Electron
Robert R. McLeod, University of Colorado http://www.allaboutcircuits.com/vol_3/chpt_3/1.html
73
• Lecture 6: Diodes
ECEN 1400 Introduction to Analog and Digital Electronics
Anodes and cathodes
Nomenclature
A
n
t
n
t
ode = Current flows i
Ca hode = Current flows ou
Current flow
Anode
Cathode
LED terminals
http://creativec0d1ng.blogspot.com/
Anode is
longer
Diode terminals
Bar on package
looks like bar
on symbol =
cathode.
http://en.wikipedia.org/wiki/File:Diode_pinout_en_fr.svg
Robert R. McLeod, University of Colorado
74
• Lecture 6: Diodes
ECEN 1400 Introduction to Analog and Digital Electronics
Why might you care?
Half-wave rectifier
Output
Vdiode
1
2
VSource
1
Idealized
operation
2
+
-
Robert R. McLeod, University of Colorado
+
V~Vsourde
-
V~0
75
• Lecture 6: Diodes
ECEN 1400 Introduction to Analog and Digital Electronics
Light emitting diodes
Symbol
Forward current turns LED on
Physics
Holes and electrons
combine to emit light.
http://en.wikipedia.org/wiki/Light-emitting_diode
Circuit use
A major consideration is limiting the current to avoid damage.
Typical forward limits are 20-30 mA and 2.5 V.
Series
Estimate that each diode
n e e d s 2 V, c a l c u l a t e
remaining voltage drop
required across resistor.
Choose I = 15 mA. A 9 V
source would then require
Small variations in
turn-on will cause one
LED to pull most of
the current.
R = (9V − 3 × 2V ) 15mA = 200Ω
http://www.kpsec.freeuk.com/components/led.htm
Half-wave rectifier
used to protect an
LED from excess
reverse bias:
http://www.allaboutcircuits.com/vol_3/chpt_3/12.html
Robert R. McLeod, University of Colorado
76
• Lecture 6: Diodes
ECEN 1400 Introduction to Analog and Digital Electronics
Quiz 6.1
Simple diode circuit
V1
+
V2
-
+
-
Q: In the two circuits above, the 12
volt supply has the opposite polarity.
V1 and V2 across the resistor are
approximately:
Left: Diode is forward
biased, so is
A: V1 ≈ 11 V, V2 ≈ 0 V
conducting with a
drop, so V ≈
B: V1 ≈ 0 V, V2 ≈ -11 V ~0.7V
12-0.7. Right: Diode
is reverse biased, so is
C: V1 ≈ 12 V, V2 ≈ 0 V
nearly an open. No
current flows, so all
drops across
D: V1 ≈ 12 V, V2 ≈ -0.7 V voltage
diode, none across
Alternatively,
E: V1 ≈ 12 V, V2 ≈ -12 V resistor.
V ≈IR=0
1
2
Robert R. McLeod, University of Colorado
77
• Lecture 6: Diodes
ECEN 1400 Introduction to Analog and Digital Electronics
Quiz 6.2
Current limiting an LED
Q: The forward voltage drop across a
blue LED is ~4V. What resistance,
R1, is needed to limit the LED current
to 20 mA?
A: R1 ≈ 100 Ω
B: R1 ≈ 200 Ω
C: R1 ≈ 400 Ω
R = (12V − 4V ) 20mA = 400Ω
D: R1 ≈ 800 Ω
E: R1 ≈ 1600 Ω
1
Robert R. McLeod, University of Colorado
78
• Lecture 6: Diodes
ECEN 1400 Introduction to Analog and Digital Electronics
Quiz 6.3
Power vs. current in an LED
Q: Consider an LED that is on
(emitting light) with 1 mA of forward
current. The current is doubled to 2
mA. The optical power now emitted
by the LED approximately…
A: Reduces by a factor of 2.
B: Increases by a factor of 2.
C: Increases by a factor of 4.
D: Is zero because the LED is
damaged.
E: There is not enough info to say.
The voltage stays roughly constant (unlike a resistor), so the power only doubles.
In a resistor, power quadruples because of the Ohm’s law IV relation.
Robert R. McLeod, University of Colorado
79