Lecture #11
Semiconductor Diodes and Basic Circuits
Outline/Learning Objectives:
• Simple circuits using ideal diode model, constant voltage drop model,
and mathematical (exponential) model.
• Use of graphical analysis, the load-line concept an iterative numerical
methods.
• Study diode rectifier, voltage-limiting, peak-detector, and clamping cir
cuits.
• Explain the temperature and breakdown behavior of a diode.
• Analyze and design basic ac/dc converters.
Lecture 11
11 - 1
Why Study Diodes and Brief Introduction to Diodes?
pn Junction Diode
The most common diode is the silicon p-n junction diode. It has non-linear I-V
characteristics which is responsible for many of its applications (see page 9 4 for 2 circuit examples).
1. Power rectifiers to convert ac to dc;
2. Signal detection in radio receivers;
3. Waveform clamp in digital circuits;
4. Photodetector; Solar cell; Light emitter etc.
It is also a principal part of other semiconductor devices,
1. in bipolar junction transistors, or
2. as the source/substrate junction of MOSFETs.
Lecture 11
11 - 2
Examples of 2 practical circuits using diodes
AM Diode Detection Circuit
IL
ac line
120Vrms
60 Hz
t
+
Diode
vS
Rectifier
t
Filter
t
+
Voltage V
Regulator L Load
t
t
Block diagram of a dc power supply
Lecture 11
11 - 3
DC Electrical Characteristics
The ideal diode model
R = 10 kΩ
ID
+
- V = 10V
+
-
VD > Von VD < Von
ID (mA)
1. Ideal Diode Model
1.0
Ideal
Diode
0.75
VD
0.5
Von = 0 V
Off
On
short
circuit
open
circuit
0.25
Von
1.0
VD
V D = 0 for I D > 0 and I D = 0 for V D < 0.
10 – 0
I D = --------------- = 1mA .
10kΩ
4
For V = -10 V, – 10V = V D – 10 I D = 0 since I D = 0 for V D < 0.
Lecture 11
11 - 4
Constant voltage drop diode model
2. Constant Voltage Drop Model
ID (mA)
R = 10 kΩ
ID
+
+
-
-
V = 10V
VD > Von VD < Von
1.0
Diode
VD
On
Off
0.75
0.5
Von
+
-
open
circuit
0.25
Von 1.0
VD
V D = V on for I D > 0 and I D = 0 for V D < V on .
10 – V on
I D = ---------------------- .
10kΩ
Lecture 11
11 - 5
Diode Current - Mathematical Model
i D = I i + I s = I s ⋅  e
qv D ⁄ nkT
– 1 .

Reverse saturation component (Is) is independent of the junction potential. Is
doubles for every 10o increase in temperature, near room temperature.
n - ideality factor (1 to 2) q - electron charge ( 1.6x10
vD - diode voltage
– 19
C)
k - Boltzmann constant  1.38x10
– 23 J 
---K
T - absolute temperature
nkT
Reverse biases: i D ≈ – I s for vD « ---------- = nV T . Zero V:
q
Forward biases: i D = I s ⋅ e
Lecture 11
qvD ⁄ nkT
iD = 0 .
for v D » nV T . (usually 4V T ≈ 0.1V )
11 - 6
Diode Temperature Coefficient
iD = Is ⋅  e

qv D ⁄ nkT
 i D
 iD
 kT
– 1 ⇒ v D = V T ⋅ ln  ----- + 1 ≈ ------- ⋅ ln  ----- .

q
 Is 
 Is

v D – V G0 – 3V T
dv D k
v D V T dI s
 i D kT 1 dI s
V
- – ------- ---- ⋅ ------- = ----- in ---- .
---------- = --- ⋅ ln  ---- – ------- ⋅ ------- = --------------------------------------T
dT
q
T
I s dT
K
 I s  q I s dT
2
3 – E G ⁄ kT
i D » I s ; I s ≈ n i and I s ≈ BT ⋅ e
.
For example, vD = 0.65 V, EG = 1.12 eV and VT = 0.025V.
v D – V G0 – 3V T
dv D
0.65 – 1.12 – 3 ( 0.025 )
mV
- = ------------------------------------------------------ = – 1.82 --------- .
---------- = --------------------------------------K
T
300
dT
EG
V G0 = -------q
Explain how it is used to make a thermometer.
Lecture 11
11 - 7
Diode Breakdown Under Reverse Bias
J = φJ + vR
( v R > 0 ).
for
wd = xn + xp =
2ε Si 1
1
----------  -------- + -------- ( φ J + v R )
q  N A N D
φJ(x)
+qND
vD < 0
Electric Field
vD = 0
-xp
xn
x
-xp
xn
x
φJ
-xp
xn
x
-qNA
Zener Breakdown - occurs only in heavily doped diods. The heavy doping
results in narrow depletion regions. Reverse bias causes tunneling between
CB and VB with I increasing rapidly. The breakdown voltage (VZ) can vary
from a few volts to 100’s of V. Zener diodes are used to maintain a constant V
for varying I’s. Used in voltage regulators. Also VZ increases as T decreases.
Lecture 11
11 - 8
Diode Breakdown Under Reverse Bias
ID
I
I
VBR VZ
Von
VD
I
+
+
V
-
-
I
RBR
-
V
Von
RZ
-
V
V
VBR Von
VZ
+
Von
Avalanche
+
Zener
Avalanche Breakdown - occurs at high voltages V > VZ. Here, carriers gain
sufficient energy from E-field to knock electron-hole pairs from covalent
bonds. Also VZ increases as T increases.
Lecture 11
11 - 9
Diode Circuit Analysis
1. Graphical analysis using load line.
2. Analysis with mathematical model of diode.
3. Simplified analysis using ideal diode model.
4. Simplified analysis using constant voltage drop model.
1. Graphical analysis using load line
1. Load-Line Analysis
1.0
R = 10 kΩ
quiescent point
(0.95 mA, 0.6V)
0.75
ID
+
- V = 10V
ID (mA)
+
-
load line
0.5
VD
0.25
Von 1.0
V = ID R + VD
2.0
VD
V
. V D = 0 ⇒ I D = ---- ; I D = 0 ⇒ V D = V
R
Quiescent point is the intersection of the diode’s I-V and the load line. This
gives the operating point of the circuit.
Lecture 11
11 - 10
PSPICE EXAMPLE
R2
R2
10k
10k
D1
D1
Dbreak
R1
Dbreak
U1
R1
+
1k
VOFF = 0
VAMPL = 2.0
FREQ = 1k
V1
-
OPAMP
+
-
1k
OUT
V1
VOFF = 0
VAMPL = 2.0
FREQ = 1k
E1
Rin
E
Rout
+
-
75
2MEG
*Libraries:
* Local Libraries :
* From [PSPICE NETLIST] section of C:\Program Files\OrcadLite\PSpice\PSpice.ini file:
.lib "nom.lib"
*Analysis directives:
.TRAN 0 10m 0 0.1u
.PROBE V(*) I(*) W(*) D(*) NOISE(*)
.INC ".\example4-SCHEMATIC1.net"
**** INCLUDING example4-SCHEMATIC1.net ****
* source EXAMPLE4
Lecture 11
11 - 11
PSPICE EXAMPLE (Cont’d)
D_D1
N00162 N00115 Dbreak
E_E1
0 N01347 N00162 0 200000
R_Rout
N01347 N00115 75
R_Rin
0 N00162 2MEG
V_V1
N00351 0
+SIN 0 2.0 1k 0 0 0
R_R2
N00162 N00115 10k
R_R1
N00351 N00162 1k
**** RESUMING example4-SCHEMATIC1-Example4Profile.sim.cir ****
.END
****
Diode MODEL PARAMETERS
************************************************************************
Dbreak
IS
RS
10.000000E-15
.1
CJO 100.000000E-15
Lecture 11
11 - 12
PSPICE EXAMPLE (Cont’d)
20V
10V
0V
-10V
0s
V(R2:2)
1ms
2ms
V(R1:1)
3ms
4ms
5ms
6ms
7ms
8ms
9ms
10ms
Time
Lecture 11
11 - 13