Diodes (RS)
-
Upload
shubham-thakur -
Category
Documents
-
view
219 -
download
2
description
Transcript of Diodes (RS)
-
1
Semiconductor
Diodes
-
Energy Band Gap in Materials
2
-
Extrinsic Semiconductors
3
N-type P-type
Typical doping level 1 part per 10 million
-
4
Semiconductor Diode
The semiconductor diode is formed by bringing p and n-type materials together (constructed from the same base Ge or Si)
Diode Symbol
-
P-N Junction (barrier) Voltage
5
-
6
Diode equation
where is the reverse saturation current
is the applied forward-bias voltage across the diode
is an which is a function of operating conditions
( 1)
ideality factor
D T
s
D
V nVsD
I
V
n
I I e
-23
T
is Boltzman constant=1.38 10 J/K
is the absolute temperature in Ke
and physical construction; it has a range between 1 &2.
(n=1 will be assumed unless otherwise noted)
V
k
T
kT
q
-19
lvins
is the electronic charge=1.6 10
For 0, and fo 0, r D TD DV nV
s sD D
q C
V VI I e I I
VT 0.26 V at room temperature
-
7
V-I characteristic of Diode
-
8
Temperature Effect
The reverse saturation current Is approximately doubles for every 10oC rise in temperature. If Is = Is1 at T = T1, then at temperature T2, Is2 is given by, Is2 = Is1x 2
(T2 T1)/10
-
9
PIV (Peak Inverse Voltage Rating)
The maximum reverse-bias potential that can be applied to the diode without damaging it or causing it to break down.
-
10
Diode Equivalent Circuit
Ideal Diode:
Piecewise Linear Model
-
11
Diode Equivalent Circuit
Simple Diode Model Typically Used:
VD
ID
VK
VK=0.7 V for Si
VK=0.3 V for Ge
VSVK I=(VS-VK)/R and VA= VS-VK
-
12
A more complicated model for a diode (Do not use this
for your circuit analysis)
Follows the actual diode characteristics better
-
What will be the states
(ON/OFF) of the two
diodes, D1 and D2?
Assume the diodes to
be ideal with a forward
voltage drop of 0.7 V.
Find the currents ID1
and ID2
Answer:
D1 ON & D2 OFF
ID1=0.953 mA ID2=0
Example 1
-
10 V
D1
D2
R1
4 K
R2
5 K
V1I1
I2 I3
V2
R3
1 K
For the circuit shown, find the
voltages V1 and V2 and the
currents I1, I2 and I3. Assume
the diodes to be ideal with a
forward voltage drop of 0.7 V.
Answer:
V1=9.3 V V2=7.53 V
I1=1.51 mA I2 =1.07 mA
I3=0.44 mA
Example 2
-
15
Zener Diode
Zener diodes are special diodes manufactured
with adequate power dissipation capabilities to
operate in the breakdown region.
Symbol:
Equivalent ckt. Approx. Eq. ckt.
-
16
V-I Characteristics of Zener Diode
-
17
Zener Regulator
Loaded Zener regulator
Vs>Vz for breakdown Rs is the current limiting resistance to limit the zener current to less than its maximum rating
Is=(Vs Vz)/Rs
Vth = voltage across zener when it is not in breakdown = Vs x RL/(Rs+RL) For breakdown, Vth>Vz
IL = VL/RL = VZ/RL Iz = Is IL Power dissipated by zener diode = VzIz Is=(Vs Vz)/Rs, Is = Iz + IL
-
18
Zener Regulator contd.
Loaded Zener regulator
Typically, for a Zener Diode, one would specify the zener voltage VZ and the maximum power dissipation PZ,max in the zener diode.
Is=(Vs Vz)/Rs, Is = Iz + IL
Additionally, we may also
specify the minimum zener
current IZ,min that must flow
through the zener diode to
provide the zener action