Single diode circuits

1. By Unsa Shakir
SINGLE DIODE CIRCUITS

2. Diode Voltages
A conducting diode has about 0.7 volts across if silicon, 0.3 volts if germanium.
To forward bias a
diode, the anode must
be more positive than
the cathode or LESS
NEGATIVE.
To reverse bias a
diode, the anode must
be less positive than
the cathode or MORE
NEGATIVE.

3. Diode Equation
iD  IS exp
qvD
nkT





1





 IS exp
vD
nVT





1






q
kT
V
nV
v
Ii
T
T
d
sD













 1exp
n ~ 1-2
VT ~ 25-26 mV

4. where
IS = reverse saturation current (A)
vD = voltage applied to diode (V)
q = electronic charge (1.60 x 10-19 C)
k = Boltzmann’s constant (1.38 x 10-23 J/K)
T = absolute temperature
n = nonideality factor (dimensionless)
VT = kT/q = thermal voltage (V)
IS is typically between 10-18 and 10-9 A, and is strongly temperature
dependent due to its dependence on ni
2. it is typically taken as 10^-14 A
for silicon and 10^-4 A for germanium. The nonideality factor is typically
close to 1, but approaches 2 for devices with high current densities. It is
generally taken as 1.

5. Analysis of Diode Circuits
DDss vRiV 
Problem
If the circuit shown below has Vss=2V and R=1kW and a diode as
shown, find the diode voltage and current at the operating point using
kirchoff’s law.
VDQ=0.639V and iDQ=1.3mA

6. Diode Voltage and Current
Calculations (Example)
Problem: Find diode voltage for diode with given specifications
Given data: IS=0.1 & ID= 300 mA
Assumptions: Room-temperature dc operation with VT= 0.025 V
Analysis:
With IS=0.1 then Vd = 3.465mV
With IS=10 then Vd = 0.738mV
With IS=0.1 then , ID= 1 mA Vd = 0.248mV

7. Diode Current for Reverse, Zero, and
Forward Bias
 Reverse bias:
 Zero bias:
 Forward bias:

iD  IS exp
vD
nVT





1





 IS 0 1  IS

iD  IS exp
vD
nVT





1





 IS 11  0

iD  IS exp
vD
nVT





1





 IS exp
vD
nVT







8. Solve the Example:
DD VRIV 
find the diode voltage and current at the operating point using
kirchoff’s law.
VDQ=0.631V and iDQ=1.3mAA

9. Example:
S
X
TXDXX
I
I
VRIVRIV ln11 
V1formA2.0
V3formA2.2


XX
XX
VI
VI Vd = 0.653V
Vd = 0.5929

10. Example

11. mAIVV
VenAkV
enAI
nnAI
eII
VIRV
DD
D
mV
V
mV
V
D
S
nkT
qV
SD
DD
D
D
D
35.4and648.0
011.1415
11.14
specified.notif(300K)uretemperatroomAssume
D1N4002afor98.1and1.14
1
015
3.51
3.51









W





















12. Id = 0.43mA
Vd = 0.636V