The document discusses single diode circuits and diode equations. It provides the following key points: - A conducting diode has a voltage drop of about 0.7V for silicon and 0.3V for germanium. To forward bias a diode, the anode must be more positive than the cathode. To reverse bias, the anode must be less positive. - The diode equation relates diode current (iD) to voltage (vD) and includes parameters like reverse saturation current (IS), temperature, and the nonideality factor. - An example circuit is solved using Kirchhoff's law to find the diode voltage (VD) and current (iD) at the operating point, calculating

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