This document discusses single diode circuits and the diode equation. It provides: 1) The diode equation that relates diode current (iD) to voltage (vD) through saturation current (IS), temperature, and other constants. 2) Typical values for the variables in the diode equation. 3) An example problem that demonstrates calculating diode voltage and current given circuit specifications by applying Kirchhoff's law and the diode equation.

1. By Unsa Shakir
SINGLE DIODE CIRCUITS

2. 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

3. 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. The nonideality factor is typically
close to 1, but approaches 2 for devices with high current densities. It is
generally taken as 1.

4. 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=0.68V and iDQ=0.93mA
Repeat for:
Vss=10V and R=10kW

5. 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
With IS=10 then
With IS=0.1 then , ID= 1 mA
V718.0)
A16-10
A4-103V)ln(10025.0(11ln 










S
I
D
I
T
nV
D
V
V603.0
D
V
V748.0
D
V

6. 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







7. Solve the Example:
DD VRIV 
find the diode voltage and current at the operating point using
kirchoff’s law.

8. Example

9. Example:
S
X
TXDXX
I
I
VRIVRIV ln11 
V1formA2.0
V3formA2.2


XX
XX
VI
VI

10. 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





















14. Diode Voltages
A conducting diode has about 0.6 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.