Activity:
 Revise all information about semiconductor materials, p- and n-type doping, p-n
junction, Ohm’s and Kirchhoff’s circuit laws on the basis of your knowledge taken
before
Contents:
1. Semiconductor diode
The semiconductor p-n junction allows the current to flow only in one direction and
blocks it in the opposite direction. In the n type semiconductor layer the movements of the
electrons and in the p type semiconductor layer the movements of the holes make the current.
Applying positive voltage on the p type layer and at the same time negative voltage on
the n type layer:
electrons enter p type layer
and
holes enter n type layer
the diode is ON
(current flows through the
junction even at low voltage
level)
Applying negative voltage on the p type layer and at the same time positive voltage
on the n type layer:
the number of the carriers
(electrons and holes) is very
limited
the diode is OFF
(the current is very small, only
10-6-10-10 A)
The device model of the semiconductor diode contains an ideal diode, a DC voltage
source (Uo) and a serial resistor (Rs).
The ideal diode is shortcut in forward direction. The value of the running current can
be arbitrarily small or large while the voltage dropout is zero. The maximum current that can
flow through the junction depends on the surface size of the chip (die). Approximately 2
A/mm2 can be the maximum current on the chip.
At reverse bias the ideal semiconductor diode has zero leakage current.
The ideal diode characteristics
The ’cut-in’ voltage (U0) of a diode is defined as UD voltage when the diode current is
one out of ten of the maximum current applied on the diode
 in case of Ge diodes 0.2 - 0.4 V
 in case of Si diodes 0.5 - 0.8 V
 in case of Schottky diodes 0.3 V
Further on we will suppose that the ’cut-in’ voltage of Si semiconductor
diodes equals to 0.6 V, but we know it is only a simple approximation.
The serial resistance of a semiconductor diode depends on manufacturing technology.
Its value is 1-10 ohm, but if we apply epitaxial layer, we can reach one out of ten of it.
Considering the facts above the voltage-current characteristic of a diode can be
approached with the following linear sections
The voltage-current characteristic of a diode approached with linear sections
A voltage-current characteristic of a real semiconductor diode changes according to
the measurements as it can be seen on the figure below:
A voltage-current characteristic of a real semiconductor diode
 Forward characteristic
At forward bias between the voltage (UD) and its current (ID) of the diode the
following empirical formula can be given:
I D  I S (e
UD
mU T
 1)
IS is the theoretical saturation current of the diode. Its value equals to 100 nA in case
of Ge diodes, and it equals about 10 pA in case of Si diodes.
The value of ’m’ varies between 1 and 2. It shows how far the features of our diodes
are from the ideal ones given by the Shockley theory.
UT is called by thermal voltage. We can determine it with the following formula
UT 
kT
q
where
k is the so called Boltzmann constant. It equals to 1.38 . 10-23 J/K or VAs/K
T is the temperature given on Kelvin scale
q is the elementary charge, its value is 1.6 . 10-19 Coulomb.
The formula mentioned above gives 26 mV in room temperature.
Since IS and UT depend on the temperature, in case of constant ID current the UD
voltage will also depend on the temperature. By increasing the temperature by degrees UD
voltage of the diode will decrease by 2-3 mV on constant ID current. Since this statement is
true on a wide range of temperature so the semiconductor diode can be used as a good
thermometer or sensor.
 Reverse characteristic
The reverse or leakage current of the diode follows the increase of the reverse voltage
on almost steady value. Theoretically it is the saturation current IS of the diode, but the real
value is much higher.
On higher temperatures this saturation current will be increased. This current will be
doubled if temperature increases 10 degrees step by step.
If the reverse voltage increases further to a defined limit, much current increase can be
experienced even if the voltage changes only a little. This voltage limit is called reverse
breakdown voltage of the diode.
The value of this reverse breakdown voltage can be adjusted in advance with the help
of the technological steps of the manufacturing. It highly depends on the concentration and
the profile of the dopant material.
The relationship between the p-n junction profile and the breakdown voltage is used to
make the so called Zener diodes. The breakdown voltage can be interpreted by tunnel effect
below 10 V, and by avalanche breakdown above 10 V. These two effects often work together
and they can hardly be separated from each other.
The value of the reverse breakdown voltage can reach 10 kV nowadays.
 Capacities of a semiconductor diode
At forward bias the diffusion capacity is accumulated in the p-n junction. The volume
of the diffusion capacity is nearly proportional to the current running through the junction and
it can even reach 10-100 nF. It is called diffusion capacity. Remove the diffusion capacity
from the p-n junction take time. So this capacity is disadvantageous if we want the diode to
switch fast.
At reverse bias the width of the p-n junction varies depending on the reverse voltage.
In this case the carriers (electrons and holes) move away from each other. So the p-n junction
is behaving like a capacity. It is called junction capacitance. Its value is about some pF. This
characteristic of the diode can be used to control the frequency of the oscillators by this
reverse voltage. These special diodes are called varicap diodes.
The different types of diodes and their symbols can be seen in the following table:
diode (general)
Schottky diode
Zener diode
Tunnel diode
varicap diode
light-emitting diode (LED)
photodiode (device sensitive to light or
other radiation)
silicon controlled rectifier
transient voltage suppression diode
Activity:
 Look for an example to apply for the different diode types above, at least one for each.
EXAMPLES:
1.1.
At room temperature what current flows through the silicon diode, the forward voltage of
which is 0.65 V?
(Data: m = 1.45; IS = 12 pA)
Solution:
Here is the following, well-known formula between the forward voltage (UD) and the current
(ID) of the silicon diode:
I D  I S  (e
UD
m U T
 1)
UT = 26 mV at room temperature, so after substituting data given above we get the following
result for the current of the diode:
I D  12 10
12
 (e
0.65V
1.4526mV
 1) A  0.369mA
On the conditions given in this case the current, 0.369 mA is expected on the diode.
1.2.
What voltage can be measured on the silicon diode, on which the flowing current is 1 mA at
20 C?
(Data: m = 1.42; IS [T =20C] = 18 pA)
Solution:
The formula between the forward voltage and the current of the diode will be solved on UD.
In this case we get:
U D  m U T  ln(
ID
I
k T
 1)  m 
 ln( D  1)
IS
q
IS
After substituting:
U D  1.42 
k  (273.15  20)
1mA
 ln(
 1)  1.42  0.02528V 17.83  0.640V
q
18 pA
The forward voltage on the diode will be 0.64 V, when the current is 1 mA and the
temperature is 20 C.
Examples for circuits:
1.A.
Define the current of the diode in the following circuit:
(D1 is a silicon semiconductor diode with U0= 0.6 V)
On all silicon diodes the voltage drops about 0.6 V at forward bias. We can measure
5 V – 0.6 V = 4.4 V
on the resistor R1. In this case the current is 4.4 V/ 1kΩ = 4.4 mA on both resistance R1 and
diode D1.
1.B.
What voltage can be expected approximately on the point P of the following circuit?
(D1 - D2 - D3 silicon semiconductor diodes with U0 = 0.6 V)
On all three diodes the voltage drops 0.6 V at forward bias.
That is why the voltage on the point P is 15 V – 0.6 V – 0.6 V – 0.6 V = 13.2 V
1.C.
What current flows through the diodes in the circuit below?
(D1 and D2 are silicon semiconductor diodes with U0= 0.6 V)
R_gen, D1, D2 and R_load are placed in serial order in the circuit. That is why the current
(ID) flowing on them is the same. So
U gen  Rgen  I D  U D1  U D 2  Rload  I D
3V  220  I D  0.6V  0.6V  860  I D
1.8V 1080  I D
I D  1.666mA
The current expected on the diode equals 1.666 mA.
Exercises for simulations:
1.a.
Assemble the following circuit on SPICE Schematics. By simulating define the value of
R_gen if the current is 1 mA on the diode D1.