Devices and Applications
Ctec 201.
Semi-Conductor Diodes
Supplement
Prepared by Mike Crompton. (Rev. 25 July 2003)
2
P.N. Junctions
Semi-conductors have a valence of 4. The only semi-conductors of concern to us at
present are carbon, silicon and germanium. Carbon we use to make resistors. Silicon and
germanium are used to make basic semi-conductor devices such as diodes and transistors.
Any atom that has a valence of 8 is considered stable and does not readily give up or
accept other electrons. An atom with a valence of 7 will readily accept another electron to
make its valence layer 8. It would therefore be an “acceptor” atom. An atom with 9
electrons in its outer layer would establish a valence layer of 8, making the extra electron
a “free” electron that it would readily give up. This makes it a “donor” atom.
This concept of “accepting” or “donating” electrons is used to create a P.N. Junction.
If we take silicon with a valence of 4 and add to it an impurity that has a valence of 3
(Aluminum, Gallium) we would end up with molecules where each atom would “see” 7
electrons in its outer (valence) layer. The atoms would be combined in such a manner as
to leave a gap or hole where an eighth electron should be. Since this hole means the
absence of an electron, and since electrons are –ve, we refer to the hole as +ve hole. Not
all the silicon atoms will join with the impurity, only some, the exact quantity would be
determined by the amount of impurity added. This process of adding impurities is known
as “Doping”. Heavily doped material has a lot of impurity, lightly doped, a little. Silicon,
which is doped with a valence of 3 impurities, is called “P type silicon” (The P being for
positive). P type silicon is an acceptor!
By adding an impurity with a valence of 5 (Arsenic, Antimony) the molecule would form
a bond with 4 silicon and 4 of the impurity allowing each atom to “see” a valence of
eight. The extra electron (the fifth from the impurity) can bond with nothing, and
therefore becomes a free electron. Since electrons are negative, we call silicon doped
with an impurity with a valence of 5, “N type silicon” It is a donator! See Fig. 1
When a piece of P type and
N Type Silicon
P Type Silicon
N type silicon are physically
Doped with Arsenic
Doped with Aluminum
joined, the junction area
or Antimony
or Gallium
becomes active as the N
type (donor) material gives
+
+
+ +
up free electrons, which
+ + +
migrate to the P type
+
+
(acceptor) material and “fill
+
in” the positive holes. This
+ve Hole
Extra Electron
creates an area where the
Fig. 1
molecules all have a valence
of eight and are completely stable, creating, in effect, a zone of resistance. The activity
continues until the zone of resistance is of such a width that the force of attraction
between free electrons (-ve) and positive holes is insufficient to overcome the resistance,
and all activity ceases. The zone of resistance has no free electrons and no +ve holes.
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Since the electrons and holes are the vehicle for current flow, called the carriers, the zone
of resistance is known as the “Depletion Zone” (Depleted of current carriers). See Fig. 2
P Type Silicon
N Type Silicon
+ + +
+ + +
+ + + +
Free electrons from N type migrate
across the junction to “Fill in” +ve
holes in P type.
+
Fig. 2
+ + +
+ + +
+ + + +
+
“Depletion Zone” formed of molecules
where each atom “Sees” 8 electrons in
its valence layer. i.e. It is an “Insulator”
or relatively high resistance.
When the activity around the junction ceases, we have a “PN Junction” device. By
attaching a wire to each end, which will allow us to connect the device into a circuit, we
have created a silicon semi-conductor “Diode”. See Fig. 3 below.
+
P
N
+ + + +
Anode
-
Pictorial representation
Cathode
Diode Symbol
Actual Diode
Fig. 3
Current (Electron) Flow
A diode is an electronic one-way street. It will allow current (electrons) to flow in one
direction only, from cathode (N type) to anode (P type).
In order to understand how this works, we must apply a voltage across the diode. This is
often referred to as “Biasing”. “Forward Biasing” a PN Junction is when we put a
positive voltage on the P type and/or a negative voltage on the N type. In the case of the
diode this would be +ve on the anode and/or –ve on the cathode. “Reverse Biasing” is the
opposite, -ve on the anode and/or +ve on the cathode. We will begin by reverse biasing
the diode. With a –ve voltage on the anode, (P type) the +ve holes are attracted to the
voltage and tend to migrate to the end of the P type material. This effectively increases
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the width of the depletion zone. In the N type material the free electrons are attracted to
the +ve voltage, and they too migrate to the end of the material, further increasing the
width of the depletion. zone. This has the effect of increasing the resistance by an amount
sufficient to prevent any current flow. See Fig. 4 below
Depletion. Zone increases with
Reverse Bias
Fig. 4
+ +
N
P
+
+
With an increase in the reverse
Reverse Bias
bias there is a further increase in
the Depletion Zone width. This
can continue until the whole
diode is depleted. Any further
increase in reverse bias will
force electron flow, but the electrons will not be free electrons. This current flow in the
reverse direction, called ‘Avalanche Current’, will destroy the diode. The Max Reverse
Bias that can be applied to any diode is one of the two most important parameters and is
specified as the Peak Inverse Voltage (P.I.V.)
-
-
+
+
When Forward Bias (+ to anode, - to cathode) is applied, the reverse of the above occurs.
The +ve holes and –ve electrons are repelled by the same polarity bias being applied to
their respective terminals. See Fig. 5.
Current (Electron) Flow
This causes a movement towards the
Fig. 5
depletion zone having a “squeezing”
+ + +
effect and reducing the resistance as
+
N
P + + +
holes and electrons (carriers) are
+ + +
forced towards and into the depletion
zone. The reduced resistance allows
+
electrons to flow through the zone.
This constitutes current flow.
Increasing the forward bias will cause further “Squeezing”, further reduction in resistance
and corresponding increase in current. Forward bias can be increased to the point where
every available electron/hole combination is being utilized. This condition is called
“Saturation”. Any further increase in forward bias will cause non-free electrons to start
moving, and will destroy the diode. To prevent this, every diode is given a limit of
maximum forward current. This is the second of the two most important parameters.
We now see that current will flow in the forward bias state but not in the reverse bias
state. We can also see that the circuit voltage (forward or reverse) will cause the depletion
zone to change, which in turn changes the resistance of the diode. This means the forward
bias resistance of the diode is inversely proportional to the circuit voltage. It also means
that the voltage across the diode will always be the same as the barrier potential. i.e. 0.7V
for Silicon 0.3V for Germanium. That is, it requires a minimum voltage of 0.7V to
overcome the initial resistance of the depletion zone barrier (0.3V for Germanium). Once
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overcome, the combination of current & resistance (V=IxR) gives us a 0.7V drop. When
circuit voltage increases, diode resistance decreases and current increases. With one
going down (R) and the other going up (I), the voltage drop remains at 0.7V. A decrease
in circuit voltage increases resistance, decreases current and the voltage drop remains at
0.7V. This is one of the most important concepts to remember in order to understand
semi conductor diodes & transistors. The voltage drop across a conducting (forward
biased) PN Junction is 0.7V (0.3V for Germanium). ALWAYS.
The diode can therefore be
regarded as an open
switch
when
reverse
biased, and a closed
switch
(virtual
short
circuit) when conducting.
Remembering that when
conducting, there will
actually be 0.7V across a
“closed switch diode” and
not 0V. See Fig. 6.
Remembering that this is a
series circuit, total current
and the resistance of the
diode can easily be
calculated.
0.7V
D1
Vsup
D1
RL
1k
RL
1k
+
10V
9.3V
Equivalent Cct
+ve Half Cycle
(Forward Bias)
Fig. 6
10V
D1
Vsup
D1
RL
1k
-
10V
RL
1k
+
Equivalent Cct
-ve Half Cycle
(Reverse Bias)
For forward bias:
ITOTAL = IRL = VRL / RL = 9.3V/1k = 0.0093A
RDIODE = VDIODE / ITOTAL = 0.7V/ .0093 = 75
For reverse bias:
RDIODE = VDIODE / ITOTAL = 10V/ 0A = 
When connected to an AC supply voltage the diode will conduct on one half cycle and
cut-off during the other half cycle. On exactly which half cycle it conducts or cuts off
will be determined by the orientation of the diode. Regardless of which way round the
diode is connected, one of the half cycles will appear across the diode (when the diode is
an “open switch” or cut off), and the other half cycle across any series components,
(when the diode is a “closed switch” or conducting). The process of “removing” one half
cycle is often referred to as Rectification, the diode being the Rectifier. This is the basis
of most power supplies that convert AC voltages to DC voltages. For a brief explanation
of this refer to Fig. 7 on the following page. An in depth explanation will be given later in
the text.
0V
6
Refer to Fig. 7a. On +ve half cycles
D1 is forward biased (closed switch)
and therefore has no voltage across it,
current flows through the circuit and
the +ve half cycles appear as a
voltage across R1.
On the –ve half
cycles, D1 is reversed biased (open
switch), no current flows therefore
there is no voltage (0V) across R1
and VSUP (the –ve half cycle) appears
across D1.
D1
Vsup
Fig.7a
RL
1k
D1
With D1 reversed in the circuit, (See
Fig. 7b) it conducts on the –ve half
cycle and cuts-off on the +ve half
cycle. This gives the opposite results,
-ve half cycles appearing across R1
and +ve across D1.
R1
Vsup
Fig.7b
Fig. 8
D1
Vout
Vin
+5V
+10V
Vin
-10V
+5.7V
-10V
Vout
RL
1k
Diodes can also be used to “clamp” a
particular point in a circuit to a given
voltage by “clipping” off any voltage,
above or below the chosen value. See
Fig. 8. Remember, that to conduct the
diode anode must be more +ve than its
cathode. If a +ve voltage was connected
to the diode’s cathode, the anode would
have to rise to +0.7V above that voltage
before the diode would conduct. In the
example shown the diode will not
conduct until it’s anode reaches 5.7V.
This means the positive 10V peak is
clipped to +5.7V while the full –ve peak
will remain. The reverse would be true
if –5V was connected to the anode; the
cathode would have to fall to –5.7V for
the diode to conduct and clip the –ve
half cycles.
It is possible to use diodes to provide the logic “OR” & “AND” functions. The output of
the logic “gates” can be logic ‘HI’ or logic ‘LO’ as desired. On the following page, both
gates are depicted and in both cases the output will be a logic Hi when the gates are
activated.
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Fig. 9 shows a diode circuit that will perform a logic “OR” function with Vout as 4.3V
representing a logic ‘Hi’, and 0V a logic Lo.
With switches A and B both open there is
no path for current to flow. Without
current flow there will be 0V across R1,
representing a logic Lo. If switch A or B
or both are closed, one or both diodes
conduct and 4.3V (5V-0.7V) appears
across R1 giving a logic 1 O/P. (As long
as the supply voltage is +5V the diodes
are actually redundant, however, they do
prevent a –ve voltage from appearing
across R1)
A
D1
B
D2
+5V
R1
Vout
Fig. 9
The circuit in Fig. 10 will perform a Logic “And” function with Vout as 5V representing
a Logic ‘Hi’ and 0.7V a logic Lo.
R1
Fig. 10
+5V
D2
D1
Vout
A
1
0
1
B
0
With switches A & B or, A or B in the 0
(right) position D1, D2 or both, have their
cathodes connected to ground and their
anodes to +5V through R1. Vout will be
0.7V as one or both diodes will conduct and
Vout is across the diodes. This will
represent a logic Lo.
With switch A and switch B in the 1
position (left), both diodes will have +5V on
their cathode (reverse bias) and will cut-off
(open switch). The full 5V supply will
appear across the diodes which represents a
logic Hi at Vout.