diodes of electric circuits
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Diode Characteristics 30
CHAPTER (3)
P-N JUNCTION DIODE CHARACTERISTICS
(3-1) I-V Relationship
The first subject to be studied in electronic devices courses is the p-n
junction diode characteristics. The p-n junction is the basic structure of
all bipolar devices. The proof of the p-n junction current is beyond the
scope of this chapter. We focus here on the device applications rather
than the p-n junction detailed physics. The I-V relationship of a p-n
junction is given by
I = IO(T) { exp (V/VT) - 1 } (3-1)
where
IO is the reverse saturation current and it is a function of
temperature,
V is the voltage applied across p-n junction,
VT is the thermal voltage (25 mV) at room temperature
( T =300OK), and
is a constant depends on the p-n junction material. =1
for Ge and = 2 for silicon.
(a) (b)
Fig. (3-1) I-V characteristics of p-n junction (a) at different
temperatures and for different materials.
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Electronic Engineering 31
Fig. (3-1a) shows a plot for the I-V characteristics given by Eq. (3-1). It
is seen that for positive values of V, (forward biased), the current
increases exponentially with the applied forward voltage. On the other
hand, for a reverse applied voltage, the diode current is almost constant
at I = - IO because the exponential term will be negative and large so it is
negligible with respect to unity. This is shown in Fig. 1(a).
(3-2) Temperature Effect
The I-V characteristics is temperature dependent because both thermal
temperature VT and reverse saturation current IO depend on T. VT is given
by
VT = k T / q (3-2)
where
k is Boltzmann constant, (= 1.381 * 10-23 J/ o K)
T is the junction absolute temperature in OK, and
q is the electronic charge (= 1.6 * 10-19 Column)
The reverse saturation current IO is related to temperature through the
relationship
IO(T) = AO T3/2 exp {-VG / ( VT)} (3-3)
where AO is a constant. Differentiating Eq. (3) with respect to
temperature and taking into account that VT is given by Eq. (2) one
obtains
dIO / dT = AO T3/2 (1/T VT) exp {-VG / ( VT)}
+ (3/2) AO T1/2 exp { -VG / ( VT)}
= AO T3/2 exp { -VG / ( VT)} {[VG / ( T VT)] + (3 /2 T )}
= IO {[VG / ( T VT)] + (3 / 2T )}
or (dIO/IO)/ dT = [VG / ( T VT)] + (3 / 2T ) (3-4)
Substituting for T = 300OK, VT = 26 mV, VG = 1.21 V and = 2 into Eq.
(3-4) one finds
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Diode Characteristics 32
(dIO/IO)/ dT = 0.08 = 8 % (3-5)
Eq. (3-5) means that the percentage change of IO (reverse saturation
current) is 8 % per degree Kelvin. For example if IO is 1 A at 300OK, it
will be 1.08 A at 301OK and (1.08*1.08) A at 302 OK and so on. But
since (1.08)9 = 2 then one can approximately write
IO(T2) = IO(T1) * 2T/9 (3-6)
where T = T2 - T1 (3-7)
where IO(T1) and IO(T2) are the reverse saturation currents at
temperatures T1 and T2 respectively. The approximation in Eq. (3-6) is
that it assumes constant reverse saturation current temperature rate for
the range between T1 and T2 equal to that at T = 300OK.
If the temperature increased, then VT increases - Eq.(2)- and
consequently the exponential term in Eq. (3-1), decreases. On the other
hand, as T increases IO increases rapidly as given by Eq. (3-3) or its
approximate relation given by Eq. (3-7).
The effect of the increase of IO due to temperature is larger than the effect
of the decrease of the exponential term due to the increase of T. Thus, the
diode current increases as temperature increases. This behavior is
illustrated in Fig. (3-1a). The sensitivity of diode current to temperature
is sometimes used as a temperature sensor. Thus the diode current (which
is passed through a resistor) is used to measure the temperature by
measuring the voltage across the resistor. The diode (when used as a
temperature sensor) should be placed in the spot where the temperature is
to be measured.
Let us now turn our attention to the effect of temperature on the voltage
applied across p-n junction V. Substituting for VT and IO from Eqs. (3-2)
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Electronic Engineering 33
and (3-3) into Eq. (3-1) and differentiating Eq. (3-1) with respect to
temperature when the current I is kept constant yields
dV
dT
V
T
d
dT
V
TTO
O G T
VI
I V V
( )
( ( / ) )1 3 2 (3-8)
Substituting for T = 300oK, VG = 1.21 V, VT = 26 mV, = 2 (Si) and V =
V= 0.6V into Eq. (3-8) one finds
dV/dT = - 2.3 mV/ 0C (3-9)
The diode material (either Ge or Si) affects the diode I-V characteristics
because the constant in Eq. (3-1) varies according to diode material. As
increases, the exponential term decreases and consequently, diode
current decreases. Since is 1 for Ge and 2 for Si then, the current is
larger for Ge diode if the applied voltage is kept constant. This is
illustrated in Fig. (3-1b).
(3-3) Piecewise Linear Model
For the diode characteristics shown in Fig. (3-2a), one can define static
resistance R which is defined at point P as
R = (V / I)at point P = V1 / I1 (3-10)
While the dynamic resistance r is determined by the relationship
r = (dV/dI)at point P (3-11)
For forward voltages, the exponential term in Eq. (3-1), is much larger
unity so one may approximate Eq. (3-1) as follows,
I = IO { exp (V/ VT)} (3-12)
Differentiating Eq. (12) with respect to V, then
dI/dV = I / VT
or
r = VT / I (3-13)
The piecewise linear model of the I-V characteristics of p-n junction is a
simplified I-V relationship. The break point is not the origin, but it is the
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Diode Characteristics 34
cut-in voltage V as shown in Fig. (3-2b). For piecewise model, the diode
is considered open circuited, for negative voltages and forward voltage
less than the cut-in voltage V. For voltages larger than V ,the diode is
considered a constant resistance Rf. The resistance Rf is relatively a small
resistance about 100 Ohms.
(a) (b)
Fig. 2 (a) Static and dynamic resistance and
(b) piecewise linear model.
The diode is said to be ideal if it is short circuited when it is forward
biased and open circuited when reverse biased. This means that an ideal
diode has V = 0 and Rf = 0. Of course such an ideal diode doesnt exist
in real world, but, sometimes such assumption is made if the applied
voltage is much larger than V and so it may be neglected. Also when
circuit resistors are much larger than Rf , one can neglect Rf.
(3-4) Transition Capacitance
When a reverse bias voltage is applied across the p-n junction, a space
charge (or depletion region) is formed around the junction. The depletion
region is a layer where there is no mobile charges. This is caused by the
reverse bias applied across the junction. This causes the majority carriers
to be away from the junction. The depletion region includes only positive
ions in the n-type region and negative ions in the p-type region as shown
in Fig. 3(a). As the reverse bias voltage increases in magnitude, the
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Electronic Engineering 35
depletion region expands on both sides of the junction. To explain this
phenomenon, assume a larger positive voltage is applied on n-type, this
(a) (b)
Fig. (3-3) (a) Depletion region when a reverse voltage V is applied
across p-n junction and (b) charge and voltage across
depletion region.
will attract more electrons towards the edge of the n-type region causing
the depletion region to expand on the n-type side. The same is applied
when a negative voltage is applied on p-type. Thus the charge Q on both
sides of the junction increases with the reverse applied voltage V. Since
Q is proportional to V, then there is a capacitance called the transition
capacitance CT defined as
CT = dQ/dV (3-14)
Thus the reverse biased p-n junction equivalent circuit is simply a
capacitance CT . This transition capacitance CT is voltage dependent. To
find this relation, one can apply Poissons equation for the depletion
region. Consider the charge density as a function of distance from an
alloy junction in which the acceptor impurity density is assumed to be
much smaller than donor concentration. Since the net charge in the
depletion region shown in Fig. (3-3b) must be zero, then
q NA WP = q ND Wn (3-15)
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Diode Characteristics 36
If NA > Wn . For simplicity, Wn is neglected and as a
result depletion region width W (= WP + Wn) will be approximately WP.
So the potential will appear across p-type region across the uncovered
acceptor ions. The relationship between potential and charge density is
given by Poissons equation,
2
2
dx
NVd
qA
O r
(3-16)
where r is the relative permittivity of the semiconductor (= 12 for Si)
and O is free space permittivity (= 8.85*10-12 F/m). The electric field is
zero outside the depletion region so, one can write
dV
dx 0 at x = 0 and at x = W WP
Also
V = 0 at x = 0
and V = VJ at x = WP
where VJ is the voltage across the p-n junction which is the difference
between the applied reverse voltage V and the built in voltage VO.
Usually V >> VO so V-VO V. Integrating Eq. (3-16) yields
dV
dx
qA
O r
N
x + k1
where k1 is a constant. But since dV/dx = 0 at x = 0, then k1 = 0. Thus
Vx = q
A
O r
N
(x2 / 2) + k2
where k2 is a constant. But since V = 0 at x = 0, then k2 = 0. As discussed
above Vx = V at x = W WP so
V = (1/2) q
A
O r
N
W2 (3-17)
The charge on the p-type region (uncovered ions) Q is given by
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Electronic Engineering 37
Q = q NA W A (3-18)
where A is the cross sectional area of the p-n junction. The transition
capacitance CT is defined as
CT = dQ / dV = q NA A (dW / dV) (3-19)
Using Eq. (3-17) one finds
dV / dW = q
A
O r
N
W (3-20)
Combining Eqs. (3-19) and (3-20) yields
CT = O r
W
A (3-21)
Substituting for W from Eq. (3-17) into Eq. (3-21) gives
CT = A q
V
A O rN 2
(3-22)
Eq. (3-22) may be rewritten as
CT = OkV
(3-23)
where kO is a constant depends on the diode manufacturing parameters
NA and area A. Thus for a given diode, kO is constant. Eq. (3-23) shows
that the diode transition capacitance CT is inversely proportional to the
square root of the reverse applied voltage. Note that CT exists only if the
applied voltage is reverse biased.
According to above mentioned analysis, a diode may be used as a
capacitor CT if a reverse biased voltage is applied across the diode. The
diode acts as a VARiable reACTOR (VARACTOR) when a reverse
biased voltage is applied across the diode. Fig. (3-4a) shows the circuit
symbol of a varactor while Fig. (3-4b) shows the dependence of CT on
the reverse applied voltage.
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Diode Characteristics 38
(a) (b)
Fig. (3-4) Varactor (a) circuit symbol and (b) C-V characteristics.
(3-5) Zener Diode
The reverse voltage characteristics of a p-n junction diode, including the
breakdown region, is plotted in Fig. (3-5a). Zener diodes are designed
with capability of enough power consumption in the breakdown region.
This device is used as voltage reference. A typical application circuit of
Zener diode is indicated in Fig. (3-5b). The source voltage V and resistor
R are selected so that, initially, the Zener diode is almost open circuited
when the voltage V is less than the Zener voltage VZ. When V exceeds
VZ the voltage across RL will remain constant at VZ while the difference
between V and VZ will be across the series resistor R. The VO - V
relationship is indicated in Fig. (3-5c). An important note is that the
Zener diode requires a minimum current IZ is required so that Zener
diode is operating at a constant voltage VZ. The output voltage remains
constant at VZ if the
(a) (b) (c)
Fig. (3-5) Zener diode (a) characteristics, (b) application circuit and
(c) input - output voltage relation of the application circuit.
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Electronic Engineering 39
input voltage V exceeds VZ. Thus Zener diode protects the load against
over voltage.
(3-6) Rectifier Circuits
Fig. (3-6) shows the basic half and full wave rectifier circuits. For the
half wave circuit, only the positive half cycle passes because of the diode
which passes only the forward current. For the bridge full
Fig. (3-6) Half wave and full wave rectifier circuits.
wave rectifier circuit, the positive half cycle passes through two opposite
diodes while the negative half cycle passes through the other opposite
two diodes. As a result the output is a rectified full wave as indicated in
Fig. (3-6).
(3-7) Clipping and Clamping Circuits
Fig. (3-7a) shows typical positive voltage clipping circuit while, Fig. (3-
7b) illustrates the output voltage of this circuit. Let us now explain the how
this circuit works?. It is assumed that the diode is ideal i.e. it is short
circuited when it is forward biased and open circuited when reverse biased.
(a) (b)
Fig. (3-7) Positive voltage clipping circuit (a) circuit and (b)
waveform.
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Diode Characteristics 40
The source voltage VS is a sinusoidal one given by
VS = Vm sin ( t) (3-24)
Since the potential at the cathode of the diode D is V -as shown in Fig.
(3-7a)- then the diode will conduct if and only if the anode voltage is
greater than V. If this condition is fulfilled, VO = V. VO cant exceed V
because the voltage across the diode is assumed zero. The difference
between VS and V will appear as a voltage across the resistor R. The
output voltage is said to be clipped at V. On the other hand, if VS is
smaller than V, diode D is reverse biased -i.e. open circuited- and
consequently no current flows in the circuit. Under this condition, the
voltage across R is zero and VO = VS as shown in Fig. (3-7b). The
maximum current Imax passes through R is
Imax. = (Vm - V) / R (3-25)
If a circuit such as that shown in Fig. 8(a) is used, a clipped output
voltage in both positive and negative sides of the sine wave is
(a) (b)
Fig. (3-8) Positive and negative voltage clipping circuit (a) circuit
diagram and (b) output voltage waveform.
obtained. The output voltage will be as shown in Fig. (3-8b). The
maximum resistor currents in both positive and negative half cycles are
[(Vm - V1) / R] and [(Vm - V2) / R] respectively.
Fig. (3-9a) shows a positive clamping circuit where the output voltage is
exactly like the input voltage but shifted by +Vm. The source voltage is a
sinusoidal one where VS is given by Eq. (3-24). Let us see
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Electronic Engineering 41
(a) (b) (c)
Fig. (3-9) Positive clamping circuit (a) circuit diagram
(b) output waveform and (c) capacitor voltage.
this circuit works?. During the first positive half cycle, the diode is
reverse biased and so no current flows in the circuit and capacitor is not
charged. During the first negative half cycle, the diode is forward biased
and the capacitor C is charged to the voltage Vm in the direction shown in
Fig. (3-9a). The capacitor charging is done through the diode forward
resistance Rf which is neglected. Thus the capacitor voltage will follow
the source voltage during charging as indicated in Fig. (3-9c). The output
voltage VO - taking into account the voltage polarities shown in Fig. (3-
9a)- may be written as
VO = VS + Vm = Vm {sin( t) + 1} (3-26)
The plot of the waveform given by Eq. (3-26), is illustrated in Fig. (3-
9b). Since sin ( t) lies between +1 and -1 then, according to Eq.(3-26)
the maximum value of VO is 2Vm , while, the minimum value of VO is
zero. The output voltage is said to be clamped at 2Vm. A negative
clamped circuit is indicated in Fig. (3-10a). In this circuit, the output
voltage is clamped at -2Vm rather than +2Vm. The output voltage
waveform is indicated in Fig. (3-10b). For the circuit in Fig. (3-10a), VO
may be written as
VO = VS - Vm = Vm {sin( t) - 1} (3-27)
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Diode Characteristics 42
It can be easily seen that VO -according to Eq. (3-27)- lies between zero
and -2Vm. The output voltage is said to be clamped at zero voltage.
Logic Gates
The diodes are used to implement the basic logic gates known as OR and
AND gates shown in Fig. (3-10a) and (3-10b) respectively. The logic
gates are characterized by two levels 5 V denoted by logic "1" and
zero voltage denoted by zero "0" and not any other level. The Tables
below are called the truth tables of OR and AND gates.
(a) (b)
Fig. (3-10) Logic gates using diodes.
A B Y
0 0 0
1 0 1
0 1 1
1 1 1
Table-1 Truth table of OR gate.
A B Y
0 0 0
1 0 0
0 1 0
1 1 1
Table-1 Truth table of AND gate.
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Electronic Engineering 43
EXERCISE (3)
Hint for all problems assume that = 8.85 * 10-12 F/m , r (Silicon) =
11.9, diode reverse saturation current (IO)= 1 A, = 2 and all diodes are
operating at room temperature (T = 3000 K) where VT = 25 mV.
1. Calculate the diode current when
a- V = 100 mV,
b- V = 250 mV and
c- V = - 150 mV.
2. (a) Calculate the reverse saturation current IO at 70O C.
(b) Calculate the temperature at which IO = 3 A.
3. If the area of a p-n junction is 10 m by 10 m and NA = 1017 cm-3 , find
(a) Transition capacitance CT if the reverse voltage is 2 V.
(b) The reverse voltage at which CT is half that of part (a).
4. Find (a) The static and dynamic resistance of a diode at V = 0.7V.
(b) Repeat part (a) if V = - 0.7V.
5. Find (a) diode current I at V = 0.7 V,
(b) diode reverse saturation current IO at T = 3500K and
(c) the temperature at which IO = 2 A.
6. For the circuit shown, R = 1 k and the current is 10 mA. Find
(a) Battery voltage V.
(b) Voltage across diode.
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Diode Characteristics 44
7. For the circuit shown, V = 3 V and CT = 25 pF when S1 is opened and S2
closed. If VS = 2 sin (t) and S1 is closed while S2 is opened
find (a) Maximum value of CT and
(Two) Minimum value of CT .
8. For the circuit shown VS = Vm sin (t). Draw VO if
(a) Vm = 10 V and V = 6V.
(b) Vm = 6 V and V = - 3 V.
(c) Find the maximum resistor current when Vm = 6 V and V = -3 V.
9. For the circuits shown draw VO if VS = 12 sin (t), R1 = 1 k, R2 = 2 k if
(a) V = 3 V and
(b) V = 6 V.
10. For the circuit shown if VS = 10 sin( t) find
(a) Maximum output voltage and
(b) Minimum output voltage.
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Electronic Engineering 45
11. Repeat problem (10) if the diode D is reversed.
12. For the circuit shown, the diodes are represented by their piecewise
linear model where V = 0.6 V, V = 10V and Rf = 20 Ohms. Find the
current in each diode if R = 1 k.
13. For the circuit shown if V = 5 V determine
(a) The current passing in the circuit I,
(b) The voltage across D1 and
(c) The voltage across D2.
14. For the circuit shown, VZ = 20 V, and R = RL = 1 k.
(a) Draw VO versus V if V varies between 0 and 20 V and
(b) Find the maximum current through R.
15. For the circuit shown in problem (14), find VO if
(a) V = 6 V,
(b) V = 10 V and
(c) V = 18 V.
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