Chapter Five P-N Junction First year class Sameer Abdul Kadhim 1 Chapter Five P-N Junction 5-1 Energy Band Diagram P-N Junction Donor impurities (N D ) are introduced into one side called N-side and accepter (N A ) into the other side called P-side. o - . . . . . . . . . . . Accepter ion Hole Donor ion Electron o - o - o - o - o - . . . . . . o - o - - - - - Depletion Region (W) x p x n Junction n p = n i 2 /p p E fp p p =N A p n =n i 2 /n n E fn n n =N D eV o E fn Ec p Ev n Ec n E fp Ev p o - o - . . Fig.(5-1) P-N junction and it's energy band diagram P-side N-side eV o
Transcript
Chapter Five P-N Junction First year class Sameer Abdul Kadhim
1
Chapter Five P-N Junction
5-1 Energy Band Diagram P-N Junction
Donor impurities (ND) are introduced into one side called N-side
and accepter (NA) into the other side called P-side.
o -
.
.
.
.
.
.
.
.
.
.
.
Accepter ion Hole
Donor ion Electron
o -
o -
o -
o -
o -
.
.
.
.
.
. o -
o -
-
-
-
-
Depletion Region (W)
xp xn
Junction
np= ni2/pp
Efp
pp=NA pn=ni2/nn
Efn
nn=ND
eVo
Efn
Ecp
Evn
Ecn Efp Evp
o -
o -
.
.
Fig.(5-1) P-N junction and it's energy band diagram
P-side N-side
eVo
Chapter Five P-N Junction First year class Sameer Abdul Kadhim
2
5-2 Equilibrium Condition
For a P-N junction when there is no external excitation (electrical,
magnetic, light,…) and net current equal zero then hole in P-side will
diffuse to N-side at xn distance, where xn is penetration of hole charge
into N-side material, and electron in N-side will diffuse to P-side in xp
distance, where xp is penetration of electron charge into P-side material.
Therefore a region from N- and P-side have no electron and have no hole
will created. This region called depletion region where it's width W= xn+
xp .As a result of diffusion of these charge, an electrical field will appear
a cross the junction. Equilibrium will be established when the field
become strong enough to restrain the process of diffusion and the net
current will equal zero. In other ward there is no charge movement across
the junction.
5-3 Potential Barrier (V0)
Potential barrier (eV0) is the electrostatic energy that prevent more
diffusion of hole from P-side to N-side and electron from N-side to P-side
as shown in Fig.(5-2).
eVo
Efn
Ecp
Evn
Ecn Efp Evp
eVo
Electron flow nn
np
pp pn
Hole flow
Fig.(5-2) Potential barrier
Chapter Five P-N Junction First year class Sameer Abdul Kadhim
1- In case of diodes connected in series we must compare E with overall
diodes voltages:
As example for circuit shown in Fig.(5-10)a .
E<3×0.7 → E= 3×Vf+ ÷÷ø
öççè
æh
´T
f0 V
VexpIR
2-In case there is more than one source we must find the equivalent
voltage and compare it with VD .
As example for the circuit shown in Fig.(5-10)b
if (E1-E2)<VD →E1-E2= Vf+ ÷÷ø
öççè
æh
´T
f0 V
VexpIR .
3- If diode not connected directly to voltage source we must find
Thevenin equivalent across diode.
As example for the circuit shown in Fig.(5-10)c we must find
RTH=R1//R2 and
ETH=2R1R
1RE
+´ .
The equivalent circuit is shown in Fig.(5-10)d. Now if ETH <VD them use
small signal analysis and:
ETH= Vf+ ÷÷ø
öççè
æh
´+T
f0TH V
VexpI)RR(
Chapter Five P-N Junction First year class Sameer Abdul Kadhim
14
Case 2: Large signal operation (Vf>VD)
As shown in Figure (5-11) when
Vf>VD (V/I) characteristic approach to
straight line, we can write the
relationship between diode voltage (Vf)
and its current (I) as Vf=VD+I×Rf
where Rf is diode forward resistance
(slop of line). In other ward diode equivalent at large signal is voltage
source (with VD voltage) connected in series with Rf resistance as
illustrated in Fig.(5-11)
Example 5-5
Find output voltage for the circuits shown below (Si diode with Rf=20Ω).
Chapter Five P-N Junction First year class Sameer Abdul Kadhim
15
Solution:
Circuit 1: E1>0.7 →large signal Vo = ( )20100
1007.010
+´- =……
Circuit 2: E2>(2×0.7) → large signal Vo = ( )40100
1004.110
+´- =……
Circuit 3:
RTH=100//100= 50 Ω and ETH=100100
10010
+´ =5>0.7→ large signal
Vo = ( )5020100
1007.05
++´- =……
Note :If two diodes connected in parallel and they was forward based
(large signal) then equivalent diode voltage is: = eqf
D
f
D RRV
RV
´÷÷ø
öççè
æ+
2
2
1
1 and
equivalent resistance is Req=Rf1 // Rf2
Exercise 2: Find output voltage for the circuit shown below.
Chapter Five P-N Junction First year class Sameer Abdul Kadhim
16
2- Reverse Region
V=-Vr → ÷÷ø
öççè
æh-
TVVr
exp << 1 → I@ -I0
Fig.(5-12) shows V/I characteristics of a reverse-biased P-N
junction. It is seen that as reverse voltage is increased from zero,
the reverse current quickly rises to its maximum or saturation value.
Keeping temperature constant as the reverse voltage is increased, I0 is
found to increase only slightly. A reverse-biased junction can be
represented by a very large resistance (Rr»MΩ) as shown in figure
below.
Chapter Five P-N Junction First year class Sameer Abdul Kadhim
17
3- Reverse Break Down Region
When the reverse voltage very high (Vr >Zener voltage Vz) that
made energy bands come closer and electrons in from p-side will
tunneling to N-side this led to made large amount of current passing from
N-side to P-side this called reverse break down current and diode can be
represented as shown in Fig.(5-13) where rz is diode resistance at break
down region (Zener resistance).
Zener voltage depend mainly on doping ratio for P- and N- sides
for conventional diode whose doping ratio is medium Vz is very high
(KV) therefore diode will damage before he reach the break down region.
5-7 Zener Diode
It is a heavily doped junction that made the energy band crossed at
relative low voltage as shown in Fig.(5-14).
Chapter Five P-N Junction First year class Sameer Abdul Kadhim
18
As shown in Fig.(5-14) high reverse voltage made the band
crossing and distance between them is narrow. Tunneling of electrons
from P-side valence band to conduction band of N-side constitutes a high
reverse current from N to P side this phenomena called Zener effect.
Zener diode used for voltage regulation.
Efp Efn
Equilibrium
Efp
Efn
Reverse Vr<Vz
eVr
Fig.(5-14) Energy band diagram Zener diode
Efp
Efn
Reverse Vr>Vz
Iz
Tunneling electrons
Heavy doping made Fermi level Fermi level very close to energy band
When small reverse voltage apply, Efn goes down and Efp goes up but energy band don't overlapped yet that made junction in reverse region
When the reverse voltage is high enough the energy band will overlapped that made electrons came from negative of battery jump across energy gap (made a tunnel through the gap) to N-side and large current will flow from N-side to P-side
N-side P-side
Chapter Five P-N Junction First year class Sameer Abdul Kadhim
19
Example 5-6
For circuit shown
below find output
voltage (at 3000K)
when input voltage (E)
is:
1- E=10V
2- E=-4V
3- E=-10volt
Solution
1- When E=10 diode forward biased (large signal model)
Output voltage = K120K1K1
7.010´
++-
=…..
2-E=-4 reverse region I=-Io=-1nA Output voltage=-1nA×1KΩ=….
3-E=-10 (>Vz)reverse break down region
Output voltage = K12K1K1
510´
+++-
=…..
Fig.(5-15) Zener diode characteristic
Chapter Five P-N Junction First year class Sameer Abdul Kadhim
20
Exercise 3: For circuit shown below find output voltage (at 3000K) when
input voltage (E) is: 1-E=10V 2-E=1V 3-E=-4 4-E=-10
5-8 Tunnel Diode (by Leo Esaki in1958)
It is a very very heavily doped P-N junction (100-104 times that of
classical diode e.g NA=ND=1020 ions/cm3) that made Fermi level lie outside
energy gap (Efn >Ecn and Efp< Evp) that made Evp>>Ecn that made energy
band structure of tunnel diode under open circuit condition shown in
Fig.(5-16).
Exercise 4: Prove that for tunnel diode
÷÷ø
öççè
æ-=
VC
DAg NN
NNkTEV ln0
Efp Efn
Ecn
Evp
Fig.(5-16)Energy band
structure of tunnel diode
under open circuit
condition
Chapter Five P-N Junction First year class Sameer Abdul Kadhim
21
Volt Ampere Characteristics of Tunnel Diode
Case 1 Reverse Biased
When a reverse voltage applied N-side Fermi level shift down Efn↓
and P-side up Efp↑ that made electron tunneling from P-side valence band
to conduction band of N-side constitutes a high reverse current from N to
P side this phenomena called Zener effect as shown in Fig.(5-17)a and
reverse characteristic is shown in Fig.(5-17)b
Case 2 Forward Biased
It is shown in Fig.(5-18) forward bias produces immediate
conduction i.e. as soon as forward bias is applied, significant current is
produced. The current quickly rises to its peak value Ip when the applied
forward voltage reaches a value Vp. When forward voltage is increased
further, diode current starts decreasing till it achieves its minimum value
called valley current Iv corresponding to valley voltage Vv. For voltages
greater than Vv , diode conduction current will start.
Efp↑ Efn↓
Fig.(5-17) Energy band structure of tunnel diode reverse biased condition
Iz
Iz
Tunneling electrons
N-side P-side
Vr>0
Vr
I
(b) (a)
Chapter Five P-N Junction First year class Sameer Abdul Kadhim
22
Explanation:
In forward region when a forward voltage applied N-side Fermi
level shift up Efn↑ and P-side down Efp↓ in other ward:
Efn= Efp + eVf
Therefore forward region can divided into four sub-regions:
a- Vf<Vp : electrons will tunnel from N-side to P-side and number of
tunneling electron increase when Vf increase until it reach maximum
value as shown in Fig.(5-19)a.
b- Vf=Vp :at this voltage Efn=Evp and Efp=Ecn that made maximum
number of electron will tunnel and current reach its peak value as
shown in Fig.(5-19)b.
Chapter Five P-N Junction First year class Sameer Abdul Kadhim
23
c- Vp <Vf< Vv: increasing of Vf made the overlapping area decrease and
number of tunneling electron decrease and forward current decreases
until it reach minimum value (Vv) as shown in Fig.(5-19)c where Vv is
the forward voltage that made Evp=Ecn that made tunneling electron
=zero as shown in Fig.(5-19)d.
Efp
Efn
Ecn
Evp
Efp
Efn
Evp
Efp
Efn
Tunneling electrons
Ecn
Tunneling electrons
No conduction electrons
Io
(a) Vf<Vp (b) Vf=Vp
(c) Vp <Vf< Vv (d) Vf= Vv
Efp
Efn conduction electrons will flow
(d) Vf> Vv
Ecn Evp No Tunneling electrons
Fig.(5-19) Forward region of tunnel diode at different forward voltage
Efn
Ecn Tunneling electrons Efp
Evp
Chapter Five P-N Junction First year class Sameer Abdul Kadhim
24
In these three region conduction current is =0 .
d- When Vf>Vv conduction current will flow and increase exponentially
(similar to conventional diode) as shown in Fig.(5-19)e.
5-9 Diode Reverse Recover Time (trr)
In most switching application a diode need (trr=ts+tp) time to switch
from forward to reverse as shown in Fig.(5-20) where:
Storage time (ts): time required for the storage charge at junction to
become zero. During this time diode is forward although a reverse
voltage is applied.
Hole life time (tp) : time required for hole (at P-side) to generate when a
negative voltage apply on P-N junction. During this time diode will
changing to reverse region.
Chapter Five P-N Junction First year class Sameer Abdul Kadhim
25
5-10 Capacitance of P-N Junction and Varactor Diode
The capacitance of diode is visualized from charge distribution in
transition region. There are two type of capacitance in the diode P-N
junction.
(a)Transition Capacitance (CT) or Space-charge Capacitance
It is dominant when junction is reversed biased where when a P-N
junction is reverse-biased, the depletion region acts like an insulator or as
a dielectric material essential for making a capacitor. The P- and N-type
regions on either side have low resistance and act as the plates. We,
therefore, have all the components necessary for making a parallel-plate
capacitor. This junction capacitance is called transition or space charge
capacitance (CT). It may be calculated by the usual formula :
WA
C r0T
ee= where A: cross section area of junction
And W is the thickness of depletion (or transition) layer which
depends on the amount of diode voltage:
( )÷÷ø
öççè
æ+
-ee=
DA
0r0
N1
N1
eVV2
W
And V is the voltage across diode terminal:
ïïî
ïïí
ì-
=
signalelforwardV
signalsmallforwardV
reverseVr
voltageappliedno
V
D
f
arg
0
Capacitance CT can be controlled with the help of applied bias. This
property of variable capacitance possessed by a reverse-biased P-N
junction is used in the construction of a device known as varicap or
varactor.
Chapter Five P-N Junction First year class Sameer Abdul Kadhim
26
This capacitance is voltage dependent as given by the relation
0
TT
VVr
1
)0(C)Vr(C
+=
Where )0(CT is the transition capacitance at zero applied voltage.
The voltage-variance capacitance of a reverse-biased P-N junction is used
in many circuits one of which is automatic
frequency control (AFC) in an FM tuner.
Other applications include self-balancing
bridge circuits, special type of amplifiers
known as parametric amplifiers and
electronic tuners in TV.
(b) Diffusion or Storage Capacitance (Cs)
This capacitive effect is present when the junction is forward-
biased. It is called diffusion capacitance to account for the time delay in
moving charges across the junction by diffusion process. Due to this fact,
this capacitance cannot be identified in terms of a dielectric and plates. It
varies directly with the magnitude of forward current because the number
of charge carriers left in depletion layer is proportional to forward
current. The capacitance assumes great significance in the operation of
devices which are required to switch rapidly from forward to reverse bias.
If Cs is large, this switch over cannot be rapid. It will delay both the
switch-on and the switch-off. This effect of Cs is variously known as
recovery time or carrier storage.
In the case of forward bias, the diode current is almost entirely due
to diffusion .If pt is hole lifetime of charge carriers, then a flow of
charge Q yields a diode current of
p
QI
t= → pIQ t´= where ÷÷
ø
öççè
æh
@T
f0 V
VexpII and
fdVdQ
Cs = =f
p
dV
dIt
Chapter Five P-N Junction First year class Sameer Abdul Kadhim
27
As shown in section (5-6) dI
dVf is called diode dynamic resistance (rd)
IV
rd Th= that made
rdCs pt=
Example 5-7
For circuit shown in Fig.(5-21) find total capacitance for the diode
at room temperature when
E=0.65volt . (Si diode
IO=1nA,tp=1psec)
From Example 5-4
Vf=0.6
rd=51.7Ω
Cs=W
´ -
7.51sec101 12
=
Example 5-8
For circuit shown in Fig.(5-21) find transition capacitance for the
diode at room temperature when E=-10volt and cross section area of
junction =0.5cm2. (Si diode re =12, NA=ND=1016 1/cm3, IO=1nA)
From Example 5-1
V0=0.31eV
E=-10 reverse biased
Vr=E-I0×R≈E =10volt
( )÷÷ø
öççè
æ+
+ee=
DA
r0r0
N1
N1
eVV2
W
W= ÷øö
çèæ
´+´´´´
-
-
1619
14
10
2
106.1
)1031.0(161085.82
WA
C r0T
ee= =
).......(5.0121085.8 14
cmCT
´´´=
-
=
Chapter Five P-N Junction First year class Sameer Abdul Kadhim
28
Tutorial Question
Q1\ A Si P-N junction at 3000K has NA= 1017 cm-3 , ND=1015 cm-3, re =12
1- Draw energy band diagram for this junction at:
a- Equilibrium
b- 0.2eV forward voltage. c- 0.7 eV reverse voltage.
2- If the junction has a circular cross section with diameter of 100mm
find transition capacitance at:
a- Equilibrium b- 7 eV reverse voltage.
Q2\ Consider a Ge P-N junction at 3000K with ND=103NA and NA
corresponding to 1 ion per 108 Ge atom (for Ge there 4.41×1022
atom/cm3) and cross section is a square with 0.1 cm length.
Calculate:
1- Transition capacitance then draw energy band diagram when a 3eV
reverse voltage were apply.
2- Number in positive charge in junction.
3- Draw the charge density for junction.
4- Resistivity of N- and P- sides .
(use re =16, Dn=99cm2/sec and Dp=47 cm2/sec).
Q3\ A Ge P-N junction at 3000K with circular cross section with diameter
of 1cm, 2Ω.cm P-side resistivity 1Ω.cm N-side resistivity find junction
capacitance when we apply 10 reveres voltage.
(use re =16, Dn=99cm2/sec and Dp=47 cm2/sec).
Q4\ Find forward voltage that made forward current =15mA for Si P-N
junction at 3000K if I0=1nA.
Chapter Five P-N Junction First year class Sameer Abdul Kadhim
29
Q5\ Find ratio between (forward current if forward voltage=0.5 volt) to
(reverse current if reverse voltage=5 volt) for Si P-N junction at 3000K.
Q6\ Find output voltage Vout for circuit shown below when :
1-E=1volt 2- E=3volt 3- E=10volt 4-E=- 10volt
Assume all diode are Si ,I0 = 1nA ,T=270C ,Rf=20Ω , Vz=5 volt ,rz=0.
Q7\ When 5 reverse voltage apply Si diode(at 3000K,NA=ND =1016cm-3)
transition capacitance was =20pF. Find transition capacitance when
reverse voltage increases to 10 volt.
Q8\ Consider a P-N junction at equilibrium with ND=103NA =17.6×103ni
.At certain temperature T it's found that potential barrier =0.327V and
depletion region width =1.15µm , if the cross section of P-N junction is a
square with 0.1 cm length , reverse saturation current =1µA and hole life
time=1psec, calculate:
1- Cut in voltage .
2- Transition capacitance if 10 reverse voltage applied.(use ϵr=10)
3- Storage capacitance if 0.25 forward voltage applied.(use ɳ=1)
4- Distance between Fermi level and valence band in N-side.
Q9\ For a Si P-N junction whose energy band diagram shown in figure
below find:
1- Resistivity for both sides. 2- Transition capacitance
Chapter Five P-N Junction First year class Sameer Abdul Kadhim
30
(Use re =12, Dn=34cm2/sec and Dp=13 cm2/sec, cross section
area=1mm2).
Q10\ Find total capacitance for the Si diode at room temperature when
forward voltage =0.6V if you know that cross section area of junction
=0.5cm2, resistivity of N-side =resistivity of P-side= 1Ω.cm, ,reverse
saturation current =1nA, hole life time =1psec. (use: Dn=34 cm2/sec
,Dp=13 cm2/sec and re =12)
Q11\ Prove that for P-N junction at equilibrium:
1- Penetration length of hole charge into N-side material is :
( )÷÷øö
ççè
æ+
´=DAA
Drp NNN
NV
ex 0
02 ee
2- Penetration length of electron charge into P-side material is:
