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Chap 3
Diodes
Ideal Diode
Terminal Characteristics
Semiconductors
Physical Operation
 Open Ckt pn junction
 Reverse bias pn junction
 Forward bias pn junction
Diode Circuits
Small signal model
Rectifiers
Clipping and clamping
The SPICE diode
C. Hutchens Chap 3 ECEN 3313 Handouts
1
Simple Diode I-V Representation
DC or Capacitance free Diode I-V models.
At higher frequencies one most include Cs


C reverse CDep or CJ “Junction”
C forward CD
C. Hutchens Chap 3 ECEN 3313 Handouts
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Diode Symbols and Ckt Representation
Ideal Math model
I D  I s ( V / nVt  1)
where n is the emission
coifficient and = 1 for Si,
Is is the saturation current, and
Vt is the thermal voltage and =
26mV at room temp
For V> 4 n Vt  100mV
VZ or Avalanche
I D  I s V / nVt
For V>> 4 n Vt
 short ckt with a voltage drop of 0.5
to 1.0V
and for V< 4Vt  100mV
ID  Is  0
C. Hutchens Chap 3 ECEN 3313 Handouts
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Diode SPICE parameters
.MODEL D1N914/125C D
+
IS = 2.25833E-15
+
RS = 1.799439
+
N = 1.067043
+
TT = 1.46E-7
+
CJO = 3.0177E-12
+
VJ = 0.4
+
M = 0.2147523
+
EG = 1.11
+
XTI = 4.799594
+
KF = 0
+
AF = 1
+
FC = 0.764906
+
BV = 94
+
IBV = 5E-6
+)
C. Hutchens Chap 3 ECEN 3313 Handouts
(
Saturation Current
Contact Resistance
Ideality factor
Fwd Transit time
Zero Bias Junction C
Built in potential
doping grading Coeff.
Breakdown Voltage
4
Simple Diode Rectifer
EX 2 Find the fraction of each cycle the
diode conducts and the peak current if vin
= 24 COS 2 f t . Note Typically f will be
in the 10 to 1000s of Hertz for this
application.
Assume vin = 24Cos 2 f t and V = 12 V,
and R =100 ohms.
 24 Cos  = 12

 Cos  =1/2

 Cos-1 1/2 = 600
 Fraction = 2 x 60/360 = 33%

 Ipeak = (24-12)/100
Other items you must consisder. If Rs =
10 ohms, and the resistance and all
diode parameters vary by +/-20% what is
the worst case peak current, peak and
continious power dissipationa? What if
vin varies by 20%.
C. Hutchens Chap 3 ECEN 3313 Handouts
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Limiters/Clippers
Observations
VO > V1
VO <V2
V2 < VO < V1
Vo > V1 D1 is a short
circuit else o.c.
Vo < V2 D1 is a short
circuit else o.c.
The polarity and
magnitude of V 1
and V2 shift the
conduction point.
Ex vin = 20V peak to peak
Sinusoid, V1 = 10 V
and V2 = -8 V
C. Hutchens Chap 3 ECEN 3313 Handouts
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LOG COMPUTATION AND THE DIODE
I D  I s ( V / nVt  1)
where n is the emission = 1
 Both IS and VT are temperature dependent
 Valid for as many as 7+ Decades
V / nV
 1) and I D 2  I S 2 ( V / nV  1)
Consider I D1  I S 2 (
t
t
I D1 / I D2   (V2 V1 ) / nVT
V2  V1  V  nVT Ln(I D1/ I D2 )  2.3nVT Log 10 (I D1/ I D2 ) = 60mV /Decade at
Rm Temp
Every Decade Increase in current results in a 60mV increase in
diode voltage.
Typical Values of Diode on voltage are 100 to
800mV.
C. Hutchens Chap 3 ECEN 3313 Handouts
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COMPUTATION AND THE DIODE
 Now ID is a function of diode area or A JS where JS is the current
density
Rewriting the previous equation
V  2.3nVT Log 10 (J S A D1 / J S A D2 )  2.3nVT Log 10 (A D1 / A D2 )
If two diodes are fabricated side by side and care the same
current but are of unequal areas the voltage difference will
be a function of the Log of the ratio their areas.
See EX 3.3
 For a single diode the FORWARD voltage shift is approx. -2mV/C0
Use Temperature measurement
 Reverse bias current -IS "Area dependent Doubles every 50C."
C. Hutchens Chap 3 ECEN 3313 Handouts
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Basic Semiconductor Concepts-Review
Conductivity is controlled by
carrier concentration
Intrinsic carrier
concentration ni = p =n =
1.5x1010 carriers/cm3 at
Room Temp.
ni  BT 3 EG / kT
where
B is a material
dependent para
8.62x10-5eV/K,
EG is the bandgap
and for Si 1.12 eV,
T is Temp. K0 equal
2730 at room temp.
Si has 5x1022 atms/cm3 vs.
ni = 1.5x1010
C. Hutchens Chap 3 ECEN 3313 Handouts
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Basic Semiconductor Concepts
Physical factors controlling pn
junction behavior.
Diffusion - Thermal
agitation coupled with
conc. (carriers/cm3)
Drift - E field associated
carrier motion.
n-type equilibrium carrier
conc. nn0  ND i.e. P
atms/cm3 free electons
p-type equilibrium carrier
conc. pp0  NA i.e. B
atms/cm3 free holes
nn0 pp0 = ni2 then
C. Hutchens Chap 3 ECEN 3313 Handouts
pp0 = ni2/ND and
nn0 = ni2/NA
10
Doping Example
Given ND = 1017/cm3 of P find nn0
pp0 at 300 K0.
ni  BT 3 E / kT where
B is a material
dependent para
8.62x10-5eV/K,
EG is the bandgap and
for Si 1.12 eV,
T is Temp. K0 equal
2730 at room temp.
G
ni 2 = (1.5x 1010 /cm3)2
nn0  ND = 1017/cm3
pp0 = ni2/ND =2.25 x 103
C. Hutchens Chap 3 ECEN 3313 Handouts
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Diffusion
Cathode
Anode
x
Consider bars of Si with the following hole and electron concentrations
Cp = -mp x and Cn = mn x "Linear"
the current (A/cm^2) can be written as
Jp = q Dp dp/dx = -qDp mp for pos. current in the x dir.
Jn = q Dn dn/dx = -q Dn mn for electrons
where D is the diffusion constant for holes ( 12cm2/s ) and
electrons ( 34cm2/s) and q is the charge on an electron of hole.
C. Hutchens Chap 3 ECEN 3313 Handouts
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Drift
vdrift =  E where  is the mobility of a charge and E is the electric field
Then considering a Si crystal with a charge density qp coulomb/cm3
Jp-drift = q pp E
hole drift current
Jn-drift = q pn E electron drift current (neg charge in neg x dir.
results in positive current)
Jtotal drift = q E(pp + pn) or using ohms Law and R=  L/A
 = 1/( q (pp + pn)) Note V/I = R =  L/A ;  = (V/L)(A/I) = E/J
x
C. Hutchens Chap 3 ECEN 3313 Handouts
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Open Ckt pn Junction
Vo = VJ, n = N
Drift (IS) and Diffusion (ID) Currents are in Equilbrium (IS = ID) for an o.c. diode
and for charge neutrality
qx p AN A  qxn AN D
x p N A  xn N D
where xp and xn are the respective depletion depths in the n and p type
materials respectively.
x n  x p  Wdep 
2 Si
q 1 / N A  1 / N D Vo
defines the depletion width.
C. Hutchens Chap 3 ECEN 3313 Handouts
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Open Ckt pn Junction
The pn work function given by;
N N 
Vo  Vt Ln A D 
 ni 
Ex Given NA = 1017 , ND = 1016 , T = 300K0 and using ni = 1.5x1010/cm3 and Si =
11.70 Fd/cm (0 = 1.04x10-12). Find Vo, W det and its componets.
Vo = 738mV
W det = 0.32um
xp = 0.03um
xn = 0.29um
C. Hutchens Chap 3 ECEN 3313 Handouts
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Rev. pn Junction-Depletion Cap.
Note C = q/RVR = VQ
or Cj =  A/t = D/W dep
Cj 
Cj0
1  VR / Vo
where
 2    N A N D  
C j 0  A  Si  

 qVo    N A  N D  
No Applied
Reverse Voltage
C. Hutchens Chap 3 ECEN 3313 Handouts
Estimating Cj - This is done by
using Cj0/2 to Cj0
i.e. VD = is 0 to 0.5V Cj = Cj0
VD = is 0 to 5V Cj = Cj0/2
Engineers to be, must be thinking
and make judgement calls
Leave the details to the
simulator
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Reverse Bias pn Junction
Applied Reverse
Voltage
Depletion Capacitance
Junction charges can be expressed as a function of either depletion region as:
qJ  qn  qx n AN D and using x p N A  xn N D and
qJ  qn  qxn A
xn  x p  Wdep
NDN A
ND  N A
where xp and xn are the respective depletion depths in the n and p type
materials respectively WHERE VR IS THE REVERSE VOLTAGE CONTROLING
W dep,
Wdep 
2 Si
q 1 / N A  1 / N D Vo  VR  .
C. Hutchens Chap 3 ECEN 3313 Handouts
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Fwd. pn Junction on the Small Signal
Forward bias small signal transconductance or resistance is defined by the
slope of the line at Qpt or the derivative of the I vs. V characteristics
evaluated at the operating point.
gd 
I
1
I
 
V I  I
rd nVt
Q
C. Hutchens Chap 3 ECEN 3313 Handouts
The idea of a QPT is
"Very Critical" !
18
Fwd. pn Junction-Diffusion Cap.
CD = T I/Vt = TT I/Vt diffusion cap
EX IF a Diode is forward biased by a ID = 0.1mA current find the small signal
ac model for a the diode the on voltage (VDQ) across the diode. Assume
a Si diode with n=N=1, TT= 27.5nS, CJ0 = 8.3pFd and Is= IS= 2fA
rd = nVt /I = 25mV/0.1mA = 250 ohms
CD = (27.5nS) (0.1mA)/25mV  110pFd
Ctotal = CD + CJ  110pFd + 8.3pFd = 118pFd Note we could just as well
have ignored CJ
VD = n Vt Ln (ID/IS) = 25mV Ln (0.1mA/2fA) = 0.640V
C. Hutchens Chap 3 ECEN 3313 Handouts
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Small Signal vs Large Signal
Typically RS= RS
should be
ignoreable until
VDQ exceeds 700
to 800mV. The
Observations
 IDQ = (V-VDQ)/R, VDQ = n Vt Ln (IDQ/IS) Transindental functions criteria is rd << RS
require iterations.
 Cblock is selected Large enough at time to appear as a short circuit
in the frequecy band if intrest.
 Vbias DC voltage to set IDQ apperas as a short circuit.
Depending on the value of R vs rd it may be possible to ignored.
Node equation across diode vd
vd (G + gd + sCj + s Cd + (vd - vin ) sCblock= 0
vd/vin = sCblock/(G+ gd + s Ctotal)  (Cblock/Ctotal){1/( 1 + gd/sCtotal)}
C. Hutchens Chap 3 ECEN 3313 Handouts
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DC vs AC (Small signal)
Observations:
 C is selected large enough to appear as a short circuit in the frequency band of the
ac generator.
 To get started quickly V is selected equal 0.6V and should be checked at times with
VD = n Vt Ln (ID/IS) = 25mV Ln (4mA/2fA) = 0.710V.

Note that rd can be ignored for all intent and purposes. What is more important is that I D not
destroy the diode or its reverse voltage not be exceeced.
C. Hutchens Chap 3 ECEN 3313 Handouts
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Zener Regulation
Example - Assume VZ = 5.1, Vsmax = 15V, Vsmin = 12V
Assume Izmin = Izmax /10
Izmax = ILMAX (Vsmax - VZ)/ (VSMIN - 0.9 VZ -0.1 VSMAX)
Note PDISS of R is very high at all times. Therefore
zener regulators are very impractical.
C. Hutchens Chap 3 ECEN 3313 Handouts
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Full Wave Rectifer
I = Cdv/dt &
f = /2 = 1/T
Given Ckt diagram and
Vp = 169 V for 120Vrms
VZ = 5V, IZ min = 5mA
Analysis of operation
Vxpeak = 169/14  12V
@ RL =  Ixmax = (Vxpeak -VZ)/R1 = (12-5)/220 = 31.8mA Sets zener power
dissapation. (This should be increased by 10% for saftey purposes)
Vxmin = (IL+ Izmin)R1 + VZ = (5/3.8K + 5mA )220 + 5 = 5.29 V Minimum
allowable voltage for zener operation.
Note proper selection of C insures that Vxmin will not be exceeded.
C> Imax dt/dv from I = Cdv/dt where Imax = IL+ Izmin  6.25mA
where worst case dv/dt = (Vxpeak - Vxmin)2f = (12-5.29)120
C > 6.25mA/{(12-5.29)(120)}  7.8uFd
C. Hutchens Chap 3 ECEN 3313 Handouts
23
Full Wave Rectifer
Observations:
 The diode conducts briefly for a small fraction of the period. See figure.
 Assuming an ideal Diode; conductions occurs during the rising edge of the Vripple signal.
 The capcitor supplies current during the falling portion of Vripple. (IC  IL)
 Only the capacitor supplies load current for the balance of the period T = 1/2f i.e. 1/(2 x 60) =
83mS
 When properly designed Vripple is small compared to Vp or Vo. i.e. 5%
IL  Vp/RL or more accurately IL  (Vp - Vripple/2)/RL
IC  CdVr/dt or after rearranging dVr = IC C dt and integration.
Vr  (IC /C) T  IC /(C2f)  (Vp/RL )/(C 2f)
C. Hutchens Chap 3 ECEN 3313 Handouts
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Voltage Doubler/Clamped Cap
Neg 1/2 cycle D1 conducts
Pos 1/2 cycle D2 conducts
How much loss occures due to the diodes?
How do we determine Pdiss require of the diodes?
How many additional diodes may be added to increase
the voltage further?
Is ripple on the capacitor a problem and if so can we solve
it?
C. Hutchens Chap 3 ECEN 3313 Handouts
25
Diode Switching
C. Hutchens Chap 3 ECEN 3313 Handouts
26
Diode SPICE parameters
.MODEL D1N914/125C D
+
IS = 2.25833E-15
+
RS = 1.799439
+
N = 1.067043
+
TT = 1.46E-7
+
CJO = 3.0177E-12
+
VJ = 0.4
+
M = 0.2147523
+
EG = 1.11
+
XTI = 4.799594
+
KF = 0
+
AF = 1
+
FC = 0.764906
+
BV = 94
+
IBV = 5E-6
+)
C. Hutchens Chap 3 ECEN 3313 Handouts
(
Saturation Current
Contact Resistance
Ideality factor
Fwd Transit time
Zero Bias Junction C
Built in potential
doping grading Coeff.
Breakdown Voltage
27
Diode Summary
Fwd
Rev
Ideal
DC or Large
Signal i.e.
fullwave
rectifier
Zener
Small signal
Small signal
dynamic
C. Hutchens Chap 3 ECEN 3313 Handouts
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