Chap 3

C. Hutchens Chap 3 ECEN 3313 Handouts 1

Chap 3

Diodes

Ideal DiodeTerminal CharacteristicsSemiconductorsPhysical Operation

Open Ckt pn junction Reverse bias pn junction Forward bias pn junction

Diode CircuitsSmall signal modelRectifiersClipping and clampingThe SPICE diode


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


Diode Symbols and Ckt RepresentationI d e a l M a t h m o d e l

I ID sV n V t ( )/ 1

w h e r e n i s t h e e m i s s i o nc o i f f i c i e n t a n d = 1 f o r S i ,I s i s t h e s a t u r a t i o n c u r r e n t , a n dV t i s t h e t h e r m a l v o l t a g e a n d =2 6 m V a t r o o m t e m p

F o r V > 4 n V t 1 0 0 m V

tnVVsD II /

F o r V > > 4 n V t

s h o r t c k t w i t h a v o l t a g e d r o p o f 0 . 5t o 1 . 0 V

a n d f o r V < 4 V t 1 0 0 m V

I ID s 0

VZ or Avalanche


Diode SPICE parameters

.MODEL D1N914/125C D (+ IS = 2.25833E-15 Saturation Current+ RS = 1.799439 Contact Resistance+ N = 1.067043 Ideality factor+ TT = 1.46E-7 Fwd Transit time+ CJO = 3.0177E-12 Zero Bias Junction C+ VJ = 0.4 Built in potential+ M = 0.2147523 doping grading Coeff.+ EG = 1.11+ XTI = 4.799594+ KF = 0+ AF = 1+ FC = 0.764906+ BV = 94 Breakdown Voltage+ IBV = 5E-6+ )


Simple Diode RectiferEX 2 Find the fraction of each cycle thediode conducts and the peak current if vin

= 24 COS 2 f t . Note Typically f will bein the 10 to 1000s of Hertz for thisapplication.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 alldiode parameters vary by +/-20% what isthe worst case peak current, peak andcontinious power dissipationa? What ifvin varies by 20%.


Limiters/Clippers

VO > V1 VO <V2 V2 < VO < V1

Observations

Vo > V1 D1 is a shortcircuit else o.c.

Vo < V2 D1 is a shortcircuit else o.c.

The polarity andmagnitude of V 1

and V2 shift theconduction point.

Ex vin = 20V peak to peakSinusoid, V1 = 10 Vand V2 = -8 V


LOG COMPUTATION AND THE DIODE

I ID sV n V t ( )/ 1 w h e r e n i s t h e e m i s s i o n = 1

B o t h I S a n d V T a r e t e m p e r a t u r e d e p e n d e n t

V a l i d f o r a s m a n y a s 7 + D e c a d e s

C o n s i d e r )1( /21 tnVV

SD II a n d )1( /22 tnVV

SD II TnVVV /)(

D2D112 I/ I

) I/ (IL o g2 .3 ) I/ L n (I D2D11 0D2D112 TT n Vn VVVV = 6 0 m V / D e c a d e a tR m T e m p

E v e r y D e c a d e I n c r e a s e i n c u r r e n t r e s u l t s i n a 6 0 m V i n c r e a s e i nd i o d e v o l t a g e .

T y p i c a l V a l u e s o f D i o d e o n v o l t a g e a r e 1 0 0 t o8 0 0 m V .


N o w I D i s a f u n c t i o n o f d i o d e a r e a o r A J S w h e r e J S i s t h e c u r r e n td e n s i t y

R e w r i t i n g t h e p r e v i o u s e q u a t i o n

) A/ (ALo g2 .3 ) AJ/ A(JLo g2 .3 D2D110D2SD1S10 TT n Vn VV

I f t w o d i o d e s a r e f a b r i c a t e d s i d e b y s i d e a n d c a r e t h e s a m ec u r r e n t b u t a r e o f u n e q u a l a r e a s t h e v o l t a g e d i f f e r e n c e w i l l

b e a f u n c t i o n o f t h e L o g o f t h e r a t i o t h e i r a r e a s .

S e e E X 3 . 3

F o r a s i n g l e d i o d e t h e F O R W A R D v o l t a g e s h i f t i s a p p r o x . - 2 m V / C 0

U s e T e m p e r a t u r e m e a s u r e m e n t

R e v e r s e b i a s c u r r e n t - I S " A r e a d e p e n d e n t D o u b l e s e v e r y 5 0 C . "

COMPUTATION AND THE DIODE


Basic Semiconductor Concepts-Review

C o n d u c t i v i t y i s c o n t r o l l e d b yc a r r i e r c o n c e n t r a t i o n

I n t r i n s i c c a r r i e rc o n c e n t r a t i o n n i = p = n =1 . 5 x 1 0 1 0 c a r r i e r s / c m 3 a tR o o m T e m p .

n B TiE k TG 3 / w h e r e

B i s a m a t e r i a ld e p e n d e n t p a r a8 . 6 2 x 1 0 - 5 e V / K ,E G i s t h e b a n d g a pa n d f o r S i 1 . 1 2 e V ,T i s T e m p . K 0 e q u a l2 7 3 0 a t r o o m t e m p .

S i h a s 5 x 1 0 2 2 a t m s / c m 3 v s .n i = 1 . 5 x 1 0 1 0


Basic Semiconductor ConceptsPhysical factors controlling pnjunction behavior.

Diffusion - Thermalagitation coupled withconc. (carriers/cm3)

Drift - E field associatedcarrier motion.

n-type equilibrium carrierconc. nn0 ND

i.e. Patms/cm3 free electons

p-type equilibrium carrierconc. pp0 NA

i.e. Batms/cm3 free holes

nn0 pp0 = ni2 then

pp0 = ni2/ND and

nn0 = ni2/NA


Doping Example

G i v e n N D = 1 0 1 7 / c m 3 o f P f i n d n n 0

p p 0 a t 3 0 0 K 0 .n B Ti

E k TG 3 / w h e r eB i s a m a t e r i a ld e p e n d e n t p a r a8 . 6 2 x 1 0 - 5 e V / K ,E G i s t h e b a n d g a p a n df o r S i 1 . 1 2 e V ,T i s T e m p . K 0 e q u a l2 7 3 0 a t r o o m t e m p .

n i 2 = ( 1 . 5 x 1 0 1 0 / c m 3 ) 2

n n 0 N D = 1 0 1 7 / c m 3

p p 0 = n i2 / N D = 2 . 2 5 x 1 0 3


Diffusion

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 ) andelectrons ( 34cm2/s) and q is the charge on an electron of hole.

Anode Cathode

x


Driftvdrift = 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 pp E hole drift current

Jn-drift = q pn E electron drift current (neg charge in neg x dir.

results in positive current)Jtotal drift = q E(p

p + pn) or using ohms Law and R= L/A

= 1/( q (pp + p

n)) Note V/I = R = L/A ; = (V/L)(A/I) = E/Jx


Open Ckt pn Junction

D r i f t ( I S ) a n d D i f f u s i o n ( I D ) C u r r e n t s a r e i n E q u i l b r i u m ( I S = I D ) f o r a n o . c . d i o d ea n d f o r c h a r g e n e u t r a l i t y

q x A N q x A Np A n D

x N x Np A n Dw h e r e x p a n d x n a r e t h e r e s p e c t i v e d e p l e t i o n d e p t h s i n t h e n a n d p t y p em a t e r i a l s r e s p e c t i v e l y .

x x W q N N Vn p d e pS i

A D o 2 1 1 / /

d e f i n e s t h e d e p l e t i o n w i d t h .

Vo = VJ, n = N


Open Ckt pn Junction

The pn work function given by;

V V LnN N

no tA D

i

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 = 738mVW det = 0.32umxp = 0.03umxn = 0.29um


Rev. pn Junction-Depletion Cap.

N o t e C = q / R

V R = V Q

o r C j = A / t = D / W d e p

CC

V Vj

j

R o

0

1 / w h e r e

C A

q V

N N

N NjS i

o

A D

A D0

2

E s t i m a t i n g C j - T h i s i s d o n e b yu s i n g C j 0 / 2 t o C j 0

i . e . V D = i s 0 t o 0 . 5 V C j = C j 0

V D = i s 0 t o 5 V C j = C j 0 / 2E n g i n e e r s t o b e , m u s t b e t h i n k i n g

a n d m a k e j u d g e m e n t c a l l sL e a v e t h e d e t a i l s t o t h e

s i m u l a t o r

No AppliedReverse Voltage


Reverse Bias pn Junction

D e p l e t i o n C a p a c i t a n c eJ u n c t i o n c h a r g e s c a n b e e x p r e s s e d a s a f u n c t i o n o f e i t h e r d e p l e t i o n r e g i o n a s :

q q q x A NJ n n D a n d u s i n g x N x Np A n D a n d x x Wn p d e p

q q q x AN N

N NJ n nD A

D A

w h e r e x p a n d x n a r e t h e r e s p e c t i v e d e p l e t i o n d e p t h s i n t h e n a n d p t y p em a t e r i a l s r e s p e c t i v e l y W H E R E V R I S T H E R E V E R S E V O L T A G E C O N T R O L I N G

W d e p ,

W q N N V Vd e pS i

A D o R 2 1 1 / / .

Applied ReverseVoltage


Fwd. pn Junction on the Small Signal

Forward bias small signal transconductance or resistance is defined by theslope of the line at Qpt or the derivative of the I vs. V characteristicsevaluated at the operating point.

gI

V r

I

nVdI I d tQ

1

The idea of a QPT is"Very Critical" !


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 signalac model for a the diode the on voltage (VDQ) across the diode. Assumea 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 wellhave ignored CJ

VD = n Vt Ln (ID/IS) = 25mV Ln (0.1mA/2fA) = 0.640V


Small Signal vs Large Signal

Observations IDQ = (V-VDQ)/R, VDQ = n Vt Ln (IDQ/IS) Transindental functions

require iterations.

Cblock is selected Large enough at time to appear as a short circuitin 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)}

Typically RS= RS

should beignoreable untilVDQ exceeds 700to 800mV. Thecriteria is rd << RS


DC vs AC (Small signal)

Observations:

C is selected large enough to appear as a short circuit in the frequency band of theac generator.

To get started quickly V is selected equal 0.6V and should be checked at times withVD = 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 ID notdestroy the diode or its reverse voltage not be exceeced.


Zener Regulation

Example - Assume VZ = 5.1, Vsmax = 15V, Vsmin = 12VAssume Izmin = Izmax /10Izmax = ILMAX (Vsmax - VZ)/ (VSMIN - 0.9 VZ -0.1 VSMAX)

Note PDISS of R is very high at all times. Thereforezener regulators are very impractical.


Full Wave Rectifer

Given Ckt diagram andVp = 169 V for 120VrmsVZ = 5V, IZ min = 5mA

Analysis of operationVxpeak = 169/14 12V

@ RL = Ixmax = (Vxpeak -VZ)/R1 = (12-5)/220 = 31.8mA Sets zener powerdissapation. (This should be increased by 10% for saftey purposes)

Vxmin = (IL+ Izmin)R1 + VZ = (5/3.8K + 5mA )220 + 5 = 5.29 V Minimumallowable 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 6.25mAwhere worst case dv/dt = (Vxpeak - Vxmin)2f = (12-5.29)120

C > 6.25mA/{(12-5.29)(120)} 7.8uFd

I = Cdv/dt &

f = /2 = 1/T


Full Wave RectiferObservations:

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)


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 increasethe voltage further?

Is ripple on the capacitor a problem and if so can we solveit?


Diode Switching


Diode SPICE parameters

.MODEL D1N914/125C D (+ IS = 2.25833E-15 Saturation Current+ RS = 1.799439 Contact Resistance+ N = 1.067043 Ideality factor+ TT = 1.46E-7 Fwd Transit time+ CJO = 3.0177E-12 Zero Bias Junction C+ VJ = 0.4 Built in potential+ M = 0.2147523 doping grading Coeff.+ EG = 1.11+ XTI = 4.799594+ KF = 0+ AF = 1+ FC = 0.764906+ BV = 94 Breakdown Voltage+ IBV = 5E-6+ )


Diode Summary

Ideal

DC or LargeSignal i.e.fullwaverectifier

Zener

Small signal

Small signaldynamic

Fwd Rev

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Chap 3