© REP 8/14/2015 EGRE224 Electronics - Diodes Page 3.1-1 Introduction The simplest and most...

© REP 04/19/23 EGRE224

Electronics - Diodes

Page 3.1-1

Introduction

The simplest and most fundamental nonlinear circuit element is the diode. It is a two terminal device like a resistor but the two terminals are not interchangeable.

We will start by describing an “ideal” diode and then look at how closely a real diode approximates the ideal situation. We will be considering silicon diodes throughout this book.

As electrical engineers we can analyze diode circuits if we have equations which describe the terminal characteristics of the device, but we need to look further and understand physically how the diode works since the diode is also the basis of the BJT and MOSFET devices we will be studying later in this course.

One of the most common uses of diode is in rectifier circuits (conversion of ac signals to dc) so we will spend some time on examples and then look at some other diode applications

We also need to look at how diode model parameters can be extracted for use in simulation programs such as SPICE. The parameters can then be used to simulate some of the application circuit examples.

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Page 3.1-2

The Ideal Diode

The diode symbol and terminal voltage and current definitions are shown to the right. The quantity VA is referred to as the Applied voltage. YOU MUST MEMORIZE THIS FIGURE!

The i-v characteristic for the ideal diode passes no current when the applied voltage (with the polarity given in the definition) is negative, and when the applied voltage is positive the diode is a perfect short circuit (zero resistance).

p n

Av

anode cathode

i

i

Av

Forward BiasReverse Bias

Reverse Bias Circuit Model Forward Bias Circuit Model

00 ivA

i Av

00 Avi

i Av

“Cut off” “ON”

The external circuit must limit the currentunder Forward Bias conditions since thediode will have no resistance

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.1-3

The diode is polarity dependent!

Forward Bias Current Limit Example (resistor limits the current)

k1

V10

k1

V10

k1

V0

V10

V10 k1

V10

V10

V0

Short circuit Open circuit

Reverse Bias Current Limit Example (diode, in cut off, limits the current)

mA1

mA0PN

NP

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Page 3.1-4

A Simple Application: The Rectifier

OvDi

Dv

Iv R

IvPv

t

IO vvDi

0Dv

Iv R

0for Iv0for Iv

0Ov0Di

ID vv

Iv R

OvPv

t

The negative half-cycle is blocked

The positive half-cycle is transmitted

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Page 3.1-5

Exercises Involving Rectification

Exercise 3.1 Sketch the transfer characteristic of simple rectifier. The transfer characteristic is vO vs. vI . We see that when vI is negative vO zero and when vI is positive vO is equal to vI

Exercise 3.2 Find the waveform of vD. Well we know that the input voltage has to divide across the diode and the resistor, so when there is no voltage across the resistor (vO =0) then it must be across the diode and vice-versa. The diode voltage will be the exact complement of the output voltage in this case.

If vI has a peak value of 10V and R=1k, find the peak value of iD and the dc component of vO.

mAR

vi PP 10

000,1

10

Vv

v

dc

dc

O

O

18.310

115

cos5

2

sin100

0

45 degreesIO vv

Iv

Ov

PvDv

t

Pv

t

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Page 3.1-6

Battery Charger Rectification Example

Example 3.1 The circuit below is used to charge a 12V battery, where vS is a sinusoid with a 24V peak amplitude. Find the fraction of each cycle during which the diode conducts, also find the peak value of the diode current and the maximum reverse-bias voltage that appears across the diode.

V12

Di

Sv

100RV12

t

V24

2

12cos24

Di

Sv

Di

cycle theof thirdoneor 1202

605.0cos 1

AVV

I Pd 12.0

100

1224

VVVV Pd 362412max

V24

maxdV

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Page 3.1-7

Another Application: Diode Logic Gates

Diodes and resistors can be used to implement digital logic functions 0V is a Low and +5V is a high In the circuit on the left below if any one of the three inputs is at +5V the output vQ will

also be at +5V and there will be a current flowing through the resistor. If all three input are zero the diodes will be cut off and the output will be grounded through the resistor. The results are summarized in the OR gate truth table next to the circuit

In the circuit on the right below, if any of the inputs are zero that diode will be on and the output will be at zero volts. If all three inputs are at +5V the diodes will be cut off and the output will be at +5V. The results are summarized in the AND gate table.

R

Av

Bv

Cv Qv

R

Av

Bv

Cv

Qv

V5Av Bv Cv Qv

0 0 0 00 0 5 50 5 0 50 5 5 55 0 0 55 0 5 55 5 0 55 5 5 5

Inputs OutputOR Gate Av Bv Cv Qv

0 0 0 00 0 5 00 5 0 00 5 5 05 0 0 05 0 5 05 5 0 05 5 5 5

Inputs Output

AND Gate

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Page 3.1-8

Simple DC Analysis of Ideal Diode Circuits

Example 3.2(a) Given the following circuit, Find the indicated values of I and V. How do we know which diodes are conducting and which are not? It might be hard to

tell, so we make an assumption (always write down your assumption), then proceed with your analysis and then check to see if everything is consistent with your initial assumption. If things are not consistent then our assumption was invalid. NOTE, this does not mean that all your work was in vain, sometimes it is just as important to prove what is incorrect as what is correct.

For now, lets assume that both diodes are conducting

k5

I1D

V10

V

2D

V10

k102DI

If D1 is on VB=0 and the output V=0 also.We can now find the current through D2

B

mAVV

ID 1000,10

0102

We can write a node equation at node B,looking at the sum of the currents

mAI

VVmAI

III kD

1

000,5

1001

52

Therefore D1 is on as assumed

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Page 3.1-9

Another Circuit

Example 3.2(b) This is the same circuit as the previous one except that the values of the two resistors have been exchanged.

Again I will assume that both diodes are on, do the analysis and check the results. Again VB=0 and V=0.

k5

I1D

V10

V

2D

V10

k10

2DI

B

mAVV

ID 2000,5

0102

We can write a node equation at node B,looking at the sum of the currents

mAI

VVmAI

III kD

1

000,10

1001

102

Not possible, thereforeassumption was wrong

mA

VVID 33.1

000,5000,10

10102

Now assume D1 is off and D2 is on

Now solving for VB we get 3.33V and I=0 since D1 is off

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Page 3.1-10

Diode Terminal Characteristics

- 4mA

- 3mA

- 2mA

- 1mA

1mA

2mA

3mA

4mA

5mA

An Analog sweep has been converted to Digital (discrete values)

(+0.2 volt increments)

VRd =

V

slope = V

1slope=

Va

I

1.0slightly negative

ForwardBias Va > 0

ReverseBias Va < 0

Rd is the dynamic (changing) resistance

BreakdownVoltage

Turn-onVoltage

from -6 to -hundreds of volts

Rd

Va

Calculations

The resistance of the diode is not constant, it depends on the polarityand magnitude of the applied voltage

The higher the doping levels of then and p sides of the diode, the lowerthe breakdown voltage.

“on” R is low

“Off”R is high

breakdown

The turn-on voltge is a function of the semiconductor used.~ 0.7V for Si and ~ 1.7V for GaAs

“closed” switch

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Page 3.1-11

Diode Analogy

A Diode can be thought of as a one-way valve (one-way street!) When no force (voltage) is applied to the valve, no current

flows

When a force (voltage) greater than a particular threshold is applied in one direction, a current can flow

When a force is applied in the opposite direction no (very little) current can flow unless the diode undergoes breakdown.

IForward

Ireverse

Breakdown

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Page 3.1-12

Determining the Polarity of a Diode

Va

I

1.0

The connections are correctVa is being applied to the p-side of the diode.

Va

I

-1.0

Reverse the connectionsto the diode, Va is being applied to the n-side of thediode

Va

Va

PN

NP

black

Curve tracer

red

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Page 3.1-13

The Forward Bias Region

Forward-bias is entered when va>0 The i-v characteristic is closely approximated by

Is, saturation current or scale current, is a constant for a given diode at a given temperature, and is directly proportional to the cross-sectional area of the diode

VT, thermal voltage, is a constant given by

K = Boltzman’s constant = 1.38 x 10-23 joules/kelvin T = the absolute temperature in kelvins = 273 + temp in C q = the magnitude of electronic charge = 1.60 x 10-19 coulomb

For appreciable current i, i >>IS, current can be approximated by

or alternatively

Va

I

1.0

p n Av

i

1eI V Tn

v

Si

q

kTV T

eI Vi Tnv

S

IVS

T

inv ln

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Page 3.1-14

The Reverse Bias Region

Reverse-bias is entered when va < 0 and the diode current becomes

Real diodes exhibit reverse currents that are much larger than IS. For instance, IS for a small signal diode is on the order of 10-14 to 10-15 A, while the reverse current could be on the order of 1 nA (10-9 A).

A large part of the reverse current is due to leakage effects, which are proportional to the junction area.

I Si

I

Va

VZK

breakdown voltage

reverse-bias region

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Page 3.1-15

The Breakdown Region

The breakdown region is entered when the magnitude of the reverse voltage exceeds the breakdown voltage, a threshold value specific to the particular diode. The value corresponds to the “knee” of the i-v curve and is denoted VZK. Z stands for Zener, which will be discussed later, and K stands for knee.

In the breakdown region, the reverse current increases rapidly, with the associated increase in voltage drop being very small.

I

Va

VZK

breakdown voltage

breakdown region

reverse-bias region

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Page 3.1-16

Conductors and Insulators

Ohm’s Law: V = I R

+

E

e-i

Conductor: small V large i R is small

Insulator: large V small i R is large ()

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Page 3.1-17

SemiconductorsTetrahedron Covalent Bonds in a Semiconductor

Figure taken from Semiconductor Devices, Physics and Technology, S. M. Sze,1985, John Wiley & Sons

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Page 3.1-18

Semiconductors (cont.)Bonds, Holes, and Electrons in Intrinsic Silicon


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Page 3.1-19

Doped SemiconductorsBonds, Holes, and Electrons in Doped Silicon


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Page 3.1-20

The Diode

n

p

p

n

B A SiO2Al

A

B

Al

A

B

Cross-section of pn -junction in an IC process

One-dimensionalrepresentation diode symbol

Figure taken from supplemental material for Digital Integrated Circuits, A Design Perspective, Jan M. Rabaey,1996, Prentice Hall

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Page 3.1-21

Carrier Motion

Carriers move due two two different mechanisms Carrier drift in response to an electric field Carriers diffuse from areas of high concentration to areas of lower concentration

Since both carrier types (electrons and holes) can be present and there are two mechanisms for each carrier there are four components to the overall current, as shown below

J J J q pE qD p

J J J q nE qD n

p p drift p diffusion p p

n n drift n diffusion n n

| |

| |

drift diffusion

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Page 3.1-22

Diffusion Current

Carriers move from areas of high concentration to low concentration

dx

dpqDJ pp

dx

dnqDJ nn

hole conc.

hole motion

current flow

negative is dx

dp

negative is dx

dn

electron conc.

electron motion

current flow

++++++

++

+++ +

++

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Page 3.1-23

Carrier Drift

Definition - Drift is the motion of a charged particle in response to an applied electric field. Holes are accelerated in the direction of the applied field Electrons move in a direction opposite to the applied field Carriers move a velocity known as the thermal velocity, th

The carrier acceleration is frequently interrupted by scattering events Between carriers Ionized impurity atoms Thermally agitated lattice atoms Other scattering centers

The result is net carrier motion, but in a disjoint fashion Microscopic motion of one particle is hard to analyze We are interested in the macroscopic movement of many, many particles

Average over all the holes or all the electrons in the sample The resultant motion can be described in terms of a drift velocity, vd

e-

E

vth

vdrift

typical value of ~ 5x106 cm/s

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Page 3.1-24

Carrier Drift, continued

Definition - Current, is the charge per unit time crossing an arbitrarily chosen plane of observation oriented normal to the direction of current flow.

Consider a p-type bar of semiconductor material, with cross-sectional area A.

The current can be written as:

We seek to directly relate J to the field. For small to moderate values of the electric field the measured drift velocity is directly

proportional to the applied field, we can write

The mobility is the constant of proportionality between the drift velocity and the electric fieldv E

cm

Vdrift p p

where is the hole mobility in 2

sec

I qpv A qpvP drift d d or in vector form J

where J is the current density, current per unit area

P drift

A+

E

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Page 3.1-25

Drift Velocity vs Electric Fieldproportionality constant is the Mobility

Some typical values for carrier mobilites in silicon at 300K and doping levels of 1015 cm-

3

100 1 103 1 104 1 105 1 1061 104

1 105

1 106

1 107

v dn( )Efield

v dp( )Efield

Efield

velocity saturation

sec480

sec1350

22

V

cm

V

cmpn

Electric Field in Volts/cm

cm/sec E

vdrift

typical value of ~ 107 cm/s

Carrier Drift Velocity

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Page 3.1-26

Abrupt Junction Formation

Junction Formation

Carrier Concentrations pp ~ Na

np0 ~ (ni )2 / Na

nn ~ Nd

pn0 ~ (ni )2 / Nd

P N

X0

pp

np0

nn

pn0

The Depletion Region represents an immobile donor impurity

(i.e. P+ ) represents an immobile acceptor impurity

(i.e. B- ) - represents a mobile electron + represents a mobile hole

+-

+- + +

+ + +

++ +

+ + ++

+ + +++

++

+

-- -- -

- --

- - - - -- - - - -

- - - - -

- - - - -

n typep type

x

+

+++ +

+ ++

+ ++

++

+++ + ++

+

--

--

-

---

-

DepletionRegion

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Page 3.1-27

The Depletion or Space Charge Region

Maximum Field (Emax )

xn

+qNd Q n = qNdxn

xQ p = -qNaxp

Electric field (x)

-qNa

+

-

xp

Abrupt depletion approximation

xn

Electrostaticpotential V(x) (Volts)

xp

Vbi

charge density

xp

xn xd = xn + xp

(Coulombs/cm-3)

(Volts/cm)

-+

++

++

electron diffusion

hole diffusion

hole drift

electron drift

---

--

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Page 3.1-28

Reverse Bias

x

pn0

np0

-W1 W20n-regionp-region

diffusion

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Page 3.1-29

Forward Bias

x

pn0

np0

-W1 W20

p n(W

2)

n-regionp-region

Lp

diffusion

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Page 3.1-30

Analysis of Forward Biased Diode Circuits

We have already looked at the ideal diode model for forward bias (short circuit). In this section we will work with a detailed model and then explore simplifying assumptions that allows us to work back towards our ideal case.

We will use a simple circuit consisting of a dc source VDD and a resistor and a diode in series. We want to determine the exact current through the circuit, ID and the exact voltage dropped across the diode VD.

If we assume that the voltage source VDD is greater than ~0.5 volts the diode will obviously be in the forward mode of operation and the current through the diode will be given by the following equation

Note we do not know the exact value of VD but we can relate it to other values in our circuit, for example we can write a Kirchhoff’s loop equation

T

D

nV

V

SD eII

R

VVI DDD

D

+VD

-

ID

VDD +

R If we assume that IS and n are known, we

have two equations and two unknowns (IS and VD) and we can solve for them by Graphical means Iterative (mathematical) means

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Page 3.1-31

Graphical (Load Line) Analysis

Our circuit has two components (not counting the voltage source), a resistor and a diode, which are connected to each other.

Each device constrains (or puts limits on) the other

Consider a toy slot car race track with a battery powered car. The car could go any where if put on a wide open surface but when placed on the track it is constrained to follow the course. The car will only be found on the course (the track constraint) and the motor will determine where on the course (car constraint). The exact position depends on both constraints

We will plot the characteristics of each device separately in the circuit as if the other device was not there and then combine our constraints for the final solution

We have already looked at the diode and its characteristic is repeated here

iD (mA)

vD (V)0 0

Since the n side of the diode is grounded the characteristic looks like our typical characteristic (already presented)

DDDD

DDD

viV

Vv

and , as

tolinked is

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Page 3.1-32

Graphical (Load Line) Analysis continued

Lets look at the resistor characteristic now

In this case one terminal of the resistor is at the voltage VDD and the other is at some unknown voltage VD at the diode

We can determine this unknown voltage (operating point) by superimposing the graphs of the expressed for diode current.

i (mA)

v (V)

0 0

R

VDD

iR (mA)

vR (V)

R

VDD

DDV

DDV

T

D

nV

V

SD eII

R

VVI DDD

D

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Page 3.1-33

The straight line is known as the load line.

The load line intersects the diode curve at point Q, the operating point. The coordinates of Q are ID and VD.

i (mA)

v (V)

0 0

R

VDD

DDV

I D

V D

Rslope

1

Q

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Page 3.1-34

Iterative Analysis

Example 3.4 Assume that the resistor in our graphical analysis circuit is 1k and VDD is 5V

The diode has a current of 1mA if it is at a voltage of 0.7 volts and the voltage drops by 0.1 volt for every decade decrease in current.

Find the current through the circuit and the exact voltage across the diode. We can start by assuming we have set up the conditions so that the voltage across the

diode is 0.7 volts, we do this so that we can do some calculations about our diode that we can use later to zero in on our actual conditions

This current is larger than the 1mA current at 0.7 volts so we conclude that the actual diode voltage will be larger than 0.7 volts. Since the relationship between the current and the voltage is exponential we can adjust our voltage estimate slightly using an equation we derived earlier relating the voltage change to the current ratio, namely

Now using this value in our original equation we get

mAR

VVI DDD

D 3.41000

7.05

74.1 then1.03.2 If :note 763.0

0.001

0.0043log1.0 so log3.2

2

121

212

nnVVV

VVI

InVVV

T

T

mAR

VVI DDD

D 237.41000

763.05

VV 762.0

0.0043

0.004237log1.0763.02

Converged toID=4.237mAVD=0.762V

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Page 3.1-35

A graphical view of the iterative analysis

i (mA)

v (V)

0 0

R

VDD

DDV0.7V

1

1.0 mA

0.763V

1

2

4.3 mA

2

0.762V 3START

4.237 mA

3END

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Page 3.1-36

Approximating the diode forward characteristic with two straight lines

The analysis of a diode circuit can be greatly simplified by approximating the exponential i-v curve with two straight lines. One line, A, has a zero slope and the second line, B, has a slope of 1/rD

The piecewise-linear model is described as follows:

,

,0

VD0D0

0

vrVviVvi

DDDD

DDD

iD (mA)

vD (V)0

B, slope =rD

1

A, slope = 0

V D0

vD

rD

ideal

iD

V D0

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Page 3.1-37

Constant-Voltage Drop Model

This model is even simpler than the piecewise-linear or battery-plus-resistance model shown on the previous slide. Here, we use a vertical straight line, B, to approximate the fast-rising part of the exponential i-v curve of the diode.

We assume that a forward-conducting diode exhibits a constant voltage drop, VD, which is approximately 0.7 V.

This model is used in the initial phases of analysis and design to give a rough estimate of circuit behavior.

iD (mA)

vD (V)0

A, horizontal

V D

B, vertical

vD

ideal

iD

V 7.0V D

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Page 3.1-38

Example

Exercise 3.16 For the circuit shown below, find ID and VD for VDD=5V and R=10kAssume that the diode has a voltage of 0.7V at 1mA current and the voltage changes by 0.1V / decade of current change.

Use (a) iteration, (b) the piecewise linear model with VD0=0.65V and rD=20, and (c) the constant voltage-drop model with VD=0.7V.

+VD

-

ID

5 V

10k

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Page 3.1-39

Example, continued

Iteration

V 663.01

0.434ln 0434.07.0

mA 434.010

663.05

V 663.043.0ln 043.07.0 Thus,

1.010 ln where,1

43.0ln 7.0

mA 43.010

7.05

V 7.0

V

I

D

D

D

TTD

D

D

V

VVV

I

V

nn V 0434.0V Tn

+VD

-5 V

10k

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Page 3.1-40

Example, continued

The piecewise-linear model

10k

5 V

+VD

- 200.65 V

V 659.002.0434.065.0

mA 434.002.0

65.05

V

I

D

D

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Page 3.1-41

Example, continued

The constant-voltage-drop model

10k

5 V

+VD

-

0.7 V

V 7.0

mA 43.010

7.05

V

D

DI

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Page 3.1-42

DC Forward Bias with an ac small signal

The DC bias level determines the ac parameters

By restricting the input signal swing to small values we can “linearize” the characteristic like we did for amplifier transfer characterisitcs

+vD(t) -

vd(t)

iD(t)

VD +DC

ac

(DC+ac)

(DC+ac)VD=0.7

VD0 vd (t)

iD (mA)

vD (V)

t

ID

Bias Point - Q

tangent at Q

dr

1slope

id (t)

0.55 0 0

1.0

0.75 0.65

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Page 3.1-43

Small Signal Analysis

If we set the ac signal to zero, the current through the diode due to the DC bias is given by

When we add in the ac small signal to the DC voltage bias the total signal is

The total (DC +ac) instantaneous current is

Which we can re-arrange to get

Substituting in the DC equation from above, we get

T

D

nV

V

SD eII

T

D

nV

v

SD eIti )(

)()( tvVtv dDD

T

dD

nV

vV

SD eIti

)(

T

d

T

D

nV

v

nV

V

SD eeIti )(

T

d

nV

v

DD eIti )(

1T

d

nV

v

T

dDD nV

vIti 1)(

T

dDDD nV

vIIti )(

dDD iIi

If we keep the amplitude of the ac signal small, such that

We can expand the exponential in an infinite series, but we find that a sufficiently accurate expression can be found using only the first two terms.

This IS the small signal approximation, valid for amplitudes less than about 10mV

We find that the total current is made up of a DC component and an ac component that is directly proportional to the small signal voltage AND the DC bias level

T

dDd nV

vIi Where

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Page 3.1-44

Small Signal Resistance (incremental resistance)

On the previous page we found

And since,

The ac small-signal resistance is inversely proportional to the DC bias current ID

In the graphical representation we find that about the Q point

DD IiD

D

d

vi

r

1

0

1DD

dD Vv

ri

D

T

dd

T

Dd

d

d

I

nV

gr

nV

Ig

v

i

1

T

dDd nV

vIi iD (mA)

vD (V)

ID

Bias Point - Q

tangent at Q

dr

1slope

0

0 VD0

The equation of the tangent line is given by:

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Page 3.1-45

The Equivalent Circuit Model for the Diode

The equation of the tangent line is a model of the diode operation for small signal changes about the bias DC point (Q point)

The total model has the components shown below

The incremental voltage across the diode is

0

1DD

dD Vv

ri

ddDD

dddDDD

ddDDD

dDDD

riVv

rirIVv

riIVv

riVv

0

0

0+vD

-

iD

rd

VD0

ideal

vD (V)

ID

Bias Point - Q

tangent at Q

dr

1slope

0

0 VD0 VD

ddd riv

di

dv

© REP 04/19/23 EGRE224


Page 3.1-46

Application of the Diode Small-Signal Model

Consider the circuit shown at the right, with combined DC and ac voltage input causing a DC and ac current. We can analyze the response of the circuit by using the diode model developed on the previous page and performing the circuit analysis

+vD=VD+vd

-

vs

iD=ID+id

VDD +

R

ddDDsDD

dddDDDsDD

ddDDdDsDD

dDDDsDD

rRiVRIvV

rRirIVRIvV

riIVRiIvV

riVRivV

0

0

0

+

vD=VD+vd

-

vs

iD=ID+id

VDD

R

rd

VD0

ideal

© REP 04/19/23 EGRE224


Page 3.1-47

Application of the Diode Small-Signal Model continued

Separate the result from the previous page into a DC response and model and an ac response and model

The small-signal analysis is done by eliminating all DC sources and replacing the diode with the small-signal resistance. Using ac voltage division of the ac signal voltage we get the small-signal voltage across the diode to be

+

VD

-

ID

VDD

R

rd

VD0

ideal

DDDD VRIV

+

vd

-

vs

id

R

rd

dds rRiv

ddDDsDD rRiVRIvV

Circuit for DC Analysis Circuit for small-signal Analysisd

dsd rR

rvv

© REP 04/19/23 EGRE224


Page 3.1-48

Power Supply Ripple Example

Example 3.6 The power supply has a 10V DC value and a 1V peak-to-peak sinusoidal ripple at a frequency of 60 Hz.

The ripple is an imperfection of the DC power supply design (we will talk about this in more detail in a later section)

Calculate the dc voltage across the diode and the magnitude of the sine-wave signal appearing across it

Assume the diode has a 0.7V drop at a current of 1mA and that the ideality factor n=2

Calculate the dc diode current by assuming VD=0.7V

Since this value is close to 1mA the diode voltage will be close to the assumed value of 0.7V. At this DC operating point we can calculate the incremental (dynamic) resistance rd as follows

The peak-to-peak small signal voltage across the diode can be found using the ac model and the voltage divider rule

This value is quite small and our use of the “small signal” model is justified

+vd

-

+V=10V+ripple

R=10k

mAID 93.0000,10

7.010

8.53

1093.0

025.023xI

nVr

D

Td

mVrR

rpeaktopeakv

d

dd 7.10

8.53000,10

8.5322

+V=10V+ripple

R=10k

rd=53.8+vd

-

© REP 04/19/23 EGRE224


Page 3.1-49

Voltage Regulation Using Diode Forward Voltage Drops

Example 3.7 The string of three diodes shown in the figure provide a voltage of about 2.1V

We want to see

a) how much of a fluctuation (percentage change in regulation) there is in the output for a 1V (10%) change in the power supply voltage

b) percentage change in regulation when there is a 1k load resistance. Assume n=2

With no load the nominal dc current is given by

Thus the dynamic resistance of each diode is

The total resistance of the diodes will be 3rd or 18.9 Using voltage division on the 1V p-p change (10%) we get

+

vo

-

10V + 1V

R=1k

RL=1kmAID 9.7

000,1

1.210

3.6

109.7

025.023xI

nVr

D

Td

mV

rR

rv

d

do 1.37

9.18000,1

9.182

3

32

© REP 04/19/23 EGRE224


Page 3.1-50

Voltage Regulation continued

When the load resistor is connected it draws current a current from the node that the diodes are connected to which reduces the dc current in the diode string.

If it is assumed that the dynamic resistance stays the same, then the output small signal change is given by

But when the dc current in the diode string is decreased the dynamic resistance changes

This leads us to

It appears that the small signal model is not entirely justified

mAIL 1.2000,1

1.2

6.8

108.5

025.023xI

nVr

D

Td

mV

RrR

Rrv

Ld

Ldo 1.49

8.25000,1

8.252

3

32

mAmAmAIII LRDiodes 8.51.29.7

mVrv do 7.399.180021.030021.0

© REP 04/19/23 EGRE224


Page 3.1-51

Diode Model for High Frequencies

The small signal model that we have developed is a resistive one and it applies for low frequencies where the charge storage is negligible.

Charge storage effects were modeled by two capacitances The diode depletion layer capacitance (Cj)

The forward biased diffusion capacitance (Cd) When we include these two capacitances in parallel

with the diode’s dynamic resistance (rd) we get the high frequency diode model shown at the right

The formulas for the model parameters are also shown at right

For high frequency digital switching applications large signal equations for Cj and Cd are used

Cjrd Cd

0for 2

0for

1

, :Point Bias DC

0

0

0

Djj

Dm

D

jj

DT

Td

D

Td

DD

VCC

V

VV

CC

IV

C

I

nVr

VI

© REP 04/19/23 EGRE224


Page 3.1-52

Small-Signal Resistance Calculation and Model

Exercise 3.20 Find the value of the diode small-signal resistance rd at bias currents of 0.1, 1, 10mA (assume n=1)

Exercise 3.21 For a diode that conducts 1 mA at a forward voltage drop of 0.7V (with n=1), find the equation of the straight line tangent at ID=1mA.

From the question above we find that rd is 25 and we can substitute in the bias point values and solve for VD0

5.2

01.0

025.01 25

001.0

025.01 250

0001.0

025.01321

D

Td

D

Td

D

Td I

nVr

I

nVr

I

nVr

VV

V

VVvr

i

D

D

DDDd

D

675.0

7.0025.0

7.025

1001.0

1

0

0

00

© REP 04/19/23 EGRE224


Page 3.1-53

Exercise 3.22, How small is small in the Small-signal model

Consider a diode with n=2 biased at a dc current of 1mA. Find the change in current as a result of changing the voltage by (a) -20mV (b) -10mV (c) -5mV (d) +5mV (e) +10mV (f) +20mV. Do the calculations using both the small-signal model and the exponential model

(a)

001.0

1

1

21

025.02

02.0

121

02.002.0

21

111

DD

DDD

nVnV

V

SnV

V

SnV

V

SDD

II

eIII

eeIeIeIII TT

D

T

D

T

D

0021

11DD

dDD

dDD Vv

rVv

rII

© REP 04/19/23 EGRE224


Page 3.1-54

Diode Characteristic in the Reverse Breakdown Region - Zener Diodes

If the Zener diode is biased in the reverse breakdown region of operation the current can fluctuate wildly about the Q point and the voltage across the diode will remain relatively unchanged

The knee current and knee voltage is usually specified on a zener diode data sheet

The incremental (dynamic) resistance in reverse breakdown is given by rZ

+VZ

-

IZ

i

v

IZT

V

VZ0

Bias Point - QZr

1slope

VZKnee VZ

Vr

Circuit symbolfor a Zener diode

IZKnee

Test current

© REP 04/19/23 EGRE224


Page 3.1-55

The Reverse Bias Zener Model

We can see from the previous page that we can model the zener diode in the breakdown region as straight line having an x (voltage) intercept at VZ0

and a slope of 1/rZ. The model is shown at the right

the reverse breakdown characteristic of a Zener diode is very steep (low resistance). For a very small change in voltage biased in the breakdown region the current changes significantly.

The zener diode can be used to absorb or buffer a load from large current changes, I.e. keep the voltage across the load approximately constant

+VZ

-

IZ

rz

VZ0

ZzZZ IrVV 0

intercept

Zr

1slope

© REP 04/19/23 EGRE224


Page 3.1-56

A Shunt Regulator

Load

+VO

-

IL

IZ

I

R

Zenerregulator

Load

+VO

-

IL

IZ

I

R

Zenerregulator

i

v

i

v

© REP 04/19/23 EGRE224


Page 3.1-57

Zener Voltage Regulation

Example 3.8 A 6.8 V Zener diode in the circuit shown below is specified to have VZ = 6.8V at IZ = 5mA and rZ =20, and IZK = 0.2mA.

The supply voltage is nominally +10V but can vary by plus or minus 1 V.

(a) Find the output VO with no load and V+ at 10V

(b) Find the value of VO resulting from the +/- 1 V change in V+

(c) Find the change in VO resulting from connecting a load resistance RL= 2 k

(d) Find the value of VO when RL =0.5 k

(e) What is the minimum value of RL for which the diode still operates in the breakdown region.

We can start by determining the value of VZ0. VZ0 is the x-axis intercept of the line tangent to the characteristic at the reverse bias operating point

+

vo

-

V+ (10V + 1V)

R = 0.5 k

RL= 1 k 6.8Vzener

VVV

IrVVIrVV

ZZ

ZzZZZzZZ

7.6 005.0208.6

00

00

© REP 04/19/23 EGRE224


Page 3.1-58

Zener example continued

With no load connected, the current through the zener diode is given by

We can now find V0, the voltage at the operating point current

Now, for a +/- 1V change in V+ can be found from

When a load resistance of 2k is connected, the load current will be approximately 6.8V/2000 or 3.4mA. This current will not be flowing through the zener diode if it is flowing through the load so the change in the zener current is -3.4mA. The corresponding change in the zener voltage (which is also the output voltage) is,

A more accurate result comes from analysis of this circuit

mArR

VVII

Z

ZZ 35.6

20500

7.6100

VrIVV ZZZ 83.602.035.67.600

mVrR

rVV

Z

Z 5.3820500

2010

mVmAIrV ZZ 684.3200

+vo

-

V+

R = 0.5 k

RL= 2 kmVV 700 rz

VZ0

© REP 04/19/23 EGRE224


Page 3.1-59

Zener Example continued

If we change the load resistance to 500 the load current would increased to 6.8V/500 = 13.6mA, but the most current we could get through the pull up resistor and still have the zener in breakdown would be (10-6.8)/500 or 6.4 mA, so we can’t approach 13.6mA unless the zener diode is off (reverse biased but not in breakdown). With the diode off we have a simple voltage divider between the pull up resistor and the load resistor.

For the zener to be at the edge of the breakdown region, the current has to be IZ=IZK=0.2mA and VZ=VZK=6.7V. At this point the current supplied through the resistor R is (9-6.7)/500 or 4.6 mA. The load current would be this current minus the current flowing through the zener to just keep it at the breakdown knee (0.2mA), or 4.6mA - 0.2mA = 4.4mA. We can now find the value of RL for to cause this

mAI

VRR

RVV

L

L

LO

10500500

10

voltagebreakdownzener thelower than is which

5500500

50010

500,10044.0

7.6LR

© REP 04/19/23 EGRE224


Page 3.1-60

Shunt Regulator

S

O

V

V

Regulation Line

L

O

I

V

Regulation Load

VS VSmax

VSmin

t

VO

t

Load+VO

-

VS

IL

IZ

I

R

Zenerregulator

reduced ripple

rz

VZ0+VO

-

RrIrR

rV

rR

RVV zL

z

zS

zZO

0

z

z

rR

r

Regulation Line

RrZRegulation Load

maxmin

min0min

LZ

ZZZS

II

IrVVR

VS

IL

IZ

I

R

© REP 04/19/23 EGRE224


Page 3.1-61

Example 3.9 Design of a Zener Shunt Regulator

It is required to design a zener shunt regulator to provide a voltage of approximately 7.5 volts. The original supply varies between 15 and 25 volts and the load current varies between 0 and 15 mA. The zener diode we have available has a VZ of 7.5 V at a current of 20 mA and its rZ is 10 .

Find the required value of R and determine the line and load regulation. Also determine the percentage change in VO corresponding to the full change in VS and IL.

)20(105.7 00 mAVVIrVV ZZZZZ VVZ 3.70

maxmin

min0min

LZ

ZZZS

II

IrVVR

383

155

)5(103.715

mAmA

mAVVR designing for Izmin=(1/3)ILmax

VmVRr

r

Z

Z /4.2538310

10Regulation Line

mAmVRrZ /7.9)383//10()//(Regulation Line

%4.3or 254.010/54.2 - Vin Change Full S VVmVVo

%2or 15.015/7.9 - Vin Change Full S VmAmAmVVo

© REP 04/19/23 EGRE224


Page 3.1-62

Temperature Effects

Temperature Coefficient (TC) is expressed in mV/degree C depends on Zener voltage and operating current For VZ<~5V the TC is typically negative and those greater than 5V the TC is positive

for certain current levels and VZ around 5V the TC and be made zero which makes a temperature insensitive supply

Another technique for making a temperature insensitive supply is to use one zener with a positive TC (say 2 mV/degree C) and a regular diode with a negative TC (say -2mV/degree C) and design a circuit in which the effects cancel

+VZ

-+VD

-

+

VD

-

© REP 04/19/23 EGRE224


Page 3.1-63

Rectifier Circuits

Dioderectifier Filter

VoltageRegulator

Load

+

-

VO

IL

t t t t t

Powertransformer

+

-

ac line

120V (rms)

60 Hz

+

-

vO

© REP 04/19/23 EGRE224


Page 3.1-64

Half-Wave Rectifier

D

Rvs

+

-

vo

+

-

Ideal

Rvs+- vo

+

-

VD0 rD

vo

vSVD00

DrR

RSlope

VS

VD0vS

t

v

vo

VD0

,0ov 0Ds Vv

,0D

DSD

o rR

RVv

rR

Rv

0Ds Vv

RrD 0DSO Vvv

SVPIV

© REP 04/19/23 EGRE224


Page 3.1-65

Full-Wave Rectifier with Center Tapped Transformer

D1

Rvs

+

-vo

+

-

vo

vSVD00

DrR

RSlope

VS

VD0vS

t

v

vo

02 DS VVPIV

D2vs

+

-

-VD0

-vS

© REP 04/19/23 EGRE224


Page 3.1-66

Full-Wave Bridge Rectifier

D4

Rvs

+

vo+ -

VS

2VD0vS

t

v

vo

0002 DSDDS VVVVVPIV

D2

-

-vS

D3

D1

© REP 8/14/2015 EGRE224 Electronics - Diodes Page 3.1-1 Introduction The simplest and most...

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Transcript of © REP 8/14/2015 EGRE224 Electronics - Diodes Page 3.1-1 Introduction The simplest and most...