Chapter 3 Diodes. Ch 3-1-

Chapter 3 Diodes

Figure 3.1

Forward-biased mode diode on

Reverse-biased mode diode off

Anode – positive terminal of diode,into which current flows Cathode – negative terminal, from which current flows

device symbol with two nodes

open ckt short ckt

3.1 Ideal Diode

3.1.1. Current-Voltage Characteristic

3.1.2: A Simple Application – The Rectifier

▪ converts AC waves in to DC …ideally with no loss.

▪ The diode blocks reverse current flow, preventing negative voltage across R.



Example 3.1 Diode Rectifier

A source (v

_S

) with peak amplitude of 24V is employed to charge a 12V dc-battery.

▪ Finding the fraction of each cycle during which the diode conducts.

▪ Finding the peak value of diode current and maximum reverse-bias voltage that appears across the diode.

The diode conducts when v

_S

exceeds 12 V

60

0











 24 cos 12

 the conduction angle is 120

⁰

(180-60-60)

A

I

_d

0 . 12

100 12 24 − =

=

The peak value of the diode current one-third of a cycle

The maximum reverse voltage across the diode occurs when v

_S

is at its negative peak

and is equal to 24 + 12 = 36 V.

OR gate AND gate

IF v

_A

= 5V applies, diode

_A

will conduct.

Then v

_Y

= v

_A

= 5V

IF any diode conducts → v

_Y

= 5V

IF v

_A

= 0V applies, diode

_A

will conduct.

Then v

_Y

= v

_A

= 0V

IF all diodes block-off→ v

_Y

= 5V

3.1.3. Another Application,Diode Logic Gates

3.2. Terminal Characteristics of Junction Diodes

Happened as v > 0

3.2.1. The Forward-Bias Region

=

_S

(

^{v V/ T}

− 1) i I e

As expressing v related to the higher conducting current

 

=  

T

 

S

v V i ln I

A 10 timeschange in i only results in 60mVchange in v.

= = 25.8

T

V kT mV

q at room temperature

I_S : saturation current V_T: thermal voltage

k : Boltzman's constant (8.62e-5 eV/K)

q : magnitude of electron charge (1.6e-19 coul)

( )

( )

( )

−

= =

=  − =



− =

1 2

2 1

/ /

1 2

( )/

2

2 1 2 1

1

2 1 2 1

and

, /

2.3

60 2.3 1

/ 0 / 1

T T

T

V V V V

S S

V V V

T

T T

I I e I I e

I e V V V I I

I

V V V I

mV

I V

l

log

n log

=

_S ^{V V/ T}

I I e

As higher voltage being applied

fully conducting region

cut-in voltage

R_diode is almost 0

▪ for negative voltages |v| is much more greater than V

_T

(25mV),

3.2.2. The Reverse-Bias Region

Happened as v < 0 i = − I e

_S ^{−v V/ T}

 

= −  

^/

  1

S v VT

i I

e

 −

_S

i I

▪ A “real” diode exhibits reverse-bias current, although small, but much larger than I

_S

.

e.g. 10

^-9

vs. 10

^-14

Amps

▪ Such increment on the reverse current is attributed to

leakage effects.

breakdown region

3.2.3. The Breakdown Region

when v < V

_ZK

(Zener-Knee Voltage)

This is normallynon-destructive.

= − _S

i I

 − _S

i I _{i = I e S} ^{− v V /}

^T

V = -V

ZK

V = -V

T

V = 10 V

T

= _S ( ^{v V /}

^T

− 1)

i I e

3.3. Modeling the Diode Forward Characteristic

A simplified diode models for diode circuit analyses:

▪ exponential model

▪ constant voltage-drop model

▪ ideal diode model

▪ small-signal (linearization) model

3.3.1 The Exponential Model

Two methods (graphical method, iterative method) can be used for solving this problem 3.3.2 Graphical Analysis Using the Exponential Model

=

^{V VD/ T}

D S

I I e

=

^DD

−

^D

D

V V

I R

Suppose V

_D

is not too small (greater than cutting voltage),

At the same time, by KVL,

load line, step #1:

Plot load line and diode characteristic step #2:

Find intersection of these two curves, it isthe operating point

▪ Pro’s

▪Intuitive

▪ Con’s

▪Poor Precision

▪Not Practical for simulation

▪ step #1:

Initially guess V

_D

.

▪ V

_D⁽⁰⁾

▪ step #2:

Use nodal / mesh analysis to solve I

_D

.

▪ step #3:

Use exponential model to update V

_D

.

▪ V

_D⁽¹⁾

= f(V

_D⁽⁰⁾

)

▪ step #4:

Repeat these steps until V

_D^(k+1)

= V

_D^(k)

.

3.3.3 Iterativel Analysis Using the Exponential Model

▪ Pro’s

▪ High Precision

▪ Con’s

▪ Not Intuitive

▪ Not Practical for Complex Analyses

▪ 10+ iterations may be required

▪ assume that voltage drop across the diode is constant.

3.3.4 The Need for Rapid Analysis

3.3.5. The Constant Voltage-Drop Model

Assumes the characteristics of I

_D

vs. V

_D

is an vertical line at the cutting voltage (0.7V)

This assumption is based on the observation that a forward-conducting diode has a voltage drop that varies in a relatively narrow range, say 0.6 to 0.8 V.

Used as medium voltages (~ 0.7V) applied

3.3.6. Ideal Diode Model

Assumes the characteristics of I

_D

vs. V

_D

is an vertical line at 0 V.

open ckt. short ckt.

Used as high voltages (>> 0.7V) applied

3.3.7. Small-Signal Model

Most frequently, the 0.7-V-drop model is utilized.

Then, for small-signal operation around the dc bias point, the diode is modeled by aresistanceequal to the inverse of the slope of the tangent to the

exponential i–v characteristic at the bias point.

There are applications in which a diode is biased to operate at a point on the forward i–v characteristic and a small ac signal is superimposed on the dc quantities.

There are applications in which a diode is biased to operate at a point on the forward i–v characteristic and a small ac signal is superimposed on the dc quantities.

▪ small-signal diode model

▪ Diode is modeled as variable resistor.

▪ defined via linearizationof exponential model.

▪ Around bias point defined by constant voltage drop model.

▪ V_D⁽⁰⁾= 0.7V

AC

Total Instantaneous Solution (v

_D.

)

Steady-State Solution (V

_D.

)

Time-Varying Solution (v

_d.

)

DC

AC

=

DC

+

▪ DC only – upper-case w/ upper-case subscript

▪ time-varying only – lower-case w/ lower-case subscript

▪ total instantaneous – lower-case w/ upper-case subscript

▪ DC + time-varying

▪ step #1:

Consider the conceptual circuit.

▪ DC voltage (V

_D

) is applied to diode

▪ arbitrary time-varying signal v

_d

is super-imposed upon V

_D

▪ DC only – upper-case w/ upper-case subscript

▪ time-varying only – lower-case w/ lower-case subscript

▪ total instantaneous – lower-case w/ upper-case subscript

▪ DC + time-varying

▪ step #2:

Define DC current

▪ step #3:

Define total instantaneous voltage (v

_D

) as composed of V

_D

and v

_d

.

▪ step #4:

Define total instantaneous current (i

_D

) as function of v

_D

.

=

^{V VD / T}

D S

I I e

= +

( ) ( )

D D d

v t V v t

=

^/

( )

^{vD VT}

D S

i t I e

▪ step #5:

Redefine i

_D

as function of both V

_D

and v

_d

.

▪ step #6:

Split this exponential in two.

▪ step #7:

In terms of DC component (I ) and time-varying voltage (v ).

( ⁺ )

=

^/

( )

^{VD vd VT}

D S

i t I e

=

^{/ /}

( )

^{T d}

D

T

D v V

V V I

D S

i t I e e

=

^/

( )

^{vD VT}

i t I e

In terms of DC component (I

_D

) and time-varying voltage (v

_d

).

=

^/

( )

^{vD VT}

D D

i t I e

▪ step #8:

Apply power series expansion

▪ step #9:

Because v

_d

/V

_T

<< 1, certain terms may be neglected.

= + + 1

²

+

³

+

⁴

+ 2! 3! 4!

x

x x x

e x







 

   

     

=  + +      +    + 

   

         

 

 

 

/

because / 1, these terms are assumed to be negligible

power series expansion o

3

f

2

1 1

( ) 1

2! 3!

d T

vd TV

d d d

D D

T T T

v V

e

v v v

i t I

V V V

= +     ( )  

d

D

D D d

T i

i t I I v

= + V

=

= ( )

1

D D d

d d

d

T d

i t I i

i v

r r V

I total instant current (i

_D

)

small-signal current (i

_d

.) small-signal resistance (r

_d

.)

Valid for for v_d< 5mV amplitude

Note this approximation is only valid for D

d

i

r

 

  

= 1

Example 3.5

▪ Consider the circuit shown over the right-hand-side for the case in which R = 10kOhm.

▪ The power supply V

⁺

has a dc value of 10V over which is super-imposed a 60Hz sinusoid of 1V peak amplitude (known as the supply ripple)

▪ Assume diode to have 0.7V drop at 1mA current.

 v

_d.

(peak) = 2.68mV

➢ Calculate both amplitude of the sine-wave signal observed across the diode.

the diode incremental resistance r_d

3.3.8. Use of Diode Forward Drop in Voltage Regulation

A Circuit whose voltage output remains stable in spite of changes in supply and load.

Example 3.6: Diode-Based Voltage Regulator

A string of three diodes is used to provide a constant voltage of 2.1V.

What is the change in this regulated voltage caused by (a) a +/- 10% change in supply voltage and

(b) connection of 1k

Ω

load resistor?

Without load

The three diodes in series will have a total incremental resistance of 9.6 Ω Corresponding to the ±10% (i.e., ±1-V) change in supply voltage,

peak-to-peak change in output voltage will be

the output voltage will change by ±9.5 mV or ±0.5%.

With 1k Ω load, it draws a current of approximately 2.1 mA.

resulting in a decrease in voltage across the diode string given by Δv_O=–2.1 × r=–2.1 × 9.6 = –20 mV A detailed calculation of the voltage change using the exponential model results in Δv = –23 mV

3.4. Operation in the Reverse Breakdown Region – Zener Diodes

▪ Under certain circumstances, diodes may be intentionally used in the reverse breakdown region.

▪ These are referred to as Zener Diodes.

knee current

3.4.1 Specifying and Modeling the Zener Diode

The manufacturer usually specifies the voltage across the zener diode V_Z at a specified test current, I_ZT.

Resistance r_z is the incremental resistance of the zener diode at operating point Q, also known as dynamic resistance

Model for the zener diode.

V

_Z

= V

_Z0

+ r

_z

I

_Z

3.4.2 Use of the Zener as a Shunt Regulator

Ex 3.7 V_Z = 6.8 V at I_Z= 5 mA, r_z = 20 Ω, and I_ZK= 0.2 mA.

(a) Find V_O with no loadand with V+ at its nominal value.

(b) Find the change in V_O resulting from the ±1-V change in V+. Note that , usually expressed in mV/V, is known as line regulation.

(c) Find the change in V_O resulting from connecting a load resistance R_Lthat draws a current I_L=1 mA, and hence find the load regulation in mV/mA.

(d) Find the change in V_O when R_L = 2 kΩ.

(e) Find the value of V_O when R_L= 0.5 kΩ.

(f) What is the minimum value of R_L for which the diode still operates in the breakdown region?

By V

_Z

= V

_Z0

+ r

_z

I

_Z Substituting V_Z = 6.8 V, I_Z = 5 mA, and r_z = 20 Ω yield V_Z0= 6.7 V (a)

(b) Thus, Line regulation = 38.5 mV/V

(c) When a load resistance RL that draws a load current IL = 1 mA is connected, the zener current will decrease by 1 mA. The corresponding change in zener voltage can be found from

Thus the load regulation is

3.4.3 Temperature Effects

➢ Temperature coefficient TC, usually expressed in mV/°C.

➢ Varies with the operating current.

➢ Zener diodes whose V

_Z

are lower than about 5 V exhibit a negative TC. On the other hand, zeners with higher voltages exhibit a positive TC.

➢ Commonly used technique for obtaining a reference voltage with low

temperature coefficient is to connect a zener diode with a positive temperature coefficient of about 2 mV/°C in series with a forward-conducting diode.

Remark :

Though simple and useful, zener diodes have lost a great deal of their popularity in recent years.

They have been virtually replaced in voltage-regulator design by specially designed integrated circuits (ICs) that perform the voltage regulation function much more effectively and with greater flexibility than zener diodes.

increase / decrease rms magnitude of AC wave via power transformer

convert full-wave AC to half-wave DC (still time-varying and periodic) employ low-pass filter to reduce wave amplitude by > 90%

employ voltage regulator to eliminate ripple supply dc load

3.5. Rectifier Circuits

▪ One important application of diode is the rectifier –

▪ Electrical device which converts alternating current (AC) to direct current (DC)

▪ One important application of rectifier is dc power supply.

3.5.1. The Half-Wave Rectifier

▪ half-wave rectifier – utilizes only alternate half-cyclesof the input sinusoid

▪ Constant voltage drop diode model is employed.

▪ current-handling capability–maximum forward current diode expected to conduct

▪ peak inverse voltage (PIV)–maximum reverse voltage expected to block w/o breakdown

Figure 3.22: Full-wave rectifier utilizing a transformer with a center-tapped secondary winding.

3.5.2. The Full-Wave Rectifier

The direction of current flowing across load never changes (both halves of AC wave are rectified).

When instantaneous source voltage is positive, D

₁

conducts while D

₂

blocks…

when instantaneous source voltage is negative, D

₂

conducts while D

₁

blocks

when instantaneous source voltage is positive, D₁ and D₂ conduct while D₃and D₄ block

when instantaneous source voltage is negative, D₁and D₂block while D₃ and D₄ conduct

▪ Pro’s

▪No need for center-tappedtransformer

▪ Con’s

▪Series connection of TWO diodeswill reduce output

3.5.3. The Bridge Rectifier

An alternative implementation of the full-wave rectifier is bridge rectifier.

3.5.4. The Rectifier with a Filter Capacitor

▪ Pulsating nature of rectifier output makes unreliable dc supply.

▪ As such, a filter capacitor is employed to remove ripple.

Fig 3.24: (a) A simple circuit used to illustrate the effect of a filter capacitor.

(b) input and output waveforms assuming an ideal diode.

▪ step #1: source voltage is positive, diode is forward biased, capacitor charges.

▪ step #2: source voltage is reverse, diode is reverse-biased (blocking), capacitor cannot discharge.

▪ step #3: source voltage is positive, diode is

forward biased, capacitor charges (maintains

voltage).

( ) ( )

( )

⁻

= −

=

output voltage for state #1

output voltage for state #2

O I D

t RC

O peak

v t v t v v t V e

circuit state #2 circuit state #1

3.5.4. The Rectifier with a Filter Capacitor

For practical application, the

converter should supply a load (i.e.

a path for capacitor discharging).

load resistor

▪ step #1: Analyze circuit state #1.

▪ When diode is forward biased and conducting.

▪ step #2: Input voltage (v

_I

) will be applied to output (v

_O

), minus 0.7V drop across diode.

=

= +

= +

: define capacitor current differentially

O L

D C L

I

D L

i v

R i i i

i C dv i dt

action

circuit state #1

= −

output voltage for state #1

O I D

v v v

circuit state #2

▪ step #3: Analyze circuit state #2.

▪ When diode is blocking and capacitor is discharging.

▪ step #4: Define KVL and KCL for this circuit.

▪ v

_O

= Ri

_L

▪ i

_L

= –i

_C

Oxford University Publishing

Microelectronic Circuits by Adel S. Sedra and Kenneth C. Smith (0195323033)

: take Laplace transform

: replace with -

: define differentially

: change sides

0

0

L C

C

C

O

O L O

O C

O O

i

O i

i

i

O

v Ri v RC dv

dt v Ri

v R C dv dt

v RC dv dt

 

=  + = 

 

= −

 

= −    

+ =

action

action

action

action

L

( ) ( ) ( )

( ) ( )

( )

( )

: take Laplace transform

tra

: seperate disalike / collect alike ter nsform of

initial

1 ( )

conditio s

n m

0 0

0

O

O

dv dt

RCs

O O O

O O

V s

O

V s RC sV s V

V s RCsV s RCV

+

+   −   =

+ =

action

action

( ) ( ) ( )

( ) ( )

( ) ( )

( )

( ) ( ) ( )

: eliminate from both sides

: solve for

: pull out C

(

1

1 )

R

1 0

0 1

1

0 1

1/

1 0

O

O

O O

O O

O O

O RC s V s

RC

s

R

V

O

C

s V s V

RC

V s V

s RC

V s V

s RC

RCs V s RCV

RC RC

 + 

 



−



 +  =

 

 

=

+

= +

+ =

action

action

action

L

( ) ( )

: take inverse Laplace

: solve

0

t RC

O O

v t V e

⁻

 

 

 

 

 

=

action

action

▪ step #5: Define output voltage for states

#1 and #2.

circuit state #2 circuit state #1

( ) ( )

( )

⁻

= −

=

output voltage for state #1

output voltage for state #2

O I D

t RC

O peak

v t v t v

v t V e

( ) ( )

( ) ⁻

=

=

output voltage for state #1

output voltage for state #2

O I

t

O peak RC

v t v t v t V e

Fig 3.25: Voltage and Current Waveforms in the Peak Rectifier Circuit WITH RC >> T.

The diode is assumed ideal.

▪ The diode conducts for a brief interval (Dt) near the peak of the input sinusoid and supplies the capacitor with charge equal to that lost during the much longer discharge interval. The latter is approximately equal to T.

▪ Assuming an ideal diode, the diode conduction begins at time t

₁

(at which the input v

_I

equals the exponentially decaying output v

_O

). Diode conduction stops at time t

₂

shortly after the peak of v

_I

(the exact value of t

₂

is determined by settling of I

_D

).

▪ During the diode off-interval, the capacitor C discharges through R causing an

exponential decay in the output voltage (v

_O

). At the end of the discharge interval, which lasts for almost the entire period T, voltage output is defined as follows – v

_O

(T) = V

_peak

– V

_r

.

▪ When the ripple voltage (V

_r

) is small, the output (v

_O

) is almost constant and equal to the peak of the input (v

_I

). the average output voltage may be defined as below…

small is

V if V

V V

V

avg =

_peak

−

_r



_{peak r}

2

) 1

(

₀

How is ripple voltage (V

_r

) defined?

▪ step #1: Begin with transient response of output during “off interval.”

▪ step #2: Note T is discharge interval.

▪ step #3: Simplify using assumption that RC >> T.

▪ step #4: Solve for ripple voltage V

_r

.

( )

is discharge interval

because , we

:

can assume...

1

1 1

solve for ripple voltage

(eq4.28)

)

(

T RC

r

t

O peak RC

peak r O

T peak r peak RC

r pea

T

RC T e T

R

k

C

T V

v t V e

V V v T

V V V e

V V T

RC

−



 −

− −

−

−

=

− =

 

−    

 

 

  

 

action

▪ step #5: Put expression in terms of frequency (f = 1/T).

▪ Observe that, as long as V

_r

<< V

_peak

, the capacitor discharges as constant current source (I

_L

).

How is conduction interval (Dt) defined?

fC I fRC

V V

^L

R I V peak

r

L

peak

⎯ ⎯ =

⎯ →

⎯



=

▪ step #1: Assume that diode conduction stops (very close to when) v

_I

approaches its peak.

▪ step #2: With this assumption, one may define expression in the following.

▪ step #3: Solve for w Dt.

cos(0

^O

)

r peak

peak

t V V

V cos( w D ) =

⁰

−

Note that peak of vI represents cos(0⁰), therefore cos(wDt) represents variation around this value

peak r

V t = 2 V D

w

As assumed, conduction interval Dt will be small when V_r<< V_peak

▪ precision rectifier – is a device which facilitates rectification of

low-voltage input waveforms.

3.6: Limiting and Clamping Circuits

▪ limits voltage output.

Fig. 3.28: General transfer characteristic for a limiter circuit

▪ passive limiter circuit

▪ has linear range

▪ has nonlinear range

▪ K < 1

▪ examples include

▪ single limiter operate in uni- polar manner

▪ double limiter operate in bi- polar manner

over linear range

outside linear range

-

constant value(s)

I O

O

I

I

I I

v Kv

L

v Kv

v L

K

L L

K v K v L

L

−

+

−

+

+

 

= 



 



 

 

 

=  

 

▪ soft vs. hard limiter

Fig 3.31: Variety of basic limiting circuits.

single limiters employ one diode

double limiters employ two diodes of opposite polarity

linear range may be controlled via string of diodes and dc sources

zener diodes may be

used to implement

soft limiting

3.6.2. The Clamped Capacitor or DC Restorer clamped capacitor, dc block, dc restorer :

a circuit which removes the dc component of an AC wave.

3.6.3: The Voltage Doubler

multiplies the amplitude of a wave or signal by two.

waveform of the voltage across D₁