EP391: Lecture 2

Diodes

Silicon

Silicon by itself is a poor conductor i.e. semi-conductor

Covalent bonding

Silicon Doping

p-doping n-doping

Arsenic: donorGallium: acceptor

Current Flow

N-Type Semiconductor

P-Type Semiconductor

Electron flow in Conduction Band

Hole flow in Valence Band

PN Junction

Diagram showing distribution of electrons and holes in a PN junction and the resulting depletion region

Potential profile across the depletion region, represented by a battery

PN Junction: Forward Bias

Forward bias PN junction showing motion of electrons and holes

Graph showing components of current

Potential across junction becomes Φ - V

PN Junction: Reverse Bias

The depletion region widens

Potential across junction becomes Φ + V

Ideal Diode: I-V characteristic

Diode Symbol

Question 1.

What is the current through the diode and the voltage across the diode for the following two circuits?

Question 2.

What is the output voltage for the following circuit? (a Rectifier)

Forward Bias Reverse Bias

Rectified Output

V across diode

Question 3. For the following circuit, if is a sinusoid with 24-V peak

amplitude, find the fraction of each cycle during which the diode conducts. Find the peak value of the diode current and the maximum reverse-bias voltage that appears across the diode.

24sin 12

30 150, or 1/3 of a cycle

24 120.12A

100The maximum revers voltage is 24+12=36V

dI

Junction Diode

There are 3 areas of operationThe forward-bias region

v > Vx

The reverse-bias regionv < Vx

The breakdown regionv < -Vzk

Vx

-Vzk

Areas Expanded

Shockley Equation: Exponential Model

I is the forward-bias current Occurs when v on the diode is positive. the “cut-in” voltage is the voltage beneath which

the current is negligible small (generally around .5V)

The current exponentially increases, and the voltage drop typically lies in a narrow range from .6V to .8V

1T

D

nV

V

s eII

Shockley Diode Equation…contd

Is is the reverse saturation current.

The saturation current is directly proportional to the cross-sectional area of the diode.

For “small-signal” diodes, the saturation current is on the order of 10e-15A.

Strongly correlated to temperature doubles for every 5˚C rise in temperature.

1T

D

nV

V

s eII

Shockley Diode Equation…contd

q

kTV

eII

T

nV

V

sT

D

Voltage Thermal is

1

-23

-19

Boltzmann's constant = 1.38x10 joules/kelvin

Absolute temperature in kelvins = 273+

the magnitude of electronic charge = 1.60x10 coulomb

k

T C

q

at room temperature (20 C), the value of is 25.2mV

We generally use 25mVT

T

V

V

Shockley Diode Equation…contd

1T

D

nV

V

s eII

n is a constant between 1 and 2 that represents variances in the material and physical structure of the diode.

Diodes made using standard integrated circuit techniques exhibit an ‘n’ close to 1.

Diodes available as two-terminal devices generally exhibit an ‘n’ closer to 2.

Also, we tend to use 1 for Ge and 2 for Si. We will use n=1 unless specified.

Reverse Bias Region

1T

D

nV

V

s eII

In the reverse-bias region, the current is theoretically

Real diodes often exhibit a much larger current due to leakages. However, the current is still quite small (nA range).

There is also a slight increase with voltage for reverse-bias current.

sII

Breakdown Region

When the voltage reaches a certain negative potential, the diode will begin conducting current. This “knee” is known as the breakdown voltage, Vzk.

The Z stands for Zener and the K for knee. We will learn more about Zener diodes later

(diodes that make use of the breakdown voltage and it’s near constant voltage/current relationship to be used in voltage regulation).

Example

A certain diode has ID = 0.1 mA for VD = 0.6 V. Assume n is unity and VT = 0.026. Compute the diode current at VD = 0.65 V.

Using the above derived equation we get ID = 0.683 mA when VD = 0.65 V

1

2

1

212

1

2

21

log3.2ln

12

1

2

21

D

DT

D

DTDD

nV

VV

nV

V

nV

V

D

D

nV

V

sDnV

V

sD

I

InV

I

InVVV

e

e

e

I

I

eIIeII

T

DD

T

D

T

D

T

D

T

D

Diode Models

Objective is to understand different diode models. Ideal-Diode ModelExponential ModelPiecewise-linear ModelConstant Voltage-drop Model

Then apply appropriate models to different circuits

Solving a Circuit: Different models

Given VDD= 5V and R=1K, find ID and VD

Assuming Ideal Diode

mAR

VI

V

DDd

D

5

0 Bias, ForwardIn

Constant Voltage-Drop Model

Problem Solving using CVD Model

Repeat previous problem using the constant voltage drop model

mAk

I

VV

D

D

3.41

7.05

7.0

Piecewise-Linear Model

Also known as the battery + resistance model

0

00

0,

,

D D D

D DD D D

D

i v V

v Vi v V

r

0 0.65V and 20D DV r

Problem Solving using PWL Model

Repeat earlier problem using the Piecewise linear model given

0 0.65V and 20D DV r

0

5 0.654.26mA

1 0.020.65 4.26 0.02 0.735V

D

D D D D

I

V V I r

Exponential Model

Assuming Exponential Model VD depend on ID, and ID depends on VD

How do we solve? Option 1. Iterative Analysis

Turns out that we can’t solve this simple little equation – it involves a:

transcendental function A function which is not an algebraic function. In other words,

a function which "transcends," i.e., cannot be expressed in terms of, algebra. Examples of transcendental functions include the exponential function, the trigonometric functions, and the inverse functions of both.

Solving Exponential Model Using Equations

Determine the current and the diode voltage for the following circuit with Vdd=5V and R=1k. Assume that the diode has a current of 1 mA at a voltage of .7 V and that its voltage drop changes by .1 V for every decade of change in current.

2 22 1

1 1

22 1

1

1 1 2 2

Iteration 1

5 0.74.3mA

1

ln 2.3 log

2.3 .1

0.1log

0.7V, 1mA, 4.3mA =0.763V

DD DD

T T

T

V VI

RI I

V V nV nVI I

nV

IV V

I

V I I V

2

Iteration 2

5 0.7634.237mA

14.237

0.763 0.1log 0.762V4.3

DI

V

Exponential Model

Option 2. Graphical Analysis

Try out yourself…

For the same circuit, find the current and the diode voltage with Vdd=5V and R=10k. Assume that the diode has a current of 1 mA at a voltage of .7 V and that its voltage drop changes by .1 V for every decade of change in current. Use (a) iteration, (b) piecewise-linear with the same parameters, (c) the constant-voltage-drop model, and (d) ideal model

(a) 0.434 mA, 0.663V; (b) 0.434mA, 0.659V; (c) 0.43mA, 0.7V; (d) 0.5 mA

Forward Bias Diode as a Regulator

Diode Regulator Design the following circuit to provide an output voltage of 2.4V. Assume

the diodes have a current of 1 mA at a voltage of .7 V and that its voltage drop changes by .1 V for every decade of change in current.

2.4V. Each diode must therefore drop .8V

The current must be 1 decade above 1mA in order

for the diode to change from .7 to .8V drops.

Thus, the current is 10mA, and the resistance must be

10 2.410m

OV

R

A, 760R

Zener Diodes

Operate in breakdown region due to their stable constant voltage

Example Part A

a) Find with no load and

with at its nominal value.

OV

V

0

0 0

0

0

6.8 20 5 , 6.7

10 6.76.346

0.5 0.02

6.7 6.346 0.02 6.827

Z Z z Z

Z Z

Zz

z

O Z Z z

V V r I

V mA V V

V VI mA

R r

V V I r V

A 6.8-V Zener diode in the circuit below is specified to have Vz=6.8V at Iz=5mA, rz=20 ohms, and Izk=0.2mA. The supply voltage is nominally 10V but can vary by +/- 1V.

Example Part B

201 38.5

500 2

Line Regulation=38.5mV/

0

V

zO

z

rV V mV

R r

b) Find the change in resulting from the 1V change in .

Note that / usually expressed in mV/V, is known

line regulation as .

O

V

V V

VV /0

Example Part C

c) Find the change in resulting from connecting a load

resistance that draws a current of 1 mA, and hence

load regulation find the in mV/mA.

O

L L

O L

V

R I

V I

The load draws a current of 1mA

from the diodes .

Lo

..

ad

20 1 20

Regulation=-20mV/mAO z ZV r I mV

Example Part D

The load current will be approximately

6.8V / 2k 3.4mA

20 3.4 68mV

This is a quick estimate as it doesn't account

for the change in the diode current.

, ,o

O z Z

O ZO OD L

z L

OD L

V r I

V VV V VI I I

R r R

V VI I I

R

10 6.740 4 100 670

500 20 2000105 710 6.762

6.762 6.827 65mV

oO Z O

z L

O O OO O O

O O

O

V V V

r R

V V VV V V

V V V

V

d) Find the change in when 2O LV R k

Example Part E

of .5k would draw a load current of

6.8 / 0.5 13.6mA. This is not possible as

the current though R is only 6.4mA. Therefore,

the Zener must be cut off.

0.510 5V

0.5 0.5

Therefore, the zener i

L

LO

L

R

RV V

R R

s not in breakdown.

e) Find the change in when 0.5O LV R k

Example Part F

L

To be at breakdown, 0.2mA

and 6.7V. In this case, the

worst-case (lowest) current through R is

9 6.74.6mA. The load current is therefore

0.54.6 0.2 4.4mA. R is therefore

6.71.523k

4.4

Z ZK

Z ZK

L

I I

V V

R

f) What is the minimum value of for which the diode

still operates in the breakdown region?LR

Halfwave Rectifier

Given Circuit in (a) Consider the piecewise-linear

model circuit (b) Draw the vo /vs transfer

characteristic (c) Draw the input and output

waveforms (d) What is the Peak-Inverse

Voltage (PIV) across the diode?

PIV=Vs

Full-Wave Rectifier Given circuit in (a) Draw the vo /vs transfer

characteristic (b) Draw the input and output

waveforms (c) What is the Peak-Inverse

Voltage (PIV) across the diode?

PIV = 2Vs – Vd

Bridge Rectifier Draw the input and output waveforms. What is the Peak-Inverse Voltage (PIV)

across the diodes? PIV = Vs

Advantages ½ the PIV of the full wave Don’t need a center-tapped transformer Only need half of the turns in the

secondary winding

2VD-2VD

Half-Wave Peak Rectifier

Half-Wave Peak Rectifier with load

Full-Wave Peak Rectifier

Diode Rectifiers

Limiter/Clipper

Hard Limiting

Soft Limiting

Examples

Draw the transfer characteristic for the following circuits.

Clamper DC Restorer

Also known as a Clamped Capacitor assume an ideal diode

The charge on the capacitor, Vc is 6 volts when the diode is conducting, and Vo is 0.

When Vi jumps to 4 volts, the diode turns off, but there is still 6 volts across the cap.

Therefore, the output is at 10V. Essentially, this circuit clamps the voltage on the bottom to 0, and moves

the waveform up. One use of this circuit is for obtaining average values and detecting “duty

cycles” based on the average value. This is useful in PWM (pulse-width modulation)

Clamper with resistor

While the output is above ground, a current must flow in R. This comes from the cap discharging (as the diode is off). This falls exponentially with CR.

When the input switches, the output switches the same amount, and then the capacitor is rapidly charged from the diode. The resulting output is just a few tenths of a volt negative.

Reverse Clamped

Voltage Doubler

Voltage Doubler Note that this circuit is

a clamp followed by a peak rectifier.

The peak-rectifier provides a voltage of -2Vp across C2.

The output voltage is thus twice the input peak voltage.

This can be extended to get larger multiples.

(Klingon Pain Stick?) Diodes and capacitors

are cool. We should try this in

lab…Or at least in a MultiSim where we can do a transient analysis