Diode Applications

Contents Diode circuit model

Diode in Series

Diode in Parallel

Half-wave rectifiers

Full-wave rectifiers

Power Supply

Line regulation

Limiter/clipper

Clamper

Diode Circuit Models

The Ideal Diode

Model

The diode is designed to allow current to flow in only one

direction. The perfect diode would be a perfect conductor in

one direction (forward bias) and a perfect insulator in the

other direction (reverse bias). In many situations, using the

ideal diode approximation is acceptable.

Example: Assume the diode in the circuit below is ideal. Determine the value

of ID if a) VA = 5 volts (forward bias) and b) VA = -5 volts (reverse bias)

+

_VA

ID

RS = 50 a) With VA > 0 the diode is in forward bias and

is acting like a perfect conductor so:

ID = VA/RS = 5 V / 50 = 100 mA

b) With VA < 0 the diode is in reverse bias and

is acting like a perfect insulator, therefore no

current can flow and ID = 0.

Series Diode Configurations

Constants

Silicon Diode: VD = 0.7 V

Germanium Diode: VD = 0.3 V

Analysis (for silicon)

VD = 0.7 V (or VD = E if E < 0.7 V)

VR = E – VD

ID = IR = IT = VR / R

4

Forward Bias

Load-Line Analysis

5

The load line plots all possible

combinations of diode current (ID)

and voltage (VD) for a given circuit.

The maximum ID equals E/R, and

the maximum VD equals E.

The point where the load line and

the characteristic curve intersect is

the Q-point, which identifies ID and

VD for a particular diode in a given

circuit.


The Ideal Diode with

Barrier Potential

This model is more accurate than the simple ideal diode

model because it includes the approximate barrier

potential voltage. Remember the barrier potential

voltage is the voltage at which appreciable current starts

to flow.

Example: To be more accurate than just using the ideal diode model include

the barrier potential. Assume V = 0.3 volts (typical for a germanium diode)

Determine the value of ID if VA = 5 volts (forward bias).

+

_VA

ID

RS = 50 With VA > 0 the diode is in forward bias and

is acting like a perfect conductor so write a

KVL equation to find ID:

0 = VA – IDRS - V

ID = VA - V = 4.7 V = 94 mA

RS 50

V+

V+


The Ideal Diode

with Barrier

Potential and

Linear Forward

Resistance

This model is the most accurate of the three. It includes a linear

forward resistance that is calculated from the slope of the linear

portion of the transconductance curve. However, this is usually

not necessary since the RF (forward resistance) value is pretty

constant. For low-power germanium and silicon diodes the RF

value is usually in the 2 to 5 ohms range, while higher power

diodes have a RF value closer to 1 ohm.

Linear Portion of

transconductance

curve

VD

ID

VD

IDRF = VD

ID

+V RF


The Ideal Diode

with Barrier

Potential and

Linear Forward

Resistance

Example: Assume the diode is a low-power diode with a

forward resistance value of 5 ohms. The barrier potential

voltage is still: V = 0.3 volts (typical for a germanium diode)

Determine the value of ID if VA = 5 volts.

+

_VA

ID

RS = 50

V+

RF

Once again, write a KVL equation for the circuit:

0 = VA – IDRS - V - IDRF

ID = VA - V = 5 – 0.3 = 85.5 mA

RS + RF 50 + 5


Values of ID for the Three Different Diode Circuit Models

Ideal Diode

Model

Ideal Diode

Model with

Barrier

Potential

Voltage

Ideal Diode

Model with

Barrier

Potential and

Linear Forward

Resistance

ID 100 mA 94 mA 85.5 mA

These are the values found in the examples on previous slides where

the applied voltage was 5 volts, the barrier potential was 0.3 volts and

the linear forward resistance value was assumed to be 5 ohms.

The Q Point

The operating point or Q point of the diode is the quiescent or no-signal

condition. The Q point is obtained graphically and is really only needed

when the applied voltage is very close to the diode’s barrier potential voltage.

The example 3 below that is continued on the next slide, shows how the Q

point is determined using the transconductance curve and the load line.

+

_VA

= 6V

ID

RS = 1000

V+

First the load line is found by substituting in different

values of V into the equation for ID using the ideal

diode with barrier potential model for the diode. With

RS at 1000 ohms the value of RF wouldn’t have much

impact on the results.

ID = VA – V

RS

Using V values of 0 volts and 1.4 volts we obtain IDvalues of 6 mA and 4.6 mA respectively. Next we will

draw the line connecting these two points on the

graph with the transconductance curve. This line is

the load line.

The Q Point

ID (mA)

VD (Volts)

2

4

6

8

10

12

0.2 0.4 0.6 0.8 1.0 1.2 1.4

The

transconductance

curve below is for a

Silicon diode. The Q

point in this example

is located at 0.7 V

and 5.3 mA.

4.6

0.7

5.3

Q Point: The intersection of the load

line and the

transconductance curve.

Series Diode Configurations

Diodes ideally behave as open circuits

Analysis

VD = E

VR = 0 V

ID = 0 A

12

Reverse Bias

Look example in monograph pages 23, 24

Parallel Configurations

13

mA 142

mA 28

D2I

D1I

mA 28.33kΩ

V .7V 10

R

DVE

RI

V 9.3R

V

V 0.7O

VD2

VD1

V

V 0.7D

V

Look example in monograph page 25

Sine Wave

The sine wave is a common type of alternating current (ac) and alternating voltage.

The time required for a sine wave to complete one full cycle is called the period (T).

Frequency ( f ) is the number of cycles that a sine wave completes in one second. The more cycles completed in

one second. The higher the frequency.

Frequency is measured in hertz(Hz)

Relationship between frequency ( f) and period (T) is:

f = 1/T

Peak-to-Peak / Average / RMS

The peak-to-peak value of a sine wave is the voltage or current from the positive peak to the negative peak.

The peak-to-peak values are represented as:

Vpp and Ipp

Where: Vpp = 2Vp and Ipp = 2Ip

The rms (root mean square) value of a sinusoidal voltage is equal to the dc voltage that produces the same amount of heat in a resistance as does the sinusoidal voltage.

Vrms = 0.707Vp

Irms = 0.707Ip

Half-Wave Rectification

16

The diode only

conducts when it is

forward biased,

therefore only half

of the AC cycle

passes through the

diode to the

output.

The DC output voltage is 0.318Vm, where Vm = the peak AC voltage.

Half-wave Rectifier

Note that the frequency stays the same

Strength of the signal is reduced

Vavg = Vp(out)/ = 0.318 x Vp(out) [31.8 % of Vp]

Vp(out) = Vp(in) – VBar

For Silicon VBar = 0.7 V

Half-wave

Rectifier

Vp(out)Vp(in)

Vavg=0.318Vm

2

PIV (PRV)

18

Because the diode is only forward biased for one-half of the AC cycle, it is

also reverse biased for one-half cycle.

It is important that the reverse breakdown voltage rating of the diode be

high enough to withstand the peak, reverse-biasing AC voltage.

PIV (or PRV) > Vm

• PIV = Peak inverse voltage

• PRV = Peak reverse voltage

• Vm = Peak AC voltage

Half-wave Rectifier - Example

• Draw the output signal– Vp(out) = Vp(in) – 0.7– Vavg = 99.3/– What happens to thefrequency?

Output:

Peak Inverse Voltage (PIV)

– The peak voltage at which the

diode is reverse biased

– In this example PIV = Vp(in)-

– Hence, the diode must be rated

for PIV = 100 V

Transformers (Review)

Transformer: Two inductors coupled together – separated by a dielectric

When the input magnetic field is changing voltage is induced on the

second inductor

The dot represents the + (voltage direction)

Applications:

Step-up/down

Isolate sources

Turns ratio (n)

n = Sec. turns / Pri. turns = Nsec/Npri

Vsec = n. Vpri

depending on value of n : step-up or step-down

Center-tapped transformer

Voltage on each side is Vsec/2

Half-wave Rectifier - Example Example:

– Assume that the input is a sinusoidal signal with Vp=156 V & T = 2

msec; assume Nsec:Npri = 1:2

– Draw the signal

– Find turns ratio;

– Find Vsec;

– Find Vout.

n = ½ = 0.5

Vsec = n.Vpri = 78 V

Vout = Vsec – 0.7 = 77.3 V

78-0.7

Full-wave Rectifier Note that the frequency is doubled

Vavg = 2Vp(out)/ = 0.636 x Vp(out)

Full-Wave Rectification

23

• Half-wave: Vdc = 0.318Vm

• Full-wave: Vdc = 0.636Vm

The rectification process can be improved by

using a full-wave rectifier circuit.

Full-wave rectification produces a greater

DC output:


24

There are 2 types of full-wave rectifiers circuit connection as below:


25

Center-Tapped Transformer Rectifier

Requires

• Two diodes

• Center-tapped transformer

VDC = 0.636Vm

Full-wave Rectifier Circuit

Center-tapped full-wave rectifier

Each half has a voltage = Vsec/2

Only one diode is forward biased at a time

The voltages at different halves are opposite of each other


Center-tapped full-wave rectifier

Each half has a voltage = Vsec/2

Only one diode is forward biased at a time

The voltages at different halves are opposite of each other


• Vout = Vsec /2 – 0.7

• Peak Inverse Voltage (PIV)

– PIV = (Vsec/2 – 0.7)- (-Vsec/2) = Vsec – 0.7

• Vout = Vsec/2 – 0.7

Assuming D2 is

reverse-biased

No current through D2

Full-wave Rectifier - Example

Assuming a center-tapped transformer

Find the turns ratio

Find Vsec

Find Vout

Find PIV

Draw the Vsec and Vout

What is the output freq? Vsec

n=1:2=0.5

Vsec=n*Vpri=25

Vout = Vsec/2 – 0.7

PIV = Vsec-0.7=24.3 V

Bridge Full-wave Rectifier Uses an untapped transformer larger Vsec

Four diodes connected creating a bridge When positive voltage

D1 and D2 are forward biased

When negative voltage D3 and D4 are forward biased

Two diodes are always in series with the load Vp(out) = Vp(sec) – 1.4V

The negative voltage is inverted

The Peak Inverse Voltage (PIV) PIV=Vp(out)+0.7


31

Bridge Rectifier

Four diodes are connected in a bridge configuration

VDC = 0.636Vm

Summary of Rectifier Circuits

32

Vm = peak of the AC voltage.

In the center tapped transformer rectifier circuit, the peak AC voltage

is the transformer secondary voltage to the tap.

Rectifier Ideal VDC Realistic VDC

Half Wave Rectifier VDC = 0.318Vm VDC = 0.318Vm – 0.7

Bridge Rectifier VDC = 0.636Vm VDC = 0.636Vm – 2(0.7 V)

Center-Tapped Transformer

RectifierVDC = 0.636Vm VDC = 0.636Vm – 0.7 V

Bridge Full-wave Rectifier - Comparison

12

0Vr

ms

Vp(2)=Peak secondary voltage ; Vp(out) Peak output voltage ; Idc = dc

load current

Make sure you understand this!

Simulation using PSPICE You can download the Orcad Pspice software

at http://www.cadence.com/products/orcad/pages/downloads.aspx#pspice.

OrCAD PCB Designer Lite DVD (All Products)

The file size is about 1GB and need to fill the form.

http://www.cadence.com/products/orcad/pages/downloads.aspx



Example on how to draw a simple circuit on PSPICESimulation on PSPICE involves 4 steps

1. Creating a new project

2. Drawing schematic diagram

3. Creating simulation profile

4. Running simulation

1. Creating a new project On the Windows start menu , All programs >>

CADENCE >> ORCAD >> ORCAD CAPTURE CIS

In the program, File >> New >> Project

A new window will appear

1. Creating a new project

In New Project window,

Enter the Name and Location

Choose “Analog or Mixed A/D”

Click “OK”

A new dialog will appear

Choose “Create a blank project”

Click “OK”

A new schematic window will appear as follows.

2. Drawing schematic diagram (Half wave rectifier)

• Place the components

• Press Place Part icon located in the upper right side of the tool palette. (shortcut key for Place Part is P)

• Choose all libraries (if you can’t find the libraries, click

Add Library and choose all libraries files in Pspice folder

• Find the parts listed below and add to the windows by left clicking.

Press “Esc” to quit the place part mode.

Parts for half wave rectifier circuit

• VSIN (voltage source)

• 1n4148 (diode)

• R (load resistor)

2. Drawing schematic diagram (Continued) Place the ground by clicking “Place Ground” on the

right tool palette (shortcut: G) and choose “0/CAPSYM”.

Arrange the parts as follows

Connect the parts with wire. (Place wire or shortcut: w)

2. Drawing schematic diagram (Continued) Change the properties of

voltage source by double clicking on each properties

VOFF=0

VAMPL = 10V

FREQ = 50Hz

3. Creating simulation profile• From the top toolbar

PSPICE>>New simulation profile

• Type in the Name, inherited from “None” and click “Create”

• Simulation Settings windows will appear• Choose “Time Domain” in

Analysis type• Set Run to time: 50ms and

Max step size 0.0001• Click OK

4. Runing Simulation Place voltage probe from

the top probe toolbar as shown in the figure.

Click on the Run PSpice icon from the toolbar.

A new simulation window will open showing the plots of two probes.

Bridge Rectifier Simulation To create a new schematic page

to simulate bridge rectifier , click on New Document button at the toolbar.

Draw the Bridge rectifier circuit as shown in the diagram as in step 2

Repeat steps 3 and 4 above guidelines to simulate the circuit.

In the simulation, use Voltage differential markers across Voltage source instead of single Voltage marker .

Bridge Rectifier Simulation (Continued) Full wave rectifier simulation results

Bridge Full-wave Rectifier - Example Assume 12 Vrms secondary voltage for the standard 120 Vrms across the

primary Find the turns ratio

Find Vp(sec)

Show the signal direction when Vin is positive

Find PIV rating

Solution:

n=Vsec/Vpri = 0.110:1

Vp(sec) = (0.707)-1 x Vrms = 1.414(12)=17 V

Vp(out) = V(sec) – (0.7 + 0.7) = 15.6 V through D1&D2

PIV = Vp(out) + 0.7 = 16.3 V

120Vr

ms

Note: Vp-Vbr ; hence, always

convert from rms to Vp

Filters and Regulators

Filters

Filters

Filters

-Ripple voltage depends on voltage variation across the capacitor

- Large ripple means less effective filter

Filters

peak-to-peak

ripple voltage

Too much ripple is bad!

Ripple factor = Vr (pp) / VDC

Vr (pp) = (1/ fRLC) x Vp(unfiltered)

VDC = (1 – 1/ fRLC) x Vp(unfiltered)

Filters

FiltersPlease refer your filter notes in the Electronics Monograph

Diode Clippers

55

The diode in a series clipper “clips”

any voltage that does not forward

bias it:•A reverse-biasing polarity

•A forward-biasing polarity less than 0.7 V (for a silicon diode)

Biased Clippers

56

Adding a DC source in

series with the clipping

diode changes the

effective forward bias of

the diode.

Parallel Clippers

57

The diode in a parallel clipper

circuit “clips” any voltage that

forward bias it.

DC biasing can be added in

series with the diode to change

the clipping level.

Summary of Clipper Circuits

58

Summary of Clipper Circuits

59

more…

Diode Limiting What is Vout?

Vout+ = Vin (RL)/(RL+R1) = 9.09

Vout- = -0.7

Forward biased when positive

Reverse biased when negative,

hence voltage drop is only -0.7

So how can we change the offset?

Diode Limiting – Changing the offset

What if we mix these together?

Positive limiter

Negative limiter

Remember:

When positive voltage reverse biased No current no clipping!

Diode Limiter When the input signal is positive D1 is reversed biased;

acting as positive limiter

Pos. Limiter

-VBIAS-0.7+VBIAS+0.7

Clampers

63

A diode and capacitor can be

combined to “clamp” an AC

signal to a specific DC level.

Biased Clamper Circuits

64

The input signal can be any type

of waveform such as sine, square,

and triangle waves.

The DC source lets you adjust

the DC camping level.

Diode Clamper

It adds a dc level

When the input voltage is negative, the capacitor is charged Initially, this will establish a positive dc

offset

Note that the frequency of the signal stays the same

RC time constant is typically much larger than 10*(Period)

Note that if the diode and capacitor are flipped, the dc level will be negative

Output:

Summary of Clamper Circuits

66

Zener Diodes

When Vi VZ

The Zener is on

Voltage across the Zener is VZ

Zener current: IZ = IR – IRL

The Zener Power: PZ = VZIZ

When Vi < VZ

The Zener is off

The Zener acts as an open circuit

67

The Zener is a diode operated

in reverse bias at the Zener

Voltage (Vz).

Zener Resistor Values

68

ZKRL I II min

min

max L

ZL

I

VR

min

max

L

Z

L

LL

R

V

R

V I

Zi

ZL

VV

RVR min

If R is too large, the Zener diode cannot conduct because the available amount of

current is less than the minimum current rating, IZK. The minimum current is

given by:

The maximum value of resistance is:

If R is too small, the Zener current exceeds the maximum current

rating, IZM . The maximum current for the circuit is given by:

The minimum value of resistance is:

Voltage-Multiplier Circuits

Voltage Doubler

Voltage Tripler

Voltage Quadrupler

69

Voltage multiplier circuits use a combination of diodes and

capacitors to step up the output voltage of rectifier circuits.

Voltage Doubler

70

This half-wave voltage doubler’s output can be calculated by:

Vout = VC2 = 2Vm

where Vm = peak secondary voltage of the transformer

Voltage Doubler

71

• Positive Half-Cycle

o D1 conducts

o D2 is switched off

o Capacitor C1 charges to Vm

• Negative Half-Cycle

o D1 is switched off

o D2 conducts

o Capacitor C2 charges to Vm

Vout = VC2 = 2Vm

Voltage Tripler and Quadrupler

72

Practical Applications

Rectifier Circuits

Conversions of AC to DC for DC operated circuits

Battery Charging Circuits

Simple Diode Circuits

Protective Circuits against

Overcurrent

Polarity Reversal

Currents caused by an inductive kick in a relay circuit

Zener Circuits

Overvoltage Protection

Setting Reference Voltages

73

DiodeCharacteristics (VRRM & IF(AV) )

References

Robert Boylestad and Louis Nashelsky. Electronic Devices and

Circuit Theory, 10e Pearson 2009

www.ccsu.edu/.../ccsu/courses/cet323/lectures/Chapter2.ppt

Documents

Diode Applications