Diode Application

WED DK2 9-10am

THURS DK6 2-4pm
 Simple Diode circuit
Series diode configuration

Characteristics
Circuit

From Kirchoff’s Voltage Law E = VD + I D R

From previous lecture (
I D = I S eVD / nVT − 1 )
Copyright ©2006 by Pearson Education, Inc.
Robert L. Boylestad
Upper Saddle River, New Jersey 07458
Electronic Devices and Circuit Theory, 9e All rights reserved.
 Load line analysis

E
When VD=0 V then ID =
R VD =0V
When ID=0 A then
VD = E I
D =0 A
Copyright ©2006 by Pearson Education, Inc.
Robert L. Boylestad
Upper Saddle River, New Jersey 07458
Electronic Devices and Circuit Theory, 9e All rights reserved.
 Example For the series diode below employing the diode
characteristic beside it, determine VDQ , IDQ and VR,

Copyright ©2006 by Pearson Education, Inc.

Robert L. Boylestad
Upper Saddle River, New Jersey 07458
Electronic Devices and Circuit Theory, 9e All rights reserved.
 E 10
ID = = = 20 mA
For load line
R V D = 0V 0 .5 k Ω
VD = E I = 10V
D =0 A
By drawing the load line on the diode characteristic we
obtained the
V DQ = 0.78V I DQ = 18.5mA
VR=IRR= IDQR=18.5mA x 1 kΩ =18.5V
Copyright ©2006 by Pearson Education, Inc.
Robert L. Boylestad
Upper Saddle River, New Jersey 07458
Electronic Devices and Circuit Theory, 9e All rights reserved.
 Diode circuit Equivalent circuit

A diode is “ON” state if the current established by the applied

sources is such that its direction matches that of the arrow in the
diode symbol and VD > 0.7V for Si, VD>0.3V for Ge , VD> 1.2V for
GaAs

Copyright ©2006 by Pearson Education, Inc.

Robert L. Boylestad
Upper Saddle River, New Jersey 07458
Electronic Devices and Circuit Theory, 9e All rights reserved.
 Diode circuit Equivalent circuit

A diode is “OFF” state if the current established by the applied

sources is reversed the direction of the arrow in the diode
symbol. Then ID=0

Copyright ©2006 by Pearson Education, Inc.

Robert L. Boylestad
Upper Saddle River, New Jersey 07458
Electronic Devices and Circuit Theory, 9e All rights reserved.
 For the diode below determine VD, VR and ID

V D = 0.7V
Using equivalent circuit and KVL

V R = E − V D = 8V − 0.7V = 7.3V
VR 7.3V
ID = IR = = = 3.32mA
R 2.2kΩ
Copyright ©2006 by Pearson Education, Inc.
Robert L. Boylestad
Upper Saddle River, New Jersey 07458
Electronic Devices and Circuit Theory, 9e All rights reserved.
 Repeat Example with the diode reverse. Thus the equivalent
circuit is

I D = 0A V R = I R R = I D R = 0V

Using equivalent circuit and KVL

V D = E − V R = 8V − 0V = 8V

Copyright ©2006 by Pearson Education, Inc.

Robert L. Boylestad
Upper Saddle River, New Jersey 07458
Electronic Devices and Circuit Theory, 9e All rights reserved.
 If the diode is biased with the voltage source less than VD,
the diode also acting like open circuit

Diode Characteristic Equivalent cct

With biasing less than 0.7V
Diode Circuit

Copyright ©2006 by Pearson Education, Inc.

Robert L. Boylestad
Upper Saddle River, New Jersey 07458
Electronic Devices and Circuit Theory, 9e All rights reserved.
 Determine Vo and ID for the series circuit below

Circuit Equivalent

Using equivalent circuit and KVL

Vo = E − V K 1 − V K 2 = 12V − 0.7V − 1.8V = 9.5V

VR 9.5V
ID = IR = = = 13.97 mA
R 680Ω

Copyright ©2006 by Pearson Education, Inc.

Robert L. Boylestad
Upper Saddle River, New Jersey 07458
Electronic Devices and Circuit Theory, 9e All rights reserved.
 Determine ID , VD and Vo for the circuit below

Since open circuit

I D = 0mA V D1 = 0V

Vo = I R R = I D R = 0 × R = 0V

And using KVL we have

V D 2 = E − V D1 − Vo = 20V − 0V − 0V = 20V

Copyright ©2006 by Pearson Education, Inc.

Robert L. Boylestad
Upper Saddle River, New Jersey 07458
Electronic Devices and Circuit Theory, 9e All rights reserved.
Determine I, V1, V2 and Vo for the series circuit below

Loop 1
Loop 2

Applying KVL in E1 + E 2 − V D 10V + 5V − 0.7V

I= = = 2.07 mA
loop1 R1 + R2 4.7 kΩ + 2.2kΩ

V1 = IR1 = 2.07 mA × 4.7 kΩ = 9.73V

V2 = IR2 = 2.07 mA × 2.2kΩ = 4.55V
And using KVL in loop 2

Vo = V2 − E 2 = 4.55V − 5V = −0.45V
Copyright ©2006 by Pearson Education, Inc.
Robert L. Boylestad
Upper Saddle River, New Jersey 07458
Electronic Devices and Circuit Theory, 9e All rights reserved.
 Determine Vo, I1, ID1 and ID2 for the parallel diode below

Since the source voltage is greater than the diode then the
current flow and the voltage across diode is 0.7V, thus Vo -0.7V

V R E − V D 10V − 0.7V
The current is I1 = = = 2.8.18mA
R R 330Ω
Since diodes are similar thus the current will be same, then
I1 28.18mA
I D1 = I D 2 = = = 14.09mA
2 2
Copyright ©2006 by Pearson Education, Inc.
Robert L. Boylestad
Upper Saddle River, New Jersey 07458
Electronic Devices and Circuit Theory, 9e All rights reserved.
 Two LEDs are used for polarity detection. Positive green and
negative red. Find R to ensure 20mA through “on” diode. Both
diodes have a reverse breakdown voltage of 3V and an
average turn-on voltage of 2V

On state

Reverse state

LEDs circuit Equivalent Turn-on Equivalent reverse

breakdown

Note: Since the turn-on voltage is 2V so it does not exceed the

reverse breakdown (3V) of the red LED. Otherwise it will damage
the red diode.
Copyright ©2006 by Pearson Education, Inc.
Robert L. Boylestad
Upper Saddle River, New Jersey 07458
Electronic Devices and Circuit Theory, 9e All rights reserved.
 Solution

Applying Ohm’s law. The current is

E − V LED 8V − 2V
I = 20mA = =
R R

Therefore
6V
R= = 300Ω
20mA

Copyright ©2006 by Pearson Education, Inc.

Robert L. Boylestad
Upper Saddle River, New Jersey 07458
Electronic Devices and Circuit Theory, 9e All rights reserved.
 What happen if we replace with LED having turn-on
voltage is 5V?

Applying Ohm’s law. The current now is

E − V LED 8V − 5V
I = 20mA = =
R R
3V
Therefore R= = 150Ω
20mA

But this time the reverse biased for red LED will be 5V and exceed
the breakdown voltage. The red LED will damage.
 Protective Measure for such case

The Si diode has a

breakdown voltage
of 20V which will
stay “off” state
when the apply
voltage is less than
20V. So will protect
the red LED.

Copyright ©2006 by Pearson Education, Inc.

Robert L. Boylestad
Upper Saddle River, New Jersey 07458
Electronic Devices and Circuit Theory, 9e All rights reserved.
 Determine the voltage Vo for the network below

The voltage across diode is the lowest one since it will “on”
first and the other still stay “off” state. Thus
Vo = 12V − 0.7V = 11.3V
Copyright ©2006 by Pearson Education, Inc.
Robert L. Boylestad
Upper Saddle River, New Jersey 07458
Electronic Devices and Circuit Theory, 9e All rights reserved.
Determine the currents I1, I2 and ID for the network below

Since R1 is // D2 then VK2 0.7V

voltage is same I1 = = = 0.212mA
R1 3.3kΩ
Applying KVL in loop 1 V2 = E − V K1 − V K 2 = 20V − 0.7V − 0.7V = 18.6V
V2 18.6V
Therefore I2 = = = 3.32mA
R2 5.6kΩ
and I D 2 = I 2 − I1 = 3.32mA − 0.212mA ≅ 3.11mA
 OR Gate

Determine Vo and I for network below

From fig. on the right apply KVL

Vo = E − V D = 10V − 0.7V = 9.3V

E − V D1 10V − 0.7V
and I= = = 9.3mA
R 1kΩ
Copyright ©2006 by Pearson Education, Inc.
Robert L. Boylestad
Upper Saddle River, New Jersey 07458
Electronic Devices and Circuit Theory, 9e All rights reserved.
 OR gate
Determine the output level for the positive logic AND gate below

Due to forward bias of D2 the output voltage is Vo= 0.7V

From fig. on the right apply KVL

E − V K 10V − 0.7V
I= = = 9.3mA
R 1kΩ
Half-Wave Rectification- Sinusoidal Input

Sinusoidal input

Forward bias

Reverse bias
For ideal rectifier the dc voltage (rms)
= 0.318Vm but the diode is conducted
after the voltage supplied is more than
0.7V as shown below so the dc voltage
will be reduced. Thus Vdc is

Vdc ≅ 0.318(Vm ) ideal

Vdc ≅ 0.318(Vm − V K ) practical

 Example; Circuit as below
a. Sketch the output vo and determine dc level of the output voltage
for ideal diode
b. What is the practical diode
c. The Vm is increase to 200V

Copyright ©2006 by Pearson Education, Inc.

Robert L. Boylestad
Upper Saddle River, New Jersey 07458
Electronic Devices and Circuit Theory, 9e All rights reserved.
 Solution

Ideal diode circuit

Vdc ≅ 0.318(Vm ) = −0.318(20V ) = −6.36V

Copyright ©2006 by Pearson Education, Inc.

Robert L. Boylestad
Upper Saddle River, New Jersey 07458
Electronic Devices and Circuit Theory, 9e All rights reserved.
(b)

Vdc ≅ 0.318(Vm − VK ) = −0.318(20V − 0.7V ) = −6.14V

Drop about 0.22V or 3.5 %

(b) Vdc ≅ 0.318(Vm ) = −0.318(200V ) = −63.6V (ideal)

Vdc ≅ 0.318(Vm − V K ) = −0.318(200V − 0.7 ) = −63.38V (practical)

Drop about 0.22V or 0.35%

 Diode Rating

The diode rating is stated as peak inverse voltage (PIV) or peak

reverse Voltage (PRV). PIV of diode must be greater than the
applied voltage otherwise the diode will damage or enter the
Zener avalanche region

PIV(rating) > Vm half –wave rectifier

Copyright ©2006 by Pearson Education, Inc.

Robert L. Boylestad
Upper Saddle River, New Jersey 07458
Electronic Devices and Circuit Theory, 9e All rights reserved.
 Full-wave rectifier

Using four diode in certain arrangement such that the

circuit are able to rectifier another half of the sinusoidal
wave.
 First half

Copyright ©2006 by Pearson Education, Inc.

Robert L. Boylestad
Upper Saddle River, New Jersey 07458
Electronic Devices and Circuit Theory, 9e All rights reserved.
 Second half

Copyright ©2006 by Pearson Education, Inc.

Robert L. Boylestad
Upper Saddle River, New Jersey 07458
Electronic Devices and Circuit Theory, 9e All rights reserved.
 ideal

Vdc = 2(0.318Vm ) = 0.636Vm

Practical diode

Vdc ≅ 0.636(Vm − V K )
 Diode Rating

PIV > Vm

Copyright ©2006 by Pearson Education, Inc.

Robert L. Boylestad
Upper Saddle River, New Jersey 07458
Electronic Devices and Circuit Theory, 9e All rights reserved.
 Center-Tapped Transformer

This is another way to get a full-wave rectification. However

the PIV > 2 Vm

Copyright ©2006 by Pearson Education, Inc.

Robert L. Boylestad
Upper Saddle River, New Jersey 07458
Electronic Devices and Circuit Theory, 9e All rights reserved.
 How this works.

1st half

2nd half

Copyright ©2006 by Pearson Education, Inc.

Robert L. Boylestad
Upper Saddle River, New Jersey 07458
Electronic Devices and Circuit Theory, 9e All rights reserved.
 PIV.

Apply KVL PIV = Vsec ondary + V R = Vm + Vm = 2Vm

Therefore PIV ≥ 2Vm

Copyright ©2006 by Pearson Education, Inc.

Robert L. Boylestad
Upper Saddle River, New Jersey 07458
Electronic Devices and Circuit Theory, 9e All rights reserved.
 Determine the output for the network below and
calculate the output dc level and the required PIV
of each diode.

Copyright ©2006 by Pearson Education, Inc.

Robert L. Boylestad
Upper Saddle River, New Jersey 07458
Electronic Devices and Circuit Theory, 9e All rights reserved.
Solution
Equivalent circuit

Redrawn network

Vo = Vi = (10V ) = 5V
1 1
From the 2nd Fig
2 2
Therefore Vdc = 0.636(Vo ) = 0.6363(5V ) = 3.18V
PIV ≥ Vm = 5V
 Rectifier Circuits

•One of the most important applications of diodes is in the

design of rectifier circuits. Used to convert an AC signal into a DC
voltage used by most electronics.
 Simple Half-Wave Rectifier

•Only lets through

positive voltages and
rejects negative
voltages

•This example assumes

an ideal diode

•What would the

waveform look like if not
an ideal diode?
 Full-Wave Rectifier
•To utilize both halves of the input sinusoid use a center-tapped
transformer…
 Bridge Rectifier
•Looks like a Wheatstone bridge. Does not require a center-
tapped transformer.

–Requires 2 additional diodes and voltage drop is double.

Peak Rectifier (filtering)
 BATTERY CHARGER

Switching
to choose
2A or 6A

Diode rectifier to change the sinusoidal wave from

transformer to develop the dc current. Some charger may
have filtering and regulator to improve dc level.
Transformer is to step down the main voltage to required
one.
 CLIPPERS.

Clippers are networks that employ diodes to “clip” away a portion

of an input signal without distorting the remaining part of the
applied waveform. The simplest form of diode clipper is one
resistor and a diode similar like half-wave rectifier. There are two
categories: series and parallel

Series clipper (diode in series)

Circuit Squared waveform Triangular waveform

 CLIPPER WITH DC SUPPLY.

Circuit

Analysis

First where the output is? In this case it is at R

Secondly see any dc supply that oppose the input signal. The system
will be “off’ state until the input voltage is grater than the diode and
the opposed voltage

Vm = Vdc + Vdiode
Output Vo = Vm − Vdc − Vdiode
For ideal case Vo = Vm − Vdc
At transition state

Vo = I R R = (0 )R = 0V
 Analysis for ideal diode

dc

After conduction

Vo = Vm − Vdc
Determine the output waveform for the sinusoidal input

From Fig. transition state will

occur at
Vi + 5V = 0V
Vi = −5V
 Transition at -5V
After transition using KVL, the peak is

Vo = Vm + 5V

Vo = 20V + 5V = 25V

The waveform is seen to be off-set by 5V

Determine the output waveform for the square wave
input as shown in the Fig.

input circuit

For 0 Æ T/2 Vo = 20V + 5V = 25V

Vo = 0V
For T/2 Æ T

output
 Parallel Clipper

Circuit

Square wave input Triangular wave input

*Here, the diode is shunted in the circuit

Determine Vo for the following network
The transition level will be at Vi = 0 since id=0

When Vi more than V=4 V , Vo will follow Vi

When Vi less than V=4V , Vo will stay at V=4V

If the diode has VK=0.7V, find Vo.

Applying KVL at transition will be

Vi + V K − V = 0V

Vi = V − V K = 4V − 0.7V = 3.3V
When Vi > 3.3V then Vo=Vi

When Vi< 3.3V then

Vo = 4V − 0.7V = 3.3V