Download - Electronic device lecture3
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Diode Application
WED DK2 9-10amTHURS DK6 2-4pm
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Robert L. BoylestadElectronic Devices and Circuit Theory, 9e
Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
Series diode configuration
CircuitCharacteristics
RIVE DD +
Simple Diode circuit
=From Kirchoff’s Voltage Law
( )1/ −= TD nVVSD eIIFrom previous lecture
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Robert L. BoylestadElectronic Devices and Circuit Theory, 9e
Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
Load line analysis
VVD
DREI
0==
AID DEV 0==
When VD=0 V then
When ID=0 A then
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Robert L. BoylestadElectronic Devices and Circuit Theory, 9e
Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
Example For the series diode below employing the diode characteristic beside it, determine VDQ , IDQ and VR,
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Robert L. BoylestadElectronic Devices and Circuit Theory, 9e
Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
mAkR
EIVV
DD
205.010
0=
Ω==
=
VEV AID D100 ==
=
VVDQ 78.0=
For load line
By drawing the load line on the diode characteristic we obtained the mAI DQ 5.18=
VR=IRR= IDQR=18.5mA x 1 kΩ =18.5V
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Robert L. BoylestadElectronic Devices and Circuit Theory, 9e
Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
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
Diode circuit Equivalent circuit
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Robert L. BoylestadElectronic Devices and Circuit Theory, 9e
Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
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
Diode circuit Equivalent circuit
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Robert L. BoylestadElectronic Devices and Circuit Theory, 9e
Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
For the diode below determine VD, VR and ID
VVD 7.0=
VVVVEV DR 3.77.08 =−=−=
mAkV
RV
II RRD 32.3
2.23.7
=Ω
===
Using equivalent circuit and KVL
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Robert L. BoylestadElectronic Devices and Circuit Theory, 9e
Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
Repeat Example with the diode reverse. Thus the equivalent circuit is
AI D 0=
VVVVEV RD 808 =−=−=
VRIRIV DRR 0===
Using equivalent circuit and KVL
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Robert L. BoylestadElectronic Devices and Circuit Theory, 9e
Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
If the diode is biased with the voltage source less than VD, the diode also acting like open circuit
Diode Circuit
Diode CharacteristicWith biasing less than 0.7V
Equivalent cct
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Determine Vo and ID for the series circuit below
Robert L. BoylestadElectronic Devices and Circuit Theory, 9e
Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
VVVVVVEV KKo 5.98.17.01221 =−−=−−=
mAVR
VII R
RD 97.13680
5.9=
Ω===
Using equivalent circuit and KVL
Circuit Equivalent
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Determine ID , VD and Vo for the circuit below
Robert L. BoylestadElectronic Devices and Circuit Theory, 9e
Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
VRRIRIV DRo 00 =×===
VVD 01 =
And using KVL we have
VVVVVVEV oDD 20002012 =−−=−−=
mAI D 0=
Since open circuit
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Determine I, V1, V2 and Vo for the series circuit below
Robert L. BoylestadElectronic Devices and Circuit Theory, 9e
Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
VVVEVVo 45.0555.422 −=−=−=
VkmAIRV 73.97.407.211 =Ω×==
And using KVL in loop 2
VkmAIRV 55.42.207.222 =Ω×==
mAkk
VVVRR
VEEI D 07.22.27.4
7.0510
21
21 =Ω+Ω
−+=
+−+
=
Loop 1 Loop 2
Applying KVL in loop1
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Determine Vo, I1, ID1 and ID2 for the parallel diode below
Robert L. BoylestadElectronic Devices and Circuit Theory, 9e
Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
mAmAIII DD 09.142
18.2821
21 ====
The current is mAVVRVE
RV
I DR 18.8.2330
7.0101 Ω
−=
−==
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
Since diodes are similar thus the current will be same, then
mAmAIII DD 09.142
18.2821
21 ====
The current is mAVVRVE
RV
I DR 18.8.2330
7.0101 Ω
−=
−==
Since diodes are similar thus the current will be same, then
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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
Robert L. BoylestadElectronic Devices and Circuit Theory, 9e
Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
LEDs circuit Equivalent Turn-on Equivalent reverse breakdown
On state
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.
Reverse state
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Solution
Robert L. BoylestadElectronic Devices and Circuit Theory, 9e
Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
Ω== 30020
6mAVR
Applying Ohm’s law. The current is
RVV
RVE
mAI LED 2820 −=
−==
Therefore
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What happen if we replace with LED having turn-on voltage is 5V?
Ω== 15020
3mAVR
Applying Ohm’s law. The current now is
RVV
RVE
mAI LED 5820 −=
−==
Therefore
But this time the reverse biased for red LED will be 5V and exceed the breakdown voltage. The red LED will damage.
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Protective Measure for such case
Robert L. BoylestadElectronic Devices and Circuit Theory, 9e
Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
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.
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Determine the voltage Vo for the network below
Robert L. BoylestadElectronic Devices and Circuit Theory, 9e
Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
VVVVo 3.117.012 =−=
The voltage across diode is the lowest one since it will “on”first and the other still stay “off” state. Thus
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Determine the currents I1, I2 and ID for the network below
mAkV
RVI 32.3
6.56.18
2
22 =
Ω==
Since R1 is // D2 then voltage is same mA
kV
RV
I K 212.03.37.0
11
2 =Ω
==
mAmAmAIII D 11.3212.032.3122
Therefore
≅−=−=
Applying KVL in loop 1 VVVVVVEV KK 6.187.07.020212 =−−=−−=
and
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OR Gate
Robert L. BoylestadElectronic Devices and Circuit Theory, 9e
Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
VVVVEV Do 3.97.010 =−=−=
From fig. on the right apply KVL
mAk
VVRVEI D 3.9
17.0101 =
Ω−
=−
=
Determine Vo and I for network below
and
I
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OR gate
From fig. on the right apply KVL
mAk
VVRVEI K 3.9
17.010
=Ω−
=−
=
Due to forward bias of D2 the output voltage is Vo= 0.7V
Determine the output level for the positive logic AND gate below
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Half-Wave Rectification- Sinusoidal Input
Sinusoidal input
Forward bias
Reverse bias
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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
( )mdc VV 318.0≅
( )Kmdc VVV −≅ 318.0
ideal
practical
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Example; Circuit as belowa. Sketch the output vo and determine dc level of the output voltage for ideal diodeb. What is the practical diodec. The Vm is increase to 200V
Robert L. BoylestadElectronic Devices and Circuit Theory, 9e
Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
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Solution
Robert L. BoylestadElectronic Devices and Circuit Theory, 9e
Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
Ideal diode circuit
( ) ( ) VVVV mdc 36.620318.0318.0 −=−=≅
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(b)
( ) ( ) VVVVVV Kmdc 14.67.020318.0318.0 −=−−=−≅
( ) ( ) VVVV mdc 6.63200318.0318.0 −=−=≅(b)
( ) ( ) VVVVV Kmdc 38.637.0200318.0318.0 −=−−=−≅
Drop about 0.22V or 3.5 %
Drop about 0.22V or 0.35%
(ideal)
(practical)
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Diode Rating
Robert L. BoylestadElectronic Devices and Circuit Theory, 9e
Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
The diode rating is stated as peak inverse voltage (PIV) or peakreverse 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
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Full-wave rectifier
Using four diode in certain arrangement such that the circuit are able to rectifier another half of the sinusoidal wave.
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Robert L. BoylestadElectronic Devices and Circuit Theory, 9e
Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
First half
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Second half
Robert L. BoylestadElectronic Devices and Circuit Theory, 9e
Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
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ideal
Practical diode
( ) mmdc VVV 636.0318.02 ==
( )Kmdc VVV −≅ 636.0
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Diode Rating
Robert L. BoylestadElectronic Devices and Circuit Theory, 9e
Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
PIV > Vm
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Center-Tapped Transformer
Robert L. BoylestadElectronic Devices and Circuit Theory, 9e
Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
This is another way to get a full-wave rectification. However the PIV > 2 Vm
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How this works.
Robert L. BoylestadElectronic Devices and Circuit Theory, 9e
Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
1st half
2nd half
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PIV.
mmmRondary VVVVVPIV 2sec =+=+=
Robert L. BoylestadElectronic Devices and Circuit Theory, 9e
Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
mVPIV 2≥
Apply KVL
Therefore
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Determine the output for the network below and calculate the output dc level and the required PIV of each diode.
Robert L. BoylestadElectronic Devices and Circuit Theory, 9e
Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
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SolutionEquivalent circuit
Redrawn network
( ) VVVV io 51021
21
===
VVPIV m 5=≥
From the 2nd Fig
Therefore ( ) VVVV odc 18.356363.0)(636.0 ===
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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.
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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?
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Full-Wave Rectifier•To utilize both halves of the input sinusoid use a center-tapped transformer…
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Bridge Rectifier•Looks like a Wheatstone bridge. Does not require a center-tapped transformer.
–Requires 2 additional diodes and voltage drop is double.
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Peak Rectifier (filtering)
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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.
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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
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CLIPPER WITH DC SUPPLY.
Circuit
Analysis
In this case it is at RFirst where the output is?
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diodedcmo VVVV −−=
diodedcm VVV +=
Output
At transition state
( ) VRRIV Ro 00 ===
dcmo VVV −=
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
For ideal case
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dcmo VVV −=
Analysis for ideal diode
dc
After conduction
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Determine the output waveform for the sinusoidal input
VVVi 05 =+
VVi 5−=
From Fig. transition state will occur at
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VVV mo 5+=
VVVVo 25520 =+=
After transition using KVL, the peak is
The waveform is seen to be off-set by 5V
Transition at -5V
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Determine the output waveform for the square wave input as shown in the Fig.
VVo 0=
VVVVo 25520 =+=For 0 T/2
For T/2 T
input circuit
output
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Parallel Clipper
*Here, the diode is shunted in the circuit
Circuit
Square wave input Triangular wave input
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Determine Vo for the following network
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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
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If the diode has VK=0.7V, find Vo.
Applying KVL at transition will be
VVVVVV Ki 3.37.04 =−=−=
VVVV Ki 0=−+
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VVVVo 3.37.04 =−=
When Vi< 3.3V then
When Vi > 3.3V then Vo=Vi
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