Diode Application WED DK2 9-10am THURS DK6 2-4pm
Simple Diode circuit Series diode configuration Circuit From Kirchoff s Voltage Law From previous lecture E = V I D D + IDR / = I V D nv T S e ( 1) Characteristics Robert L. Boylestad Electronic Devices and Circuit Theory, 9e Copyright 2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.
Load line analysis When V D =0 V then When I D =0 A then I V D D = = E R E V I D D =0V =0 A Robert L. Boylestad Electronic 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 V DQ, I DQ and V R, Robert L. Boylestad Electronic Devices and Circuit Theory, 9e Copyright 2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.
For load line I V D = = E R E VD = 0V = 10 0.5kΩ = 10V = 20 ma D ID = 0 A By drawing the load line on the diode characteristic we obtained the V DQ = 0. 78V V R =I R R= I DQ R=18.5mA x 1 kω =18.5V I DQ = 18. 5mA Robert L. Boylestad Electronic Devices and Circuit Theory, 9e Copyright 2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 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 V D > 0.7V for Si, V D >0.3V for Ge, V D > 1.2V for GaAs Robert L. Boylestad Electronic Devices and Circuit Theory, 9e Copyright 2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 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 I D =0 Robert L. Boylestad Electronic 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 V D, V R and I D V D = 0. 7V Using equivalent circuit and KVL VR = E VD = 8 V 0.7V = 7. 3V VR 7.3V I D = I R = = = 3. 32mA R 2.2kΩ Robert L. Boylestad Electronic 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 I D = 0A VR = I R R = I D R = 0V Using equivalent circuit and KVL VD = E VR = 8 V 0V = 8V Robert L. Boylestad Electronic 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 V D, the diode also acting like open circuit Diode Circuit Diode Characteristic With biasing less than 0.7V Equivalent cct Robert L. Boylestad Electronic Devices and Circuit Theory, 9e Copyright 2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.
Determine V o and I D for the series circuit below Circuit Equivalent Using equivalent circuit and KVL Vo = E VK1 VK 2 = 12V 0.7V 1.8V = 9. 5V VR 9.5V I D = I R = = = 13. 97mA R 680Ω Robert L. Boylestad Electronic Devices and Circuit Theory, 9e Copyright 2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.
Determine I D, V D and V o for the circuit below Since open circuit I D = 0mA V D 1 = 0V Vo = I R R = I D R = 0 R = 0V And using KVL we have VD2 = E VD1 Vo = 20V 0V 0V = 20V Robert L. Boylestad Electronic Devices and Circuit Theory, 9e Copyright 2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.
Determine I, V 1, V 2 and V o for the series circuit below Loop 1 Loop 2 Applying KVL in loop1 And using KVL in loop 2 I = E 1 + E R + 1 2 V R 2 D 10V + 5V = 4.7kΩ + V = IR = 2.07mA 4.7kΩ 9. 73V 1 1 = V = IR = 2.07mA 2.2kΩ 4. 55V 2 2 = V o = V E = 4.55V 5V = 0. 45V 2 2 0.7V 2.2kΩ = 2.07mA Robert L. Boylestad Electronic Devices and Circuit Theory, 9e Copyright 2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.
Determine V o, I 1, I D1 and I D2 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 V o -0.7V The current is I 1 = V R R = E V R D = 10 V 0.7 V 330 Ω 2.8.18 ma Since diodes are similar thus the current will be same, then I 1 28.18 ma I D 1 = I D 2 = = = 14. 09 ma 2 2 Robert L. Boylestad Electronic Devices and Circuit Theory, 9e Copyright 2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 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. Robert L. Boylestad Electronic Devices and Circuit Theory, 9e Copyright 2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.
Solution Applying Ohm s law. The current is I = 20mA = E VLED 8V 2V = R R Therefore R 6V = = 300Ω 20mA Robert L. Boylestad Electronic Devices and Circuit Theory, 9e Copyright 2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.
What happen if we replace with LED having turn-on voltage is 5V? Applying Ohm s law. The current now is I = 20mA Therefore E VLED 8V 5V = = R R 3V 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. Robert L. Boylestad Electronic Devices and Circuit Theory, 9e Copyright 2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.
Determine the voltage V o 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 V o = 12 V 0.7V = 11. 3V Robert L. Boylestad Electronic Devices and Circuit Theory, 9e Copyright 2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.
Determine the currents I 1, I 2 and I D for the network below Since R 1 is // D 2 then voltage is same Applying KVL in loop 1 Therefore and VK2 0.7V I1 = = = 0.212mA R1 3.3kΩ V2 = E VK VK = 20V 0.7V 0.7V = 18. 6V 1 2 V2 18.6V I 2 = = = 3. 32mA R2 5.6kΩ I D = I I = 3.32mA 0.212mA 3. 11mA 2 2 1
OR Gate Determine V o and I for network below I From fig. on the right apply KVL Vo = E VD = 10 V 0.7V = 9. 3V and I = E V R D1 = 10V 0.7V 1kΩ = 9.3mA Robert L. Boylestad Electronic Devices and Circuit Theory, 9e Copyright 2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.
OR gate Determine the output level for the positive logic AND gate below Due to forward bias of D 2 the output voltage is V o = 0.7V From fig. on the right apply KVL I = E V R K 10V 0.7V = 1kΩ = 9.3mA
Half-Wave Rectification- Sinusoidal Input Sinusoidal input Forward bias Reverse bias
For ideal rectifier the dc voltage (rms) = 0.318V m 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 V dc is V 0. 318 V ( ) dc V m dc 0. 318 ( V V ) m K ideal practical
Example; Circuit as below a. Sketch the output v o and determine dc level of the output voltage for ideal diode b. What is the practical diode c. The Vm is increase to 200V Robert L. Boylestad Electronic Devices and Circuit Theory, 9e Copyright 2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.
Solution Ideal diode circuit V ( V ) = 0.318( 20V ) = 6. V dc 0.318 m 36 Robert L. Boylestad Electronic Devices and Circuit Theory, 9e Copyright 2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.
(b) V ( V V ) = 0.318( 20V 0.7V ) = 6. V dc 0.318 m K 14 Drop about 0.22V or 3.5 % (b) V ( V ) = 0.318( 200V ) = 63. V dc 0.318 m 6 (ideal) V ( V V ) = 0.318( 200V 0.7) = 63. V dc 0.318 m K 38 (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) > V m half wave rectifier Robert L. Boylestad Electronic Devices and Circuit Theory, 9e Copyright 2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 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 Robert L. Boylestad Electronic Devices and Circuit Theory, 9e Copyright 2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.
Second half Robert L. Boylestad Electronic Devices and Circuit Theory, 9e Copyright 2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.
ideal dc ( 0.318V m ) 0. Vm V = 2 = 636 Practical diode V dc 0. 636 ( V V ) m K
Diode Rating PIV > V m Robert L. Boylestad Electronic Devices and Circuit Theory, 9e Copyright 2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.
Center-Tapped Transformer This is another way to get a full-wave rectification. However the PIV > 2 V m Robert L. Boylestad Electronic Devices and Circuit Theory, 9e Copyright 2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.
How this works. 1 st half 2 nd half Robert L. Boylestad Electronic Devices and Circuit Theory, 9e Copyright 2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.
PIV. Apply KVL Therefore PIV = V 2V sec ondary + VR = Vm + Vm = PIV 2V m m Robert L. Boylestad Electronic Devices and Circuit Theory, 9e Copyright 2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.
Determine the output for the network below and calculate the output dc level and the required PIV of each diode. Robert L. Boylestad Electronic Devices and Circuit Theory, 9e Copyright 2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.
Solution Equivalent circuit Redrawn network From the 2 nd Fig 1 1 Vo = Vi = 5 2 2 PIV Vm = 5V ( 10V ) = V Therefore V = 0.636( V ) = 0.6363( 5V ) = 3. V dc o 18
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 centertapped 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 Output V = V + V V m o = V m dc V dc diode V diode For ideal case V o = V m V dc At transition state ( 0 ) R = V Vo = I R R = 0
Analysis for ideal diode dc After conduction V o = V m V dc
Determine the output waveform for the sinusoidal input From Fig. transition state will occur at V i + 5 V = 0V V i = 5V
After transition using KVL, the peak is Vo = Vm + 5V Transition at -5V V o = 20 V + 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 For 0 T/2 For T/2 T V o = 20 V + 5V = 25V V o = 0V circuit 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 V i = 0 since i d =0 When V i more than V=4 V, V o will follow V i When V i less than V=4V, V o will stay at V=4V
If the diode has V K =0.7V, find V o. Applying KVL at transition will be Vi + VK V = 0V Vi = V VK = 4 V 0.7V = 3. 3V
When V i > 3.3V then V o =V i When Vi< 3.3V then V o = 4 V 0.7V = 3. 3V