# اريد مساعدة ارجوكم



## ahmad05541 (23 أبريل 2010)

السلام عليكم و رحمة الله :
انا عندي امتحان الاسبوع القادم في مادة دوائر الكترونية 1
الامتحان في الدروس الاتية :
clipping circuits
camping circuits
DC-voltage doubler
DC-voltage Quadrupler
diode application in rectifier
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
هالمواضيع كلها على الدايود ....
و انا لم اجد اي كتاب على هذه الدروس ..
ارجوكم اريد مرجع ادرس فيه ..
و جزاكم الله الف خير و احسان ..
انا في انتظار الرد 
والسلام عليكم و رحمة الله


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## ahmad05541 (23 أبريل 2010)

ارجوكم ساعدوني 
انا في حاجة ماسة الى كتاب عن الدايود


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## فائق حمادي (23 أبريل 2010)

ahmad05541 قال:


> السلام عليكم و رحمة الله :
> انا عندي امتحان الاسبوع القادم في مادة دوائر الكترونية 1
> الامتحان في الدروس الاتية :
> clipping circuits
> ...


 

*Very simple just write in Google tool bar the words and you will find subjects, or visit Wikipedia.com and you will find reasonable *


*information*


*Good Luck*​


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## ahmad05541 (25 أبريل 2010)

انا انتضر الرد 
ارجوكم انا امتحاني قرب 
انا اناشد كل المختصين


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## ahmad05541 (2 مايو 2010)

انا مستغرب كل هالخبراؤ و الناس المختصين و ما فيش حد قدر ايساعدني ..
ليش يا شباب ؟؟؟هل لاني جديد في المنتدى ؟؟؟انا في حاجة للمساعدة ..
الحمد لله الامتحان تأجل ..
و انا بإنتضار الرد


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## فائق حمادي (2 مايو 2010)

ahmad05541 قال:


> انا مستغرب كل هالخبراؤ و الناس المختصين و ما فيش حد قدر ايساعدني ..
> ليش يا شباب ؟؟؟هل لاني جديد في المنتدى ؟؟؟انا في حاجة للمساعدة ..
> الحمد لله الامتحان تأجل ..
> و انا بإنتضار الرد


 
See details below and the PDF
*[SIZE=+4]Diode Applications[/SIZE]*

*[SIZE=+1]See nice reference on Diode Applications from the Georgia State University, Hyperphysics project.[/SIZE]*
​*Transfer Characteristic*





*Rectification ("frequency shifting")*

*[SIZE=+1]Typical power supply applications[/SIZE]* ​





*[SIZE=+1]Half-Wave Rectification[/SIZE]* 
*"Figure shows a half-wave rectifier circuit. The signal is exactly the top half of the input voltage signal, and for an ideal diode does not depend at all on the size of the load resistor.*







*"The rectified signal is now a combination of an AC signal and a DC component. Generally, it is the DC part of a rectified signal that is of interest, and the un-welcomed AC component is described as ripple. It is desirable to move the ripple to high frequencies where it is easier to remove by a low-pass filter.* 
*"When diodes are used in small-signal applications - a few volts - their behaviour is not closely approximated by the ideal model because of the PN turn-on voltage. The equivalent circuit model can be used to evaluate the detailed action of the rectifier under these conditions. During the part of the wave when the input is positive but less than the PN turn-on voltage, the model predicts no loop current and the output signal voltage is therefore zero. When the input exceeds this voltage, the output signal becomes proportional to, or about 0.6 V lower than the source voltage."* (source)​*Op Amp solution to PN turn-on problem*



(source)​

*[SIZE=+1]Half-wave rectifier with filter capacitor or peak detector[/SIZE]*​





*[SIZE=+1]Full-Wave Retification[/SIZE]*
*[SIZE=+1]Version 1 - Center-Tap Full-Wave Rectifier[/SIZE]*








*(source)*​
*[SIZE=+1]Version 2 - Bridge Full-Wave Rectifier[/SIZE]* 
*"The diode bridge circuit shown Ö is a full-wave rectifier. The diodes act to route the current from both halves of the AC wave through the load resistor in the same direction, and the voltage developed across the load resistor becomes the rectified output signal. The diode bridge is a commonly used circuit and is available as a four-terminal component in a number of different power and voltage ratings."*








(source)​












*[SIZE=+1]Go to Diode Bridge Modules for a collection of pdf data sheets on many integrated diode bridges.[/SIZE]*​ *Op Amp solution to PN turn-on problem*



(source)​*Split Power Supply* 
*"Often a circuit requires a power supply that provides negative voltage as well as positive voltage. By reversing the direction of the diode and the capacitor (if it is polarized), the half-wave rectification circuit with low-pass filter provides a negative voltage. Similarly, reversing the direction of the diodes and capacitor in the full-wave rectified supply produces a negative voltage supply. A split power supply is shown in figure Ö" *​



(source)​*A Variety of Other Applications:*



*[SIZE=+1]Clamp[/SIZE]* 
*[SIZE=+1]Also called a "dc restorer" in Sedra & Smith[/SIZE]*
*"When a signal drives an open-ended capacitor the average voltage level on the output terminal of the capacitor is determined by the initial charge on that terminal and may therefore be quite unpredictable. Thus it is necessary to connect the output to ground or some other reference voltage via a large resistor. This action drains any excess charge and results in an average or DC output voltage of zero*.







*"A simple alternative method of establishing a DC reference for the output voltage is by using a diode clamp as shown in figure Ö. By conducting whenever the voltage at the output terminal of the capacitor goes negative, this circuit builds up an average charge on the terminal that is sufficient to prevent the output from ever going negative. Positive charge on this terminal is effectively trapped."* *(source)*​



(source)​



*[SIZE=+1]Clippers/Limiters[/SIZE]* ​



*"A diode clipping circuit can be used to limit the voltage swing of a signal. Figure Ö shows a diode circuit that clips both the positive and negative voltage swings to references voltages."*​




*(source)*​*[SIZE=+1]Limiting with ordinary diodes:[/SIZE]*​



*(source)*​*[SIZE=+1]Limiting with Zener diodes:[/SIZE]*​



*[SIZE=+1]Multiplier[/SIZE]* 

*[SIZE=+1]Doubler - Version 1: A dc restorer followed by a peak detector[/SIZE]*










*(source)*​
*[SIZE=+1]Doubler - Version 2[/SIZE]* 
*"A voltage multiplier circuit is shown in figureÖ. We can think of it as two half-wave rectifier circuits in series. During the positive half-cycle one of the diodes conducts and charges a capacitor. During the negative half-cycle the other diode conducts negatively to charge the other capacitor. The voltage across the combination is therefore equal to twice the peak voltage. In this type of circuit we have to assume that the load does not draw a significant charge from the capacitors."*
*(source)*​












*(source)*​

*[SIZE=+1]Tripler[/SIZE]*




*(source)*​

*[SIZE=+1]For more on multipliers see:[/SIZE]* *[SIZE=+1]Cockroft-Waton Diode Voltage Multipliers[/SIZE]*​
*[SIZE=+1]Diode Logic[/SIZE]* ​






*"To the left (above) you see a basic Diode Logic OR gate. We'll assume that a logic 1 is represented by +5 volts, and a logic 0 is represented by ground, or zero volts. In this figure, if both inputs are left unconnected or are both at logic 0, output Z will also be held at zero volts by the resistor, and will thus be a logic 0 as well. However, if either input is raised to +5 volts, its diode will become forward biased and will therefore conduct. This in turn will force the output up to logic 1. If both inputs are logic 1, the output will still be logic 1. Hence, this gate correctly performs a logical OR function.*​
*"To the right (above) is the equivalent AND gate. We use the same logic levels, but the diodes are reversed and the resistor is set to pull the output voltage up to a logic 1 state. For this example, +V = +5 volts, although other voltages can just as easily be used. Now, if both inputs are unconnected or if they are both at logic 1, output Z will be at logic 1. If either input is grounded (logic 0), that diode will conduct and will pull the output down to logic 0 as well. Both inputs must be logic 1 in order for the output to be logic 1, so this circuit performs the logical AND function."* *(source)*​*[SIZE=+1]Standby Voltage[/SIZE]* ​






*(source)*​*[SIZE=+1]Gate[/SIZE]* ​



*(source)*​*Mixer Circuits*

*[SIZE=+1]Consider the following circuit:[/SIZE]*​



*[SIZE=+1]Using the Shockley Diode Equation we can write[/SIZE]*​



*[SIZE=+1]This is a pretty complicated expression, but we can get a reasonable and useful result with a bit algebra and analysis. Let us assume that there are dc biases on the signals (usual case) so that[/SIZE]*​



*[SIZE=+1]where



are fluctuations around the bias values



.[/SIZE]* 
*[SIZE=+1]With much algebra we can show that[/SIZE]* 



*[SIZE=+1]and[/SIZE]*



*[SIZE=+1]where[/SIZE]*




*[SIZE=+1]The important point is that the output has components[/SIZE]*



​


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