Diode schematic symbol, Electrical Engineering

Assignment Help:

Diode Schematic Symbol

The diode in figure is a unidirectional device. Electron current just only flows in one direction, against the arrow, corresponding to forward bias. The cathode, bar, of the diode symbol corresponds to N-type semiconductor. The anode, arrow, corresponds to the P-type semiconductor. To memorize this relationship, Not-pointing (bar) on the symbol corresponds to N-type semiconductor. Pointing (arrow) corresponds to P-type.

If a diode is forward biased like in Figure above (a), current will get increase slightly as voltage is increased from 0 V. In the case of a silicon diode a computable current flows while the voltage approaches 0.6 V at (c). As greater the voltage is increases past 0.6 V, current increases significantly after the knee. Raising the voltage well beyond 0.7 V may result in high enough current to destroy the diode. The forward voltage, VF, is a characteristic of the semiconductor: 0.6 to 0.7 V for silicon, 0.2 V for germanium, some volts for Light Emitting Diodes (LED). The forward current ranges from a few mA for point contact diodes to 100 mA for small signal diodes to tens or thousands of amperes for power diodes.

960_Diode Schematic Symbol.png

Figure: (a) Forward biased PN junction, (b) Corresponding diode schematic symbol (c) Silicon Diode I vs V characteristic curve.

Just only the leakage current of the intrinsic semiconductor flows, if the diode is reverse biased. This is planned to the left of the origin in diagram above(c). This current will just only be as high as 1 µA for the most intense conditions for silicon small signal diodes. This current does not increase noticeably with increasing reverse bias until the diode breaks down. At breakdown, the current get increases so significantly that the diode will be destroyed unless a high series resistance limits current. We generally select a diode with a higher reverse voltage rating than as compared to any applied voltage to prevent this. Silicon diodes are commonly available with reverse break down ratings of 50, 100, 200, 400, 800 V and higher. It is probable to fabricate diodes with a lower rating of a few volts for utilize as voltage standards.

We earlier mentioned that the reverse leakage current of within a µA for silicon diodes was because of conduction of the intrinsic semiconductor. This is the leakage which can be described by theory. Thermal energy generates few electron hole pairs that conduct leakage current until recombination. In concrete practice this predictable current is only part of the leakage current. Much of the leakage current is because of surface conduction, related to the lack of cleanliness of the semiconductor surface. Both of the leakage currents increase with growing temperature, approaching a µA for small silicon diodes.

Since the p-type material is now associated to the negative terminal of the power supply, 'holes' within the P-type material are pulled away from the junction, causing the width of the depletion zone to increase. Likewise, because the N-type region is linked to the positive terminal, the electrons will as well be pulled away from the junction. Hence the depletion region widens, and does so much increasingly with increasing reverse-bias voltage. This get increases the voltage barrier causing a high resistance to the flow of charge carriers thus permitting minimal electric current to cross the p-n junction. The increase in resistance of the p-n junction results in the junction to behave like an insulator. This is significant for radiation detection since if current was able to flow; the charged particles would just dissipate into the material. The reverse bias makes sure that charged particles are capable to create it to the detector system.


Related Discussions:- Diode schematic symbol

Matrix converters, i need tutor for teaching me matrix converter based upfc...

i need tutor for teaching me matrix converter based upfc modelling and building using matlab

Active-mode npn transistors in circuits, Active-mode NPN transistors in cir...

Active-mode NPN transistors in circuits Figure:  Structure and use of NPN transistor. Arrow according to schematic. The figure opposite is a schematic presentation

Add the contents of register - add instruction , Add the Contents of Regist...

Add the Contents of Register   This  instruction is used to add the contents  of register R with  the contents  of accumulator  and result  of operation is stored  in the accumu

Explain working of digital-to-analog converters, Q. Explain working of Digi...

Q. Explain working of Digital-to-Analog Converters? Digital-to-Analog (D/A) Converters For the results of digital computations to be used in the analog world, it becomes n

ELECTRO MAGNETIC INDUCTION, If aluminium and copper pipes are of same lengt...

If aluminium and copper pipes are of same length and diameter ... same magnet dropped through them...in copper it takes more time to come out of other end, i myself have done this.

Calculate the antenna aperture area, Q. Amicrowave relay systemuses two ide...

Q. Amicrowave relay systemuses two identical horn antennas mounted on towers spaced 40 km apart. If f c = 6 GHz, and each relay hop has L = 60 dB, calculate the antenna aperture a

Show common emitter output characteristics, Q. A Silicon Transistor Whose C...

Q. A Silicon Transistor Whose Common Emitter Output Characteristics Are Shown  In Dig Is Used In The Circuit With V cc =22.5v, R c =5.6k, R e =1k, R 2 =10k, And R 1 =90

Radio hardware, Filters 1. You need to design a lowpass filter with cutoff ...

Filters 1. You need to design a lowpass filter with cutoff frequency Fc= 1MHz. a. What is the minimum filter order required for 30 dB rejection (-30 dB gain) of 10 MHz? b. What is

Show the classifications of data network, Q. Show the Classifications of Da...

Q. Show the Classifications of Data Network? Classifications of Data Network: Data Networks are categorized according to their geographical coverage: • Wide area networks

Determine the total energy loss, Determine the total energy loss: Two ...

Determine the total energy loss: Two capacitors C 1 = 50 μF and C 2 = 100 μF are connected in parallel across 250 V supply. Determine the total energy loss. Figure

Write Your Message!

Captcha
Free Assignment Quote

Assured A++ Grade

Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!

All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd