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Electronic circuits and components * Course Index * Introduction * About the Author * About this Course * Feedback * Course Navigation * How to use this Course * TINA * Locktronics Fundamentals * Introduction: Fundamentals * Units and Multiples * Electricity * Electronic Principles * Electrical Circuits * Alternating Current * Assessment: Fundamentals Passive Components * Introduction: Passive Components * Resistors * Capacitors * Inductors * Transformers * Batteries, Fuses, Lamps and Switches * Assessment: Passive Components Semiconductors * Introduction: Semiconductors * Diodes ## Operation ## Forward and Reverse Bias ## Diode Characteristics ## Worksheet: Diode Characteristics ## Zener Diodes ## Examples: Zener Diodes ## Worksheet: Zener Diodes ## Light Emitting Diodes * Transistors * Logic Gates * Assessment: Semiconductors Passive Circuits * Introduction: Passive Circuits * Series and Parallel Connections * Kirchoff's Laws * Potential and Current Dividers * Passive Time Variant Circuits * Assessment: Passive Circuits Active Circuits * Introduction: Active Circuits * Power Supply Circuits * Operational Amplifier Circuits * Transistor Amplifier Circuits * 555 Timer Circuits * Assessment: Active Circuits Parts Gallery * Introduction: Parts Gallery * Passive Component Images * Semiconductor Images * Other Images * Quizzes |
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Examples: Zener Diodes<^< Zener Diodes | Course Index | Worksheet: Zener Diodes >^>(:nl:) >>frame round<< Example 1: A 5.1 V zener diode is connected in series with a 680 Ω resistor across a 12 V DC supply. Determine the current flowing in the diode. >><<
>>id=example1 class=question margin=10px frame round<< The voltage dropped across the resistor will be (12 - 5.1) = 6.9 V The current flowing in the diode will be the same as that flowing through the resistor. Thus I = V / R = 6.9 V / 680 Ω = 10.1 mA >><< >>frame round<< Example 2: An LED has an operating voltage of 2 V and is designed to operate with a current of 15 mA. Determine the value of series resistance required if the LED is to operate from a 5 V supply. >><<
>>id=example2 class=question margin=10px frame round<< The voltage dropped across the series resistor will be (5 - 2) = 3 V The current in the resistor will be the same as that flowing through the LED, i.e. 15 mA. Thus R = V / I = 3 V / 15 mA = 200 Ω. >><< (:nl:)(:table style="clear:both":)
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