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Door phone intercom
The digital door phone intercom is a device which can be used for communication between two people separated by a distance so that one can determine exactly who it is before access can be granted.
|language || ||english
|wordcount || ||7888 (cca 22 pages)
|contextual quality || ||N/A
|language level || ||N/A
|price || ||free
|sources || ||4
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Preview of the essay: Door phone intercom
CHAPTER ONE DOOR PHONE INTERCOM 1.1 Introduction The digital door phone intercom is a device which can be used for communication between two people separated by a distance so that one can determine exactly who it is before access can be granted. 1.2 Objectives The purpose of this project is: ➢ To design and construct a device which can allow easy communication between two parties separated by a distance before establishing who it is so that they can be allowed. ➢ To construct a quality, cheap and affordable door phone intercom which can meet the market requirement 1.3 Specifications Supply voltage 9V d.c Frequency 50HZ Transmission distance 60M 1.4 BLOCK DIAGRAM [pic] Functions of each block Power supply The door phone intercom uses a 9V dc power supply to both the pre-amplifier and the power amplifier. The self DC bias is used in allowing for variations in transistor current gain. This can be obtained by stepping down the 240V a.c. to 9V d.c. Pre-amplifier ...
... a half – cycle, there is a half- wave signal with an average value of ; - [pic] = 477.47mA Efficiency Advantage of push-pull class as amplifiers is the high efficiency. This advantage overrides the difficulty of biasing he class. Ad. Push – pull amplifiers to eliminate cross over distortion. Efficiency = Pout X 100% = 4.49652 x 100 5.7296 = 78.48% Input resistance The complementary push – pull configuration of class AB amplifiers is in effect two counter followers. Therefore, Ri = Bac(re + Re) Since Re = Rn Rin = Bac(re + Rn) Let re = 5Ω and Bac = 50 Rin = 50 (5 + 4) = 450Ω (ideal situation) Rin = R1//R2 = 1 x 10³ x 1 x 10³ 1 x10³ + 1 x 10³ = 500Ω Coupling capacitors Capacitors C1 and C2 Xcl = 1 2πFc.l Let fc = 150HZ And Xc l = 1 2πFc.l = 1 2π x 1 x 10³ x 15 = 1.06µF (nearest preferred value 1µF) Since Q1 and Q2 are matched, hence C1 = C2 = 1µF
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The writer reminds me of my physics teacher who like socrates gently guide us through a very technical subject.