zener diode

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Presentation Transcript

Slide 1: 

LP3 1 Unlike ordinary diodes they are used in the breakdown region Zener diodes are used to get reasonably regulated dc voltage when the input voltage and load resistance vary This is achieved by passing controlled current through the zener diode in the breakdown region The nominal breakdown voltage VZ is specified as the zener voltage (example 4.7 V zener, 8.2 V zener) Symbol looks like Zener Diodes

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LP3 2 Example1 on Zener Diode Voltage Regulator VSS = 24V R = 1.2 k RL= 6 k; 1.2 k Find IS and vL using the zener characteristics

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LP3 3 Thevenin voltage Example1 on Zener Diode Voltage Regulator(contd..) Compute the Thevenin equivalent of the previous circuit with the zener diode as the load Thevenin Resistance k We can then write VT +RTiD+vD = 0 and find out vD,, iD using the zener diode characteristics vL = vD and IS = vL /RL + iD

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LP3 4 Load Line : vD = -VT -RTiD Example1 on Zener Diode Voltage Regulator(contd..)

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LP3 5 Answer to Example1 on Zener Diode Voltage Regulator vL = 10V IS = vL /RL + iD = 10/6 +10 mA = 11.67mA vL = 9.5 V IS = vL /RL + iD = 9.5/1.2 +5 mA = 12.92mA Note that load resistance change has caused the output voltage to change by 0.5V due to non-ideal reverse characteristics of the zener diode

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LP3 6 Example 2 on Zener Diode Voltage Regulator (ideal-characteristics) Find v0 for iL= i) 0 ii) 30 iii) 80 mA

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LP3 7 Example 2 on Zener Diode Voltage Regulator The characteristics shown for the zener diode is the ideal characteristics- one without any zener bulk resistance. Hence the zener will be able to provide ideally regulated voltage unless the load current demand exceeds (15- v0)/100 = 50 mA. When iL  50 mA the zener absorbs the differential (excess) current to maintain the load voltage at 10 volts. When iL  50 mA the zener loses its regulating capacity(since it cannot generate current!) and the output voltage starts drooping. Then v0 = 15-100 iL.

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LP3 8 Answer to Example 2 on Zener Diode Voltage Regulator i) 10V, ii) 10V, iii) 7 V

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LP3 9 Wave-shaping Circuits Necessary to shape waveforms Many examples can be found in transmitters and receivers in TV or radar or equipment control and protection circuits Uses diodes and zener diodes for clipping and clamping (shaping) input

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LP3 10 A Clipper Circuit (Assume ideal diode)

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LP3 11 A Clipper Circuit (Practical Implementation)

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LP3 12 A Clamp (adds dc offset to ac) Circuit Assumption: RC (2)/

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LP3 13 In the positive half cycle C gets charged through D to 10V (peak of sinewave + 5 V) with the straight plate of C at a higher potential. D Clips the output to a maximum of -5V. In the negative half cycle D is reverse biased. The output can reach a minimum of –15V (-VC + negative peak of sinewave). How Does A Clamp Circuit Work?

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LP3 14 Application of Clamp Circuit –A Voltage-Doubler Assumption: R1C1 or R1C2 (2)/

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LP3 15 In the negative half cycle C1 gets charged through D2 to peak of sinewave (say Vm) with the right plate of C1at a higher potential. D1is reverse biased since D2 conducts. In the positive half cycle D1 is conducting and D2 is reverse biased. Thus C2 gets charged to the maximum of Vm+ Vmsin(t) or 2Vm with the top plate of C2 at a higher potential How does a Voltage-Doubler works

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LP3 16 Other types of Diodes LED (Light Emitting Diodes) They emit light when forward biased. Typical forward voltage drop for these diodes are 1.5 - 2.5V for currents between 10 – 20 mA. Application areas: Instrument/Equipment panel indicators, Seven Segment Displays. Photo Diodes Get forward biased when light falls on them. Application areas: Optocouplers for electrical isolation. Schottky diodes, Ultrfast recovery diodes Application areas:Power Electronics

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LP3 17 Shockley Equation An equation named after the inventor of Transistor Describes approximately the Diode curves like the one shown below Is an important equation to be used in Transistor Analysis

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LP3 18 Shockley Equation(2) Shockley equation: Where IS is the saturation current,VT is the thermal voltage,n is the emission coefficient that takes values between 1 and 2, k is Boltzmann’s constant ( k = 1.38*10-23 J/K) ,q is the magnitude of electrical charge on an electron (k = 1.60*10-19 J/K), T is p-n junction temperature in degree Kelvin. For example, at T =3000K, IS = 10-14 A, VT  0.026 V

Slide 19: 

LP3 19 Plus and minuses of Shockley Equation(2) Fairly accurate with diode near the forward biased region Not a good predictor of IS which happens to much larger in magnitude Does not predict reverse breakdown Usually simpler models for diodes are useful