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Published byAlyson Marsh Modified over 9 years ago
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Diode Diode is the simplest semiconductor device. It’s a two-terminal device
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Basic operation Ideal diode conducts current in only one direction and acts like open in the opposite direction
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Characteristics of an ideal diode: Conduction region
- the voltage across the diode is zero - the current near infinite - the diode acts like short
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Characteristics of an ideal diode: Non-conduction region
Non-conduction region - all of the voltages is across the diode - the current is zero - the diode acts like open
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Semiconductor materials
Two types of materials used in development of semiconductor are Silicon(Si) and Germanium(Ge) Doping is a process to add impurity(such as Antimony(Sb) or Boron(B)) to Si or Ge to make two types of semiconductor materials: n- type material: make Si(or Ge) more negative (has “free” electrons) p- type material: make Si(or Ge) more positive (has “holes”) Joining n-type and p-type materials makes a p-n junction diode
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P-n junction diode “electrons” in n-types material migrate across the junction to p-type material and forms a so-called “depletion region” around the junction.
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No bias condition
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Reverse bias condition
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Forward bias condition
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Actual diode characteristics
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Zener region
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Resistance levels DC or Static resistance AC or Dynamic resistance Average AC resistance
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DC or Static resistance
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AC or Dynamic resistance
Forward bias region: The resistance depends on the amount of current(ID) in the diode. The voltage across the diode is fairly constant(26 mA for 25 C). rB ranges from a typical 0.1 Ohms for high power device to 2 Ohms for low power, general purpose diodes. Reverse bias region: The resistance is infinite. The diode acts like an open.
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AC or Dynamic resistance(cont’d)
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Average AC resistance
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Diode specification sheets
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Other types of diodes Zener diode Light emitting diode (LED) Diode arrays
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Zener diode
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Light emitting diode (LED)
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Diode arrays
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Diode equivalent circuits
Piecewise linear model Simplified model Ideal device
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Piecewise linear model
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Simplified model
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Ideal device
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Determine VDQ, IDQ and VR if
(a) E = 20 V, R = 1 kOhms (b) E = 10 V, R =10 kOhms
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(a) E=20V, R = 1k: (b) E=10V, R = 10k: E = VD + VR = VD + RID
Try VD = 0.5 ID = (E - VD)/R = 19.5/1000 = 19.5 mA From the graph, at ID = 19.5 mA, VD = 0.9 V Try VD = 0.9 ID = (E - VD)/R = 19.1/1000 = 19.1 mA Therefore, ID = mA, VD = 0.9, VR = V (b) E=10V, R = 10k: Try VD = 0.5 ID = (E - VD)/R = 9.5/10000 = 0.95 mA From the graph, at ID = 0.95 mA, VD = 0.55 V Try VD = 0.55 ID = (E - VD)/R = 9.45/10000 = 0.94 mA Therefore, ID = 0.94 mA, VD = 0.55, VR = V
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Load-line analysis
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Example: Simple diode circuit
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Example: Simple diode circuit
Find VDQ, IDQ, and VR if (a) E = 10 Volts, R = 1 kOhms (b) E = 10 Volts, R = 2 kOhms (c) Repeat (a) using the approximate equivalent model for the Si diode. (d) Repeat (b) using the approximate (e) Repeat (a) using the ideal diode model. (f) Repeat (b) using the ideal diode model.
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Example: (a) E = 10 Volts, R = 1 kOhms
Y-intercept = E/R = 10 mA Slope = -1/ R = 1 mA/V From the load-line: IDQ = 9.15 mA VDQ = 0.82 V VR = IDQ x R = 9.15 V
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Example: (b) E = 10 Volts, R = 2 kOhms
Y-intercept = E/R = 5 mA Slope = -1/ R = mA/V From the load-line: IDQ = 4.6 mA VDQ = 0.78 V VR = IDQ x R = 9.2 V
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