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© Electronics ECE 1231 Recall-Lecture 3 Current generated due to two main factors Drift – movement of carriers due to the existence of electric field Diffusion – movement of carriers due to gradient in concentrations

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© Electronics ECE 1231 Recall-Lecture 3 Introduction of PN junction Space charge region/depletion region Built-in potential voltage V bi Reversed biased pn junction no current flow Forward biased pn junction current flow due to diffusion of carriers. V bi

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Analysis of PN Junction Diode in a Circuit

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© Electronics ECE 1231 CIRCUIT REPRESENTATION OF DIODE i vDvD The I -V characteristics of the ideal diode. V = 0V Conducting state Reverse bias Reverse biased open circuit Conducting state short circuit

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© Electronics ECE 1231 CIRCUIT REPRESENTATION OF DIODE – Piecewise Linear Model i vDvD VV Conducting state Reverse bias V D = V for diode to turn on. Hence during conducting state: = VV Represented as a battery of voltage = V Reverse biased open circuit

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© Electronics ECE 1231 i vDvD VV Conducting state Reverse bias CIRCUIT REPRESENTATION OF DIODE – Piecewise Linear Model V D ≥ V for diode to turn on. Hence during conducting state: Reverse biased open circuit = VV Represented as a battery of voltage = V and forward resistance, r f in series rf rf + V D -

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© Electronics ECE 1231 Diode Circuits: DC Analysis and Models Example Consider a circuit with a dc voltage V PS applied across a resistor and a diode. Applying KVL, we can write, or, The diode voltage V D and current I D are related by the ideal diode equation: (I S is assumed to be known for a particular diode) Equation contains only one unknown, V D :

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Why do you need the Piecewise Linear Model?

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© Electronics ECE 1231 Diode Circuits: Direct Approach Question Determine the diode voltage and current for the circuit. Consider I S = 10 -13 A. and ITERATION METHOD

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© Electronics ECE 1231 Diode Circuits: Using Models Example Determine the diode voltage and current using a piecewise linear model. Assume piecewise linear diode parameters of V f = 0.6 V and r f = 10 Ω. Solution: The diode current is determined by:

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© Electronics ECE 1231 DIODE DC ANALYSIS Find I and V O for the circuit shown below if the diode cut in voltage is V = 0.7V I = 0.2325mA Vo = -0.35V +VO-+VO- I

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© Electronics ECE 1231 Find I and V O for the circuit shown below if the diode cut in voltage is V = 0.7V +VO-+VO- I I = 0.372mA Vo = 0.14V

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© Electronics ECE 1231 Example 2 a) Determine I D if V = 0.7V R = 4k b) If V PS = 8V, what must be the value of R to get I D equal to part (a)

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DIODE AC EQUIVALENT

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© Electronics ECE 1231 ●Sinusoidal Analysis The total input voltage v I = dc V PS + ac v i i D = I DQ + i d v D = V DQ + v d I DQ and V DQ are the DC diode current and voltage respectively.

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© Electronics ECE 1231 Current-voltage Relation The relation between the diode current and voltage can be written as: V DQ = dc quiescent voltage v d = ac component The -1 term in the equation is neglected. The equation can be written as: Diode Circuits: AC Equivalent Circuit If v d << V T, the equation can be expanded into linear series as: The DC diode current Is:

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© Electronics ECE 1231 i D = I D [ 1 + v d /V T ] i D = I D + I D v d / V T = I D + i d where i d = I D v d / V T using Ohm’s law: I = V/R hence, i d = v d / r d compare with i d = I D v d / V T which reveals that r d = V T / I D CONCLUSION: During AC analysis the diode is equivalent to a resistor, r d

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© Electronics ECE 1231 DC equivalentAC equivalent rdrd idid I DQ V DQ = V

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© Electronics ECE 1231 Example 1 Analyze the circuit (by determining V O & v o ). Assume circuit and diode parameters of V PS = 5 V, R = 5 kΩ, V γ = 0.6 V & v i = 0.1 sin ωt V DQ = V I DQ

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© Electronics ECE 1231 rdrd idid

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DC ANALYSIS DIODE = MODEL 1,2 OR 3 CALCULATE DC CURRENT, I D CALCULATE r d AC ANALYSIS DIODE = RESISTOR, r d CALCULATE AC CURRENT, i d

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© Electronics ECE 1231 EXAMPLE 2 Assume the circuit and diode parameters for the circuit below are V PS = 10V, R = 20k , V = 0.7V, and v i = 0.2 sin t. Determine the current, I DQ and the time varying current, i d

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