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Published byLeon Poole Modified over 9 years ago
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Recall-Lecture 4 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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Recall-Lecture 4 Vbi Introduction of PN junction
Space charge region/depletion region Built-in potential voltage Vbi Reversed biased pn junction no current flow Forward biased pn junction current flow due to diffusion of carriers. Vbi
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PN Junction Diode The basic PN junction diode circuit symbol, and conventional current direction and voltage polarity. The graphs shows the ideal I-V characteristics of a PN junction diode. The diode current is an exponential function of diode voltage in the forward-bias region. The current is very nearly zero in the reverse-bias region.
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PN Junction Diode Temperature Effects
Both IS and VT are functions of temperature. The diode characteristics vary with temperature. For silicon diodes, the change is approximately 2 mV/oC. Forward-biased PN junction characteristics versus temperature. The required diode voltage, V to produce a given current decreases with an increase in temperature.
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Analysis of PN Junction Diode in a Circuit
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CIRCUIT REPRESENTATION OF DIODE
The I -V characteristics of the ideal diode. i vD Reverse bias Conducting state V = 0V Reverse biased open circuit Conducting state short circuit
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CIRCUIT REPRESENTATION OF DIODE – Piecewise Linear Model
vD Reverse bias Conducting state Reverse biased open circuit V VD = V for diode to turn on. Hence during conducting state: = V Represented as a battery of voltage = V
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CIRCUIT REPRESENTATION OF DIODE – Piecewise Linear Model
vD Reverse bias Conducting state Reverse biased open circuit V VD ≥ V for diode to turn on. Hence during conducting state: = V Represented as a battery of voltage = V and forward resistance, rf in series rf VD
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Diode Circuits: DC Analysis and Models
Example Consider a circuit with a dc voltage VPS applied across a resistor and a diode. Applying KVL, we can write, or, The diode voltage VD and current ID are related by the ideal diode equation: (IS is assumed to be known for a particular diode) Equation contains only one unknown, VD:
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Why do you need to use these models?
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Diode Circuits: Direct Approach
Question Determine the diode voltage and current for the circuit. Consider IS = A. and ITERATION METHOD
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Diode Circuits: Using Models
Example Determine the diode voltage, current and the power dissipated by the diode using piecewise linear model. Assume piecewise linear diode parameters of V = 0.6 V and rf = 10 Ω. Solution: The diode current is determined by: Power dissipated: VD x ID = x 2.19 = 1.36 mW
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DIODE DC ANALYSIS Find I and VO for the circuit shown below if the diode cut in voltage is V = 0.7V + VO - I I = mA Vo = -0.35V
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Find I and VO for the circuit shown below if the diode cut in voltage is V = 0.7V
+ VO - I = 0.372mA Vo = 0.14V
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Example 2 Determine ID if V = 0.7V
R = 4k b) If VPS = 8V, what must be the value of R to get ID equal to part (a)
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DC Load Line A linear line equation ID versus VD
Obtain the equation using KVL
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V 2 Use KVL: 2ID + VD – 5 = 0 ID = -VD + 5 = - VD + 2.5
The value of ID at VD = V V 2 Use KVL: 2ID + VD – 5 = 0 ID = -VD = - VD + 2.5
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DIODE AC EQUIVALENT
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Sinusoidal Analysis The total input voltage vI = dc VPS + ac vi
iD = IDQ + id vD = VDQ + vd IDQ and VDQ are the DC diode current and voltage respectively.
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Diode Circuits: AC Equivalent Circuit
Current-voltage Relation The relation between the diode current and voltage can be written as: VDQ = dc quiescent voltage vd = ac component The -1 term in the equation is neglected. The equation can be written as: If vd << VT, the equation can be expanded into linear series as: The DC diode current Is:
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iD = ID [ 1 + vd/VT] iD = ID + ID vd / VT = ID + id where id = ID vd / VT using Ohm’s law: I = V/R hence, id = vd / rd compare with id = ID vd / VT which reveals that rd = VT / ID CONCLUSION: During AC analysis the diode is equivalent to a resistor, rd
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IDQ VDQ = V rd id DC equivalent AC equivalent
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Example 1 VDQ = V IDQ Analyze the circuit (by determining VO & vo ).
Assume circuit and diode parameters of VPS = 5 V, R = 5 kΩ, Vγ = 0.6 V & vi = 0.1 sin ωt VDQ = V IDQ
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rd id
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CALCULATE DC CURRENT, ID CALCULATE AC CURRENT, id
DC ANALYSIS AC ANALYSIS DIODE = MODEL 1 ,2 OR 3 CALCULATE rd DIODE = RESISTOR, rd CALCULATE DC CURRENT, ID CALCULATE AC CURRENT, id
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EXAMPLE 2 Assume the circuit and diode parameters for the circuit below are VPS = 10V, R = 20k, V = 0.7V, and vi = 0.2 sin t. Determine the current, IDQ and the time varying current, id
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