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

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

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.

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.

Analysis of PN Junction Diode in a Circuit

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

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

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 -

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:

Why do you need to use these models?

Diode Circuits: Direct Approach Question Determine the diode voltage and current for the circuit. Consider IS = 10-13 A. and ITERATION METHOD

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 = 0.622 x 2.19 = 1.36 mW

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 = 0.2325mA Vo = -0.35V

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

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)

DC Load Line A linear line equation ID versus VD Obtain the equation using KVL

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 + 5 = - VD + 2.5

DIODE AC EQUIVALENT

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.

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:

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

IDQ VDQ = V rd id DC equivalent AC equivalent

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

rd id

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

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