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 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.

Analysis of PN Junction Diode in a Circuit

CIRCUIT REPRESENTATION OF DIODE – Ideal Model Reverse-bias VD = - VS Forward-bias I-V characteristics of ideal model

EXAMPLE: Determine the diode voltage and current in the circuit using ideal model for a silicon diode. Also determine the power dissipated in the diode.

CIRCUIT REPRESENTATION OF DIODE – Piecewise Linear Model Reverse-bias VD = - VS Forward-bias I-V characteristics of constant voltage model

EXAMPLE: Determine the diode voltage and current in the circuit (using constant voltage model) for a silicon diode. Also determine the power dissipated in the diode. Consider the cut-in voltage V = 0.65 V.

CIRCUIT REPRESENTATION OF DIODE – Piecewise Linear Model Reverse-bias VD = - VS Forward-bias I-V characteristics of piecewise model

EXAMPLE: Determine the diode voltage and current in the circuit using piecewise linear model for a silicon diode. Also determine the power dissipated in the diode. Consider the cut-in voltage V = 0.65 V and the diode DC forward resistance, rf = 15 Ω.

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. VPS = IDR + VD 5 = (2 x 103) (10-3) [ e ( VD / 0.026) – 1 ] + VD VD = 0.619 V And ID = 2.19 mA ITERATION METHOD

PIECEWISE LINEAR MODEL 1 PIECEWISE LINEAR MODEL 2 IDEAL MODEL PIECEWISE LINEAR MODEL 1 PIECEWISE LINEAR MODEL 2 DIRECT APPROACH Diode voltage VD 0V 0.65 V 0.652 V 0.619 V Diode current ID 2.5 mA 2.175 mA 2.159 mA 2.19 mA

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.

The DC diode current IDQ in term of diode voltage VDQ Total voltage Total current VDQ = DC voltage vd = ac component The DC diode current IDQ in term of diode voltage VDQ If vd << VT , the equation can be expanded into linear series as:

During AC analysis the diode is equivalent to a resistor, rd Therefore, the diode current-voltage relationship can be represented as The relationship between the AC components of the diode voltage and diode current is Or, Where, During AC analysis the diode is equivalent to a resistor, rd

IDQ VDQ = V + - rd id DC equivalent AC equivalent

Example 1 DC Current DC Output voltage Analyze the circuit (by determining VO & vo ). Assume circuit and diode parameters of VPS = 10 V, R = 5 kΩ, Vγ = 0.6 V & vi = 0.2 sin ωt DC Current DC Output voltage

vi vi

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 VPS = 10V, R = 20k, V = 0.7V, and vi = 0.2 sin t (V). Assume the circuit and diode parameters for the circuit below are VPS = 10V, R = 20k, V = 0.7V, and vi = 0.2 sin t (V). Determine the current, IDQ and the time varying current, id ANSWERS IDQ = 0.465 mA Id = 9.97 sin t (µA)