RecallLecture 4 l Current generated due to two

Recall-Lecture 4 l Current generated due to two main factors l l Drift – movement of carriers due to the existence of electric field Diffusion – movement of carriers due to gradient in concentrations © Electronics ECE 1231

Recall-Lecture 4 l Introduction of PN junction l Space charge region/depletion region l Built-in potential voltage Vbi l Reversed biased pn junction l no current flow © Electronics ECE 1231

Forward-Biased pn Junction Ø +ve terminal is applied to the p-region of the pn junction and vice versa. Ø Direction of the applied electric field EA is the opposite as that of the E-field in the space-charge region. Ø The net result is that the electric field in the space-charge region lower than thermal equilibrium value causing diffusion of charges to begin again. Ø The diffusion process continues as long as VD is applied. Ø Creating current in the pn junction, i. D. © Electronics ECE 1231

Ideal Current-Voltage Relationship Current ID equation of a pn junction diode: IS = the reverse-bias saturation current (for silicon 10 -15 to 10 -13 A) VT = thermal voltage (0. 026 V at room temperature) n = the emission coefficient (1 ≤ n ≤ 2) © Electronics ECE 1231

Ideal Current-Voltage Relationship Example Determine the current in a pn junction diode. Consider a pn junction at T = 300 K in which IS = 1. 4 x 10 -14 A and n = 1. 1 Find the diode current for v. D = +0. 75 V and v. D = -0. 75 V. Very small current © Electronics ECE 1231

PN Junction Diode Ø The basic PN junction diode circuit symbol, and conventional current direction and voltage polarity. Ø The graphs shows the ideal I-V characteristics of a PN junction diode. Ø The diode current is an exponential function of diode voltage in the forward-bias region. Ø The current is very nearly zero in the reverse-bias region. © Electronics ECE 1231

Analysis of PN Junction Diode in a Circuit

CIRCUIT REPRESENTATION OF DIODE – Ideal Model VD = - V S Reverse-bias I-V characteristics of ideal model Forward-bias © Electronics ECE 1231

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. © Electronics ECE 1231

CIRCUIT REPRESENTATION OF DIODE – Piecewise Linear Model VD = - V S Reverse-bias Forward-bias © Electronics I-V characteristics of constant voltage model ECE 1231

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. © Electronics ECE 1231

CIRCUIT REPRESENTATION OF DIODE – Piecewise Linear Model VD = - V S Reverse-bias I-V characteristics of piecewise model Forward-bias © Electronics ECE 1231

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. © Electronics ECE 1231

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 ITERATION METHOD And ID = 2. 19 m. A © Electronics ECE 1231

IDEAL MODEL PIECEWISE LINEAR MODEL 1 PIECEWISE LINEAR MODEL 2 DIRECT APPROACH Diode voltage VD 0 V 0. 652 V 0. 619 V Diode current ID 2. 5 m. A 2. 175 m. A 2. 159 m. A 2. 19 m. A © Electronics ECE 1231

DC Load Line l l l A linear line equation ID versus VD Obtain the equation using KVL © Electronics ECE 1231

The value of ID Use KVL: 2 ID + VD – 5 = 0 ID = -VD + 5 = - VD + 2. 5 2 © Electronics VD 2 ECE 1231

EXAMPLE A diode circuit and its load line are as shown in the figure below Design the circuit when the diode is operating in forward bias condition. Determine the diode current ID and diode forward resistance rf in the circuit using a piecewise linear model. Consider the cut-in voltage of the diode, Vγ = 0. 65 V. © Electronics ECE 1231

ID R + V D – 5 = 0 ID = -VD/R + 5/ R = 2. 5 m. A R = 2 kΩ © Electronics at VD = 0. 7 V ID = (5 – 0. 7) / 2 = 2. 15 m. A Now, VD = Vγ + ID rf 0. 7 = 0. 65 + 2. 15 rf rf = 0. 05 / 2. 15 m. A = 23. 3 Ω ECE 1231

DIODE DC ANALYSIS l Find I and VO for the circuit shown below if the diode has the value of V = 0. 7 V I + VO - I = 0. 2325 m. A © Electronics VO = -0. 35 V ECE 1231

MULTIPLE DIODE CIRCUITS

Problem-Solving Technique: Multiple Diode Circuits 1. Assume the state of the diode. a. b. If assumed on, VD = V If assumed off, ID = 0. 2. Analyze the ‘linear’ circuit with assumed diode states. 3. Evaluate the resulting state of each diode. 4. If any initial assumptions are proven incorrect, make new assumption and return to Step 2. © Electronics ECE 1231

Example 1 The figure shows a multiple diode circuit. If each diode cutin voltage is Vγ= 0. 7 V, determine i. D 1, i. D 2 and v. O +5 V = 5 k Let’s assume both diodes are on KVL at L 1: 5 i. D 2 + 0. 7 + 10 (i. D 2+ i. D 1) – 5 = 0 5 i. D 2 + 10 (i. D 2+ i. D 1) = 9. 3 ---(1) KVL at L 2: 0. 7 + 10 (i. D 2+ i. D 1) – 5 = 0 10 (i. D 2+ i. D 1) = 4. 3 replace in (1) 0 V = 10 k -5 V 5 i. D 2 + 4. 3 = 9. 3 i. D 2 = 1 m. A Hence, Using 10 (i. D 2+ i. D 1) = 4. 3 so i. D 1 = 0. 43 – 1 = -0. 57 m. A a negative diode current means that the diode is actually OFF. Therefore, we need to calculate again, but now we know that diode D 1 is actually off which means that i. D 1 = 0 © Electronics ECE 1231

D 1 is off and D 2 is on +5 V = 10 k KVL at L 1: 5 i. D 2 + 0. 7 + 10 (i. D 2) – 5 = 0 5 i. D 2 + 10 (i. D 2) = 9. 3 i. D 2 = 0. 62 m. A KVL at L 3: vo – 5 + 5 i. D 2 = 0 vo = 5 – 3. 1 = 1. 9 V = 5 k -5 V © Electronics ECE 1231