# PNJunction Diode Characteristics Forward Bias External battery makes

• Slides: 17

PN-Junction Diode Characteristics Forward Bias --- External battery makes the Anode more positive than the Cathode --- Current flows in the direction of the arrow in the symbol. Reverse Bias --- External battery makes the Cathode more positive than the Anode --- A tiny current flows opposite to the arrow in the symbol. ECE 442 Power Electronics 1

Graphical PN-Junction Diode V-I Characteristic Forward Bias Region Reverse breakdown ECE 442 Power Electronics 2

Mathematical Approximation ECE 442 Power Electronics 3

Ideal PN Junction Diode V-I Characteristic Forward Bias – Short Circuit Reverse Bias – Open Circuit ECE 442 Power Electronics 4

Diode Reverse Recovery Time ta is the time to remove the charge stored in the depletion region of the junction tb is the time to remove the charge stored in the bulk semiconductor material ECE 442 Power Electronics 5

Reverse Recovery Characteristics Soft Recovery Reverse recovery time = trr = ta+tb Peak Reverse Current = IRR = ta(di/dt) ECE 442 Power Electronics 6

Reverse Recovery Characteristics Abrupt Recovery Reverse recovery time = trr = ta+tb Peak Reverse Current = IRR = ta(di/dt) ECE 442 Power Electronics 7

Series-Connected Diodes • Use 2 diodes in series to withstand higher reverse breakdown voltage. • Both diodes conduct the same reverse saturation current, Is. ECE 442 Power Electronics 8

Diode Characteristics • Due to differences between devices, each diode has a different voltage across it. • Would like to “Equalize” the voltages. ECE 442 Power Electronics 9

Series-Connected Diodes with Voltage Sharing Resistors ECE 442 Power Electronics 10

Series-Connected Diodes with Voltage Sharing Resistors ECE 442 Power Electronics 11

Series-Connected Diodes with Voltage Sharing Resistors • Is = Is 1+IR 1 = Is 2+IR 2 • IR 1 = VD 1/R 1 • IR 2 = VD 2/R 2 = VD 1/R 2 • • Is 1+VD 1/R 1 = IS 2+VD 1/R 2 Let R = R 1 = R 2 Is 1 + VD 1/R = Is 2 +VD 2/R VD 1 + VD 2 = Vs ECE 442 Power Electronics 12

Example 2. 3 • Is 1 = 30 m. A, Is 2 = 35 m. A • VD = 5 k. V • (a) – R 1=R 2=R=100 kΩ, find VD 1 and VD 2 • (b) – Find R 1 and R 2 for VD 1=VD 2=VD/2 ECE 442 Power Electronics 13

Example 2. 3 (a) ECE 442 Power Electronics 14

Example 2. 3 (a) simulation ECE 442 Power Electronics 15

Example 2. 3 (b) ECE 442 Power Electronics 16

Example 2. 3 (b) simulation ECE 442 Power Electronics 17