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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.
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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
Graphical PN-Junction Diode V-I Characteristic Forward Bias Region Reverse Bias Region Reverse breakdown ECE 442 Power Electronics
Mathematical Approximation ECE 442 Power Electronics
Ideal PN Junction Diode V-I Characteristic Forward Bias – Short Circuit Reverse Bias – Open Circuit ECE 442 Power Electronics
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
Reverse Recovery CharacteristicsSoft Recovery Reverse recovery time = trr = ta+tb Peak Reverse Current =IRR = ta(di/dt) ECE 442 Power Electronics
Reverse Recovery CharacteristicsAbrupt Recovery Reverse recovery time = trr = ta+tb Peak Reverse Current = IRR = ta(di/dt) ECE 442 Power Electronics
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
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
Series-Connected Diodes with Voltage Sharing Resistors ECE 442 Power Electronics
Series-Connected Diodes with Voltage Sharing Resistors ECE 442 Power Electronics
Series-Connected Diodes with Voltage Sharing Resistors • Is = Is1+IR1 = Is2+IR2 • IR1 = VD1/R1 • IR2 = VD2/R2 = VD1/R2 • Is1+VD1/R1 = IS2+VD1/R2 • Let R = R1 = R2 • Is1 + VD1/R = Is2 +VD2/R • VD1 +VD2 = Vs ECE 442 Power Electronics
Example 2.3 • Is1 = 30mA, Is2 = 35mA • VD = 5kV • (a) – R1=R2=R=100kΩ, find VD1 and VD2 • (b) – Find R1 and R2 for VD1=VD2=VD/2 ECE 442 Power Electronics
Example 2.3 (a) ECE 442 Power Electronics
Example 2.3 (a) simulation ECE 442 Power Electronics
Example 2.3 (b) ECE 442 Power Electronics
Example 2.3 (b) simulation ECE 442 Power Electronics