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PN-Junction Diode Characteristics

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Presentation on theme: "PN-Junction Diode Characteristics"— Presentation transcript:

1 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

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

3 Mathematical Approximation
ECE 442 Power Electronics

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

5 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

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

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

8 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

9 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

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

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

12 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

13 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

14 Example 2.3 (a) ECE 442 Power Electronics

15 Example 2.3 (a) simulation
ECE 442 Power Electronics

16 Example 2.3 (b) ECE 442 Power Electronics

17 Example 2.3 (b) simulation
ECE 442 Power Electronics


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