pn junction

1. ECE 442 Power Electronics 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.

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

3. ECE 442 Power Electronics 3
Mathematical Approximation
D
T
V
ηV
D s
I =I (e -1)

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

5. 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

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

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

8. 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.

9. 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.

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

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

12. 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

13. 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

14. ECE 442 Power Electronics 14
Example 2.3 (a)

s1
s2
1 2
D D1 D2
D2 D D1
D1 D2
s1 s2
D
D1 S2 S1
-3 -3
D1
D2 D D1
I = 30mA
I = 35mA
R =R =R =100kΩ
-V = -V - V
V = V - V
V V
I + =I +
R R
V R
V = + (I -I )
2 2
5kV 100k
V = + (35Χ10 -30Χ10 ) = 2750Volts
2 2
V = V - V = 5kV - 2750 = 2250Volts

15. ECE 442 Power Electronics 15
Example 2.3 (a) simulation
D1
DIODE_VIRTUAL*
D2
DIODE_VIRTUAL**
R1
100kOhm
R2
100kOhm
U1
DC 1MOhm
-2.727k V
+
-
U2
DC 1MOhm
-2.273k V
+
-
V1
5000 V

16. ECE 442 Power Electronics 16
Example 2.3 (b)
s1
s2
D
D1 D2
D1 D2
s1 s2
1 2
D2 1
2
D1 1 s2 s1
1
2 -3 -3
2
I = 30mA
I = 35mA
V
V = V = = 2.5kV
2
V V
I + =I +
R R
V R
R =
V -R (I -I )
R =100kΩ
2.5kVΧ100kΩ
R =
2.5kV -100kΩΧ(35Χ10 -30Χ10 )
R =125kΩ

17. ECE 442 Power Electronics 17
Example 2.3 (b) simulation
D1
DIODE_VIRTUAL*
D2
DIODE_VIRTUAL**
R1
100kOhm
R2
125kOhm
U1
DC 1MOhm
-2.500k V
+
-
U2
DC 1MOhm
-2.500k V
+
-
V1
5000 V