Download Diode Characteristics

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.
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Graphical
PN-Junction Diode V-I Characteristic
Forward Bias Region
Reverse Bias Region
Reverse
breakdown
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Mathematical Approximation
ID =Is (e
ECE 442 Power Electronics
VD
ηVT
-1)
3
Ideal PN Junction Diode V-I Characteristic
Forward Bias – Short Circuit
Reverse Bias – Open Circuit
ECE 442 Power Electronics
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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
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Reverse Recovery Characteristics
Soft Recovery
Reverse recovery time = trr = ta+tb
Peak Reverse Current = IRR = ta(di/dt)
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Reverse Recovery Characteristics
Abrupt Recovery
Reverse recovery time = trr = ta+tb
Peak Reverse Current = IRR = ta(di/dt)
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Series-Connected Diodes
• Use 2 diodes in series
to withstand higher
reverse breakdown
voltage.
• Both diodes conduct
the same reverse
saturation current, Is.
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Diode Characteristics
• Due to differences
between devices,
each diode has a
different voltage
across it.
• Would like to
“Equalize” the
voltages.
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Series-Connected Diodes with
Voltage Sharing Resistors
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Series-Connected Diodes with
Voltage Sharing Resistors
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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
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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
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Example 2.3 (a)
Is1 = 30mA
Is2 = 35mA
R1 = R 2 = R = 100kΩ
-VD = -VD1 - VD2
VD2 = VD - VD1
VD1
VD2
Is1 +
= Is2 +
R
R
VD R
VD1 =
+ (IS2 -IS1 )
2 2
5kV 100k
VD1 =
+
(35Χ10-3 - 30Χ10-3 ) = 2750Volts
2
2
VD2 = VD - VD1 = 5kV - 2750 = 2250Volts
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Example 2.3 (a) simulation
R1
100kOhm
+
U1
-2.727k V DC 1MOhm
-
D1
DIODE_VIRTUAL*
V1
5000 V
R2
100kOhm
+
-
U2
-2.273k V DC 1MOhm
D2
DIODE_VIRTUAL**
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Example 2.3 (b)
Is1 = 30mA
Is2 = 35mA
VD
= 2.5kV
2
V
V
Is1 + D1 = Is2 + D2
R1
R2
VD1 = VD2 =
R2 =
VD2R1
VD1 - R1(Is2 -Is1 )
R1 = 100kΩ
2.5kVΧ100kΩ
2.5kV -100kΩΧ(35Χ10-3 - 30Χ10-3 )
R2 = 125kΩ
R2 =
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Example 2.3 (b) simulation
R1
100kOhm
+
U1
-2.500k V DC 1MOhm
-
D1
DIODE_VIRTUAL*
V1
5000 V
R2
125kOhm
+
-
U2
-2.500k V DC 1MOhm
D2
DIODE_VIRTUAL**
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