MBR2535CT Vishay, MBR2535CT Datasheet - Page 3

• Manufacturer Part Number: MBR2535CT
• Description: DIODE SCHOTTKY 35V 15A TO-220AB
• Manufacturer: Vishay

Specifications of MBR2535CT

• Voltage - Forward (vf) (max) @ If: 820mV @ 30A
• Current - Reverse Leakage @ Vr: 200µA @ 35V
• Current - Average Rectified (io) (per Diode): 15A
• Voltage - Dc Reverse (vr) (max): 35V
• Diode Type: Schottky
• Speed: Fast Recovery =< 500ns, > 200mA (Io)
• Diode Configuration: 1 Pair Common Cathode
• Mounting Type: Through Hole
• Package / Case: TO-220-3 (Straight Leads)
• Product: Schottky Diodes
• Peak Reverse Voltage: 35 V
• Forward Continuous Current: 30 A
• Max Surge Current: 150 A
• Configuration: Dual Common Cathode
• Forward Voltage Drop: 0.82 V
• Maximum Reverse Leakage Current: 200 uA
• Operating Temperature Range: - 65 C to + 150 C
• Mounting Style: Through Hole
• Lead Free Status / RoHS Status: Contains lead / RoHS non-compliant
• Reverse Recovery Time (trr): -
• Lead Free Status / RoHS Status: Contains lead / RoHS non-compliant, Lead free / RoHS Compliant
• Other names: *MBR2535CT; MBR2535CTIR; VS-MBR2535CT; VS-MBR2535CTIR; VS-MBR2535CTIRL; VSMBR2535CTIR

Document Number: 94291
Revision: 13-Aug-08
Fig. 1 - Maximum Forward Voltage Drop Characteristics
100
10
1
0.01
0.2
0.1
10
0.00001
1
V
0.4
FM
- Forward Voltage Drop (V)
0.6
(thermal resistance)
(Per Leg)
0.0001
Single pulse
Fig. 4 - Maximum Thermal Impedance Z
0.8
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1000
100
1.0
T
T
T
J
J
J
= 150 °C
= 125 °C
= 25 °C
Schottky Rectifier, 2 x 15 A
0
Fig. 3 - Typical Junction Capacitance vs.
1.2
0.001
t
1
- Rectangular Pulse Duration (s)
10
Reverse Voltage (Per Leg)
V
1.4
R
- Reverse Voltage (V)
20
D = 0.75
D = 0.50
D = 0.33
D = 0.25
D = 0.20
0.01
T
J
= 25 °C
30
thJC
Characteristics (Per Leg)
40
0.0001
0.001
0.1
0.01
100
0.1
10
1
0
Fig. 2 - Typical Values of Reverse Current vs.
50
Vishay High Power Products
Notes:
1. Duty factor D = t
2. Peak T
MBR25..CTPbF Series
5
T
J
T
= 125 °C
J
= 75 °C
10
Reverse Voltage (Per Leg)
V
R
T
J
P
J
- Reverse Voltage (V)
= P
DM
1
15
T
= 25 °C
J
DM
= 150 °C
t
1
20
x Z
t
2
1
/t
thJC
2
25
+ T
C
T
30
J
10
= 100 °C
T
J
35
= 50 °C
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40
45
3