MBRD1035CTL ON Semiconductor, MBRD1035CTL Datasheet - Page 4

• Manufacturer Part Number: MBRD1035CTL
• Description: DIODE SCHOTTKY 35V 5A DPAK
• Manufacturer: ON Semiconductor
• Series: SWITCHMODE™r

Specifications of MBRD1035CTL

• Voltage - Forward (vf) (max) @ If: 470mV @ 5A
• Current - Reverse Leakage @ Vr: 2mA @ 35V
• Current - Average Rectified (io) (per Diode): 5A
• Voltage - Dc Reverse (vr) (max): 35V
• Diode Type: Schottky
• Speed: Fast Recovery =< 500ns, > 200mA (Io)
• Diode Configuration: 1 Pair Common Cathode
• Mounting Type: Surface Mount
• Package / Case: DPak, TO-252 (2 leads+tab), SC-63
• Lead Free Status / RoHS Status: Contains lead / RoHS non-compliant
• Reverse Recovery Time (trr): -

1000
100
8.0
7.0
6.0
5.0
4.0
3.0
2.0
1.0
10
0
0
0
* Reverse power dissipation and the possibility of thermal runaway must be considered when operating this device under any
reverse voltage conditions. Calculations of T
T
This graph displays the derated allowable T
where r(t) = Rthja. For other power applications further calculations must be performed.
J
(50% DUTY CYCLE)
may be calculated from the equation:
20
SQUARE WAVE
Figure 5. Current Derating Per Leg
Figure 7. Capacitance Per Leg
I
5
I
dc
I
I
pk
pk
pk
pk
/I
/I
/I
/I
o
o
o
o
40
T
= 20
= p
= 10
= 5
L
V
, LEAD TEMPERATURE (°C)
R
, REVERSE VOLTAGE (VOLTS)
10
60
freq = 20 kHz
80
15
100
20
J
T
J
due to reverse bias under DC conditions only and is calculated as T
120
J
therefore must include forward and reverse power effects. The allowable operating
T
r(t) = thermal impedance under given conditions,
Pf = forward power dissipation, and
Pr = reverse power dissipation
= 25°C
J
http://onsemi.com
= T
Jmax
140
25
− r(t)(Pf + Pr) where
4
125
105
115
4.0
3.5
3.0
2.5
2.0
1.5
1.0
0.5
95
85
75
65
0
0
0
Figure 6. Forward Power Dissipation Per Leg
I
pk
/I
o
Figure 8. Typical Operating Temperature
1.0
= 20
5
I
pk
I
/I
O
o
, AVERAGE FORWARD CURRENT (AMPS)
= 10
V
2.0
R
R
qJA
, DC REVERSE VOLTAGE (VOLTS)
10
I
pk
/I
Derating Per Leg *
= 48°C/W
o
= 5
3.0
R
qJA
15
I
pk
= 67.5°C/W
/I
4.0
o
R
= p
qJA
(50% DUTY CYCLE)
R
20
SQUARE WAVE
qJA
= 2.43°C/W
5.0
= 84°C/W
J
25
= T
6.0
R
qJA
Jmax
= 25°C/W
30
7.0
− r(t)Pr,
dc
8.0
35