IRF7341QTRPBF International Rectifier, IRF7341QTRPBF Datasheet - Page 7

MOSFET N-CH DUAL 55V 8-SOIC

IRF7341QTRPBF

Manufacturer Part Number
IRF7341QTRPBF
Description
MOSFET N-CH DUAL 55V 8-SOIC
Manufacturer
International Rectifier
Series
HEXFET®r
Datasheet

Specifications of IRF7341QTRPBF

Fet Type
2 N-Channel (Dual)
Fet Feature
Logic Level Gate
Rds On (max) @ Id, Vgs
50 mOhm @ 5.1A, 10V
Drain To Source Voltage (vdss)
55V
Current - Continuous Drain (id) @ 25° C
5.1A
Vgs(th) (max) @ Id
1V @ 250µA
Gate Charge (qg) @ Vgs
44nC @ 10V
Input Capacitance (ciss) @ Vds
780pF @ 25V
Power - Max
2.4W
Mounting Type
Surface Mount
Package / Case
8-SOIC (3.9mm Width)
Lead Free Status / RoHS Status
Lead free / RoHS Compliant
Other names
IRF7341QTRPBFCT

Available stocks

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Part Number
Manufacturer
Quantity
Price
Part Number:
IRF7341QTRPBF
Manufacturer:
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Quantity:
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Company:
Part Number:
IRF7341QTRPBF
Quantity:
2 306
www.irf.com
0.001
0.01
100
0.1
10
140
120
100
1
1.0E-06
80
60
40
20
0
Fig 16. Maximum Avalanche Energy
25
Duty Cycle = Single Pulse
0.01
0.05
0.10
Starting T J , Junction Temperature (°C)
50
1.0E-05
Vs. Temperature
75
TOP
BOTTOM 10% Duty Cycle
I D = 5.1A
Fig 15. Typical Avalanche Current Vs.Pulsewidth
100
1.0E-04
Single Pulse
125
1.0E-03
150
175
tav (sec)
1.0E-02
Notes on Repetitive Avalanche Curves , Figures 15, 16:
(For further info, see AN-1005 at www.irf.com)
1. Avalanche failures assumption:
2. Safe operation in Avalanche is allowed as long asT
3. Equation below based on circuit and waveforms shown in
4. P
5. BV = Rated breakdown voltage (1.3 factor accounts for
6. I
7. ∆T
Purely a thermal phenomenon and failure occurs at a
not exceeded.
Figures 12a, 12b.
T
t
D = Duty cycle in avalanche = t
Z
temperature far in excess of T
every part type.
avalanche pulse.
voltage increase during avalanche).
av
av =
thJC
D (ave)
jmax
= Allowable avalanche current.
=
Average time in avalanche.
(D, t
Allowable rise in junction temperature, not to exceed
(assumed as 25°C in Figure 15, 16).
= Average power dissipation per single
av
P
) = Transient thermal resistance, see figure 11)
D (ave)
1.0E-01
Allowed avalanche Current vs
avalanche
assuming ∆ Tj = 25°C due to
avalanche losses
= 1/2 ( 1.3·BV·I
I
E
av
AS (AR)
= 2DT/ [1.3·BV·Z
1.0E+00
= P
D (ave)
jmax
pulsewidth,
av
av
. This is validated for
·f
) = DT/ Z
·t
th
av
]
1.0E+01
thJC
tav
jmax
1.0E+02
7
is

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