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2SK2241 参数 Datasheet PDF下载

2SK2241图片预览
型号: 2SK2241
PDF下载: 下载PDF文件 查看货源
内容描述: [Power Field-Effect Transistor, 7A I(D), 450V, 0.95ohm, 1-Element, N-Channel, Silicon, Metal-oxide Semiconductor FET]
分类和应用: 局域网晶体管
文件页数/大小: 50 页 / 635 K
品牌: SANKEN [ SANKEN ELECTRIC ]
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Avalanche energy capability of MOS FET
1. What is avalanche energy capability ?
When a MOS FET is used for high-speed switching, the
inductive load and wiring inductance may cause a counter
electromotive voltage at cutoff that the device cannot
withstand.
Avalanche energy capability is the non-clamped ability
to withstand damage expressed as energy. As long as the
energy applied to the device at cutoff is within the
guaranteed avalanche energy capability, the device will not
be damaged even if the Drain-Source voltage exceeds the
capability.
For example, a Drain-Source voltage that is within the
guaranteed capability when electrically stationary may
exceed the limit at startup or cutoff. Usually, a snubber
circuit or similar surge absorbing circuit is used to keep the
Drain-Source voltage within the guaranteed capability.
Sanken MOS FETs, however, do not require this kind of
protective circuit because the avalanche energy capability
is guaranteed. Sanken MOS FETs enable the number of
parts to be reduced, saving board area.
* Consult the engineering department of Sanken when
planning to use MOS FETs in avalanche mode.
Sanken MOS FETs feature guaranteed
avalanche energy capability.
3. Temperature derating for E
AS
The E
AS
Value in the specifications is guaranteed when
the channel temperature Tch is 25ºC. Since the E
AS
Value
drops as the channel temperature rises, derating
depending on the temperature is necessary.
Fig.B shows the derating curve for single avalanche
energy capability. This is the derating curve of E
AS
and the
channel temperature (Tch (start)) immediately before the
avalanche occurs in the product, with the E
AS
value
(maximum rating) at 25ºC as 100%.
For example, if the product temperature is 50ºC, the E
AS
value is derated to 64% of the value at 25ºC.
Fig. B
E
AS
— Tch (start)
100
I
Lp
= I
D
max
E
AS
(normalized) (%)
80
60
40
2. E
AS
calculation method
If the current in an inductive load L is I
LP
at the moment
when the MOS FET is cut off, E
AS
can be expressed as
follows:
V
DSS
1
........................... 1
E
AS
=
• L • I
Lp2
2
V
DSS
– V
DD
* V
DD
: Supply voltage
20
0
25
50
75
100
125
150
Tch (start) (ºC)
4. Continuous avalanche energy capability
This section explains the derating method for continu-
ous avalanche.
Considering continuous avalanche as the repetition of a
single avalanche, the safe operating area (SOA) is deter-
mined using the derating curve shown in Fig. B.
Calculate the energy and Tch (start) of avalanche in the
worst condition and determine SOA using the calculated
data and the derating curve shown in Fig.B. The tempera-
ture rise due to avalanche should not cause the channel
temperature to exceed the maximum rating.
The following is an example of determining SOA judg-
ment by calculation when a MOS FET enters a transient
avalanche state at power-on then changes to a stationary
state.
Supposing that the waveform is as shown in Fig. C until
the MOS FET changes to the stationary state, calculate the
start loss and switching (turn-on/off) Ioss. To simplify the
calculation, the average loss Pa and the last two wave-
forms are used for approximation. (Fig. D)
First, calculate the channel temperature Tch ( ) at time
( ) where the temperature condition is severest.
If the Tch ( ) value is within the maximum rating, there
is no problem as far as the temperature is concerned.
If the value of L is not known in an actual circuit, E
AS
can also be calculated from the actual voltage and current
waveforms as follows:
E
AS
= Ps •t ............................................................
2
* Ps: Surge power
* t: Surge time
The following calculation is used to determine E
AS
where the voltage and current shown in Fig.A are applied
to the MOS FET in a circuit.
Integrate the overlapping section of I
D
and V
DS
to calcu-
late I
D
•V
DS
•dt. When the I
D
waveform is triangular, E
AS
will be as follows:
E
AS
=
Fig. A
1
•10(A) • 550(V) • 10(µs) = 27.5 (mJ)
2
10A
0
I
D
550V
(V
DSS
)
V
DS
0
10µs
4