Vbb disconnect with energized inductive
load
4
3 IN1
high
Vbb
1
OUT1
6 IN2 PROFET
ST
5
OUT2
7
GND
2
BTS612N1
with an approximate solution for RL > 0 Ω:
EAS=
2IL·R· LL·(Vbb
+
|VOUT(CL)|)·
ln
(1+
IL·RL
|VOUT(CL)|
)
Maximum allowable load inductance for
a single switch off (both channels parallel)
L = f (IL ); Tj,start = 150°C,TC = 150°C const.,
Vbb = 12 V, RL = 0 Ω
L [mH]
1000
Vbb
Normal load current can be handled by the PROFET
itself.
100
Vbb disconnect with charged external
inductive load
4
3 IN1
Vbb
1
10
high
OUT1
6 IN2 PROFET
D
OUT2
ST
5
GND
7
2
Vbb
If other external inductive loads L are connected to the PROFET,
additional elements like D are necessary.
Inductive Load switch-off energy
dissipation
E bb
E AS
IN
Vbb
ELoad
PROFET OUT
=
ST
EL
GND
L
{Z L RL
ER
1
2
3
4
5
6
7
8
IL [A]
Energy stored in load inductance:
EL = 1/2·L·I2L
While demagnetizing load inductance, the energy
dissipated in PROFET is
EAS= Ebb + EL - ER= VON(CL)·iL(t) dt,
Semiconductor Group
9
2003-Oct-01