# IGBT Modules - Technologies, Driver and Application (Second Edition) - page 240

228
GE
G
min ,G
C
L 2
R
Σ
Eq. 6.7
Σ
L
G
: Sum of thegate lead inductances (L
G
+L
Gon
or L
G
+ L
Goff
) [H]
Resolving
o I
peak
results in:
min ,G
GE
min ,G
GE
peak
R
U 74.0
R
U
e
2 I
⋅ =
Eq. 6.8
e: Euler's number 2.71828
Consequently for large L
G
the value of R
G
(actually R
Gon
) has to be increased to avoid
snap-off behaviour of the freewheeling diode.
If different gate resistors R
Gon
and R
Goff
are used, the required peak current for the
selection of the driver stage has to be calculated for the lower resistor value. It should
also be noted that some driver IC provide different peak currents for the turn-on and
turn-off phases of the IGBT. In this case, the peak currents for both switching events
have to be calculatedand thedriver IC has tobe selected accordingly.
Fig. 6.26
Driver stagewith parasitic elements
If the peak current capability of the driver stage is not sufficient for the target application
an additional booster can be placed between the output of the driver and the gate
resistor of the IGBT. When using BJTs, for example, for the booster stage, the
calculation sequence is as follows:
Calculation of the required peak current using
Different R
Gon
and R
Goff
aswell as anoptional C
G
have to be considered.
Selection of suitable npn- and pnp-BJTs and setting themaximum current per
booster stage.When using only one booster stage, this current corresponds to
the previously calculated peak current. However, more than one booster stage
may be used, which can be connected in parallel. Each of them has to provide
a current resulting from the calculated peak current divided by the amount of
booster stages.
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