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12
Applications Information
Eliminating Negative IGBT Gate Drive
To keep the IGBT firmly off, the
HCPL- 5120 has a very low
maximum VOL specification of
0.5 V. The HCPL- 5120 realizes
this very low VOL by using a
DMOS transistor with 1 Ω
(typical) on resistance in its
pull down circuit. When the
HCPL- 5120 is in the low state,
the IGBT gate is shorted to the
emitter by Rg + 1 Ω. Minimizing
Rg and the lead inductance from
the HCPL- 5120 to the IGBT
gate and emitter (possibly by
mounting the HCPL- 5120 on a
small PC board directly above
the IGBT) can eliminate the
need for negative IGBT gate
drive in many applications as
shown in Figure 25. Care should
be taken with such a PC board
design to avoid routing the
IGBT collector or emitter traces
close to the HCPL- 5120 input
as this can result in unwanted
coupling of transient signals
into the HCPL- 5120 and
degrade performance. (If the
IGBT drain must be routed near
the HCPL- 5120 input, then the
LED should be reverse- biased
when in the off state, to prevent
the transient signals coupled
from the IGBT drain from
turning on the HCPL- 5120.)
Selecting the Gate Resistor (Rg) to
Minimize IGBT Switching Losses.
Step 1: Calculate Rg Minimum from
the IOL Peak Specification.
The IGBT and Rg in Figure 26
can be analyzed as a simple RC
circuit with a voltage supplied
by the HCPL- 5120.
(VCC - V
EE - V
OL)
Rg = –––––––––––––––––
IOLPEAK
(VCC – VEE – 2V)
= ––––––––––––––––––
IOLPEAK
(15 V + 5 V – 2V)
= –––––––––––––––––––
2.5 A
= 7.2
Ω
≈ 8 Ω
The VOL value of 2 V in the
previous equation is a
conservative value of VOL at the
peak current of 2.5A (see Figure
6). At lower Rg values the
voltage supplied by the HCPL-
5120 is not an ideal voltage
step. This results in lower peak
currents (more margin) than
predicted by this analysis. When
negative gate drive is not used
VEE in the previous equation is
equal to zero volts
Step 2: Check the HCPL-5120 Power
Dissipation and Increase Rg if
Necessary.
The HCPL- 5120 total power
dissipation (PT) is equal to the
sum of the emitter power (PE)
and the output power (PO):
PT = PE + PO
PE = IF
•
VF
•
Duty Cycle
PO = PO(BIAS) + PO (SWITCHING)
= ICC
•
(VCC - V
EE) + ESW(Rg,
Qg)
•
f
For the circuit in Figure 26
with IF (worst case) = 18 mA,
Rg = 8 Ω, Max Duty Cycle = 80%,
Qg = 500 nC, f = 20 kHz and TA
max = 125°C:
PE = 18 mA
•
1.8 V
•
0.8 = 26 mW
PO = 4.25 mA
•
20 V + 1.0 µJ
•
20
kHz
= 85 mW + 20 mW
= 105 mW
< 112 mW (PO(MAX) @ 125
°
C
= 250 mW - 23
°
C
•
6 mW/°C)
The value of 4.25 mA for ICC in
the previous equation was
obtained by derating the ICC
max of 5 mA (which occurs at -
55°C) to ICC max at 125°C.
Since PO for this case is less
than PO(MAX), Rg of 8 Ω is
appropriate.
Figure 25. Recommended LED Drive and Application Circuit
+ HVDC
3-PHASE
AC
- HVDC
0.1 µF VCC = 18 V
1
3
+
2
4
8
6
7
5
270 Ω
CONTROL
INPUT
Rg
Q1
Q2
74XXX
OPEN
COLLECTOR
_
+5 V