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12
Applications Information
Eliminating Negative IGBT Gate Drive
To keep the IGBT rmly o , the HCPL-5120 has a very
low maximum VOL speci cation of 0.5 V. The HCPL-5120
realizes this very low VOL by using a DMOS transistor
with 1 (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 . 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 elimi-
nate the need for negative IGBT gate drive in many ap-
plications 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 sig-
nals into the HCPL-5120 and degrade performance. (If
the IGBT drain must be routed near the HCPL-5120 in-
put, then the LED should be reverse-biased when in the
o 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 Speci cation.
The IGBT and Rg in Figure 26 can be analyzed as a simple
RC circuit with a voltage supplied by the HCPL-5120.
(VCC - VEE - VOL)
Rg = –––––––––––––––––
IOLPEAK
(VCC – VEE – 2V)
= ––––––––––––––––––
I
OLPEAK
(15 V + 5 V – 2V)
= –––––––––––––––––––
2.5 A
= 7.2Ω ≈ 8Ω
The VOL value of 2 V in the previous equation is a con-
servative 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 - VEE) + ESW(Rg , Qg ) f
For the circuit in Figure 26 with IF (worst case) = 18 mA,
Rg = 8 , Max Duty Cycle = 80%, Qg = 500 nC, f = 20 kHz
and TA max = 125C:
PE = 18 mA 1.8 V 0.8 = 26 mW
PO = 4.25 mA 20 V + 1.0J 20 kHz
= 85 mW + 20 mW
= 105 mW
< 112 mW (PO(MAX) @ 125C = 250 mW - 23C 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 -55C) to ICC max at 125C.
Since PO for this case is less than PO(MAX) , Rg of 8 is ap-
propriate.
+ 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
Figure 25. Recommended LED Drive and Application Circuit