Selecting the Gate Resistor
(Rg) to Minimize IGBT
Switching Losses.
Step 1: Calculate Rg Minimum
From the IOL Peak Specifica-
tion. The IGBT and Rg in Figure
26 can be analyzed as a simple
RC circuit with a voltage supplied
by the HCPL-3150.
The VOL value of 2 V in the pre-
vious equation is a conservative
value of VOL at the peak current
of 0.6 A (see Figure 6). At lower
Rg values the voltage supplied by
the HCPL-3150 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.
PT = PE + PO
•
•
PE = IF V Duty Cycle
F
PO = PO(BIAS) + PO (SWITCHING)
•
= ICC (VCC - VEE)
•
+ ESW(RG, QG) f
For the circuit in Figure 26 with IF
(worst case) = 16 mA, Rg =
30.5 Ω, Max Duty Cycle = 80%,
Qg = 500 nC, f = 20 kHz and TA
max = 90°C:
(V – VEE - V )
Rg≥ ––C–C–––––––––O–L––
IOLPEAK
•
•
PE = 16 mA 1.8 V 0.8 = 23 mW
(V – VEE - 1.7 V)
= ––C–C–––––––––––––
IOLPEAK
Step 2: Check the HCPL-3150
Power Dissipation and
•
PO = 4.25 mA 20 V
•
+ 4.0 µJ 20 kHz
= 85 mW + 80 mW
= 165 mW
Increase Rg if Necessary. The
HCPL-3150 total power dissipa-
tion (PT) is equal to the sum of
the emitter power (PE) and the
output power (PO):
(15 V + 5 V - 1.7 V)
= ––––––––––––––––––
0.6 A
> 154 mW (PO(MAX) @ 90°C
= 250 mW−20C 4.8 mW/C)
= 30.5 Ω
•
HCPL-3150
+5 V
1
2
3
4
8
V
= 15 V
CC
+ HVDC
270 Ω
0.1 µF
+
–
7
Rg
Q1
= -5 V
3-PHASE
AC
CONTROL
INPUT
6
5
V
EE
–
+
74XXX
OPEN
COLLECTOR
Q2
- HVDC
Figure 26. HCPL-3150 Typical Application Circuit with Negative IGBT Gate Drive.
PE
PO Parameter
Description
Supply Current
Positive Supply Voltage
Negative Supply Voltage
Parameter
Description
LED Current
LED On Voltage
ICC
VCC
VEE
IF
VF
Duty Cycle
Maximum LED
Duty Cycle
ESW(Rg,Qg)
Energy Dissipated in the HCPL-3150 for each
IGBT Switching Cycle (See Figure 27)
f
Switching Frequency
1-207
HP [ HEWLETT-PACKARD ]
