HCPL3150 HP [Agilent(Hewlett-Packard)], HCPL3150 Datasheet - Page 11

no-image

HCPL3150

Manufacturer Part Number
HCPL3150
Description
0.5 Amp Output Current IGBT Gate Drive Optocoupler
Manufacturer
HP [Agilent(Hewlett-Packard)]
Datasheet

Available stocks

Company
Part Number
Manufacturer
Quantity
Price
Part Number:
HCPL3150
Manufacturer:
FAIRCHILD/仙童
Quantity:
20 000
Part Number:
HCPL3150SD
Manufacturer:
FAIRCHILD/仙童
Quantity:
20 000
Figure 26. HCPL-3150 Typical Application Circuit with Negative IGBT Gate Drive.
Selecting the Gate Resistor
(Rg) to Minimize IGBT
Switching Losses.
Step 1: Calculate Rg Minimum
From the I
tion. The IGBT and Rg in Figure
26 can be analyzed as a simple
RC circuit with a voltage supplied
by the HCPL-3150.
CONTROL
COLLECTOR
Parameter
Duty Cycle
INPUT
Rg
+5 V
74XXX
P
V
OPEN
I
= ––––––––––––––––
= ––––––––––––––––––
= 30.5
F
E
F
(V
–––––––––––––––
(V
(15 V + 5 V - 1.7 V)
OL
CC
CC
270
0.6 A
Peak Specifica-
– V
– V
I
I
OLPEAK
OLPEAK
LED On Voltage
Maximum LED
Description
LED Current
EE
EE
Duty Cycle
- V
- 1.7 V)
1
2
3
4
OL
)
HCPL-3150
The V
vious equation is a conservative
value of V
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 V
is equal to zero volts.
Step 2: Check the HCPL-3150
Power Dissipation and
Increase Rg if Necessary. The
HCPL-3150 total power dissipa-
tion (P
the emitter power (P
output power (P
P
O
E
OL
SW
EE
Parameter
T
) is equal to the sum of
value of 2 V in the pre-
(Rg,Qg)
in the previous equation
V
V
I
OL
CC
CC
EE
8
7
6
5
f
0.1 µF
at the peak current
O
):
+
+
E
) and the
V
V
CC
EE
Energy Dissipated in the HCPL-3150 for each
Rg
= 15 V
= -5 V
IGBT Switching Cycle (See Figure 27)
Q1
Q2
Negative Supply Voltage
Positive Supply Voltage
Switching Frequency
P
P
P
P
P
For the circuit in Figure 26 with I
(worst case) = 16 mA, Rg =
30.5 , Max Duty Cycle = 80%,
Qg = 500 nC, f = 20 kHz and T
max = 90 C:
Supply Current
E
O
T
E
O
Description
= P
= 16 mA 1.8 V 0.8 = 23 mW
= I
= 4.25 mA 20 V
= P
= I
= 85 mW + 80 mW
= 165 mW
F
CC
E
O(BIAS)
+ P
V
+ 4.0 J 20 kHz
> 154 mW (P
= 250 mW 20C 4.8 mW/C)
+ E
(V
F
CC
Duty Cycle
O
SW
3-PHASE
+ P
- V
+ HVDC
- HVDC
(R
AC
EE
O (SWITCHING)
G
)
, Q
G
O(MAX)
) f
1-207
@ 90 C
A
F

Related parts for HCPL3150