ACPL-H342-560E Avago Technologies US Inc., ACPL-H342-560E Datasheet - Page 6

• Manufacturer Part Number: ACPL-H342-560E
• Description: Plastic Optocoupler, LF+T/R
• Manufacturer: Avago Technologies US Inc.
• Series: -r
• Type: Gate Driver, Miller Clampr

Specifications of ACPL-H342-560E

• Voltage - Isolation: 3750Vrms
• Input Type: DC
• Voltage - Supply: 15 V ~ 30 V
• Operating Temperature: -40°C ~ 105°C
• Mounting Type: Surface Mount
• Package / Case: 8-SOIC (0.268", 6.81mm Width)
• Lead Free Status / Rohs Status: Lead free / RoHS Compliant

Table 5. Electrical Specifications (DC)
Unless otherwise noted, all typical values are at T
fications are at Recommended Operating Conditions (T
15 to 30 V, V
6
Parameter
High Level Peak Output Current
Low Level Peak Output Current
High Output Transistor RDS(ON)
Low Output Transistor RDS(ON)
Clamp Output Peak Current
Clamp Pin Threshold
Clamp Output Transistor
RDS(ON)
High Level Output Voltage
High Level Output Voltage
Low Level Output Voltage
High Level Supply Current
Low Level Supply Current
Threshold Input Current
Low to High
Threshold Input Voltage
High to Low
Input Forward Voltage
Temperature Coefficient of
Input Forward Voltage
Input Reverse Breakdown
Voltage
Input Capacitance
UVLO Threshold
UVLO Hysteresis
EE
= Ground).
Symbol
I
I
R
R
I
V
R
V
V
V
I
I
I
V
V
'V
BV
C
V
V
UVLO
OH
OL
CLAMP
CCH
CCL
FLH
DS,OH
DS,OL
tCLAMP
DS,CLAMP
OH
OH
OL
FHL
F
UVLO+
UVLO-
IN
R
F
/'T
HYS
A
Min.
-0.5
-2.0
0.5
2.0
1.0
V
0.5
0.8
1.2
5
10.0
9.0
CC
-2.0
A
= 25°C, V
Typ.
-1.2
2.7
2.6
0.8
2.5
2.3
0.8
V
V
0.07
1.68
2.0
1.5
1.55
-1.7
70
12.3
10.7
1.4
CC
CC
A
= -40 to 105°C, I
-0.80
CC
- V
Max.
5.0
2.0
2.0
0.25
2.5
2.5
4.0
1.95
13.5
12.0
EE
= 30 V, V
Units
A
A
A
A
:
:
A
V
:
V
V
V
mA
mA
mA
V
V
mV/°C
V
pF
V
F(ON)
EE
= 7 to 16 mA, V
= Ground; all Minimum/Maximum speci-
Test Conditions
V
V
V
V
I
I
V
I
I
I
I
R
C
R
C
R
C
I
I
f = 1 MHz, V
V
OH
OL
CLAMP
O
O
O
F
R
g
g
g
O
O
O
O
O
g
g
g
O
= 10 mA
= 100 PA
= -100 mA
= 0 mA , I
= 100 mA
= 10:,
= 25 nF, I
= 10:,
= 25 nF, I
= 10:,
= 25 nF, V
= V
= V
= V
= V
= 2.0A
= V
> 5 V, I
= -2.0A
CC
CC
EE
EE
EE
= 1.5 A
+ 2.5V
+ 15V
+ 2.5
– 4
– 15
F
= 10 mA
F
F
F
F
O
=10 mA
= 10 mA,
= 0 mA
= 0 V
F(OFF)
> 5 V
= -3.6 to 0.8 V, V
Fig.
3, 4, 23
6, 7, 24
8
9
14, 16,
27
15,16,
28
2, 4, 25 6, 7
1
5, 7, 26
10, 11
12, 13,
29
22
30
Note
5
2
5
2
2
CC
=