30
MAX. I , I
vs. GATE RESISTANCE
/ V = 25 V / 5 V
ON OFF
(Average switching energy
supplied to HCPL-316J per
The HCPL-316J total power
dissipation (P ) is equal to the
(V
CC2 EE2
4
T
cycle vs. R plot);
sum of the input-side power (P )
and output-side power (P ):
O
G
I
3
2
3. Compare the input and output
power dissipation calculated in
step #2 to the maximum
P = P + P
T
I
O
1
I
I
(MAX.ꢀ
OFF
recommended dissipation for
the HCPL-316J. (If the
maximum recommended level
has been exceeded, it may be
necessary to raise the value of
P = I
* V
CC1
I
CC1
0
P = P
O
+ P
O,SWTICH
O(BIAS)
(MAX.ꢀ
ON
-1
-2
-3
= I
* (V
–V ) +
CC2
E
CC2 EE
* f
SWITCH
SWITCH
R to lower the switching
G
0
20 40 60 80 100120140160180 200
where,
power and repeat step #2.)
Rg (Ωꢀ
P
= steady-state power
O(BIAS)
dissipation in the HCPL-316J
due to biasing the device.
Figure 76. Typical Peak I
and I
OFF
As an example, the total input
and output power dissipation can
be calculated given the following
conditions:
ON
Currents vs. Rg (for HCPL-316J
Output Driving an IGBT Rated at
600 V/100 A.
P
= transient power
O(SWITCH)
dissipation in the HCPL-316J
due to charging and discharging
power device gate.
• I
~ 2.0 A
SWITCHING ENERGY vs. GATE RESISTANCE
(V / V = 25 V / 5 V
ON, MAX
CC2
EE2
• V
= 18 V
CC2
9
8
7
6
5
4
3
2
• V = -5 V
EE
• f
= 15 kHz
CARRIER
E
= Average Energy
SWITCH
dissipated in HCPL-316J due to
switching of the power device
over one switching cycle
(µJ/cycle).
Step 1: Calculate R minimum
G
from I peak specification:
OL
Ess (Qg = 650 nCꢀ
To find the peak charging l
OL
assume that the gate is initially
charged the steady-state value of
f
= average carrier signal
SWITCH
1
0
V
. Therefore apply the
EE
frequency.
following relationship:
0
50
100
150
200
For R = 10.5, the value read
G
Rg (Ωꢀ
from Figure 77 is E
=
SWITCH
Figure 77. Switching Energy Plot for
Calculating Average Pswitch (for
HCPL-316J Output Driving an IGBT
Rated at 600 V/100 A).
R =
G
6.05 µJ. Assume a worst-case
[V @650 µA – (V +V )]
OH
OL
EE
average I = 16.5 mA (which is
CC1
I
OL,PEAK
given by the average of I
and
CC1H
I
I
). Similarly the average
= 5.5 mA.
CC1L
= [V
– 1 – (V + V )]
OL EE
CC2
CC2
I
OL,PEAK
P = 90.8 mW < 150 mW
I
P = 16.5 mA * 5.5 V = 90.8 mW
(abs. max.) ➱ OK
I
18 V – 1 V – (1.5 V + (-5 V))
2.0 A
P = P
+ P
P = 217.3 mW < 400 mW
O
O(BIAS)
O,SWITCH
O
(abs. max.) ➱ OK
= 10.25 Ω
= 5.5 mA * (18 V – (–5 V)) +
6.051 µJ * 15 kHz
≈ 10.5 Ω (for a 1% resistor)
Therefore, the power dissipation
absolute maximum rating has not
been exceeded for the example.
= 126.5 mW + 90.8 mW
= 217.3 mW
(Note from Figure 76 that the
real value of I may vary from
OL
the value calculated from the
simple model shown.)
Please refer to the following
Thermal Model section for an
explanation on how to calculate
the maximum junction
temperature of the HCPL-316J
for a given PC board layout
configuration.
Step 3: Compare the
calculated power dissipation
with the absolute maximum
values for the HCPL-316J:
Step 2: Calculate total power
dissipation in the HCPL-316J:
For the example,
AGILENT [ AGILENT TECHNOLOGIES, LTD. ]
