Quick Gate Drive Design Example using ACPL-H312/K312
The total power dissipation (PT) is equal to the sum of the
LED input-side power (PI) and detector output-side power
(PO) dissipation:
PT = PI + PO
PI = I
F(ON) ,max
* V
F,max
where,
I
F(ON),max
= 16mA (Table 4)
V
F,max
= 1.8V (Table 5)
PO = PO(BIAS) + PO(SWTICH) = I
CC2
* (V
CC2
–V
EE
) + ∆V
GE
*
Q
G
* f
SWITCH
where,
PO(BIAS) = steady-state power dissipation in the driver
due to biasing the device.
PO(SWITCH) = power dissipation in the driver due to charg-
ing and discharging of power device gate capacitances.
I
CC2
= Supply Current to power internal circuity = 3.0mA
(Table 5)
∆V
GE
= V
CC2
+ |V
EE
| = 18 – (-5V) = 23V (Application exam-
ple)
Q
G
= Total gate charge of the IGBT or MOSFET as described
in the manufacturer speci cation = 240nC (approxima-
tion of 100A IGBT which can be obtained from IGBT data-
sheet)
f
SWITCH
= switching frequency of application = 10kHz
Similarly using the maximum supply current I
CC2
= 3.0
mA.
PI = 16 mA * 1.8 V = 28.8mW
PO = PO(BIAS) + PO(SWITCH)
= 3.0 mA * (18 V – (–5 V)) + (18V + 5V) * 240nC * 10 kHz
= 69mW + 55.2mW
= 124.2 mW
Using the given thermal resistances and thermal model
formula in this datasheet, we can calculate the junction
temperature for both LED and the output detector. Both
junction temperature should be within the absolute maxi-
mum rating. For this application example, we set the am-
bient temperature as 78 ºC and use the high conductivity
thermal resistances.
LED junction temperature,
T1 = (R
11
* P
1
+ R
12
* P
2
) + T
a
= (311 * 28.8 + 111 * 124.2) + 78
= 22.7 + 78 = 100.7 ºC
Output IC junction temperature,
T2 = (R
21
x P
1
+ R
22
x P
2
) + T
a
= (111 * 28.8 + 168 * 124.2) + 78
= 24 + 78 = 102 ºC
In this example, both temperature are within the maxi-
mum 125C. If the juntion temperature is higher than the
maximum junction temperature rating, the desired speci-
cation must be derated according.
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AV02-0821EN - June 28, 2011
