acpl-h312 Avago Technologies, acpl-h312 Datasheet - Page 17

• Manufacturer Part Number: acpl-h312
• Description: 2.5 Amp Output Current Igbt Gate Driver Optocoupler With Low Icc & Uvlo In Stretched So8
• Manufacturer: Avago Technologies

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
where,
I
V
PO = PO(BIAS) + PO(SWTICH) = I
Q
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
(Table 5)
∆V
ple)
Q
in the manufacturer specification = 240nC (approxima-
tion of 100A IGBT which can be obtained from IGBT data-
sheet)
f
Similarly using the maximum supply current I
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
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AV02-0821EN - June 16, 2008
F(ON),max
CC2
SWITCH
F,max
G
G
GE
* f
= Total gate charge of the IGBT or MOSFET as described
= Supply Current to power internal circuity = 3.0mA
F(ON) ,max
= V
SWITCH
= 1.8V (Table 5)
= switching frequency of application = 10kHz
CC2
= 16mA (Table 4)
+ |V
* V
EE
F,max
| = 18 – (-5V) = 23V (Application exam-
CC2
* (V
CC2
–V
EE
) + ∆V
CC2
= 3.0
GE
*
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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 80 ºC and use the high conductivity
thermal resistances.
LED junction temperature,
T1 = (R
Output IC junction temperature,
T2 = (R
In this example, both temperature are within the maxi-
mum 125°C. If the juntion temperature is higher than the
maximum junction temperature rating, the desired speci-
fication must be derated according.
= (399 * 28.8 + 164.1 * 124.2) + 80
= 31.9 + 80 = 111.9 ºC
= (164.1 *28.8 + 251 * 124.2) + 80
= 35.9 + 80 = 115.9 ºC
11
21
* P
x P
1
1
+ R
+ R
12
22
* P
x P
2
2
) + T
) + T
a
a