12
Selecting the Gate Resistor (R
g
) for HCPL-3020
Step 1: Calculate R
g
minimum from the I
OL
peak specication. The IGBT and R
g
in Figure 17 can be analyzed as a
simple RC circuit with a voltage supplied by the HCPL-3020.
R
g
≤ V
CC
– V
OL
I
OLPEAK
= 24 - 1
0.4
= 57.5 Ω
The V
OL
value of 1 V in the previous equation is the V
OL
at the peak current of 0.4 A. (See Figure 4).
Step 2: Check the HCPL-3020 power dissipation and increase R
g
if necessary. The HCPL-3020 total power dissipation
(P
T
) is equal to the sum of the emitter power (P
E
) and the output power (P
O
).
P
T
= P
E
+ P
O
P
E
= I
F
• V
F
• Duty Cycle
P
O
= P
O(BIAS)
+ P
O(SWITCHING)
= I
CC
• V
CC
+ E
SW
(R
g
;Q
g
) • f
= (I
CCBIAS
+ K
ICC
• Q
g
• f) • V
CC
+ E
SW
(R
g
;Q
g
) • f
where K
ICC
• Q
g
• f is the increase in I
CC
due to switching and K
ICC
is a constant of 0.001 mA/(nC*kHz). For the circuit
in Figure 17 with I
F
(worst case) = 10 mA, R
g
= 57.5 Ω, Max Duty Cycle = 80%, Q
g
= 100 nC, f = 20 kHz and T
AMAX
=
85°C:
P
E
= 10 mA • 1.8 V • 0.8 = 14 mW
P
O
= [3 mA + (0.001 mA/nC • kHz) • 20 kHz • 100 nC] • 24 V + 0.3mJ • 20 kHz
= 126 mW < 250 mW (P
O(MAX)
) @ 85°C
The value of 3 mA for I
CC
in the previous equation is the max. I
CC
over entire operating temperature range.
Since P
O
for this case is less than P
O(MAX)
, R
g
= 57.5 Ω is alright for the power dissipation.
Figure 18. Energy dissipated in the HCPL-3020 and HCPL-0302
and for each IGBT switching cycle.
Esw – ENERGY PER SWITCHING CYCLE – µJ
0
0
Rg – GATE RESISTANCE – Ω
100
1.5
20
4.0
40
1.0
60 80
3.5
Qg = 50 nC
Qg = 100 nC
Qg = 200 nC
Qg = 400 nC
3.0
2.0
0.5
2.5
