Datasheet

19
Selecting the Gate Resistor (Rg) to Minimize IGBT
Switching Losses. (Discussion applies to HCPL-3120,
HCPL-J312 and HCNW3120)
Step 1: Calculate Rg Minimum from the I
OL
Peak Specica-
tion. The IGBT and Rg in Figure 26 can be analyzed as a
simple RC circuit with a voltage supplied by the HCPL-
3120.
(V
CC
– V
EE
- V
OL
)
Rg ≥ ———————
I
OLPEAK
(V
CC
– V
EE
- 2 V)
= ———————
I
OLPEAK
(15 V + 5 V - 2 V)
= ———————
2.5 A
= 7.2 Ω @ 8 Ω
The V
OL
value of 2 V in the previous equation is a con-
servative value of V
OL
at the peak current of 2.5A (see
Figure 6). At lower Rg values the voltage supplied by
the HCPL-3120 is not an ideal voltage step. This results
in lower peak currents (more margin) than predicted by
this analysis. When negative gate drive is not used V
EE
in
the previous equation is equal to zero volts.
Step 2: Check the HCPL-3120 Power Dissipation and
Increase Rg if Necessary. The HCPL-3120 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
PE = I
F
• V
F
· Duty Cycle
P
O
= P
O(BIAS)
+ P
O (SWITCHING)
= I
CC
• (V
CC
- V
EE
)+ E
SW
(R
G
, Q
G
) • f
Figure 26. HCPL-3120 typical application circuit with negative IGBT gate drive.
+ HVDC
3-PHASE
AC
- HVDC
HCPL-3120 fig 26
0.1 µF
V
CC
= 15 V
1
3
+
–
2
4
8
6
7
5
HCPL-3120
Rg
Q1
Q2
V
EE
= -5 V
–
+
270 Ω
+5 V
CONTROL
INPUT
74XXX
OPEN
COLLECTOR
For the circuit in Figure 26 with I
F
(worst case) =
16 mA, Rg = 8 Ω, Max Duty Cycle = 80%, Qg = 500 nC,
f = 20 kHz and T
A
max = 85 °C:
P
E
= 16 mA • 1.8 V • 0.8 = 23 mW
P
O
= 4.25 mA • 20 V + 5.2 µ J • 20 kHz
= 85 mW + 104 mW
= 189 mW > 178 mW (P
O(MAX)
@ 85°C
= 250 mW-15C*4.8 mW/C)
The value of 4.25 mA for I
CC
in the previous equation was
obtained by derating the I
CC
max of 5 mA (which occurs
at -40°C) to I
CC
max at 85C (see Figure 7).
Since P
O
for this case is greater than P
O(MAX)
, Rg must be
increased to reduce the HCPL-3120 power dissipation.
P
O(SWITCHING MAX)
= P
O(MAX)
- P
O(BIAS)
= 178 mW - 85 mW
= 93 mW
P
O(SWITCHINGMAX)
E
SW(MAX)
= ———————
f
93 mW
= ———— = 4.65 µW
20 kHz
For Qg = 500 nC, from Figure 27, a value of E
SW
= 4.65 µW
gives a Rg = 10.3 Ω.