HCPL-3150 Datasheet, PDF (15/22 Pages) Agilent(Hewlett-Packard) – 0.5 Amp Output Current IGBT Gate Drive Optocoupler

+5 V
1
CONTROL
270 Ω
INPUT
2
74XX
OPEN
COLLECTOR
3
GND 1
+5 V
6
270 Ω
CONTROL
INPUT
7
74XX
OPEN
COLLECTOR
8
GND 1
HCPL-315J
16
0.1 µF
15
14
FLOATING
SUPPLY
V_CC = 18 V
+
–
Rg
11
0.1 µF
10
9
V_CC = 18 V
+
–
Rg
Figure 25b. Recommended LED Drive and Application Circuit (HCPL-315J).
+ HVDC
3-PHASE
AC
- HVDC
HCPL-3150 fig 25b
Selecting the Gate Resistor (Rg) to Minimize IGBT Switching Losses.
Step 1: Calculate Rg Minimum From the I_OL Peak Specifica­tion. The
IGBT and Rg in Figure 26 can be analyzed as a simple RC
circuit with a voltage supplied by the HCPL-3150/315J.
 R g ^≥
(V_CC – V_EE - V_OL)
I_{OLPEAK
    =}
(

V_CC
–
V_EE - 1.7
I_OLPEAK
V)
_{    = (}15
V
+5V-
0.6 A
1.7
V)
= 30.5 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 0.6 A (see
Figure 6). At lower Rg values the voltage supplied by the
HCPL-3150/315J 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-3150/315J Power Dissipation and Increase Rg
if Necessary. The HCPL-3150/315J 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 = P + P _{O O(BIAS) O (SWITCHING)}
   = I_CC^•(V_CC - V_EE) + E_SW(R_G, Q_G)^•f
For the circuit in Figure 26 with I_F (worst case) = 16 mA,
Rg = 30.5 Ω, Max Duty Cycle = 80%,Qg = 500 nC, f = 20
kHz and T_Amax = 90°C:
P_E = 16 mA^•1.8 V^•0.8 = 23 mW
P_O = 4.25 mA^•20 V + 4.0 µJ^•20 kHz
   = 85 mW + 80 mW
   = 165 mW > 154 mW (P_O(MAX) @ 90°C
   = 250 mW−20C^•4.8 mW/C)
15