HCPL-5120 Datasheet, PDF (12/16 Pages) Agilent(Hewlett-Packard) – 2.0 Amp Output Current IGBT Gate Drive Optocoupler

Applications Information
Eliminating Negative IGBT Gate Drive
To keep the IGBT firmly off, the
HCPL- 5120 has a very low
maximum V_OL specification of
0.5 V. The HCPL- 5120 realizes
this very low V_OL by using a
DMOS transistor with 1 Ω
(typical) on resistance in its
pull down circuit. When the
HCPL- 5120 is in the low state,
the IGBT gate is shorted to the
emitter by R_g + 1 Ω. Minimizing
R_g and the lead inductance from
the HCPL- 5120 to the IGBT
gate and emitter (possibly by
mounting the HCPL- 5120 on a
small PC board directly above
the IGBT) can eliminate the
need for negative IGBT gate
drive in many applications as
shown in Figure 25. Care should
be taken with such a PC board
design to avoid routing the
IGBT collector or emitter traces
close to the HCPL- 5120 input
as this can result in unwanted
coupling of transient signals
into the HCPL- 5120 and
degrade performance. (If the
IGBT drain must be routed near
the HCPL- 5120 input, then the
LED should be reverse- biased
when in the off state, to prevent
the transient signals coupled
from the IGBT drain from
turning on the HCPL- 5120.)
Selecting the Gate Resistor (R_g) to
Minimize IGBT Switching Losses.
Step 1: Calculate R_g Minimum from
the I_OL Peak Specification.
The IGBT and R_g in Figure 26
can be analyzed as a simple RC
circuit with a voltage supplied
by the HCPL- 5120.
R_g
=
–⁽–^V–^C–^C––^-––^V–^E–^E––^-––^V–^O–^L–⁾
I_OLPEAK
= –⁽–^V–^C–^C––^–––^V––^E–^E––^–––²â€“^V–⁾–
I_OLPEAK
(15 V + 5 V – 2V)
= –––––––––––––––––––
2.5 A
= 7.2Ω ≈ 8 Ω
The V_OL value of 2 V in the
previous equation is a
conservative value of V_OL at the
peak current of 2.5A (see Figure
6). At lower R_g values the
voltage supplied by the HCPL-
5120 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-5120 Power
Dissipation and Increase R_g if
Necessary.
The HCPL- 5120 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 - V_EE) + ESW(R_g,
Q_g) • f
For the circuit in Figure 26
with I_F (worst case) = 18 mA,
R_g = 8 Ω, Max Duty Cycle = 80%,
Q_g = 500 nC, f = 20 kHz and T_A
max = 125°C:
P_E = 18 mA•1.8 V•0.8 = 26 mW
P_O = 4.25 mA•20 V + 1.0 µJ • 20
kHz
= 85 mW + 20 mW
= 105 mW
< 112 mW (P_O(MAX) @ 125°C
= 250 mW - 23°C • 6 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 -
55°C) to I_CC max at 125°C.
Since P_O for this case is less
than P_O(MAX), R_g of 8 Ω is
appropriate.
+5 V
1
270 Ω
2
CONTROL
3
INPUT
74XXX
4
OPEN
COLLECTOR
8
0.1 µF
7
6
5
Figure 25. Recommended LED Drive and Application Circuit
+_ ^{V CC = 18 V}
Rg
Q1
Q2
+ HVDC
3-PHASE
AC
- HVDC
12