ISL6140 Datasheet, PDF (9/19 Pages) Intersil Corporation – Negative Voltage Hot Plug Controller

ISL6140, ISL6150
Applications: Overcurrent
CORRECT
INCORRECT
TO SENSE
AND V_EE
CURRENT
SENSE RESISTOR
FIGURE 5. SENSE RESISTOR
Physical layout of R_{1 SENSE} resistor is critical to avoid
the possibility of false overcurrent occurrences. Since it
is in the main input-to-output path, the traces should
be wide enough to support both the normal current,
and up to the overcurrent trip point. Ideally trace
routing between the R₁ resistor and the ISL6140 and
ISL6150 (pin 4 (V_EE) and pin 5 (SENSE) is direct and
as short as possible with zero current in the sense lines
(see Figure 5).
There is a short filter (3µs nominal) on the
comparator; current spikes shorter than this will be
ignored. Any longer pulse will shut down the output,
requiring the user to either power-down the system
(below the UV voltage), or pull the UV pin below its
trip point (usually with an external transistor).
If current pulses longer than the 3µs are expected, and
need to be filtered, then an additional resistor and
capacitor can be added. As shown in Figure 29, R₇ and
C₃ act as a low-pass filter such that the voltage on the
SENSE pin won’t rise as fast, effectively delaying the
shut-down. Since the ISL6140/ISL6150 has essentially
zero current on the SENSE pin, there is no voltage drop
or error associated with the extra resistor. R₇ is
recommended to be small, 100Ω is a good value.
The delay time is approximated by the added RC time
constant, modified by a factor relative to the trip point
(see Equation 2).
t = –R*C*In [1 - (V(t) - V(t₀ ) ) ⁄ (V_i – V(t₀))]
(EQ. 2)
where V(t) is the trip voltage (nominally 50mV); V(t₀)
is the nominal voltage drop across the sense resistor
before the overcurrent condition; V_i is the voltage drop
across the sense resistor while the overcurrent is
applied.
For example: a system has a normal 1A current load,
and a 20mΩ sense resistor, for a 2.5A overcurrent. It
needs to filter out a 50µs current pulse at 5A.
Therefore:
V(t) = 50mV (from spec)
V(t₀) = 20mV (V = IR = 1A*20mΩ)
V_i = 100mV (V = IR = 5A*20mΩ)
If R₇ = 100Ω, then C₃ is around 1µF.
Note that the FET must be rated to handle the higher
current for the longer time, since the IC is not doing
current limiting; the RC is just delaying the overcurrent
shutdown.
Applications: OV and UV
The UV and OV input pins are high impedance, so the
value of the external resistor divider is not critical with
respect to input current. Therefore, the next
consideration is total current; the resistors will always
draw current, equal to the supply voltage divided by
the total of R4 + R5 + R6; so the values should be
chosen high enough to get an acceptable current.
However, to the extent that the noise on the power
supply can be transmitted to the pins, the resistor
values might be chosen to be lower. A filter capacitor
from UV to VEE or OV to UV is a possibility, if certain
transients need to be filtered. (Note that even some
transients which will momentarily shut off the gate
might recover fast enough such that the gate or the
output current does not even see the interruption).
Finally, take into account whether the resistor values
are readily available, or need to be custom ordered.
Tolerances of 1% are recommended for accuracy. Note
that for a typical 48V system (with a 36V to 72V
range), the 36V or 72V is being divided down to
1.223V, a significant scaling factor. For UV, the ratio is
roughly 30x; every 3mV change on the UV pin
represents roughly 0.1V change of power supply
voltage. Conversely, an error of 3mV (due to the
resistors, for example) results in an error of 0.1V for
the supply trip point. The OV ratio is around 60. So the
accuracy of the resistors comes into play.
The hysteresis of the comparators (20mV nominal)
is also multiplied by the scale factor of 30 for the UV
pin (30 * 20mV = 0.6V of hysteresis at the power
supply) and 60 for the OV pin (60*20mV = 1.2V of
hysteresis at the power supply).
With the three resistors, the UV equation is based on
the simple resistor divider:
1.223 = V_UV*(R₅ + R₆)/(R₄ + R₅ + R₆) or
V_UV = 1.223 (R₄ + R₅ + R₆)/(R₅ + R₆)
Similarly, for OV:
1.223 = V_OV*(R₆)/(R₄ + R₅ + R₆) or
V_OV = 1.223 (R₄ + R₅ + R₆)/(R₆)
Note that there are two equations, but 3 unknowns.
Because of the scale factor, R₄ has to be much bigger
than the other two; chose its value first, to set the
current (for example, 50V/500kΩ draws 100µA), and
then the other two will be in the 10kΩ range. Solve the
two equations for two unknowns. Note that some
iteration may be necessary to select values that meet
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FN9039.4