ISL6140 Intersil Corporation, ISL6140 Datasheet - Page 8

• Manufacturer Part Number: ISL6140
• Description: Negative Voltage Hot Plug Controller
• Manufacturer: Intersil Corporation

Applications: Over-Current
Physical layout of R1 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
over-current trip point. Ideally trace routing between the R1
resistor and the ISL6140/ISL6150 (pin 4 (VEE) 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.
t = - R * C * ln [1 - (V(t) - V(t
where V(t) is the trip voltage (nominally 50mV); V(t
nominal voltage drop across the sense resistor before the
over-current condition; V
sense resistor while the over-current is applied.
For example: a system has a normal 1A current load, and a
20mΩ sense resistor, for a 2.5A over-current. It needs to
filter out a 50µs current pulse at 5A. So,
V(t) = 50mV (from spec)
V(t
V
If R7 = 100Ω, then C3 is around 1µF.
i
CORRECT
= 100mV (V = IR = 5A * 20mΩ)
0
) = 20mV (V = IR = 1A * 20mΩ)
To SENSE
and V
EE
FIGURE 5. SENSE RESISTOR
SENSE RESISTOR
i
CURRENT
is the voltage drop across the
0
)) / (V
8
INCORRECT
i
- V(t
0
))]
0
) is the
ISL6140, ISL6150
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 over-current 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 30 times; 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
V
Similarly, for OV:
1.223 = V
V
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 the requirement, and are also readily available
standard values.
The three resistors (R4, R5, R6) is the recommended
approach for most cases. But if acceptable values can’t be
UV
OV
= 1.223 (R4 + R5 + R6)/(R5 + R6)
= 1.223 (R4 + R5 + R6)/(R6)
UV
OV
* (R5 + R6)/(R4 + R5 + R6) or
* (R6)/(R4 + R5 + R6) or