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ISL6140CBZ Datasheet(PDF) 9 Page - Intersil Corporation |
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ISL6140CBZ Datasheet(HTML) 9 Page - Intersil Corporation |
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9 / 19 page  9 FN9039.4 Applications: Overcurrent 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 overcurrent trip point. Ideally trace routing between the R1 resistor and the ISL6140 and 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 (see Equation 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; Vi 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) Vi = 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 = VUV*(R5 + R6)/(R4 + R5 + R6) or VUV = 1.223 (R4 + R5 + R6)/(R5 + R6) Similarly, for OV: 1.223 = VOV*(R6)/(R4 + R5 + R6) or VOV = 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 CORRECT TO SENSE CURRENT SENSE RESISTOR INCORRECT AND VEE FIGURE 5. SENSE RESISTOR t R*C*In [1 - (V(t) - V(t0)) Vi Vt0 () – () ⁄ ] – = (EQ. 2) ISL6140, ISL6150 |
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