Datasheet LT1086 (Analog Devices) - 9

LT1086 Series
U U W U APPLICATIONS INFORMATION
plugging and unplugging in the system will not generate the power supply may need to be cycled down to zero and current large enough to do any damage. brought up again to make the output recover. The adjustment pin can be driven on a transient basis ±
Ripple Rejection
25V, with respect to the output without any device degradation. Of course as with any IC regulator, exceeding For the LT1086 the typical curves for ripple rejection the maximum input-to-output voltage differential causes reflect values for a bypassed adjust pin. This curve will be the internal transistors to break down and none of the true for all values of output voltage. For proper bypassing protection circuitry is functional. and ripple rejection approaching the values shown, the impedance of the adjust pin capacitor at the ripple fre- D1 1N4002 quency should equal the value of R1, (normally 100Ω to (OPTIONAL) 120Ω). The size of the required adjust pin capacitor is a function of the input ripple frequency. At 120Hz the adjust pin capacitor should be 13µF if R1 = 100Ω; at 10kHz only VIN IN LT1086 OUT VOUT + 0.16 ADJ R1 C µF is needed. OUT 150µF + For circuits without an adjust pin bypass capacitor the CADJ R2 10µF LT1086 • AI01 ripple rejection will be a function of output voltage. The output ripple will increase directly as a ratio of the output
Overload Recovery
voltage to the reference voltage (VOUT/VREF). For ex- ample, with the output voltage equal to 5V and no adjust Like any of the IC power regulators, the LT1086 has safe pin capacitor, the output ripple will be higher by the ratio area protection. The safe area protection decreases the of 5V/1.25V or four times larger. Ripple rejection will be current limit as input-to-output voltage increases and degraded by 12dB from the value shown on the LT1086 keeps the power transistor inside a safe operating region curve. Typical curves are provided for the 5V and 12V for all values of input-to-output voltage. The LT1086 devices since the adjust pin is not available. protection is designed to provide some output current at all values of input-to-output voltage up to the device
Output Voltage
breakdown. The LT1086 develops a 1.25V reference voltage between When power is first turned on, as the input voltage rises, the output and the adjust terminal (see Figure 1). By the output follows the input, allowing the regulator to start placing resistor R1 between these two terminals, a con- up into very heavy loads. During the start-up, as the input stant current is caused to flow through R1 and down voltage is rising, the input-to-output voltage differential through R2 to set the overall output voltage. Normally this remains small, allowing the regulator to supply large current is chosen to be the specified minimum load output currents. With high input voltage, a problem can current of 10mA. Because IADJ is very small and constant occur wherein removal of an output short will not allow the when compared with the current through R1, it repre- output voltage to recover. Older regulators such as the sents a small error and can usually be ignored. For fixed 7800 series also exhibited this phenomenon, so it is not voltage devices R1 and R2 are included in the device. unique to the LT1086. The problem occurs with a heavy output load when the V IN LT1086 OUT V input voltage is high and the output voltage is low, such as IN OUT + ADJ 10µF immediately after a removal of a short. The load line for VREF R1 TANTALUM I such a load may intersect the output current curve at two ADJ 50µA points. If this happens there are two stable output operat- R2 R2 ( ) VOUT = VREF 1 + + IADJ R2 ing points for the regulator. With this double intersection R1 1086 • F01
Figure 1. Basic Adjustable Regulator
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