Datasheet LT1940, LT1940L (Analog Devices) - 8

LT1940/LT1940L
U U W U APPLICATIO S I FOR ATIO
The maximum input voltage is determined by the absolute (VOUT/VIN < 0.5), there is a minimum inductance required maximum ratings of the V to avoid subharmonic oscillations. See AN19. The discus- IN and BOOST pins and by the minimum duty cycle DC sion below assumes continuous inductor current. MIN = 0.15: V The current in the inductor is a triangle wave with an INMAX = (VOUT + VD)/DCMIN – VD + VSW. average value equal to the load current. The peak switch This limits the maximum input voltage to ~14V with current is equal to the output current plus half the peak-to- VOUT = 1.8V and ~19V with VOUT = 2.5V. Note that this is peak inductor ripple current. The LT1940 limits its switch a restriction on the operating input voltage; the circuit will current in order to protect itself and the system from tolerate transient inputs up to the absolute maximum overload faults. Therefore, the maximum output current rating. For the LT1940L, the maximum input voltage is 7V. that the LT1940 will deliver depends on the current limit, the inductor value, and the input and output voltages. L is
Inductor Selection and Maximum Output Current
chosen based on output current requirements, output A good first choice for the inductor value is: voltage ripple requirements, size restrictions and effi- L = (V ciency goals. OUT + VD)/1.2 where V When the switch is off, the inductor sees the output D is the voltage drop of the catch diode (~0.4V) and L is in voltage plus the catch diode drop. This gives the peak-to- µH. With this value the maximum load current will be ~1.4A, independent of input voltage. The inductor’s peak ripple current in the inductor: RMS current rating must be greater than your maximum ∆IL = (1 – DC)(VOUT + VD)/(L • f) load current and its saturation current should be about where f is the switching frequency of the LT1940 and L is 30% higher. To keep efficiency high, the series resistance the value of the inductor. The peak inductor and switch (DCR) should be less than 0.1Ω. Table 1 lists several current is vendors and types that are suitable. I Of course, such a simple design guide will not always SWPK = ILPK = IOUT + ∆IL/2. result in the optimum inductor for your application. A To maintain output regulation, this peak current must be larger value provides a slightly higher maximum load less than the LT1940’s switch current limit ILIM. ILIM is at current, and will reduce the output voltage ripple. If your least 1.8A at low duty cycle and decreases linearly to 1.5A load is lower than 1.4A, then you can decrease the value of at DC = 0.8. The maximum output current is a function of the inductor and operate with higher ripple current. This the chosen inductor value: allows you to use a physically smaller inductor, or one with IOUTMAX = ILIM – ∆IL/2 = 1.8A • (1 – 0.21 • DC) – ∆IL/2 a lower DCR resulting in higher efficiency. Be aware that if the inductance differs from the simple rule above, then the If the inductor value is chosen so that the ripple current is maximum load current will depend on input voltage. There small, then the available output current will be near the are several graphs in the Typical Performance Character- switch current limit. istics section of this data sheet that show the maximum One approach to choosing the inductor is to start with the load current as a function of input voltage and inductor simple rule given above, look at the available inductors, value for several popular output voltages. Also, low and choose one to meet cost or space goals. Then use inductance may result in discontinuous mode operation, these equations to check that the LT1940 will be able to which is okay, but further reduces maximum load current. deliver the required output current. Note again that these For details of maximum output current and discontinuous equations assume that the inductor current is continuous. mode operation, see Linear Technology Application Discontinuous operation occurs when IOUT is less than Note 44. Finally, for duty cycles greater than 50% ∆IL/2 as calculated above. 1940fa 8