Datasheet LT1375, LT1376 (Analog Devices) - 10

LT1375/LT1376
U U W U APPLICATIONS INFORMATION
pin when output voltage is low. The equivalent circuitry is formula assumes continuous mode operation, implying shown in Figure 2. Q1 is completely off during normal that the term on the right is less than one-half of IP. operation. If the FB pin falls below 1V, Q1 begins to conduct current and reduces frequency at the rate of (VOUT)(VIN−VOUT) approximately 5kHz/µA. To ensure adequate frequency IOUT(MAX) = I − foldback (under worst-case short-circuit conditions), the Continuous Mode P (2L)(f)(VIN) external divider Thevinin resistance must be low enough to pull 150µA out of the FB pin with 0.6V on the pin (RDIV ≤ 4k). The net result is that reductions in frequency and For the conditions above and L = 10µH, current limit are affected by output voltage divider imped- ance. Although divider impedance is not critical, caution ( )5(8− )5 should be used if resistors are increased beyond the I = 1 4 . 4 − OUT MAX ( ) ⎛ −5⎞ ⎛ 3⎞ suggested values and short-circuit conditions will occur 2⎝10 ⎠⎝500•10 ⎠(8) with high input voltage. High frequency pickup will in- crease and the protection accorded by frequency and = 1 4 . 4 − 0.19 = 1 2 . 5A current foldback will decrease. At VIN = 15V, duty cycle is 33%, so IP is just equal to a fixed 1.5A, and IOUT(MAX) is equal to:
MAXIMUM OUTPUT LOAD CURRENT
Maximum load current for a buck converter is limited by 5 15 5 ( )( − ) the maximum switch current rating (I 1 5 . − 1 5 . 0.33 1.17A P) of the LT1376. ⎛ 5⎞ ⎛ 3⎞ This current rating is 1.5A up to 50% duty cycle (DC), 2 10 500 • 10 15 ⎝ ⎠ ⎝ ⎠ ( ) = − = − decreasing to 1.35A at 80% duty cycle. This is shown graphically in Typical Performance Characteristics and as Note that there is less load current available at the higher shown in the formula below: input voltage because inductor ripple current increases. I This is not always the case. Certain combinations of P = 1.5A for DC ≤ 50% I inductor value and input voltage range may yield lower P = 1.65A – 0.15 (DC) – 0.26 (DC)2 for 50% < DC < 90% available load current at the lowest input voltage due to DC = Duty cycle = VOUT/VIN reduced peak switch current at high duty cycles. If load Example: with VOUT = 5V, VIN = 8V; DC = 5/8 = 0.625, and; current is close to the maximum available, please check maximum available current at both input voltage ex- ISW(MAX) = 1.64 – 0.15 (0.625) – 0.26 (0.625)2 = 1.44A tremes. To calculate actual peak switch current with a Current rating decreases with duty cycle because the given set of conditions, use: LT1376 has internal slope compensation to prevent cur- rent mode subharmonic switching. For more details, read V V ( −V ) Application Note 19. The LT1376 is a little unusual in this OUT IN OUT I I SW PEAK OUT ( ) = + regard because it has nonlinear slope compensation which 2 L f V ( )( )( )IN gives better compensation with less reduction in current limit. For lighter loads where discontinuous operation can be used, maximum load current is equal to: Maximum load current would be equal to maximum switch current for an infinitely large inductor, but with ( )2( )( )( ) finite inductor size, maximum load current is reduced by I f L V P OUT I one-half peak-to-peak inductor current. The following OUT(MAX) = Discontinuous mode 2 V V V ( )( − ) OUT IN OUT 13756fd 10