Datasheet LT1959 (Analog Devices) - 9

LT1959
U U W U APPLICATIONS INFORMATION
The internal circuitry which forces reduced switching finite inductor size, maximum load current is reduced by frequency also causes current to flow out of the feedback one-half peak-to-peak inductor current. The following 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 0.7V, Q1 begins to conduct current and reduces frequency at the rate of (VOUT)(VIN−VOUT) approximately 2kHz/µA. To ensure adequate frequency IOUT(MAX) = I − P foldback (under worst-case short-circuit conditions), the Continuous Mode (2L)(f)(VIN) external divider Thevinin resistance must be low enough to pull 150µA out of the FB pin with 0.3V on the pin (RDIV ≤ 2k). The net result is that reductions in frequency and For the conditions above and L = 3.3µH, 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 = 4 3 . − OUT MAX ( )  −  6  3 suggested values and short-circuit conditions will occur 23 3. •10 500•10 (8) with high input voltage. High frequency pickup will = 4 3 . − 0 5 . 7 = 3 7 . 3 A increase and the protection accorded by frequency and current foldback will decrease. At VIN = 15V, duty cycle is 33%, so IP is just equal to a fixed 4.5A, 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 (IP) of the LT1959. 4 5 . −  −    This current rating is 4.5A up to 50% duty cycle (DC), 2 3 3 . • 10 6 500 • 103 15     ( ) decreasing to 3.7A at 80% duty cycle. This is shown = − = graphically in Typical Performance Characteristics and as 4 5 . 1 0 . 1 3 4 . 9A shown in the formula below: Note that there is less load current available at the higher IP = 4.5A for DC ≤ 50% input voltage because inductor ripple current increases. IP = 3.21 + 5.95(DC) – 6.75(DC)2 for 50% < DC < 90% This is not always the case. Certain combinations of inductor value and input voltage range may yield lower DC = Duty cycle = VOUT/VIN available load current at the lowest input voltage due to Example: with VOUT = 5V, VIN = 8V; DC = 5/8 = 0.625, and; reduced peak switch current at high duty cycles. If load I current is close to the maximum available, please check SW(MAX) = 3.21 + 5.95(0.625) – 6.75(0.625)2 = 4.3A maximum available current at both input voltage ex- Current rating decreases with duty cycle because the tremes. To calculate actual peak switch current with a LT1959 has internal slope compensation to prevent cur- given set of conditions, use: rent mode subharmonic switching. For more details, read Application Note 19. The LT1959 is a little unusual in this V V ( −V ) regard because it has nonlinear slope compensation which OUT IN OUT I I ( ) = + gives better compensation with less reduction in current SW PEAK OUT 2 L f V ( )( )( )IN limit. Maximum load current would be equal to maximum switch current for an infinitely large inductor, but with 9