Electronic Components Datasheet Search 

LT1959 Datasheet(PDF) 9 Page  Linear Technology 

LT1959 Datasheet(HTML) 9 Page  Linear Technology 
9 / 24 page 9 LT1959 APPLICATIONS INFORMATION The internal circuitry which forces reduced switching frequency also causes current to flow out of the feedback pin when output voltage is low. The equivalent circuitry is shown in Figure 2. Q1 is completely off during normal operation. If the FB pin falls below 0.7V, Q1 begins to conduct current and reduces frequency at the rate of approximately 2kHz/ µA. To ensure adequate frequency foldback (under worstcase shortcircuit conditions), the external divider Thevinin resistance must be low enough to pull 150 µA out of the FB pin with 0.3V on the pin (RDIV ≤ 2k). The net result is that reductions in frequency and current limit are affected by output voltage divider imped ance. Although divider impedance is not critical, caution should be used if resistors are increased beyond the suggested values and shortcircuit conditions will occur with high input voltage. High frequency pickup will increase and the protection accorded by frequency and current foldback will decrease. MAXIMUM OUTPUT LOAD CURRENT Maximum load current for a buck converter is limited by the maximum switch current rating (IP) of the LT1959. This current rating is 4.5A up to 50% duty cycle (DC), decreasing to 3.7A at 80% duty cycle. This is shown graphically in Typical Performance Characteristics and as shown in the formula below: IP = 4.5A for DC ≤ 50% IP = 3.21 + 5.95(DC) – 6.75(DC)2 for 50% < DC < 90% DC = Duty cycle = VOUT/VIN Example: with VOUT = 5V, VIN = 8V; DC = 5/8 = 0.625, and; ISW(MAX) = 3.21 + 5.95(0.625) – 6.75(0.625)2 = 4.3A Current rating decreases with duty cycle because the LT1959 has internal slope compensation to prevent cur rent mode subharmonic switching. For more details, read Application Note 19. The LT1959 is a little unusual in this regard because it has nonlinear slope compensation which gives better compensation with less reduction in current limit. Maximum load current would be equal to maximum switch current for an infinitely large inductor, but with finite inductor size, maximum load current is reduced by onehalf peaktopeak inductor current. The following formula assumes continuous mode operation, implying that the term on the right is less than onehalf of IP. IOUT(MAX) = Continuous Mode For the conditions above and L = 3.3 µH, I A OUT MAX ( ) − =− () − () () =− = 43 58 5 2 3 3 10 500 10 8 43 057 3 73 63 . .• • .. . At VIN = 15V, duty cycle is 33%, so IP is just equal to a fixed 4.5A, and IOUT(MAX) is equal to: 45 515 5 2 3 3 10 500 10 15 45 101 3 49 63 . .• • .. . − () − () () =− = − A Note that there is less load current available at the higher input voltage because inductor ripple current increases. This is not always the case. Certain combinations of inductor value and input voltage range may yield lower available load current at the lowest input voltage due to reduced peak switch current at high duty cycles. If load current is close to the maximum available, please check maximum available current at both input voltage ex tremes. To calculate actual peak switch current with a given set of conditions, use: II VV V Lf V SW PEAK OUT OUT IN OUT IN ( ) =+ − () ()( )( ) 2 IP − () − () ()( )( ) VV V Lf V OUT IN OUT IN 2 
