Datasheet LT1374 (Analog Devices) - 10

FabricanteAnalog Devices
Descripción4.5A, 500kHz Step-Down Switching Regulator
Páginas / Página32 / 10 — APPLICATIONS INFORMATION. MAXIMUM OUTPUT LOAD CURRENT. CHOOSING THE …
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APPLICATIONS INFORMATION. MAXIMUM OUTPUT LOAD CURRENT. CHOOSING THE INDUCTOR AND OUTPUT CAPACITOR

APPLICATIONS INFORMATION MAXIMUM OUTPUT LOAD CURRENT CHOOSING THE INDUCTOR AND OUTPUT CAPACITOR

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LT1374
U U W U APPLICATIONS INFORMATION MAXIMUM OUTPUT LOAD CURRENT
Note that there is less load current available at the higher input voltage because inductor ripple current increases. Maximum load current for a buck converter is limited by This is not always the case. Certain combinations of the maximum switch current rating (IP) of the LT1374. inductor value and input voltage range may yield lower This current rating is 4.5A up to 50% duty cycle (DC), available load current at the lowest input voltage due to decreasing to 3.7A at 80% duty cycle. This is shown reduced peak switch current at high duty cycles. If load graphically in Typical Performance Characteristics and as current is close to the maximum available, please check shown in the formula below: maximum available current at both input voltage IP = 4.5A for DC ≤ 50% extremes. To calculate actual peak switch current with a IP = 3.21 + 5.95(DC) – 6.75(DC)2 for 50% < DC < 90% given set of conditions, use: 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 For lighter loads where discontinuous operation can be Current rating decreases with duty cycle because the used, maximum load current is equal to: LT1374 has internal slope compensation to prevent current mode subharmonic switching. For more details, read Ap- plication Note 19. The LT1374 is a little unusual in this regard IOUT(MAX) = because it has nonlinear slope compensation which gives Discontinuous mode better compensation with less reduction in current limit. Example: with L = 1.2µH, VOUT = 5V, and VIN(MAX) = 15V, 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 one-half peak-to-peak inductor current. The following formula assumes continuous mode operation, implying that the term on the right is less than one-half of IP. The main reason for using such a tiny inductor is that it is physically very small, but keep in mind that peak-to-peak I inductor current will be very high. This will increase output OUT(MAX) = Continuous Mode ripple voltage. If the output capacitor has to be made larger to reduce ripple voltage, the overall circuit could actually For the conditions above and L = 3.3µH, wind up larger.
CHOOSING THE INDUCTOR AND OUTPUT CAPACITOR
For most applications the output inductor will fall in the range of 3µH to 20µH. Lower values are chosen to reduce physical size of the inductor. Higher values allow more At VIN = 15V, duty cycle is 33%, so IP is just equal to a fixed output current because they reduce peak current seen by 4.5A, and IOUT(MAX) is equal to: the LT1374 switch, which has a 4.5A limit. Higher values also reduce output ripple voltage, and reduce core loss. Graphs in the Typical Performance Characteristics section show maximum output load current versus inductor size and input voltage. A second graph shows core loss versus inductor size for various core materials. 1374fd 10