Datasheet LT1934, LT1934 (Analog Devices) - 9
| Fabricante | Analog Devices |
| Descripción | Micropower Step-Down Switching Regulators in ThinSOT |
| Páginas / Página | 20 / 9 — APPLICATIONS INFORMATION. Table 2. Inductor Vendors. VENDOR. PHONE URL. … |
| Formato / tamaño de archivo | PDF / 245 Kb |
| Idioma del documento | Inglés |
APPLICATIONS INFORMATION. Table 2. Inductor Vendors. VENDOR. PHONE URL. PART. SERIES. COMMENTS. Input Capacitor
Línea de modelo para esta hoja de datos
- LT1934EDCB#PBF LT1934EDCB#TRMPBF LT1934EDCB#TRPBF LT1934EDCB-1#PBF LT1934EDCB-1#TRMPBF LT1934EDCB-1#TRPBF LT1934ES6#TRMPBF LT1934ES6#TRPBF LT1934ES6-1#TRMPBF LT1934ES6-1#TRPBF LT1934IDCB#PBF LT1934IDCB#TRMPBF LT1934IDCB#TRPBF LT1934IDCB-1#PBF LT1934IDCB-1#TRMPBF LT1934IDCB-1#TRPBF LT1934IS6#PBF LT1934IS6#TRMPBF LT1934IS6#TRPBF LT1934IS6-1#PBF LT1934IS6-1#TRMPBF LT1934IS6-1#TRPBF
Versión de texto del documento
LT1934/LT1934-1
APPLICATIONS INFORMATION Table 2. Inductor Vendors VENDOR PHONE URL PART SERIES COMMENTS
Murata (404) 426-1300 www.murata.com LQH3C Small, Low Cost, 2mm Height Sumida (847) 956-0666 www.sumida.com CR43 CDRH4D28 CDRH5D28 Coilcraft (847) 639-6400 www.coilcraft.com DO1607C DO1608C DT1608C Würth (866) 362-6673 www.we-online.com WE-PD1, 2, 3, 4 Electronics the average inductor current equals the load current, the The inductor must carry the peak current without satu- maximum load current is: rating excessively. When an inductor carries too much current, its core material can no longer generate ad- IOUT(MAX) = IPK – ΔIL/2 ditional magnetic fl ux (it saturates) and the inductance where IPK is the peak inductor current and ΔIL is the drops, sometimes very rapidly with increasing current. peak-to-peak ripple current in the inductor. The ripple This condition allows the inductor current to increase current is determined by the off time, tOFF = 1.8μs, and at a very high rate, leading to high ripple current and the inductor value: decreased overload protection. ΔIL = (VOUT + VD) • tOFF/L Inductor vendors provide current ratings for power induc- I tors. These are based on either the saturation current or PK is nominally equal to ILIM. However, there is a slight delay in the control circuitry that results in a higher peak on the RMS current that the inductor can carry without current and a more accurate value is: dissipating too much power. In some cases it is not clear which of these two determine the current rating. Some data IPK = ILIM + 150ns • (VIN – VOUT)/L sheets are more thorough and show two current ratings, These expressions are combined to give the maximum one for saturation and one for dissipation. For LT1934 ap- load current that the LT1934 will deliver: plications, the RMS current rating should be higher than the load current, while the saturation current should be IOUT(MAX) = 350mA + 150ns • (VIN – VOUT)/L – 1.8μs higher than the peak inductor current calculated above. • (VOUT + VD)/2L (LT1934) I
Input Capacitor
OUT(MAX) = 90mA + 150ns • (VIN – VOUT)/L – 1.8μs • (VOUT + VD)/2L (LT1934-1) Step-down regulators draw current from the input sup- The minimum current limit is used here to be conservative. ply in pulses with very fast rise and fall times. The input The third term is generally larger than the second term, capacitor is required to reduce the resulting voltage ripple so that increasing the inductor value results in a higher at the LT1934 and to force this switching current into output current. This equation can be used to evaluate a tight local loop, minimizing EMI. The input capacitor a chosen inductor or it can be used to choose L for a must have low impedance at the switching frequency to given maximum load current. The simple, single equation do this effectively. A 2.2μF ceramic capacitor (1μF for the rule given above for choosing L was found by setting LT1934-1) satisfi es these requirements. ΔIL = ILIM/2.5. This results in IOUT(MAX) ~0.8ILIM (ignoring If the input source impedance is high, a larger value ca- the delay term). Note that this analysis assumes that the pacitor may be required to keep input ripple low. In this inductor current is continuous, which is true if the ripple case, an electrolytic of 10μF or more in parallel with a 1μF current is less than the peak current or ΔIL < IPK. ceramic is a good combination. Be aware that the input 1934fe 9
APPLICATIONS INFORMATION Table 2. Inductor Vendors VENDOR PHONE URL PART SERIES COMMENTS
Murata (404) 426-1300 www.murata.com LQH3C Small, Low Cost, 2mm Height Sumida (847) 956-0666 www.sumida.com CR43 CDRH4D28 CDRH5D28 Coilcraft (847) 639-6400 www.coilcraft.com DO1607C DO1608C DT1608C Würth (866) 362-6673 www.we-online.com WE-PD1, 2, 3, 4 Electronics the average inductor current equals the load current, the The inductor must carry the peak current without satu- maximum load current is: rating excessively. When an inductor carries too much current, its core material can no longer generate ad- IOUT(MAX) = IPK – ΔIL/2 ditional magnetic fl ux (it saturates) and the inductance where IPK is the peak inductor current and ΔIL is the drops, sometimes very rapidly with increasing current. peak-to-peak ripple current in the inductor. The ripple This condition allows the inductor current to increase current is determined by the off time, tOFF = 1.8μs, and at a very high rate, leading to high ripple current and the inductor value: decreased overload protection. ΔIL = (VOUT + VD) • tOFF/L Inductor vendors provide current ratings for power induc- I tors. These are based on either the saturation current or PK is nominally equal to ILIM. However, there is a slight delay in the control circuitry that results in a higher peak on the RMS current that the inductor can carry without current and a more accurate value is: dissipating too much power. In some cases it is not clear which of these two determine the current rating. Some data IPK = ILIM + 150ns • (VIN – VOUT)/L sheets are more thorough and show two current ratings, These expressions are combined to give the maximum one for saturation and one for dissipation. For LT1934 ap- load current that the LT1934 will deliver: plications, the RMS current rating should be higher than the load current, while the saturation current should be IOUT(MAX) = 350mA + 150ns • (VIN – VOUT)/L – 1.8μs higher than the peak inductor current calculated above. • (VOUT + VD)/2L (LT1934) I
Input Capacitor
OUT(MAX) = 90mA + 150ns • (VIN – VOUT)/L – 1.8μs • (VOUT + VD)/2L (LT1934-1) Step-down regulators draw current from the input sup- The minimum current limit is used here to be conservative. ply in pulses with very fast rise and fall times. The input The third term is generally larger than the second term, capacitor is required to reduce the resulting voltage ripple so that increasing the inductor value results in a higher at the LT1934 and to force this switching current into output current. This equation can be used to evaluate a tight local loop, minimizing EMI. The input capacitor a chosen inductor or it can be used to choose L for a must have low impedance at the switching frequency to given maximum load current. The simple, single equation do this effectively. A 2.2μF ceramic capacitor (1μF for the rule given above for choosing L was found by setting LT1934-1) satisfi es these requirements. ΔIL = ILIM/2.5. This results in IOUT(MAX) ~0.8ILIM (ignoring If the input source impedance is high, a larger value ca- the delay term). Note that this analysis assumes that the pacitor may be required to keep input ripple low. In this inductor current is continuous, which is true if the ripple case, an electrolytic of 10μF or more in parallel with a 1μF current is less than the peak current or ΔIL < IPK. ceramic is a good combination. Be aware that the input 1934fe 9