Datasheet LT8490 (Analog Devices) - 35
| Manufacturer | Analog Devices |
| Description | High Voltage, High Current Buck-Boost Battery Charge Controller with Maximum Power Point Tracking (MPPT) |
| Pages / Page | 42 / 35 — applicaTions inForMaTion |
| File Format / Size | PDF / 707 Kb |
| Document Language | English |
applicaTions inForMaTion
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LT8490
applicaTions inForMaTion
• With RFBOUT2 set at 20kΩ and a desired Stage 2 voltage • Using the standard value resistors calculated above, the limit of 14.2V, the top output feedback resistor, RFBOUT1, VX3, N1 and N2 checking equations yield the following: is calculated according to the following equation: VX3 = 14.31V ⎡ ⎤ R ⎛ ⎞ N FBOUT1 = RFBOUT2 • V ⎢ S2 • 1.241 ⎜ − 0.128⎟−1⎥Ω 1 = 1.22 ⎣ ⎝1.211 ⎠ ⎦ N2 = 0.804 ⎡ ⎛ ⎞ ⎤ = 20k • 14.2 • 1.241 ⎢ ⎜ − 0.128⎟−1⎥Ω • In order to find a resistor combination that yields VX3 ⎣ ⎝1.211 ⎠ ⎦ closer to the desired 14.2V, RFBOUT2 is increased to the = 234,684Ω next higher standard value and the above calculations are repeated. Choose RFBOUT1 = 237kΩ which is the closest standard value resistor. • Iterations of the previous step are performed that include adjustments to RFBOUT1, RDACO1 and RDACO2 • Following the calculation of RFBOUT1, solve for RDACO1, until the following standard value feedback resistors RDACO2 and CDACO according to the following formulas: were chosen: R R FBOUT1 • RFBOUT2 • 0.833 R DACO2 = Ω FBOUT1 = 274kΩ R ⎛ ⎞ ⎜ FBOUT2 • VS2 • 1.241⎟−R R ⎝ 1.211⎠ FBOUT2 −RFBOUT1 FBOUT2 = 23.2kΩ 234,684 • 20k • 0.833 RDACO1 = 26.1kΩ = Ω ⎛ ⎞ RDACO2 = 124kΩ ⎜20k • 14.2 • 1.241⎟− 20k − 234,684 ⎝ 1.211⎠ CDACO = 0.082µF = 107,556Ω where: Choose RDACO2 = 107kΩ which is the closest standard VX3 = 14.27V value resistor. N1 = 1.22 RDACO1 = (0.2 • RDACO2) Ω N2 = 0.805 = 0.2 • 107,556Ω • With the output feedback network determined, use = 21,511Ω VMAX and solve for the input resistor feedback network Choose R according to the following formulas: DACO1 = 21.5kΩ which is the closest standard value resistor. ⎡ ⎛ ⎞⎤ ⎢1 4.47V +⎜ ⎟⎥ C 1 ⎢ ⎝ VMAX − 6V ⎠⎥ DACO = F R 500 •R FBIN1 = 100k • Ω DACO1 ⎢ ⎛ ⎞⎥ ⎢1 5.593V + 1 ⎜ ⎟⎥ ⎣ ⎝ V = F MAX − 6V ⎠⎦ 500 • 21,511 ⎡ ⎤ 1 ⎛ 4.47V ⎞ ⎢ + ⎜ ⎟ = 93nF ⎝53V ⎥ = 100k • − 6V ⎠ ⎢ ⎥Ω ⎢1 ⎛ 5.593V ⎞ +⎜ ⎟⎥ ⎣⎢ ⎝53V − 6V ⎠⎦⎥ = 97,865Ω 8490fa For more information www.linear.com/LT8490 35 Document Outline Features Applications Typical Application Description Absolute Maximum Ratings Order Information Pin Configuration Electrical Characteristics Typical Performance Characteristics Pin Functions Block Diagram Operation Applications Information Package Description Revision History Typical Application Related Parts
applicaTions inForMaTion
• With RFBOUT2 set at 20kΩ and a desired Stage 2 voltage • Using the standard value resistors calculated above, the limit of 14.2V, the top output feedback resistor, RFBOUT1, VX3, N1 and N2 checking equations yield the following: is calculated according to the following equation: VX3 = 14.31V ⎡ ⎤ R ⎛ ⎞ N FBOUT1 = RFBOUT2 • V ⎢ S2 • 1.241 ⎜ − 0.128⎟−1⎥Ω 1 = 1.22 ⎣ ⎝1.211 ⎠ ⎦ N2 = 0.804 ⎡ ⎛ ⎞ ⎤ = 20k • 14.2 • 1.241 ⎢ ⎜ − 0.128⎟−1⎥Ω • In order to find a resistor combination that yields VX3 ⎣ ⎝1.211 ⎠ ⎦ closer to the desired 14.2V, RFBOUT2 is increased to the = 234,684Ω next higher standard value and the above calculations are repeated. Choose RFBOUT1 = 237kΩ which is the closest standard value resistor. • Iterations of the previous step are performed that include adjustments to RFBOUT1, RDACO1 and RDACO2 • Following the calculation of RFBOUT1, solve for RDACO1, until the following standard value feedback resistors RDACO2 and CDACO according to the following formulas: were chosen: R R FBOUT1 • RFBOUT2 • 0.833 R DACO2 = Ω FBOUT1 = 274kΩ R ⎛ ⎞ ⎜ FBOUT2 • VS2 • 1.241⎟−R R ⎝ 1.211⎠ FBOUT2 −RFBOUT1 FBOUT2 = 23.2kΩ 234,684 • 20k • 0.833 RDACO1 = 26.1kΩ = Ω ⎛ ⎞ RDACO2 = 124kΩ ⎜20k • 14.2 • 1.241⎟− 20k − 234,684 ⎝ 1.211⎠ CDACO = 0.082µF = 107,556Ω where: Choose RDACO2 = 107kΩ which is the closest standard VX3 = 14.27V value resistor. N1 = 1.22 RDACO1 = (0.2 • RDACO2) Ω N2 = 0.805 = 0.2 • 107,556Ω • With the output feedback network determined, use = 21,511Ω VMAX and solve for the input resistor feedback network Choose R according to the following formulas: DACO1 = 21.5kΩ which is the closest standard value resistor. ⎡ ⎛ ⎞⎤ ⎢1 4.47V +⎜ ⎟⎥ C 1 ⎢ ⎝ VMAX − 6V ⎠⎥ DACO = F R 500 •R FBIN1 = 100k • Ω DACO1 ⎢ ⎛ ⎞⎥ ⎢1 5.593V + 1 ⎜ ⎟⎥ ⎣ ⎝ V = F MAX − 6V ⎠⎦ 500 • 21,511 ⎡ ⎤ 1 ⎛ 4.47V ⎞ ⎢ + ⎜ ⎟ = 93nF ⎝53V ⎥ = 100k • − 6V ⎠ ⎢ ⎥Ω ⎢1 ⎛ 5.593V ⎞ +⎜ ⎟⎥ ⎣⎢ ⎝53V − 6V ⎠⎦⎥ = 97,865Ω 8490fa For more information www.linear.com/LT8490 35 Document Outline Features Applications Typical Application Description Absolute Maximum Ratings Order Information Pin Configuration Electrical Characteristics Typical Performance Characteristics Pin Functions Block Diagram Operation Applications Information Package Description Revision History Typical Application Related Parts