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LT3574 Datasheet(PDF) 16 Page - Linear Technology |
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LT3574 Datasheet(HTML) 16 Page - Linear Technology |
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16 / 26 page ![]() LT3511 16 3511fa Example: NPS < 150V – 72V – 40V 15V + 0.5V NPS < 2.45 NPS = 2 The choice of turns ratio is critical in determining output power as shown earlier in the Output Power section. At this point, a third winding can be added to the transformer to drive the BIAS pin of the LT3511 for higher efficiencies. Choose a turns ratio that sets the third winding voltage to regulate between 3.3V and 6V for maximum efficiency. Choose a third winding ratio to drive BIAS winding with 5V. (Optional) Example: NTHIRD NS = VTHIRD VOUT = 5V 15V = 0.33 The turns ratio of the transformer chosen is as follows NPRIMARY: NSECONDARY: NTHIRD = 2:1:0.33. Step 2: Calculate maximum power output at minimum VIN. POUT(VIN(MIN)) = η • VIN(MIN) • IIN = η • VIN(MIN) • D • IPEAK • 0.5 D = VOUT + VF ( ) •NPS VOUT + VF ( ) •NPS + VIN(MIN) η = Efficiency = ~75% IPEAK = Peak switch current = 0.26A Example: D = 0.46 POUT(VIN(MIN)) = 1.62 IOUT(VIN(MIN)) = POUT(VIN(MIN))/VOUT = 0.11A The chosen turns ratio satisfies the output current re- quirement of 100mA. If the output current was too low, the minimum input voltage could be adjusted higher. The turns ratio in this example is set to its highest ratio given switch voltage requirements and margin for leakage in- ductance voltage spike. Step 3: Determine primary inductance, switching fre- quency and saturation current. Primary inductance for the transformer must be set above a minimum value to satisfy the minimum off time requirement. LPRI ≥ tOFF(MIN) • NPS • VOUT + VF ( ) IPEAK(MIN) tOFF(MIN) = 400ns IPEAK(MIN) = 55mA Example: LPRI ≥ 400ns • 2 • 15 + 0.5 ( ) 0.055 LPRI ≥ 225µH In addition, primary inductance will determine switching frequency. fSW = 1 tON + tOFF = 1 LPRI •IPEAK VIN + LPRI •IPEAK NPS • VOUT + VF ( ) IPEAK = VOUT •IOUT • 2 η • VIN • D Example: Let’s calculate switching frequency at our nominal VIN of 48V. D = 15 + 0.5 ( ) • 2 15 + 0.5 ( ) • 2+ 48 = 0.39 IPEAK = 15V • 0.1A • 2 0.75 • 48V • 0.39 = 0.21A Let’s choose LPRI = 350μH. Remember, most transform- ers specify primary inductance with a tolerance of ±20%. fSW = 256kHz Finally, the transformer needs to be rated for the correct saturation current level across line and load conditions. In the given example, the worst-case condition for switch current is at minimum VIN and maximum load. applicaTions inForMaTion |
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