Datasheet LT1316 (Analog Devices) - 8
| 制造商 | Analog Devices |
| 描述 | Micropower DC/DC Converter with Programmable Peak Current Limit |
| 页数 / 页 | 16 / 8 — APPLICATIONS INFORMATION. Discontinuous Mode Operation. Design Example 2. … |
| 文件格式/大小 | PDF / 313 Kb |
| 文件语言 | 英语 |
APPLICATIONS INFORMATION. Discontinuous Mode Operation. Design Example 2. Design Example 1
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LT1316
U U W U APPLICATIONS INFORMATION Discontinuous Mode Operation
2(I 2(10mA) 2. I OUT) PEAK = = = 58mA A boost converter with a high V 1 – DC 1 – 0.654 OUT:VIN ratio operates with a high duty cycle in continuous mode. For duty cycles 3. Find L exceeding the LT1316’s guaranteed minimum specifica- tion of 0.73, the circuit will need to be designed for VOUT – VIN + VD ) discontinuous operation. Additionally, very low peak cur- L = t 0.4(I OFF PEAK) rent limiting below 50mA may necessitate operating in this mode unless high inductance values are acceptable. When 5 – 2 + 0.4 = ) )2µs operating in discontinuous mode, a different equation 0.4(58mA) governs available output power. For each switch cycle, the = 293µH inductor current ramps down to zero, completely releas- ing the stored energy. Energy stored in the inductor at any 4. Find RSET resistor time is equal to 1/2 LI2. Because this energy is released each cycle, the equation for maximum power out is: V Overshoot = IN 300ns ) )L P 2 OUT(MAX) = 1/2L(IPEAK )f 2 = = 1.8mA ) 1 ) 330µH Where f = IPEAK(L) + tOFF V Find R IN – VSAT SET from Figure 3 for 58mA – 1.8mA = 56.2mA RSET ≈ 47k When designing for very low peak currents (< 50mA), the inductor size needs to be large enough so that on-time is
Design Example 2
a least 1µs. On-time can be calculated by the equation: Requirements: VIN = 3.3V, VOUT = 28V and ILOAD = 5mA. I 1. Find duty cycle: On-Time = PEAK • L ) (VIN – VSAT) V DC = OUT – VIN + VD ) 28 – 3.3 + 0.4 = = 0.89 ) 28 – 0.2 + 0.4 where V VOUT – VSAT + VD SAT = 0.2V. Also, at these low current levels, current overshoot due to Because duty cycle exceeds LT1316 minimum specifi- power transistor turn-off delay will be a significant portion cation of 73%, the circuit must be designed for discon- of peak current. Increasing inductor size will keep this to tinuous operation. a minimum. 2. Find POUT(MAX)
Design Example 1
Multiply POUT by 1.4 to give a safe operating margin Requirements: V P IN = 2V, VOUT = 5V and ILOAD = 10mA. OUT(MAX) = POUT(1.4) = (5mA)(28V)(1.4) = 0.196W 1. Find duty cycle 3. Set the on-time to the data sheet minimum of 3.4µs and find L V DC = OUT – VIN + VD ) 5 – 2 + 0.4 = = 0.654 ) (t 2)(V V ON IN – VSAT)2 OUT – VSAT + VD 5 – 0.2 + 0.4 L = 2POUT(MAX)(tON + tOFF) Because duty cycle is less than the LT1316 minimum (3.4µs2)(3.3 – 0.2)2 specification (0.73), the circuit can be designed for = = 52µH continuous operation. 2(0.196W)(3.4µs + 2µs) 8
U U W U APPLICATIONS INFORMATION Discontinuous Mode Operation
2(I 2(10mA) 2. I OUT) PEAK = = = 58mA A boost converter with a high V 1 – DC 1 – 0.654 OUT:VIN ratio operates with a high duty cycle in continuous mode. For duty cycles 3. Find L exceeding the LT1316’s guaranteed minimum specifica- tion of 0.73, the circuit will need to be designed for VOUT – VIN + VD ) discontinuous operation. Additionally, very low peak cur- L = t 0.4(I OFF PEAK) rent limiting below 50mA may necessitate operating in this mode unless high inductance values are acceptable. When 5 – 2 + 0.4 = ) )2µs operating in discontinuous mode, a different equation 0.4(58mA) governs available output power. For each switch cycle, the = 293µH inductor current ramps down to zero, completely releas- ing the stored energy. Energy stored in the inductor at any 4. Find RSET resistor time is equal to 1/2 LI2. Because this energy is released each cycle, the equation for maximum power out is: V Overshoot = IN 300ns ) )L P 2 OUT(MAX) = 1/2L(IPEAK )f 2 = = 1.8mA ) 1 ) 330µH Where f = IPEAK(L) + tOFF V Find R IN – VSAT SET from Figure 3 for 58mA – 1.8mA = 56.2mA RSET ≈ 47k When designing for very low peak currents (< 50mA), the inductor size needs to be large enough so that on-time is
Design Example 2
a least 1µs. On-time can be calculated by the equation: Requirements: VIN = 3.3V, VOUT = 28V and ILOAD = 5mA. I 1. Find duty cycle: On-Time = PEAK • L ) (VIN – VSAT) V DC = OUT – VIN + VD ) 28 – 3.3 + 0.4 = = 0.89 ) 28 – 0.2 + 0.4 where V VOUT – VSAT + VD SAT = 0.2V. Also, at these low current levels, current overshoot due to Because duty cycle exceeds LT1316 minimum specifi- power transistor turn-off delay will be a significant portion cation of 73%, the circuit must be designed for discon- of peak current. Increasing inductor size will keep this to tinuous operation. a minimum. 2. Find POUT(MAX)
Design Example 1
Multiply POUT by 1.4 to give a safe operating margin Requirements: V P IN = 2V, VOUT = 5V and ILOAD = 10mA. OUT(MAX) = POUT(1.4) = (5mA)(28V)(1.4) = 0.196W 1. Find duty cycle 3. Set the on-time to the data sheet minimum of 3.4µs and find L V DC = OUT – VIN + VD ) 5 – 2 + 0.4 = = 0.654 ) (t 2)(V V ON IN – VSAT)2 OUT – VSAT + VD 5 – 0.2 + 0.4 L = 2POUT(MAX)(tON + tOFF) Because duty cycle is less than the LT1316 minimum (3.4µs2)(3.3 – 0.2)2 specification (0.73), the circuit can be designed for = = 52µH continuous operation. 2(0.196W)(3.4µs + 2µs) 8