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LM3405AXMK データシート(PDF) 9 Page - Texas Instruments |
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LM3405AXMK データシート(HTML) 9 Page - Texas Instruments |
9 / 32 page LM3405A www.ti.com SNVS508C – OCTOBER 2007 – REVISED MAY 2013 BOOST FUNCTION Capacitor C3 and diode D2 in Figure 13 are used to generate a voltage VBOOST. The voltage across C3, VBOOST - VSW, is the gate drive voltage to the internal NMOS power switch. To properly drive the internal NMOS switch during its on-time, VBOOST needs to be at least 2.5V greater than VSW. A large value of VBOOST - VSW is recommended to achieve better efficiency by minimizing both the internal switch ON resistance (RDS(ON)), and the switch rise and fall times. However, VBOOST - VSW should not exceed the maximum operating limit of 5.5V. When the LM3405A starts up, internal circuitry from VIN supplies a 20mA current to the BOOST pin, flowing out of the BOOST pin into C3. This current charges C3 to a voltage sufficient to turn the switch on. The BOOST pin will continue to source current to C3 until the voltage at the feedback pin is greater than 123mV. There are various methods to derive VBOOST: 1. From the input voltage (VIN) 2. From the output voltage (VOUT) 3. From a shunt or series zener diode 4. From an external distributed voltage rail (VEXT) The first method is shown in the Simplified Block Diagram of Figure 13. Capacitor C3 is charged via diode D2 by VIN. During a normal switching cycle, when the internal NMOS power switch is off (TOFF) (refer to Figure 14), VBOOST equals VIN minus the forward voltage of D2 (VD2), during which the current in the inductor (L1) forward biases the catch diode D1 (VD1). Therefore the gate drive voltage stored across C3 is: VBOOST - VSW = VIN - VD2 + VD1 (1) When the NMOS switch turns on (TON), the switch pin rises to: VSW = VIN – (RDS(ON) x IL) (2) Since the voltage across C3 remains unchanged, VBOOST is forced to rise thus reverse biasing D2. The voltage at VBOOST is then: VBOOST = 2VIN – (RDS(ON) x IL) – VD2 + VD1 (3) Depending on the quality of the diodes D1 and D2, the gate drive voltage in this method can be slightly less or larger than the input voltage VIN. For best performance, ensure that the variation of the input supply does not cause the gate drive voltage to fall outside the recommended range: 2.5V < VIN - VD2 + VD1 < 5.5V (4) The second method for deriving the boost voltage is to connect D2 to the output as shown in Figure 15. The gate drive voltage in this configuration is: VBOOST - VSW = VOUT – VD2 + VD1 (5) Since the gate drive voltage needs to be in the range of 2.5V to 5.5V, the output voltage VOUT should be limited to a certain range. For the calculation of VOUT, see OUTPUT VOLTAGE section. Figure 15. VBOOST Derived from VOUT Copyright © 2007–2013, Texas Instruments Incorporated Submit Documentation Feedback 9 Product Folder Links: LM3405A |
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