有关升压开关电源的分析

橄榄之家

Question:

Assuming ideal components, consider this boost converter:                                                                         1. Draw the current waveforms at the input (I1) and output (I2) of the boost converter. Determine whether the currents are linear or exponential and explain why.                                                 2. What is the voltage of a boost converter�s output when it is placed in shutdown mode (transistor is turned off)?                                                                         Our Answer:                                                 1. When the boost converter's switch closes, the current through the inductor charges per the equation IL = 1/L ∫Vdt (assuming the switch resistance is small). Since the input voltage is a constant DC level, the inductor current ramps up linearly over time.                                                 When the switch opens, the inductor voltage reverses polarity (and boosts the output voltage in the process) in an attempt to keep the inductor current constant at the moment of switching. Because the inductor's voltage polarity has reversed, the inductor current must decrease over time, based on the formula above. You might think that this current ramps down in an exponential fashion, given that it discharges through the load resistor. But actually, because the output capacitor keeps the output voltage relatively constant, the output resistance is close to zero, so the discharge current, like the charge current, is linear.                                                 Note that because the current at the converter's input ramps up and down (instead of pulsing up and down sharply), filtering the input source is fairly easy.                                                 The current through the diode, I2, shoots up from zero when the switch opens. It quickly reaches the level of the inductor current and follows that current down until the switch closes, at which point it falls again to zero. Filtering the output of a boost converter is more difficult than filtering its input because of these fast rising and falling edges.                                                 See the application note: Circuit Tradeoffs Minimize Noise in Battery-Input Power Supplies.                                                                         2. Because of the DC path between the input voltage and the boost converter's output, the output voltage doesn't go to zero volts during shutdown. Instead, it hovers at a diode drop below the input voltage. Thus the circuit never goes into shutdown because current is continually drawn from the input voltage by the load. This is a problem when that input voltage is supplied by a battery.                                                 To fix this problem, you can add a MOSFET in series with the output. It switches off when the boost converter goes into shutdown. The figure below shows one such approach.                                                 Here the MAX810 power-on reset turns off the MOSFET when it detects the reduced output voltage that results when the boost converter goes into shutdown.                                                 See the application note: A Smart Solid-State Fuse Cures Boost-Converter Ills.                                                                         Additional Note: Strictly speaking, the current waveforms are truly linear only if you neglect parasitic circuit resistances and assume the output capacitor is infinite. In reality, the "linear" ramps are the beginning of an exponential, but with an asymptote so far away that you don't see any curvature.                                                 When you see a charging ramp with a slope that is not linear:                                                         If the slope decreases with time, it probably means there is parasitic resistance somewhere.                                If it increases with time, it means the inductor is running into saturation.