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对于BOOST同样的功率下,DCM的优点电感电流纹波更小,可以用相对更小的电感。但由于每个周期电感电流都不会降到零,所以在下管二极管都会导通,SW的开关波形会有震荡,需要考虑EMI以及SW过压等问题。同时如果需要设计补偿的话,DCM会相对简单些。
芯片有TPS40210,TPS61040等。
Hi shibin
A detailed description of differences between CCM and DCM for you.
Discontinuous conduction can result in a decrease the size of the inductor. This is a desirable result in many designs. The inductor is selected based on L = V(dt/dI). From this equation it can be seen that larger the change in current (dt/dI) the smaller the size of the inductor. The current required to keep an inductor continuous is the ripple current dI divided by 2. So as you decrease L, dI will increase thus causing a higher minimum current to keep the inductor continuous.
Another benefit is when voltage mode control is used, the control to output (Vout/Vc) transfer function will have a single pole response. This can make compensating the feedback loop easier. There is a detailed explanation of this that can be found in Unitrode seminar book SEM-300 topic 2 Closing the Feedback Loop by Lloyd Dixon.
Some drawbacks of discontinuous mode DCM: Power supplies designed with DCM have a tendency to be noisier than power supplies designed for continues conduction mode (CCM). This is because when the inductor swings to zero current it has a tendency to ring with parasitic capacitance. Also the peak to peak current is higher, resulting in a higher RMS content, and will require a lareger EMI filter and may influence sensitive areas of the design.
The higher peak to peak current can also increase the losses, since many losses are resistive and are calculated by Power = Isquared x R, so the losses increase with the square of the current. Higher peak to peak currents also can result in larger output ripple currents than designs that use continuous conduction.
To maintain the same output voltage ripple, the design that uses DCM will require a much larger output capacitance with lower equivalent series resistance (ESR) as compared to the CCM design. The lower ESR and larger capacitance is required to suppress the larger ripple current.