6V-20V to 12V Step Up Down Converter Boost Buck Voltage Regulator Module for Car Screen, Monitor Camera, Fan, Water Pump, Motor, Router, etc(2A)

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As told at the beginning of this section, the converter operates in discontinuous mode when low current is drawn by the load, and in continuous mode at higher load current levels.

The 50A Buck-Boost DC-DC Converter is a DC-DC converter for charging a 12/24V service battery in vehicles with an intelligent dynamo. Switching converters (such as buck converters) provide much greater power efficiency as DC-to-DC converters than linear regulators, which are simpler circuits that dissipate power as heat, but do not step up output current. The term T V i L {\displaystyle {\frac {TV_{\text{i}}}{L}}} is equal to the maximum increase of the inductor current during a cycle; i. Where I L ¯ {\displaystyle {\overline {I_{\text{L}}}}} is the average value of the inductor current. while in the Off-state, the inductor is connected to the output load and capacitor, so energy is transferred from L to C and R.

Disconnect the source and use the inertia of the current in the inductor to provide more current than the source delivers ("off" in fig.

The only difference in the principle described above is that the inductor is completely discharged at the end of the commutation cycle (see waveforms in figure 4). If the switch is opened while the current is still changing, then there will always be a voltage drop across the inductor, so the net voltage at the load will always be less than the input voltage source. That means that the current I L {\displaystyle I_{\text{L}}} is the same at t = 0 {\displaystyle t=0} and at t = T {\displaystyle t=T} (figure 4). If the switch is closed again before the inductor fully discharges (on-state), the voltage at the load will always be greater than zero.As can be seen in figure 4, t on = D T {\displaystyle t_{\text{on}}=DT} and t off = ( 1 − D ) T {\displaystyle t_{\text{off}}=(1-D)T} . However, since the inductor doesn't allow rapid current change, it will initially keep the current low by dropping most of the voltage provided by the source. the current at the limit between continuous and discontinuous mode is I o lim = V i T 2 L D ( 1 − D ) = I o lim 2 | I o | D ( 1 − D ) {\displaystyle \scriptstyle I_{o_{\text{lim}}}={\frac {V_{i}\,T}{2L}}D\left(1-D\right)={\frac {I_{o_{\text{lim}}}}{2\left|I_{o}\right|}}D\left(1-D\right)} .