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Topic: Step Down Buck Converter with MP1584 (Read 5210 times) previous topic - next topic


What is the advantage of using a MP1584 compared to a LM2596?

When I look at a breakout board with a MP1584 it looks unfamiliar (no capacitor, no coil)

(1) the MP1584
(2) is that a coil (4,7┬ÁH)?
(3) are those capacitors?


I would assume #2 is the coil.   If you are curious you can trace-out the connections (an Ohmmeter can help), draw the schematic, and compare to the manufacturer's recommended circuit shown on the datasheet.  

The coil will measure a short (or nearly  zero Ohms) with an Ohmmeter.

When I look at a breakout board with a MP1584 it looks unfamiliar (no capacitor, no coil)
The black part above the IC that you didn't mark is probably the output capacitor.


May 04, 2015, 11:25 pm Last Edit: May 04, 2015, 11:47 pm by Wawa
The big square part is the inductor. The small black part is the schottky rectifier diode.
The brown parts are ceramic smoothing capacitors (~4.7-10uF?) across DC input and output.
Electrolytic caps are bad at the higher switching frequencies these supplies use (~1Mhz)
The general advantage of these micro supplies is better efficiency at low loads.


May 05, 2015, 09:15 am Last Edit: May 05, 2015, 09:15 am by Zealot
What exactly is this "switching frequency"?


May 05, 2015, 09:59 am Last Edit: May 05, 2015, 10:01 am by Wawa
A switchmode DC/DC converter uses a switch and inductor to convert from one voltage to another.
The switch (mosfet) works on a high frequency, so inductors and capacitors can be small.
First generation- and bigger supplies work on ~100Khz. Small newer ones upto ~2Mhz.
A switching supply is usually more efficient than a common lineair regulator.
e.g. to convert 12volt to 5volt/1A, ~7watt is lost in a lineair regulator, and ~1watt is lost in a switching regulator.


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