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I'm going to call the lower input voltage range but less efficiency circuit, circuit A. I'm going to call the more efficient configuration, circuit B. Note that both circuits are actually the same, only some component values are different: The inductor L1: 0.33 uH (circuit A) vs 6.8 uH (circuit B) The RC network connected to the Vc pin These two changes ...


2

When in doubt, read the datasheet, apply some circuit design knowledge, and maybe go dig up some applications notes. The SG3525 has become a standard part, so there may even be books written around it. If you look in the datasheet, you'll see that the error amplifier gain is stated as a transconductance, with a range of \$1.1\mathrm{mS} < G_m < 1.5\...


1

The "typical" 0.1µF capacitor you refer to is used to 'kill' very short pulses (in the oreder of nanoseconds) of low current (in the order of tens of milliamperes), as they occur when a logic IC switches its output state. The LM2596 is a step-down converter. All it does is generating one current pulse after the other, but the operating frequency is around ...


1

According to datasheet section 9.1.1 about input capacitor, it prevents large voltage transients at the input, and provides instantaneous current for the switching.


1

You you can use a ultra low power voltage monitoring IC. This IC or similar IC will keep the MCU at reset state even down till 0.7 V of \$V_{IN}\$ . As soon as the Voltage drops below the internal reference, the MCU will be held under reset. This will make sure that MCU will work normally when the input voltage is back in its valid range.


1

A larger step up ratio is less efficient because the power stage has to handle much more current through the inductor and mosfet. This results in higher switching losses in the mosfet and higher conduction losses in the mosfet and inductor. At light loads, the quiescent current of the regulator becomes significant. So efficiency suffers. This does not ...


1

First, the filters have two different purposes, unless this were a true two-wire device, and the AOZ1284 is not. An input filter provides a smoothed voltage source to the buck converter. and it prevents high-frequency harmonics (RF noise) from getting into the supply line. An output filter eliminates ripple and noise from the output line. Second, the ...


1

Idea My current approach is as follows: Here when the current exceeds the limit set by the variable resistor (the one between Vref2 and GND), the output of OPAMP will clamp to ground, so the led will illuminate which in turn will decrease the Vout, thus the output current. Solution I've successfully implemented the following solution:


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