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Think about the Huge low ESR cap that you are charging .What limits the peak current in Q2 ? Also think about the discharge current peaks in Q1 .Is your hysteric current sensing fast enough on R1 ?


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The sensor you talk about is a simple current-controlled voltage source. You can place it by invoking >place>controlled sources>Current Controlled Voltage source: However, if it does the job of sensing the current, I prefer to resort to a current-controlled current source (press F) that I direct to a 1-ohm resistance. That way, I can easily add another ...


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I am also new to these things, so I thought my viewpoint would be helpful for others who are new. At a glance it looks like your buck converter is only constant-voltage, not constant-current. So the current from the battery rushing to the device can burn it out immediately, as it has no current regulation. I think this is what the other answer meant by ...


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Your power source can source (dc converter or the power supply feeding it) can supply a limited power (P=V*I). So in example if a power supply is set on 1W then at 10V it can source 100mA. If a 1k resistor is placed it consumes Pload = V²/R = 0.1W which is less than the power supply can provide and you get all the current you need for it. Now if you have a ...


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Buck/boost off of 3x AA's, or buck from 4x AA's. If you read the datasheet off of a common AA battery you will see that more than half of it's life is below 1.1V. If you have an LDO or a buck converter you simply will miss out on most of the battery life. Throw out the 9v idea unless you have extremely low power draw, 9v's are notorious for their internal ...


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The choice on Vin beeing 4.5 (3x1.5) or 9V mostly depends on how efficient your voltage converter is. If it only "burns" the over voltage one would go with the closest input voltage that is higher then the output voltage. Also have in mind that the "burning" regulator needs a voltage that is Vout + 1V to operate correctly (Might even be higher then 1V). If ...


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The MPPT converter will probably have a significantly higher power draw for its own circuitry that can waste more than any gains from power tracking - even the 5-10mA quiescent current of the LM2596 might be worse than the gains over a simple low quiescent LDO. Since in this case there is a significant voltage difference between your panel and the load (3....


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The phone can detect what the charger can deliver using sensing resistors placed on the charger port data (DP/DM) lines, and limit its current draw accordingly. Difficulty: there's different standards that are in use: USB, Sony, Apple. More about that here: https://www.maximintegrated.com/en/design/technical-documents/tutorials/5/5801.html


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Your idea won't work. The phone will require a certain minimum current - probably 500 mA, since this should be available on any USB port - and if the voltage collapses the charger will shut off, the voltage will rise back to 5 V and the cycle will repeat. I'd like to limit the current that the phone can draw to around 500 mA. I thought of adding a ...


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You're on the right track. Take the next step and consider using a 48V source, following emerging practice in data center technology. More here: https://blog.se.com/datacenter/2018/05/24/12v-vs-48v-the-rack-power-architecture-efficiency-calculator-illustrates-energy-savings-of-ocp-style-psus/ 48V ATX power units are available from various sources in the ...


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I'm inferring that LED stability is important for your application. The ripple will definitely affect the output with a noise spectrum of harmonics. My suggestion? Use both. Use DC-DC to make the primary voltage, followed by an LDO (perhaps wired as a current regulator?) to post-regulate the ripple.


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You could add a power lowpass filter to remove the fundamental and higher frequencies. There are a ton of examples and calculators if googled. The simplest may be a series resistor of a low value (maybe 1 Ohm) and a parallel capacitor of say, 1000µF. The resistor drops a little power while operating, but the cap conducts AC, causing the AC component (ripple) ...


2

The ripple on a buck converter will have a negligible effect on LED intensity, and will be far to high a frequency to be visible in any case. Note that such a converter could also be set up to operate as a constant current source (also with negligible, high frequency ripple), providing greater control of the intensity over all conditions.


3

As the name implies, a SMPS relies on the principle of transferring energy from input to the output cyclically, and this is accomplished via a switching device and a corresponding driver. Thus, there will always be high frequency components (main switching frequency + harmonics + higher order frequency components) at the output of your power supply, but it ...


2

To turn off that FET, the gate driver Vout needs to be up at 24 volts. This circuit does not provide that.


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