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I'm using an LM2596 buck converter set to 5.0V to power a bunch of stuff among which is a smart phone.

My total consumption is around 1A excluding the phone. I'm not really sure what will happen if the phone tries to draw more than 1A, I suppose the voltage will collapse and nothing will get damaged but I'd rather not take any chances especially in today's circumstances.

I'd like to limit the current that the phone can draw to around 500mA. I thought of adding a resistor in series but I'm not sure how would that work as it will create a voltage drop as well as that's 2.5W it will probably create a lot of heat.

What would be an acceptable way to limit the current?

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    \$\begingroup\$ the proper way to limit the current drawn by the phone would be to plug the phone into a different charger \$\endgroup\$
    – jsotola
    Mar 21, 2020 at 21:44
  • \$\begingroup\$ @jsotola sadly that's the only 5V power source I have and I have to manage - paraphrased. \$\endgroup\$
    – php_nub_qq
    Mar 21, 2020 at 21:45
  • \$\begingroup\$ There's two things we like to see on SE: having self-researched the problem first, and having a problem with more than $10 at stake, i.e. that isn't trivial. You have a $5 answer: go down to the local gas station and buy a USB charging block. Mains or 12V they have both. Another $5 solution is a second LM 2596 module fed off the same source as the first. Your "resistor" proposal makes me doubt what research you have done. Hint: USB has a protocol for negotiating current draw limits. \$\endgroup\$ Mar 21, 2020 at 21:51
  • \$\begingroup\$ learn.adafruit.com/minty-boost/icharging?embeds=allow \$\endgroup\$ Mar 21, 2020 at 22:45

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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 resistor in series but I'm not sure how would that work as it will create a voltage drop as well as that's 2.5W it will probably create a lot of heat.

You're not thinking this through correctly. The only way you can dissipate 2.5 W in the series resistor with 0.5 A is by having a voltage drop across it, \$ V_R = \frac P I = \frac {2.5}{0.5} = 5 \ \text V \$ which leaves you with 0 V for the phone. Obviously this isn't going to work!

What would be an acceptable way to limit the current?

The phone should do this itself if it is unable to negotiate a higher current from the charger. Some phones look for certain voltages on the D+ and D- lines to indicate the capability of the power source. These may or may not be present on your PSU.

I'd be surprised if you don't have a USB socket somewhere around you - laptop, back of TV, hi-fi, etc. - that will allow you to keep plug in your devices separately.


From the comments:

I also thought about using a transistor for example with a gain of 100, apply 5 mA to the base, get 500 mA on the collector, would that work out?

Nope. Any current limiter will cause voltage drop when the demand exceeds the current limit. You can have constant voltage (which is what the phone expects) or constant current but not both.

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  • \$\begingroup\$ It's a mobile setup, and the phone is used as a display screen. There is no data lines, I've soldered the power lines of a male micro USB directly to the converter's output. I can't use a different charger because there is logic that determines when the converter gets powered. \$\endgroup\$
    – php_nub_qq
    Mar 21, 2020 at 22:03
  • \$\begingroup\$ I also thought about using a transistor for example with a gain of 100, apply 5mA to the base, get 500mA on the collector, would that work out? \$\endgroup\$
    – php_nub_qq
    Mar 21, 2020 at 22:16
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    \$\begingroup\$ See the update. \$\endgroup\$
    – Transistor
    Mar 21, 2020 at 22:19
  • \$\begingroup\$ But I'm a little confused. This will probably hurt your brain as it's most likely taught in electronics 101 yet I still can't have a clear understanding in my head. Say we start at 5.5V with a series resistor of 1 ohm, that will leave 5V for the phone and limit to 500mA. However if the phone tries to draw more the voltage will collapse, right? Why is that not happening with, say, diodes? If you increase the resistance the current decreases no matter how much current the diode wants? And yes I mean LED. \$\endgroup\$
    – php_nub_qq
    Mar 21, 2020 at 22:33
  • \$\begingroup\$ Make that a new question and we'll clear that up separately. \$\endgroup\$
    – Transistor
    Mar 21, 2020 at 22:37
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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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