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This question already has an answer here:

I have a DC voltage source (between 5 and 10V) from which I want to charge cell phone with a USB cable. I believe that using a LM317 or similar IC for charging cell phones is not a good idea as most cell phones are coming with Lithium-ion batteries, which requires constant current phase first, then constant voltage phase. I have seen MAX1555 IC but it gives only 4.2 V as regulated output. If I connect a VBUS (usb cable) from MAX155 chip to iPhone, my iPhone may not charge because it's less than 5 V.

Can anyone give some insight?

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marked as duplicate by Ignacio Vazquez-Abrams, PeterJ, JRE, Daniel Grillo, placeholder Apr 23 '16 at 15:36

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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You will be charging the phone through its USB port which takes 5 V from the USB host. The battery charger and logic is built into the phone.

Make a 5 V supply and leave the figuring out of the charge to the phone.

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  • \$\begingroup\$ Thank you. It's very easy then. Another question. When phone is fully charged, the current drawn from the LM317 is closed to zero? I want to tun off the power to the LM317 when phone is fully charged. Is there anyway that I can figure out the phone is charged fully? \$\endgroup\$ – user72035 Apr 22 '16 at 20:01
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You want to charge the cell phone through USB, right ? You don't charge the battery of the phone directly (taking the battery out of the phone) ? So you don't have to care about the constant current thing. This is taking care of by the circuitry within the phone. Just provide 5V to the USB cable, that's all.

A LM317 would do. Or a 7805, or any regulator, as long as it can provide the appropriate current.

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  • \$\begingroup\$ Thank you. It's very easy then. Another question. When phone is fully charged, the current drawn from the LM317 is closed to zero? I want to tun off the power to the LM317 when phone is fully charged. Is there anyway that I can figure out the phone is charged fully? \$\endgroup\$ – user72035 Apr 22 '16 at 20:00
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    \$\begingroup\$ Yes, the current will be close to zero. If you want to know that, you can add a small resistor before the input of the regulator and check that there is not voltage drop at this point. But I think this would deserve its own question if you need details. \$\endgroup\$ – dim Apr 22 '16 at 20:05
  • \$\begingroup\$ Thanks again. Any idea how much current would LM317 draw from input when phone is fully charged? I did not find input quiescent current of LM317 when no load is connected. \$\endgroup\$ – user72035 Apr 22 '16 at 21:02
  • \$\begingroup\$ Quiescent current for LM317 is the current through the ADJ pin (specified in the datasheet), so <100µA \$\endgroup\$ – dim Apr 22 '16 at 21:07
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If you use lm317, set its output to 5.5v, high but safe voltage for phone. This reduces voltage drop across the linear regulator and the associated power loss.

Most phones draw close to 1 amp during charging, thus even a 3 volt drop across the regulator makes it considerably hot if no heatsink is used.

After the phone is fully charged it will still draw fluctuating current if the phone is connected to the mobile network. So, it won't be anywhere near zero.

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