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I'm trying to use a DC-DC converter to charge a mobile phone.

This is the converter....TRACO 12v-5v converter

The converter is relatively low power, (5w) and can supply a maximum of 1amp. Testing with both a USB voltmeter and a multimeter indicates that the device is correctly wired and producing 5v.

However, when I plug in my iPhone (or tablet or any other mobile device) there's no sign of any actually charging taking place. (I would normally expect to see a little lightening strike zig-zag symbol appear on the battery)

Is this because there is a minimum current requirement to power the mobile device? Is the device likely to be not charging at all? or is it probably charging ok, but at a very slow rate?

I know that if I were charging a lead-acid battery at a very low current then the battery would still charge, albeit very slowly.

If there is a minimum current requirement for my mobile device, what's it likely to be? I know I can buy higher current 5v supplies, but would prefer to know exactly what I need before hand rather than just buying DC-DC converters on a trial and error basis.

UPDATE - I have left the 'data +' and 'data -' unconnected, and simply connected ground to ground, and Vcc to 5v. My understanding (perhaps wrong) is that a USB device should be able to charge itself with this configuration. It's only if attempting to fast-charge a device 'data +' and 'data -' need to be connected to anything.

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  • \$\begingroup\$ How does your charger connector look like? \$\endgroup\$
    – Huisman
    May 4, 2019 at 8:57
  • \$\begingroup\$ It's a USB 'A' connector \$\endgroup\$ May 4, 2019 at 9:02
  • \$\begingroup\$ Did you check this USB A receptable with another stripped USB A plug if the 5 volt is still there? \$\endgroup\$
    – Huisman
    May 4, 2019 at 9:12
  • \$\begingroup\$ Yes - I've measured the voltage on the red/black wires on a stripped down USB lead, I've also used a USB voltage tester \$\endgroup\$ May 4, 2019 at 9:37
  • \$\begingroup\$ you need resistors on the USB fro the phone to recognise it as a charger \$\endgroup\$ May 4, 2019 at 10:14

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