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Does anyone know how much power out we can get from the USB port in a DSLR camera. Is it possible to get an output like 3.7v 1500mAh battery can supply from a DSLR?

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    The information which camera you are using might come handy. On the other hand, what are you trying to achieve? – Alexander von Wernherr May 24 at 5:33
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    mAh are a unit of energy, not power. – Peter Taylor May 24 at 6:29
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    So you're trying to build something like a Nikon WU-1a? That product's existence proves feasibility with some Nikon cameras, although it leaves the question of whether the lack of support for earlier Nikon DSLRs was a technical or a sales/marketing decision. – Peter Taylor May 24 at 7:18
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    vtc b/c This is a question about USB specs and capabilities that only incidentally involves camera equipment. – xiota May 24 at 7:28
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    This is an X→Y problem. The true question should be, "How can I power a device that requires 3.7V using the internal battery of any number of different DSLRs?" – Michael C May 24 at 11:44
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The cameras often have a "type-B" USB connector (the "peripheral" side) and these aren't supposed to provide current (they are the side that normally draws current).

On addition the batteries in a camera aren't that big, you couldn't draw significant current from them without impacting the battery life.

  • How is Nikon WU-1a powered? – xiota May 24 at 7:58
  • Some recent cameras (like the Nikon) have a mini-A/B so they can be both sides. But this isn't universal (yet). My 70D only has a mini-B (and so has the 80D it seems) but then they have WiFi built-in :) – xenoid May 24 at 8:04
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Does anyone know how much power out we can get from the USB port in a DSLR camera.

For the vast majority of DSLR cameras, the answer is: "None."

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    "Most cameras introduced in the last couple years" is still a pretty small minority of the vast majority of DSLR cameras produced, or even of DSLRs currently in service. – Michael C May 24 at 13:12
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    Absolutely. I added your point that when I moved my comment to its own answer. – scottbb May 24 at 13:25
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Most cameras have historically operated in USB Device mode (which can, but is not required to, receive power from the host), not USB Host mode (which can, but is not required to, supply power to the connected device).

There is a third mode, USB OTG (on-the-go), which was created to allow for devices to act as both hosts (to power peripherals) and devices (to act as a peripheral to hosts controlling them). USB OTG is not strictly a software feature — the USB controller (or in the case of system-on-a-chip, the SOC chip) must support that mode of operation.

If I had to guess, most cameras introduced in the last couple years probably have the ability (in the chipset) to operate in USB OTG (on-the-go), meaning they can be either a host or a device. But that doesn't mean that manufacturers expose that functionality, or even turn it on, for anything but 1st-party devices. And to emphasize, that would still be a tiny minority of the cameras currently owned and in use by photographers.

It may be possible to reverse-engineer the chipset to figure out how to enable it, if it is even present. But that is far beyond the scope of Photo-SE. The camera industry is particularly guarded with their specs and interfaces, rarely releasing such information to the public. Probably the most reverse-engineered DSLRs are Canons. You can check out the CHDK — Canon Hack Developer Kit project. They have the most knowledge and information about Canon internals.

  • Thanks for the detailed explanation scottbb. :) – Tharanga May 24 at 15:30

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