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I have very little knowledge about power draw / conversions etc. I would like to power a Kinect v2 Sensor from a USB Power bank. I saw the USB C and QC3.0 on this powerbank has between 9 and 12V output. I have read although the Kinect sensor is 12V 2.67A it only draws 16w per hour according to this user. So I would like to know can I splice up a USB-C cable (using red & black wires) to a wire that connects to my Kinect to Windows Adapter Any advice would be great thanks.

The goal is to have a mobile scanner with at least 3 to 4 hours battery life.

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    \$\begingroup\$ "watts per hour" is not a sensible unit. Also, implementing QC3.0 to get 12V from that charger is not trivial, to my knowledge. \$\endgroup\$
    – Hearth
    Commented Oct 24, 2019 at 17:37

3 Answers 3

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has between 9 and 12V output

It's not that easy. To get power from that bank, you need to speak the USB-C PD protocol. Otherwise, you'd only be getting "normal" VUSB of 5V at low current capability.

That's a pretty complicated protocol, and requires you to afterwards handle quite a bit of power, so that's why you'd typically do that by buying a special chip to talk to the power source.

Frankly, considering the level of understanding for electrical units, this is probably nothing you'll achieve on your own. You'd be better off just buying a cheaper 5V power bank and buying a (good) 5V -> 12V step up converter.

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    \$\begingroup\$ QC3.0 and USB-C PD are two different standards. \$\endgroup\$
    – Hearth
    Commented Oct 24, 2019 at 18:19
  • \$\begingroup\$ Thanks for the comments. So I have to connect something inline between the USB C output and the input on the Kinect cable?. Something like this? amazon.co.uk/LAOMAO-Step-up-Supply-Module-Adjustable/dp/… So I have to trick the battery to push out 12V \$\endgroup\$
    – Kevin T
    Commented Oct 24, 2019 at 18:44
  • \$\begingroup\$ @Hearth yes, and the product page mentions the latter in the description: USB Type-C PD Input &Output】The Bidirectional Type-C port with PD and QC 3.0 Quick Charge protocol both supports fast in and out. Up to 2A input and 3A output with USB Type-C port achieves 9-12V high voltage output that makes charging faster. Use your Original Type-C to Type-C cable to charge your New 2017/2016/2015 MacBook, Nintendo Switch.etc. \$\endgroup\$
    – Marcus Müller
    Commented Oct 24, 2019 at 18:45
  • \$\begingroup\$ @Hearth MaxBooks (I think) are PD devices, so I infered that while the statement is ambiguous w.r.t. QC, it's clear that PD for charging from the bank is implemented. \$\endgroup\$
    – Marcus Müller
    Commented Oct 24, 2019 at 18:46
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    \$\begingroup\$ @KevinT I'm honestly not a fan of Amazon-sourced electronic components - Amazon marketplace is prone to hosting scams, and the device you selected can not deliver the current you want at 12V, almost certainly. Get something with a datasheet. There's many cheap electronics distributors in the UK; try uk.farnell.com. \$\endgroup\$
    – Marcus Müller
    Commented Oct 24, 2019 at 18:51
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If the voltage needed is between 9V and 12V you could use an 11.1V RC car/plane battery, but the Kinect might need too much current and the battery would be low in a couple of hours.

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As my understanding power bank circuit uses a bi-directional buck-boost converter. When you connect for charging, it boosts the voltage to 12V and charge battery. When you connect mobile for charging, the circuit steps down the 12V into 5V. So you cannot draw 12V from power bank USB output.

What you can do is, you can solder wires from the internal battery which is not really recommended because you will break the Li-Ion protection mechanism and accidents might occur.

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    \$\begingroup\$ no, that contradicts the feature set that the powerbank has: fast out at 9–12 V, it says. \$\endgroup\$
    – Marcus Müller
    Commented Oct 24, 2019 at 17:52
  • \$\begingroup\$ Shuvangkar Shuvo - Welcome :-) Unfortunately there are a few errors in your answer. I'll highlight one. You said: "you cannot draw 12V from power bank USB output". Not true - on some power banks, you can. Notice that the question specifies a power bank with QC3.0. This Wikipedia article shows some basics (you can research more) about Quick Charge versions, including Quick Charge 3.0 and how the device and power source (the power bank in this case) can negotiate various voltages from a QC3.0 power source, including 12V. \$\endgroup\$
    – SamGibson ♦
    Commented Oct 24, 2019 at 18:24
  • \$\begingroup\$ @SamGibson, Thank for the correction. I experience the non-quick charging version of the power bank. The specification says "both the host providing power and the device must support it"- That means the device and host(power bank) should have communication for quick charge support confirmation. This implies that normally the USB provides 5V. Upon device request, it negotiates the charge voltage. \$\endgroup\$
    – Shuvangkar Das
    Commented Oct 24, 2019 at 19:30
  • \$\begingroup\$ @ShuvangkarShuvo - "Upon device request, it negotiates the charge voltage." Yes, that's what I said, the QC protocol involves a negotiation. The point is that the power bank specified in the question can output 12V if the correct QC negotiation occurs. Your answer states the opposite, without any caveats. If you meant to say: "you cannot draw 12V from [a normal, non-QC or similar] power bank USB output" then you would be correct - but that answer would not be helpful for this question, since it is a QC-compatible power bank here. Answers should apply to the specific question. Thanks. \$\endgroup\$
    – SamGibson ♦
    Commented Oct 24, 2019 at 20:16
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    \$\begingroup\$ @KevinT - "I can get an inline chip to negotiate a 12V output" Yes, boards (and ICs) to do that QC3.0 negotiation and switch the power bank to 12V output do exist. But, I'm not saying they will work with that power bank in your specific case - if I had the time to do all the research, I would have posted a well-researched answer :-) Specifically, can that power bank output 2.67A at 12V? I don't know. The Amazon page mentions 3A output, but doesn't specify the o/p voltage. Hence more research (or existing knowledge) is needed, but I spent my Stack Exchange time today on other questions. \$\endgroup\$
    – SamGibson ♦
    Commented Oct 24, 2019 at 21:26

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