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I've bought a bunch of dlink wifi cameras for my home. These all come with 5V wall wart psus connecting to the camera using what appears to be a simple 2 core cable terminated in a micro USB socket.

There are, of course, no power outlets close to where I would like to position the cameras. So I have to extend the cables and, rather than terminating each in a separate plug, I was wondering if I could simply run a 5V pair supplied by a beefier psu, possibly with battery backup, around the house and hang all the cameras in parallel off that.

iirc though, proper usb devices do some kind of power negotiation at start up, which, presumably, would prevent them being used in parallel. How might I determine if these cameras are simply using the microUSB as a convenient small physical plug, or if there's anything special going on?

This article (https://www.medo64.com/2016/11/100-ma-is-a-myth/) suggests I might be able to just ignore the whole thing anyway. Is that realistic?

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    \$\begingroup\$ If the cables from the supplied PSUs only have two cores, then there is no way for the devices to negotiate power with the supply/host and will draw as much current as they need and hope the host can supply it. You can connect them in parallel, but leave the data pins unconnected. \$\endgroup\$ – Oskar Skog May 23 at 9:56
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    \$\begingroup\$ Additionally to Oskar's answer: Make sure your cables are thick enough to avoid voltage drop over the cables. \$\endgroup\$ – Bora May 23 at 10:07
  • \$\begingroup\$ fwiw - MA is definitely a myth. mA usually is:-) \$\endgroup\$ – Russell McMahon May 23 at 13:45
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It is highly likely that you can use another 5V power supply.** - either for individual cameras or for a group of cameras.

  • Using a 2 wire power pair only - while it would be possible to customise a device/supply pair so that they would only work together* this would be extremely unusual, and D -Link are not known for making it hard to use their equipment in this manner (unlike some other suppliers).
    [* eg the device could signal the supply by modulating the current drawn - and/or the supply could wake up the device by voltage signalling. Doable but extremely unlikely. As two examples only: Apple go out of their way to make it hard for other people to interface with their products, and some Dell systems do. ]

What is the camera model?
What is the camera current draw mean/peak?

If operating over a significant distance, voltage drop can be an issue, especially at low voltages and higher currents. Several means of addressing this exist.

  • As Boa suggested, use adequately thick wires.
    "How thick" is needed depends on current draw and circuit length.

  • Using a supply voltage at the upper tolerable limit for the device will allow close devices to "survive" the voltage and allow greater distance to the end device before voltage is too low.

  • Adding a capacitor at the feed point for each camera will almost certainly allow lower voltage operation that if not used. A 100 uF electrolytic at say 10V would help and larger or much larger would not be amiss.
    A good quality 100 uF 10V aluminium electrolytic capacitor can be bought for $US0.13 each in 10 quantity
    and a [1000 uF 10V capacitor(https://www.digikey.com/product-detail/en/panasonic-electronic-components/ECA-1AM102/P5127-ND/244986) for $US0.30 in 10 quantity

  • Adding a regulator at each camera or close proximity group of cameras and using a somewhat higher voltage power supply allows for voltage drop in the cable.
    Regulators are not complex but do need a small degree of 'design'.
    Some regulators have more demanding input and output capacitor requirements than others.
    Unless specifically intended as "low drop out regulators" a regulator will often drop typically 1.5 to 3 volts across the regulator in normal operation. In this application you may thus need 5V supply + say 2V wire drop + 2V say regulator "dropout voltage" = 9V or more feed voltage. An LDO (low dropout" regulator will have well under 1 V dropout voltage in normal operation - under 0.1V in some devices.

    As an example only - the LD1117 has aa dropout voltage of up to 1 V, max current of 800 mA, and costs around $US0.75 in 10 quantity.


Fire risk:

Note that at the currents and voltages concerned fire is not a major risk if sensible construction methods are used - but also not a nonexistent one. At say 10V feed x 5A supply you have 50 Watts available. You can definitely set things alight at that power level if you are clever enough to short the supply in a close-to-supply portion of the circuit. Common sense should avoid this.

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  • \$\begingroup\$ Thanks Russell for that most comprehensive answer! May I ask how a capacitor at the feed point help? \$\endgroup\$ – Ian May 24 at 14:13
  • \$\begingroup\$ @Ian Devices MAY draw a constant current but it is usual for current drain to vary - perhaps due to changes in focus or aperture or ... . As the feed circuit has resistance which causes voltage drop (V=IR) and as their are multiple devices on the same power circuit, changes of current drain for any one will cause voltage fluctuations for all. For thos at the far end of the circuit this is mre likely to cause them to malfunction due to low voltage. Such malfunctions will first start to occur on current peaks - the average current may be adequate but the peak may drop voltage too low. ... \$\endgroup\$ – Russell McMahon May 24 at 14:52
  • \$\begingroup\$ A capacitor acts as a local reservoir, supplying current when the circuit cannot - for short periods. | A second closely related effect is noise passed along the circuit. This may come from other devices or from induced signals from radio equipment machinery, ... . A capacitor close to a device offers lower impedance than when further away (less resistance and inductance in the intervening circuit. Proximity to the device both absorbs noise it generates better abd filters out noise from other devices better. \$\endgroup\$ – Russell McMahon May 24 at 14:55

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