Wired power from old 26V AC system (solved)

[stevemur]

Posting this here just in case others have the same issue. Please note that this is NOT an officially sanctioned solution by Arlo and may void your warranty (and a lawyer would probably say there's even some fire risk involved if the buck converter fails), but it's working well for me here and I thought I'd help others out. 

I'm a happy new owner of an Arlo Pro 2 system. I've fully replaced a low-fidelity, wired analog closed circuit system that was wired into our house. Arlo adds numerous improvements, including a much better remote app viewing experience, higher fidelity in both daylight and nighttime, motion detection and more. 

The problem: I don't love relying upon the camera's battery option; it's a hassle to keep replacing these batteries. I also hate the idea of dangling wires from these cameras to the nearest outlet.

I really wanted to make use of the EXISTING power that those analog cameras had, and in checking the transformer, I noticed it was sending 26V AC to power these cameras via a twisted pair line. 

What I ended up doing was installing a 24V AC to 5V DC buck converter at the endpoint of these cameras, soldering in a female USB A connector. Works!

The buck converter I used is this one:

https://www.amazon.com/SMAKN%C2%AE-Converter-Voltage-Supply-Waterproof/dp/B00RE6QN4U

The female USB A connectors are these:

https://www.sparkfun.com/products/9011

(note you can buy them for much less in bulk)

The pinout diagram of USB A can be found with a simple Google -- basically the rightmost pin is 5V+ and the leftmost pin is Ground. The center two pins are for data and are not required. 

Be sure to check each connection with a multimeter before connecting in the camera. 

In my particular case, I let it run overnight and checked the heat dissipation before installation, and I've wrapped fireproof insulation around these devices just in case. But the devices are not even warm to the touch. 

DISCLAIMER: Your decision is your own. Simply posting this here in hopes that it might be helpful to others who are comfortable with basic electronics and soldering. 

[AncientGeek]

Fingers crossed.  The USB spec for PD cables is over 600 pages long with a crap load of protocols and handshaking rules.  Hopefully, the Arlo cameras will just accept your analog power and run with it.

[stevemur]

I didn’t wire in either of the data pins, as they aren’t used in this scenario. Data travels wirelessly from the camera to the base station. The vast majority of the USB spec is about the data format over the wire, which is unused.

[stevemur]

To be clear, I leave the battery in the camera, and the 5V DC trickle is enough to keep the battery charged and the camera powered.

[AncientGeek]

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@stevemur wrote:
I didn’t wire in either of the data pins, as they aren’t used in this scenario. Data travels wirelessly from the camera to the base station. The vast majority of the USB spec is about the data format over the wire, which is unused.

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I am referencing purely the USB “power delivery” (PD) spec which is 547 pages long.  Not the data specs.  Power delivery is supposed to be negotiated and I assume there are safety protocols included for disconnects.  If it works for you, that is great.  But USB isn’t really an analog power bus from what I’ve researched.

[stevemur]

Another entirely different suggestion for people reading this post:

Find the transformer which is supplying power to your old AC 24V cameras. Unplug it. Disconnect the cameras. Mount 5V DC power at the central source.

However, there will be voltage drop based on the resistance of your wires and the distance (Google some voltage drop calculators to find out just how much.) I didn’t try this method but it should be possible. Always test with a multimeter before connecting the cameras. I’ve read that the tolerance on acceptable voltage is quite narrow.

Feel free to chime in on this thread with your results.

[Dannybear]

@stevemur,
I think your suggestion to minimise the number of 5vdc supplies with this method would work if the cables to each camera are short and feed directly to each camera from the supply as you state. I can see an issue if cameras are daisy chained along the cable as the voltage drop would be incremental and likely to be under voltage eventually. If you were to compensate by increasing the supply and then you loose a camera on the chain it could cause over voltage issues to other cameras.

[AncientGeek]

The best answer is an Arlo approved POE adapter.  Which doesn’t exist other than for the Q Plus.  POE uses 48 volts.  Higher voltages tend to experience lower loss over distance, so dropping to a lower voltage at the end of the line is best.  That’s one reason electric power is distributed in this manner.  

The native Arlo power adapters provide dual voltage.  5V and 9V DC.  Elsewhere in this community others have said the cameras will run on 5V, but use 9V for charging.  I haven’t read the Arlo specs, so I don’t know firsthand.

Of course, you also need the correct amperage at the end of each line as well.  I can’t see daisy chaining ever being successful unless the wire was of suffice to gauge to carry “n” multiples of the amperage needed to power each of the “n” devices on the circuit plus the loss from the beginning to the end.  Usually these small gauge cables are sized to carry the load for a single device at the end of the cable over a specified maximum distance.

[Dannybear]

To be accurate with determining the voltage drop over a long length of cable you would additionally need to know the current draw of the camera in amperes (Amps). As the camera’s operation will vary, so will the current required and hence the voltage drop will also. This could mean the camera works ok during the day but dies when the IR leds turn on at night as the voltage drops below the camera’s required specifications. As ancientgeek has recommended, use the individual 5VDC converter as close to the camera end of the cable as possible.

[AncientGeek]

https://www.calculator.net/voltage-drop-calculator.html?material=copper&wiresize=52.96&voltage=5&pha...

You can enter the specifics here and see the loss over distance.