sorry for the silly question but what adapter can be used to plug the rb5009 using the 2-pin terminal? are ther any simple dc adpater pluggable to an outlet available?
Pretty much any 12-57VDC power supply can be used.
Direct connections like that one are commonly used in telco and ISP installations, such as in a central office, carrier hut, or outdoor cabinet, where the equipment is powered by batteries via some means of DC distribution, like a fuse or breaker panel.
On the bench, you might hook it up to a variable power supply while testing.
In a home or small lab, you might have two normal power supplies (like those that come with it) or a combination of a power brick and a battery.
OK I see. Thanks for the answer! now i will have to look to a way connect when we have four of themn in 1 u . i wish people from mokkrotik would have put more place for it. and an easy way unplug. the raxk tool could be improved.
Well I am more looking for DC adapter or multi-dc outlets if it exists. A multi-dc outlet would allow me to plug it to the UPS easily.
I used this 48V 3A power supply: https://www.amazon.com/dp/B07G21P3KD
Coupled with these DC splitters: https://www.amazon.com/dp/B07J28SX6Y
The power supply goes into one splitter, which feeds to two more splitters, and each splitter goes to a pair of RB5009’s in the rack (two on the left, two on the right).
I also use a CRS328 that provides redundant power to the 5009’s, but it means the POE-in port only goes up to 1Gbps instead of 2.5Gbps.
I have gotten 2.5Gbps links to work through Tycon POE-INJ-1000-WT’s https://www.tyconsystems.com/poe-inj-1000-wt, which would work with the RB5009’s POE-in port.
Thanks for the links this is very helpful! Interresting you’re using the CRS328. You’re only using to power or also to extend the network ? Is it powering the 4 rb5009? What’s the schema.
The CRS328 provides power to a number of radios, routers, and switches at the site. For some of them, it’s only providing power (or redundant power) and the Ethernet port is disabled. On the RB5009’s it’s because I’m using the 10G ports for data instead.
You can also get those as 4+ ports.
But the four way splitter can’t reach all four when racked up using their rack kit.
one thing annoying using the 4 rb5009 in 1u design is the position of the 2-pin input. There isn’t a lot of room to put a cable for the one at the bottom right (front view)… Would have been better to have them in front…
I can’t stand those bricks so picked up a Mean Well DIN power supply (HDR-30-24) and a rack mountable DIN rail (WatchfulEyE DIN Rail).
Taking the PoE path means only the power delivery device needs UPS power.
A nice bonus falls out of this if you were using the fiber port on the 5009 as the LAN core uplink. While you might be tempted to reject this idea since you can’t do PoE over fiber, nothing stops you from doing both copper and fiber. Simply set the RSTP priority value for the copper link high (confusingly meaning low-priority) so it’s mainly used for power delivery, then set the fiber port’s priority to a lower value so it’s always the main link while it’s up.
In addition to centralizing your power distribution, this scheme means you get backup links between the core and the routers, and best of all, you’re likely to wire it in such a way that if the fiber link goes down, the same cause won’t take down the copper link, because they’re presumably taking different paths.
The 2-pin terminal on the RB5009 can be powered using a variety of different power sources, including a DC power supply, a battery, or a solar panel. It is important to make sure that the power source you choose is capable of providing the correct voltage and current to power the device properly. Additionally, you should be sure to read the documentation for the RB5009 to ensure that the power source you choose is compatible with the device.
this is true. Though it would let so much ports unused if I used the 24 ports POE? Especially if use the POE-OUT version of the rb5009…
I was thinking to use such things like 2 Ubiquity Edgepower with power failover or something equivalent wth the COSEL KRE-20A and the Mean Well DDR-240 to split in 4 feeds. But It seems that using 4 routers imply to either use the DC jack or POE. I don’t see an easy way to plug electryicity in such case with the 2-pin terminal?
There are 8-port PoE-out switches in the MikroTik line. The newest is the CSS610-8P, though you’ll lose RouterOS support by going that way. The CRS112-8P solves that, but it’s much older tech.
probably a stupid questio, but what DC output would be more efficient:
- 48 VDC, current range ~0-5A, peak 7.5A
- 24 VDC, current range ~0-10A, peak 15A
I think the first one would be more efficient? I have 54 VDC on input and max 150W (or 300W it depends) . Also what cable diameter did you use?
For longer power supply wires go for higher voltage. It decreases current and it’s current times wire resistance equals power loss.
For shorter runs, go for lower voltage. Supply wire power losses will be negligible. However: modern electronic devices require relatively low voltages for actual operation (usually 5V or less) and DC-DC downconverters tend to have higher conversion losses when voltage difference is higher. Which means down conversion losses from 12V will be lower than losses from 57V.
Source of wisdom written above.
All else being equal, yes.
it’s current times wire resistance equals power loss.
Sorry, but you’ve jumbled several concepts. Current (I) times wire resistance (R) equals voltage loss (V), not power loss (P). V=IR.
Rearranged to solve for I — I=V/R — that equation says that if you change the voltage (e.g. the 24 to 48 in the question up-thread) across a fixed resistance, you know the resulting difference in current.
You then plug that result into a different equation — P=I²R — to get the power loss from the lower current across that fixed resistance.
A much simpler way of saying all of this is that, for a fixed resistance, power loss goes up as the voltage loss squared. That formulation doesn’t get you proper units, for which you need the equations, but for questions like this, it tells you what you want to know: doubling the voltage will cut the power loss to ¼ its prior value, all else being equal.
DC-DC downconverters tend to have higher conversion losses when voltage difference is higher.
I’d need to see data on that.
A more reliable relationship is that converter efficiency has an “umbrella” shape, low at low currents, low at really high currents, and peaking at some ideal load current in the middle.
The application to this question is that it’s possible for the lower-voltage power supply to be more efficient overall because it’s operated in its sweet spot. You’d have to dig into the data sheet curves to find out. The equations above only talk about ideal physics, not practical engineering applications.
Source of wisdom written above.
Try this one, written by yours truly.
Apart from nitpicking about exact power loss formula (you’re right, power loss is not linearly proportional to current, it’s proportional to square of current … but my point was intended to be that voltage loss is proportional to line resistance, I typed wrong word)…
The article I linked at the end of my previous post includes some very informative tables and charts. So if you disagree with those, please write what you consider wrong. Your article doesn’t even touch efficiency of DC-DC converters (which are, as you surely very well know, a completely different beast than AC-AC transformers) while the article I linked is all about DC-DC converters.