Hi,

I'll be honest - I have no clue what I'm doing when it comes to calculating voltages, current etc. I'm looking to power a POE hat for a Raspberry Pi with a POE injector but don't know which specs I need for the injector.

The POE hat accepts 37-57V DC and outputs 5V/2.5A DC. I don't know how many amps it accepts.
The POE injector I'm looking at provides 56V/1.1A DC.

I'm thinking that as the current provided by the injector is only 1.1A and the POE hat outputs 2.5A, the injector isn't providing nearly enough current, but for all I know there could be some conversion possible which ramps up the current by sacrificing voltage.

Any help would be appreciated because I'm utterly clueless!

 

Remember that power equals Volts time Amps (P = V * A) and that is what you work with here.

The POE Hat has a switching power supply which should have an efficiency of at least 80%.
For the maximum 5V/2.5A output (12.5W) the input current for the 56V from the POE injector would be 12.5W/56V * 1/80% = .279A.

Thus the POE injector would be only operating at only about 25% of its 1.1A capacity, max.

Alternately you could look at just power.
The input power to the POE would be output power divided by the efficiency, or 12.5w/0.8 = 15.6W.
The maximum output power of the injector is 56V * 1.1A = 61.6W.
The injector is thus operating at 15.6W/61.6W *100 =25.4% of its maximum (same as calculated above).

 

Not being familiar with the Raspberry Pi I'm not sure what you're asking but it sound like some sort of DC-DC power converter is involved?

If the input is 56V@1.1A, with approx 80% efficiency, then that would give approximately 5V@7A, more than enough?

I'm sure you'll get a more definitive answer.

 

Remember that power equals Volts time Amps (P = V * A) so that is what you work with.

The POE Hat has a switching power supply which should have an efficiency of at least 80%.
For the maximum 5V/2.5A output (12.5W) the input current for the 56V from the POE injector would be 12.5W/56V * 1/80% = .279A.

Thus the POE injector would be only operating at only about 25% of its 1.1A capacity, max.

Thanks for the quick reply crutschow. Would you mind verifying that my logic here is correct?

Let's say I'm looking to use the Cat5 cable at https://docs-emea.rs-online.com/webdocs/13eb/0900766b813eb191.pdf.

10 mΩ/m is the transfer impedance per length at 10MHz. I'm not sure where the frequency value comes from in my situation.
Assuming it's 10MHz, the resistance over 20m, for example, would be 0.02Ω.

Ω = V / A - now I'm stuck. I'm not sure how I'd calculate the voltage and current drop over a distance of 20m from here because both the current and voltage will drop.

Ideally, I'd like to know roughly the maximum length of the cable before I need to look at a POE injector with a greater current/voltage output.

 

Not being familiar with the Raspberry Pi I'm not sure what you're asking but it sound like some sort of DC-DC power converter is involved?

If the input is 56V@1.1A, with approx 80% efficiency, then that would give approximately 5V@7A, more than enough?

I'm sure you'll get a more definitive answer.

I'm focusing on powering the POE hat using a POE injector, not the pi with the hat. The hat is already compatible with the pi and is directly mounted so I have no concerns about its ability to provide enough voltage/current. I just need to make sure that it receives enough from the POE injector.

 

10 mΩ/m is the transfer impedance per length at 10MHz. I'm not sure where the frequency value comes from in my situation.
Assuming it's 10MHz, the resistance over 20m, for example, would be 0.02Ω.

You want the DC resistance of the cable, not the AC impedance (which can have a higher resistance due to the skin effect).

Ω = V / A - now I'm stuck. I'm not sure how I'd calculate the voltage and current drop over a distance of 20m from here because both the current and voltage will drop.

Yes, they both change, but not enough to cause a significant error if you assume the current doesn't change significantly.
(Actually the current would increase slightly to maintain the same power input to the hat).

So, if the cable resistance is 0.02Ω, then for the .279A of current I calculated for the hat input, the voltage drop would be .279A * .02Ω = 5.6mV, which is negligible.
(Actually it may be double that if the wire resistance is just for one wire and doesn't include the return wire, but that would still be negligible).

 

You want the DC resistance of the cable, not the AC impedance (which can have a higher resistance due to the skin effect).
Yes, they both change, but not enough to cause a significant error if you assume the current doesn't change significantly.
(Actually the current would increase slightly to maintain the same power input to the hat).

So, if the cable resistance is 0.02Ω, then for the .279A of current I calculated for the hat input, the voltage drop would be .279A * .02Ω = 5.6mV, which is negligible.
(Actually it may be double that if the wire resistance is just for one wire and doesn't include the return wire, but that would still be negligible).

Ok, this is sounding promising. So the current required for the hat is so low that the voltage drop is tiny as well.

So if I can afford a voltage drop of 19V (56 provided - 37 required), I can afford a drop of 19,000mV.

19,000mV / 0.279A = 68,100Ω

Let's say the DC resistance is 10 mΩ/m.

68,100Ω is 68,100,000mΩ, which would be accumulated over 6,810,000 metres? That's not right.

 

That's not right.

No it's not.
Your calculator needs new batteries.
You need to carry the m in mV.
It's 19,000mV/.279A = 68,100mΩ = 68.1Ω

 

No it's not.
Your calculator needs new batteries.
You need to carry the m in mV.
It's 19,000mV/.279A = 68,100mΩ = 68.1Ω

That sounds more like it, but why is the m not carried in .279A * .02Ω = 5.6mV?

 

That sounds more like it, but why is the m not carried in .279A * .02Ω = 5.6mV?

Carried where?
.279 * .02 = .0056 = 5.6mV.

Sounds like you need to brush up on how number prefixes work.

 

Carried where?
.279 * .02 = .0056 = 5.6mV.

Sounds like you need to brush up on how number prefixes work.

I have to agree; I've been doing software and web development for years with minimal mathematical involvement so that part of my brain must've withered away. I need to remember that whether it's mV or V, that's the exact same value but just presented in a different way.

So now 19,000mV / 0.279A = 68,100mΩ, meaning at a resistance of 10mΩ/m, I can lose 6,810 metres' worth of voltage. 7km from just a little POE injector still seems unrealistic.

 

the resistance over 20m, for example, would be 0.02Ω.

No. It's 0.2Ω.

 

No. It's 0.2Ω.

Ah of course, so 681m?

That still seems like a long distance.