How can we power -48 VDC device

[dufferdev]

Hi,

I want to power a wifi device using solar power. But I see the datasheet which has input voltage -48 VDC.

I am confused on how to power this system using battery ?

Any suggestions are welcome.

Thanks a ton...

 

[f95toli]

Use 4 12V batteries connected in series

 

[dufferdev]

Use 4 12V batteries connected in series

Thanks for your quick reply will try this...

 

[Fish4Fun]

Just to be clear...which data sheet specifies 48V? Is it the WiFi device or the solar panel?

Fish

 

[dufferdev]

Just to be clear...which data sheet specifies 48V? Is it the WiFi device or the solar panel?

Fish

The -48VDC is specified in the datasheet of the DEVICE and not the solar panel.

 

[Fish4Fun]

48Vdc is a lot for a typical consumer WiFi device, but ok. The steps to calculating your system should go like this:

First Determine:

System Voltage = 48V
System Current = ??
System Power Consumption = 48 * I
Number of Hours of Operation Per Day = ??

For now I will Assume:
I = 1A
H = 24 Hours

From here we can determine the minimum battery size:

24 hours * 1A = 24Ah

Next we need to know something about the amount of full sun in the place the device will be located on the Winter Equinox. I will assume this is 4 hours.

You will need to charge the batteries back to at least 24Ah each day. I will assume an 80% charging efficiency, so:

(24Ah / 0.80) * 48V = 1440W / 4hours = 360W

You would need a minimum panel size of 360W @ 48V.

Next you need to evaluate how critical the device is to determine and calculate how long you might want the device to withstand adverse weather conditions. Let's say the device is highly critical, and you want to ensure there are NO interruptions for up to 7 consecutive days of 100% cloud cover.

So, our minimum battery size would be 24Ah * 7days = 168Ah

To charge our 168Ah battery bank, we would want to ensure the batteries could be fully recharged in no more than 4 days of full sun, so:

(168Ah/0.80) + (4 Days * 24Ah) / (4 Days * 4 Hours) = 19.125Ah * 48V = 918W

So, for this level of protection you would need a minimum of 1kW of panels and 210Ah of storage. If you plan on using lead acid batteries, you would want to ensure that you never discharged them more than 50%, so you might use two parallel banks of 4 series connected 210Ah batteries (total of 8 batteries). If you plan on using Lithium Ion batteries rated @ 3.6V & 2.5Ah, you would use 84 parallel connected banks of 14 series batteries (1176 batteries).

The cost of the system is extremely sensitive to actual power consumption and the fault tolerance duration.

For Zero Fault Tolerance, 14 * 2.5Ah Lithium Ion Batteries in series charged by a pair of 180W solar panels might be built for less than $2,000. For the 7 day fault tolerance system, the cost would certainly exceed $15,000. Cutting the power consumption by 25% could easily cut the cost of the system by 50%.

Hope that helps,

Fish

 

[dufferdev]

48Vdc is a lot for a typical consumer WiFi device, but ok. The steps to calculating your system should go like this:

First Determine:

System Voltage = 48V
System Current = ??
System Power Consumption = 48 * I
Number of Hours of Operation Per Day = ??

For now I will Assume:
I = 1A
H = 24 Hours

From here we can determine the minimum battery size:

24 hours * 1A = 24Ah

Next we need to know something about the amount of full sun in the place the device will be located on the Winter Equinox. I will assume this is 4 hours.

You will need to charge the batteries back to at least 24Ah each day. I will assume an 80% charging efficiency, so:

(24Ah / 0.80) * 48V = 1440W / 4hours = 360W

You would need a minimum panel size of 360W @ 48V.

Next you need to evaluate how critical the device is to determine and calculate how long you might want the device to withstand adverse weather conditions. Let's say the device is highly critical, and you want to ensure there are NO interruptions for up to 7 consecutive days of 100% cloud cover.

So, our minimum battery size would be 24Ah * 7days = 168Ah

To charge our 168Ah battery bank, we would want to ensure the batteries could be fully recharged in no more than 4 days of full sun, so:

(168Ah/0.80) + (4 Days * 24Ah) / (4 Days * 4 Hours) = 19.125Ah * 48V = 918W

So, for this level of protection you would need a minimum of 1kW of panels and 210Ah of storage. If you plan on using lead acid batteries, you would want to ensure that you never discharged them more than 50%, so you might use two parallel banks of 4 series connected 210Ah batteries (total of 8 batteries). If you plan on using Lithium Ion batteries rated @ 3.6V & 2.5Ah, you would use 84 parallel connected banks of 14 series batteries (1176 batteries).

The cost of the system is extremely sensitive to actual power consumption and the fault tolerance duration.

For Zero Fault Tolerance, 14 * 2.5Ah Lithium Ion Batteries in series charged by a pair of 180W solar panels might be built for less than $2,000. For the 7 day fault tolerance system, the cost would certainly exceed $15,000. Cutting the power consumption by 25% could easily cut the cost of the system by 50%.

Hope that helps,

Fish

Hello Fish,

Thanks for your support and help.

However, I am clear on the solar calculation.

I want to know how to generate minus (- 48VDC) which is stated in the wifi device datasheet ?

 

[Fish4Fun]

It might be best if you posted the data sheet to determine what the -48V reference is to, but in general, a DC device has two supply inputs. Choosing which is "ground" is fairly arbitrary. If you select the positive terminal of a 48V battery as "ground", then the negative terminal, by definition is -48V. If you select the negative terminal as "ground" then the positive terminal is, by definition +48V. If there is some Voltage Reference, and relative to that reference you need -48V, then you simply connect the positive terminal of the battery to that reference point.

Fish

 

[dufferdev]

It might be best if you posted the data sheet to determine what the -48V reference is to, but in general, a DC device has two supply inputs. Choosing which is "ground" is fairly arbitrary. If you select the positive terminal of a 48V battery as "ground", then the negative terminal, by definition is -48V. If you select the negative terminal as "ground" then the positive terminal is, by definition +48V. If there is some Voltage Reference, and relative to that reference you need -48V, then you simply connect the positive terminal of the battery to that reference point.

Fish

Thanks for your reply fish, I have attached the datasheet here

 

[Fish4Fun]

The protocol they are using is

PoE, Power Over Internet, you can read about it here:

http://en.wikipedia.org/wiki/Power_over_Ethernet

The power supply you need will likely be included with the unit, and it will be similar to this:

http://www.tranzeo.com/products/docs/TR-SUR-001.pdf

In that PDF it states in a footnote @ the bottom:

Use TR-SUR-48 POE for 48VDC operation.

As long as you use their power supply, you should be able to power it easily from any 48V battery source.

Fish

 

[Fish4Fun]

The protocol they are using is

PoE, Power Over Internet, you can read about it here:

http://en.wikipedia.org/wiki/Power_over_Ethernet

The power supply you need will likely be included with the unit, and it will be similar to this:

http://www.tranzeo.com/products/docs/TR-SUR-001.pdf

In that PDF it states in a footnote @ the bottom:

Use TR-SUR-48 POE for 48VDC operation.

As long as you use their power supply, you should be able to power it easily from any 48V battery source.

Fish

 

[Studiot]

PoE, Power Over Internet,

I wish. Can I have some too?

Interestingly 48 volts is the standard phantom power supply for stage microphones.

 

[Fish4Fun]

I wish. Can I have some too?

Interestingly 48 volts is the standard phantom power supply for stage microphones.

ooops...PoE => Power Over Ethernet

I got the link right, by my dyslexic brain transposed Internet for Ethernet...

Fish

 

[MATLABdude]

[...]Interestingly 48 volts is the standard phantom power supply for stage microphones.

-48V is also the DC supply voltage on phonelines. I wonder where else this might be used?

To the OP, if you Google for solar POE, there are various [strike]hobbiest[/strike] hobbyist project pages and commercialish devices that may help with that aspect of the problem.

EDIT: Fixed that for me...

 

[davenn]

-48V is also the DC supply voltage on phonelines. I wonder where else this might be used?...

also on telephone systems in NZ and Australia and its almost universally used on much RF telecomm's equip world wide and has been since Adam was a boy ;)

Dave

 

[sophiecentaur]

Telephone exchanges used to have vast banks of -48V battery supply and backup, on float charge, for the 'old fashioned' switching and signalling circuits. Using the same supply voltage for other electronic equipment would make a lot of sense as you'd have a built in UPS for the electronics. I can't think that -48V would be a good idea under any other circs where 12V and 5V are the most common power supply voltages. All the internal devices would use these lower voltages, after all.
It makes me wonder why not use a different, more convenient WIFI modem?