Portable solar powered power bank

Hi,
the reason I am posting this here is this not an assignment given to me, this is a project I chose to do as my degrees final year project.

Abstract:
I am going to design and build a portable solar powered device which is able to charge a USB device and drive a 4w fan. It also includes lithium ion batteries to store energy in case the direct illumination is not present.
Objectives:
*Portable ( foldable pv panel)
*Efficient ( combination of MPPT and PMW)
*keep the fan running in full speed and charge USB devices at the same time.(Ideally)
* Prevent the battery from getting over-charged and over-discharged.
*Satisfy USB devices charging specifications.(assuming be able to charge a mobile phones 3500mah battery within 1 hours in full illumination)
*relatively low cost.

I would appreciate it if you guys help me to find out where should I start in calculating . what steps should I take. what information's do I need. Components have to carefully calculated so u have to keep assumption minimal. In the end of the day this is a device that I eventually need to build and it has to work.
P.S: attached a diagram of the system.

First consider your energy balance; here we shall talk about power.

What is the total (boilerplate) energy requirement of the load?

What is the voltage and the AH capacity of the battery?

How much margin you want to keep?

What is the PV cell capacity?

Let us say the load is a 4W fan; keep some margin and make it 5W.

The battery capacity is 3.7Vx3.5A say 15W

Your solar panel must supply say 20W under typical light levels (whatever that may mean!!); if you want to be conservative add 10% to the angels.

First step should be the solar panels and they need to be portable. You decide on the panels first and then move on to regulator.

You have to run as fast as you can just to stay where you are.
If you want to get anywhere, you'll have to run much faster.
-- Lewis Carroll

I have made some progress since last night. basically worked out the battery and the PV module. would appreciate it if you take a look.

Battery: Need to choose what device needed to be charged first in order to choose the appropriate battery size and PV panel respectively. I assume, load (the device needed to charged is a phone) is 3500 mAh with 3.7v. PV panel: To calculate the energy content of the battery:


E=V*I*T
3.7*3500mAh=12.95Wh ~ 13Wh

Considering the Li-ion battery loss which is close to 20%:

E= 13Wh/0.8=16.25 Wh~ 17Wh

The Lithium ion battery is the most suitable option here as it has a higher rated power, more compact and also Peukart effect is negligible in this type of batteries.

Candide component: Lithium Ion Battery pack 4400 mAh 2 x 2200 mAh 3.7V akku
(**broken link removed**)


PV module: There are three main factors to consider in order to choose a solar panel:
How much energy will your appliance(s) use over a period?
How much energy can your battery store?
How much energy can a Solar panel generate over a period?

The amount of energy the load uses over a period has been already addressed in previous section. Consequently, a battery which can facilitate the same amount of energy and current has been chosen. Therefore, the PV module has to produce at least 17Wh power.
PV module sizing: The fact that different PV module sizes generate different amount of power, must be taken into consideration. The peak Watt produced (Wp) not only depends on the size but also the climate of site location (Radiation angel). In other words, the Total Watt-Peak-Rating (Wp) would be on its maximum when the sunlight and absorbing surface (radiation incident) are preopercular.
The Relationship between S module and S horizontal is as followed:

SMODULE=├ Shorizontal*sin⁡(α+β) ┤/Sinα



As mentioned earlier solar electricity generation depends on many factors such as Incident angel, sun elevation, longitude and latitude of the site, etc.
Below two tables which demonstrate annual estimated solar electricity generated in (Birmingham, UK), have been compared to each other. Two tables below, one with the incident angel of 39˚(optimum) and the other one 90˚ (window mounted) have been compared to each other.









Ed = panel generation factor.




Referring to information provided in the table on the left (optimum, south facing). Watt-peak rating needed for PV module would be the total Watt-hours per day needed from the PV modules divided by daily solar radiation:

Wp=needed PV module energy/ Ed
Wp=17Wh/1.03=16.50Wh
Now to calculate the number of PV modules:
number of PV modules=Wp/(PV module rated Wp)
number of PV modules=16.50Wh/21Wh=0.78 ~ 1
However, this result only represents the outdoor use, as the results have been calculated using the optimum solar radiation angel (39˚). By going through a similar procedure, the unit’s feasibility in a different environment would be tested. In order to replicate a window mounted stationary PV module (indoor use, 90˚, south facing), the table on the right has been used.

Wp=17Wh/1.14Wh=14.91Wh
number of PV modules=14.91Wh/21Wh=0.71 ~ 1

The above calculations examined the daily solar radiation in December which is the lowest in the year (worst case scenario). The results show that even under mentioned circumstances only one PV panel is needed in both situations.
Calculations above show that only one 21Wh foldable PV panel would satisfy the project needs in all conditions.

Candid Component: 21W, 5V Foldable Solar Panel.



I am not sure about the solar incident radiation. The PFG I used was in kWh/m2/day. I simply ignored the k and used it. in formula otherwise it wouldn't have worked!

Baxxxter said:

*Satisfy USB devices charging specifications.(assuming be able to charge a mobile phones 3500mah battery within 1 hours in full illumination)

Click to expand...

You should know that there are numerous USB specifications out there with vastly different charging specs. Most USB chargers aren't capable of meeting your desired spec... but anyways, you're basically asking for approximately 20W max.

1kW/m2 of solar irradiance is a commonly cited number for "peak sun hours," i.e. the highest peak irradiance you can expect on a "typical" day (depending on where you are....). Typical PV efficiency is around 15%. So for every 1W of power you want during peak sun hours, you need about 0.007m2 of PV area. For 20W, you will need 0.134m2. And this is before factoring other factors including converter efficiency, irradiance angle, etc...

Thanks I will look into these matters.