Any Electrical Engeiners Want to Help an Idiot? Version 2!

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Hello readers and electrical engineer type people! A week ago, I posted this thread: https://forum.teksyndicate.com/t/any-electrical-engeiners-want-to-help-an-idiot/96632

It was regarding charging some Sanyo 16650 Lithium cells that I acquired from a laptop battery that I took apart. I had asked how I would go along with charging the batteries, and I got great answers and a lot of help and learned quite a bit.

This week, I took apart one of my portable battery banks, which contains 4 Samsung 18650 cells. From a bit of research, the only difference that I could find that the Samsung Cells hold more power (300mah difference)

Link for Samsung Cells: http://lygte-info.dk/review/batteries2012/Samsung%20ICR18650-28A%202800mAh%20(Purple)%20UK.html
AND http://www.ebay.ca/itm/SAMSUNG-ICR-18650-28A-2800-mAh-with-Solder-Lug-/262041777822?hash=item3d02e8369e:g:F-4AAOSwg3FUc4N3

Link for Sanyo Cells: http://www.illumn.com/16650-sanyo-ur16650zta-4-35v-2500mah-flat-top.html
AND http://www.ebay.ca/itm/2PCS-Sanyo-UR16650ZTA-Li-Ion-Rechargeable-Battery-Batteries-2500mah-3-7V-/281843162051?hash=item419f295bc3:g:wKQAAOSwl9BWM1wg

From looking at the 4 different website that I checked out, it seems like both the Samsung and Sanyo batteries charge at between 4.2 and 4.35V, and both have an end discharge voltage of 2.5v (which I think means that it won't go any lower than 2.5v when empty? Correct me if i'm wrong). From what i've read so far, the Samsung and Sanyo cells are very similar, so I was wondering if I could replace the 4 Samsung cells which are inside my portable battery bank, with the Sanyo cells. Of course, i'd have to wire things identically or I might fry/blow something up.

That also leads me to my second thing... I don't have a voltmeter lol, however, I do have a soldering iron (60w), soldering wick, and a soldering pump! If I were to Google the polarity of each side of the Samsung and Sanyo cells, and find out what each colored cable on the PCB means of the logicboard controlling the charging/discharging of the batteries, would that be good enough?

Here are pictures of my portable battery bank, taken apart (album): http://imgur.com/a/pcTVV

Heres a picture of my Sanyo Cells from the laptop battery: http://imgur.com/miDGVox

Will it be possible for me to remove the 4 Samsung Cells, and replace them with my 4 Sanyo Cells (and wired correctly), and charge the 4 Sanyo Cells just fine? Also, would it be possible to charge all 8 batteries at once, without causing any of them to be overcharged/burning down my house?

Thank you to everyone who understands what I just wrote above and can help me! I really appreciate it! :D

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If it was not suggested in the previous thread I would suggest you look into a YouTube channel by the name of Rinoa Super Genius. He is always doing videos with 18650 cells, I enjoy them a lot but I am not a Electrical Engineer.

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Already am subbed to him! He's a really cool person.

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Bump?

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Well sir your in luck. I am an electrical engineer. First off let me start by correcting you here on what an end discharge voltage value is or an EDV

What an end discharge voltage is the voltage in which a battery has hit when roughly 92~96 percent of its energy is used up. This can differ based on the type of battery. For example, The end-of-discharge voltage for lead acid type storage medium is 1.75V/cell; a nickel metal hydride battery would be about 1.00V/cell; and most Li-ions and lithium polymer batteries is roughly 3.00V/cell. At this level, What this value essentially means is the voltage would drop rapidly if the discharge were to continue. passed this point. If you were to plot such a model on a graph it would be a very steep exponential/logarithmic decay. Another correction I am going to make for you. By Convention the wire color tells you a lot. The red wire is for Positive Voltage. The Black wire is for negative voltage. The Yellow or green wire tells you that it is a common ground rail wire

Second off lets look at the two batteries you are comparing. I would say the answer to your question is yes you can do what you are trying to do but I would advise caution. You would likely need a slightly different regulator to accompany the slightly higher charger and nominal voltage rating of the samsung cells. Assuming you can get your hands on some white pages or technical documents that involve similar cells and can aquire the information the regulator. Go right ahead.

Third .. Uhmm looking at your pictures you have wired the cells very differently from the sanyos. See the sanyos are in what we call a series configuration which increases the voltage of the bank. Your samsungs are in parrellel which keeps the voltage at spec (with the battery cells) but increases your capacity. You need to match the configuration of the sanyos or you will not have enough volts to power the laptop.

Fourth off. In the unlikely event that you cannot get technical pages. SInce they are so close and within margin of error you are likely not to notice anything but reduced capacity on the samsungs as you are not reaching nominal charge at 4.2 volts.

Assuming you do everything correctly you will not have any issues. The internal regulator circuit will handle all of the quirks. These batteries are of the same chemical composition with the samsungs being slightly larger. So in reality there should not be a difference in charging characteristics. Since this is your project I will need you to find me battery curves for each of the batteries and specs and white pages and what not so we can do an apples to apples comparison

As helpful as your original post was I need more details .. i,e why you intend on doing this. What is the purpose. The make and model of the device being powered.. yada yada. You know what I am getting at

From what I see the samsungs are smaller and provide a more power dense package. I could see you making a more power dense pack and doing some more interesting configurations to increase capacity using a hybrid series-parallel configuration but more on that later

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GET ONE .. !!

seriously not something to lol at

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What I plan to do with these batteries is use them for a flashlight. Recently, I ordered a 10W LED from ebay (9-12v, 900mah), and was planning on using these batteries to power the LED. My thought was to just have it so that 3 batteries will be in series (so 11.1v, and approx 900mah, I hope?) and just solder that directly to the LED, and also solder in a switch to allow me to turn on and off the LED.

I did know the Sanyo cells were in series, so what I had done is cut them apart, and left a bit of the metal connecting the batteries, and then added solder to the metal pieces hanging off of the battery. So this is how it turned out: http://imgur.com/h01pqT5 I copied the layout of how the Samsung cells were arranged, and began soldering. I charged them for about 30 minutes, and the only thing getting hot was the logicboard controlling the charging of the batteries, and the batteries did hold a charge and were able to also deliver power to a USB device. I let them charge for around 8 hours, and the battery indicator on the logicboard is indicating they are full, and if I quickly short circuit the batteries, they lot out quite a big spark, so it seems like it worked fine, without burning down my house.

My guess as to what EDV is was completely wrong, and thank you so much for correcting that lol. As for the voltmeter, I plan to purchase one of those soon because I want to learn a lot more about circuits and such, and what each individual component does in a circuit (ex, what a capacitor does... I still don't know and couldn't really find a simple thing telling me about it. Does it reduce volts, reduce amps, smooths the power and converts AC to DC? I dunno), and eventually be able to diagnose what goes wrong on a logicboard/motherboard and be able to replace the faulty component (if it isn't a BGA, of course. I don't have a BGA rework station and won't get one for a long time lol).

In my chemistry class, we learned about how electrolytic cells and galvanic cells have reverse polarity or something like that (learned it about 3 months ago, so I sorta forgot it) and that an anode and cathode are the complete opposite between the two different cells. Would power coming out of the mains and out of the battery have the same anode, and the same cathode? Also, which is which? The cathode is the one receiving the electrons or sending them to the anode? It really confuses me.

I know I sorta replied to your long but very helpful paragraph in a not-so-organized order, sorry, but I hope it answers some of the questions you had for me, and also gives you an idea as to what I want to do, how little my knowledge is at this point, and how much I want to learn.

P.S Don't make fun of my soldering job. I'm also a noob at that, and the solder didn't want to stick to the metal so I sorta added a whole lot to get it to temporarily work.

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Draw the design up. This is what you should do before committing to any prototype or design work. Also do you have part numbers for that LED? There are a lot better more reputable suppliers for things you are getting. Head over to digi key or mauser electronics... You can get anything you need.

Yes this is true between the two different cell types their polarities are reversed. To answer your question. by convention an anode is an electrode through which a current flows into a polarized electrical device. You can use the mnemonic device ACID to remember this. Anode current into device. While a cathode is the exact opposite in most conventional current situations.

Thats definitely something for another thread. I could explain here but it would detract from your project as each component behaves differently whether its in AC or DC. Also things that smooth power perform power factor correction on an AC source before rectifying it into a DC source. They can also smooth power in DC switching conditions or pulsing (pwm) conditions. Like i said definitely stuff for another thread..

Use the quote feature like I just did. Highlight the text you are responding to and then a quote button should appear above. Youll get the hang of it

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If you can't get solder to stick to those tabs you should probably use more heat and or clean the surface with acetone or somesuch.
Please treat these batteries like they will burst into flames any moment, regardless if you're around them or not.
Be very careful about what voltages they're at, above 4.2 you enter the dangerzone™ and above 2.5 (or whatever the battery is speced for) you start degrading the batteries capacity and reliability.

mah (milliampere hours) is 1/1000th of a ampere houre, which is 1 ampere over 1 hour, or 2 ampere over 30 mins, you get the idea.
Which means your 2800mah samsung cells can be discharged with a constant current of 2.8 ampere over 1 hour before they're empty.
To figure out your total energy in your battery pack or cell is to take the average voltage, which is about 3.7v for your cells and multiply it by the ampere hours, 2.8ah*3.7v=10.36wh, which stands for watt hours.

If you take two of your 2.8ah batteries and connect them in parallel you get twice the amp hours but the same voltage, connect them in series and you get twice the voltage but the same amp hours, total energy doesn't change.

Multimeter is a must by the way.

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So this is the LED that I purchased from eBay: http://www.ebay.ca/itm/301862404632?_trksid=p2057872.m2749.l2649&var=600672650387&ssPageName=STRK%3AMEBIDX%3AIT
It seems like some generic LED, probably also has shitty quality, but because i'm only doing this experiment to learn, I don't really mind.

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This above is going to be my test circuit (hopefully). I know that the LED only allows power to travel one way. Everything will be connected via small jumper cables on my first couple of trys, to reduce the amount of heat I put onto the cells and LED by soldering. Underneath the LED, will place a headsink that I got from some Intel cooler, and connect the 2 via thermal paste. I heard these LED's get pretty hot and need a heatsink.

After I do this experiment, I will advance to finding some sort of tubular object, like a can of Pringles, purchase a higher output LED, more batteries, and add a small 12V fan somewhere inside the pringles can to cool the whole thing, and add something to limit the amount of amps that gets sent to the fan (it's only around 2 watts, so really low amps), so I think i'll need a resistor? I'll have to search up what reduces amps to a device.

As for charging the batteries, I plan to purchase some cheap charging circuit on ebay which can charge LiPo batteries, and every time I want to charge the battery, just use jumper cables to the charging circuit and the battery, to reduce the amount of soldering needed, maybe even find something to charge all of the batteries at once, instead of disconnecting each, every time I want to charge the whole thing.

Thanks! I didn't know how to use this before, but saw others do it lol

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I don't want to apply to much heat to them because i'm afraid it may damage the batteries or may cause them to explode. I also do not have acetone. I'm going to try to use sandpaper on the metal strips, and hope that allows for the solder to stick better.

Regarding all of that milliampere hour math and stuff, I really did not know about any of that! Thank you so much for all of that! :D

I also plan to get a multimeter.

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@CynicRF So i'm trying to make a circuit with a motor (electric fan at12v, 0.18a, 2.16w) , 3 cells in series (11.1v, 2.8a I think), a 10w LED, and I know I need a resistor for the motor, but have no clue which one to use and how to get the value. I'm stuck. Do I also need a step-up transformer to make the 11.1v to 12v? I have no clue. This is what I have so far.

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You don't need a resistor in series with the fan.
Fully charged your 3 batteries give out 12.6v when in series like that.
Your fan is going to work between 14v and 6v if its a PC fan, but speed will depend on the voltage.

You can't use a step up transformer with DC, only AC.

1000ohm resistor is probably fine for your LED, depending on the type of course.

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Here is a nice discharge curve of a 18650 style battery.

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I have some experience making high powered LED flashlights. (I'm also an engineer)

I can tell you this.

You can't just hook the LED's right up to the battery supply. They will not last long. Although these LEDs do have some internal resistance, it drops the hotter they get. If your batteries can provide the current, they LEDs will have a thermal run-away and fry. What you need is a current driver. It limits the current through the LEDs to the level they were intended to be driven by and therefore prolong the life of your LEDs immensely.

Here is a driver I've used in the past. http://www.ledsupply.com/led-drivers/buckpuck-dc-led-drivers
You can get them in various current ratings.

Also note that the driver does have a small drop-out voltage so you'll need to take that into consideration when you setup your circuit.

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You can totally hook up LEDs directly (kinda) to a battery via a current limiting resistor.
If you cannot squeeze more current through the resistor at a given voltage than the LED can handle you're fine. Implying you're only using a indicator LED of course.

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These are high powered LEDs. Your talking about 1amp of current. Using a resistor to limit the current is very impractical. This "resistor" would basically need to be a 10w incandescent light bulb cause it would need to be very low resistance and be able to dissipate quite a lot of wasted power.

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10w led, didn't catch that, only thought it was a indicator.
Yeah hes going to need a CC LED driver for that.

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@Castellorizon The reason I wanted to add a resistor to the PC fan is because I thought that applying 11.1-12v and 2.8A of power (2800mah battery, I think that's how it works?) would kill the fan very quickly and spin it very fast because it would be giving it 33watts. I wanted to give it 2.16watts (it's recommended amount of power).

I did not know that, so thank you :p

Regarding your battery curve chart, I did not know that a LiPo battery would deliver close to 4.2v at it's highest. I thought that the voltage stayed constant. Because of it delivering 4.2v per cell, that would be 12.6v, which is a bit over the voltage that the LED is meant to handle.

Regarding ampere, if, for example, I were to discharge 3 battery's to 3.7ish volts, connect them in series, and connected it straight to the LED, would the LED receive 11.1 volts and 2.8m, or will the LED only take as much power as it needs, so approx 11.1v at approx 0.9a?

I was planning to add a heatsink and a PC fan to the heatsink to reduce the amount of heat going to the LED's.

The one that you linked is over $10, and because I am only a student, and I want to do this project to learn about electricity, and don't really care about these LED's/cells lasting months, I don't mind if they burn out after a week or two. The cheapest LED current driver that I found was $5 CAD, and only could handle 12v 1a, which is what the LED recommends, but I want to make this project as cheap as possible, and make/solder as many of the items as I can at home.

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@eidolonFIRE @Castellorizon thanks for taking over. I dont have the time this week to really get a solid reply in here but I may tonight. Ill try to chime in here and there @TheAlmightyBaconLord

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