Battery charger: more power or more current?

See http://batteryuniversity.com/learn/article/charging_lithium_ion_batteries. Looks like B would be the best.

Clem

Hi Enrico,

Perhaps you can show me that I am wrong but I do not believe you can make this type of choice. For a given battery increasing the charge voltage will increase the current. Your parameters, Charge Condition A and Charge Condition B, can not both exist for a single specific battery. Also, just as a side observation, when we typically refer to power it is not just the amps but the product of the volts and amps ( VA ) which is the definition of Watts. If I misunderstood your question please clarify and send it back.

John

PS since neither A nor B has a voltage potential above the nominal voltage of the battery neither one would actually charge the 7.5 Volt battery.

What you say is something I tend to agree. But I am not so expert of battery charger so my doubt is if this shorten the battery life. Usually what I see - I mean logic suggest is that the B would be the best, also following the article you suggest. It is true what jw0752 suggests that the charge potential is not higher than the one declared on the battery. But it is also true usually what I see is that the chargers not always erogate higher voltage but lower ... The doubt remain.

Hi John, thanks for the advice. In this case the problem will be only the power charger, nothing else. As for as the literature I have read in the case of the battery the equation W = VxA (that we all know very well) seems not so applicable. As a matter of fact the battery charge uses the current to reverse an electrochemical reaction and this makes the difference. Don't you agree?

BTW, you have not misunderstood my question, maybe it's me that I don't know some aspect of this matter.

Enrico, The "LI ion 7.5V" sounds like a "two-in-series" configuration of (about) 3.7V Lithium ion batteries. Those batteries are pretty picky about what they tolerate. Go outside the allowed-parameters of the battery and you're in trouble. (Possible exploading batteries).

The "1500mA" could be one of three things. I'm guessing the last option:

1) Max discharge current.

2) Max charge current.

3) You meant mAh instead of mA.

Now, assuming the battery is 2S (two in series) battery, charging it with 4.3V will mean... discharging! A single cell may be discharged to about 3.0V. Below that is really bad for the battery! Don't do that!

6.5V is "almost empty". Charging the battery to "almost empty" seems of little use to me.

As these lithium ion batteries are "picky", you really should get a proper lithium ion charger. I'm guessing that your 7.4V nominal battery can be charged to 8.4V MAXIMUM. Get a proper lithium battery charger. They might say "7.4V" or "8.4V" on the outside, but they mean the same thing (They both mean: "For a 7.4V nominal batery which can be charged to 8.4V maximum").

If you have an adapter that says: "6.5V 1400mA" that adapter could very well provide more than 8.4V nominal when unloaded. This means that it will "fast-charge" your battery to "almost empty" and then continue to slow-charge until your battery has had too much. THAT's what makes not-specially-for-li-ion chargers not suitable for li-ion batteries.

IF you really need your battery charged today, your special charger is on its way,  and you have the time to baby-sit it, THEN you could connect it up to the "6.5V nominal" adapter. Keep measuring the voltage on the battery every 5 minutes until it reaches 8V. Then stop the charge. (you'll be at about 90% full. and I'll be guessing, probably more than 24H into the charge process!!!....)

The "LI ion 7.5V" sounds like a "two-in-series" configuration of (about) 3.7V Lithium ion batteries. Those batteries are pretty picky about what they tolerate. Go outside the allowed-parameters of the battery and you're in trouble. (Possible exploading batteries).

It sounds something like to me too. This is a test battery and it does no matter if it has a short life, what I want to see is if the theory works fine. So don't worry about the battery itself. It's for a test only.

The "1500mA" could be one of three things. I'm guessing the last option:

1) Max discharge current.

2) Max charge current.

3) You meant mAh instead of mA.

Sorry for my mistake in the original question post. It is true, I was meaning mAh not mA.

Now, assuming the battery is 2S (two in series) battery, charging it with 4.3V will mean... discharging! A single cell may be discharged to about 3.0V. Below that is really bad for the battery! Don't do that!

That's true. I have also found documentation about this on the document that clem57 linked in a previous replay. The specific page refers to the LiIon 3.6 V  batteries but the principle for charging / discharging remain the same.

6.5V is "almost empty". Charging the battery to "almost empty" seems of little use to me.

I agree and the concept is correct. But I apologise but I don't understand why "charged" (or similar) means "almost empty" Please can you detail me about this ?

As these lithium ion batteries are "picky", you really should get a proper lithium ion charger. I'm guessing that your 7.4V nominal battery can be charged to 8.4V MAXIMUM. Get a proper lithium battery charger. They might say "7.4V" or "8.4V" on the outside, but they mean the same thing (They both mean: "For a 7.4V nominal batery which can be charged to 8.4V maximum").

As a matter of fact no, the nominal battery is only 7.4 V 1500 mAh and it is explicitly written that should not be charged with more than 7.4V See the image below.

If you have an adapter that says: "6.5V 1400mA" that adapter could very well provide more than 8.4V nominal when unloaded. This means that it will "fast-charge" your battery to "almost empty" and then continue to slow-charge until your battery has had too much. THAT's what makes not-specially-for-li-ion chargers not suitable for li-ion batteries.

It is true that - as occur in all the  power adapers I saw - when unloaded the value is higher than when there is a load device.

About the battery charger, I have used a couple of LM350 with a circuit like the one in the following schematic:

I can't see the point of the second LM350 - your charger looks to me to be a 125mA current source and this is a potentially deadly way to charge a lithium ion battery.

If the battery is trickle charged to too high a voltage it will form metallic lithium internally and eventually explode or catch fire (read up on it.)

The normal way to charge a lithium ion battery is to charge at a reasonable current (often C/5 but in some case up to C (where C is 10x the current the battery could deliver for 10 hours if fully charged)).

The battery is charged to V1 at this high current and then the current is reduced and the battery charged until the voltage reaches V2. Typical values for V1 and V2 are 3.9 and 4.2 V but they vary and you MUST use the correct values for the battery and you must take into account the fact that two cells in series will not charge identically. Good chargers measure and even balance the voltage per cell.

You must never overcharge a Lithium Ion or Lithium Polymer battery  - some have built in over voltage protection but you should not rely on it.

Batteries will last longer if you don't charge all the way to the maximum voltage but they will hold less charge .

It doesn't hurt if you charge up slowly.

You need precision voltage measurement to get the best out of the battery.

MK

Think of the battery as a "fuel tank". But it is a bit of a "complex" fuel tank. It has mechanisms inside to pump the fuel  in and out. Now, over time in a real fuel tank you get some "gunk" at the bottom. That means that completely running the tank empty may feed the "gunk" into your motor ruining your motor. In the case of the lithium battery, it is the "mechanisms inside the tank" that get ruined if you drain it too much. Similarly overfilling the fuel tank may damage it and the mechanisms inside.

So lithium ion batteries have strict "min" and "max" fuel levels (voltages).

So, when a lithium ion cell is below 3.0V, the "mechanism" inside the cell starts to degrade. The cell is damaged.

Thinking of the cell as a fuel tank can show you some other characteristics of the lithium ion cell. If the fuel level is low, (below say 3.6V) the fuel tank is very narrow. Adding just a little fuel will raise the level of the fuel surface a lot. Then the tank goes wider. It takes a lot of fuel to take the tank through 3.7V to 3.8V. The tank is very wide at that level. Then it is narrower again, but not as narrow as at the bottom.

So, when you charge your battery to 6.5V, each cell is at 3.25V, and you can take out only very little fuel before you reach the minimum level of 3.0V.

Even though your battery clearly states maximum charging voltage: 7.4V, I think that's a typo, and they mean 8.4V. Still, everybody, including me should warn you to heed the advice printed on the battery.

The reason for me thinking that the printing on the battery is wrong, is that it also states: 7.4V / 1500mAh / 11.1Wh. That last piece of info means that the battery has an average voltage over the full "100% charged to 0% charged" of about 11.1Wh/1.5Ah = 7.4V. This is not possible if the fully charged state is 7.4V, and it is common for 8.2 or 8.4V fully charged batteries.

So SOMETHING on your label is wrong. I'm guessing the max charge voltage. But if my guess is wrong, the battery might go BOOM when charged to 8.4V....

I'm pretty good at "reading schematics". But especially when I'm doing it for fun, the person making the schematic has to make an effort to make a schematic readable. This means: Putting the GND signal near the bottom (or at least near the bottom of each sub-part). Your schematic is very difficult to read.

It might represent a charging circuit current, and voltage limited at the top. Nice. That's the way to charge a Lithium battery. But the specs for the end-voltage are very tight. This cannot be easily achieved with old fashioned components like the LM350. There are special modern chips that will provide exactly the right voltage for Lithium Ion cells.

When charging a battery, you force a certain amount of current through the battery. The battery will determine the voltage at which this happens. When that voltage reaches a certain value, you switch to forcing a specific voltage on the battery, and the battery will determine the current at which this happens. In both (actually ALL) cases the power is simply voltage times current.

Your original question can be answered. A higher current will always mean a higher power. And higher power will always mean more current. But if the current is more than is allowed by the battery, it might go BOOM.

So: more power, lower current is impossible. And lower voltage more power is also impossible.

Oh. Wait! You're saying "power" when you mean current. Aaargh! When asking technical questions, it becomes very difficult if you use the wrong terms. The 4.3V, 2000mA supply can provide more current than allowed by your battery. Moreover, at 4.3V no charging will occur as the voltage is too low. The 1400mA of the 6.5V supply is close to the maximum of your battery, but it will barely charge it with that current. If the powersupply is badly regulated, it will continue to charge your battery above 6.5V, but at a very, very slow rate.

So..... Hook your battery up to the 12V of a 600W ATX powersupply and... you can put a lot of power into the battery. But you'll exceed the maximum current, maximum voltage, and your battery will not last one cycle.

If you want to make your own circuit to charge this battery,

go to the LM 317 Calculator and figure out the resistor value for 7.4V. This will be 1240 Ohms. Now if instead of searching for a 1240 ohm resistor, you find a 1200 ohm resistor and place a 100 ohm potmeter in series, you will be able to adjust your regulator from 7.2 to 7.7V. Adjust for precisely 7.4V. Use a Vin of about 10-15V.

The LM317 will currentlimit at close to 1.5A, close to the maximum of your battery. That takes care of the current limit. Officially you're probably going over the limit of the battery. The '317 is rated for 1.5A nominal, so the current limit is a bit higher, I'd feel a bit more comfortable using a 1.0A nominal regulator, but I don't know a type number by heart. Ah! google for 1A adjustable voltage regulator leads to LM2941. The spec says MAX 1.6A, so it is unlikely to go far over the currentlimit of your battery. The 1200 ohm resistor is "too big" if you stay with the 240 ohm reference resistor. (the minimum adjustable voltage will be 7.6V nominal). Maybe use a 500 (or 470) ohm pot and a 1k resistor. But your adjustment range will be big so fine-tuning to exactly 7.4V will be difficult.

Michael, I've never heard of the two voltages charging method. Except for: slow charge (C/20) to 3.0V and then normal CC/CV up to 4.2V. Also, the battery clearly states maximum charging voltage 7.4V, so I would not recommend going above that. (even though I think it is a typo).

Ok, first of all thank you for your advice. I think that this part of the project is possible just thanking the suggestions of the community.

First of all the facts.

1. I have found the suggestion of the second LM350 as a feedback regulator to block the overcharge issue. And it seems working testing the current during the various phases. The other thing I noted is that the charger section of the circuit does not heat at all. Maybe that this second LM350 is put wrong or redundant.
2. Where you read +9V it's only a silly mistake because I have not in this moment the Eagle correct schematic component The charger applies 12 Vcc at 1.4A DC
3. The battery receives 8.4 V from the charger. All the values were read during charge time (about three hours just to have a reference parameter) using the BitScope Micro and Chart software..
4. To stress the situation I have powered the Raspi with the battery until it spontaneously power-off. Then the battery shows 0V
5. After a charge cycle (until now are 3) the battery reach the nominal value of 7.4 V. In this moment it is powering the device again, later I can see how much time the device will work (with WiFi active)

One obscure point is how much time (average) I should expect the PI will work with the battery. Should investigate on it.

A new obscure point - not related to this specific test battery - is how can I calculate the battery deliverable current in 10 hours.

It is really bad for a lithium battery to reach 0.0V.

If the voltage goes below 3V per cell (6V on your battery) it will lose capacity every minute it stays like that. You need to charge it back to within 6 to 7.4V as soon as possibe. But the maximum current is now: 75mA. I don't know if your battery goes BOOM or if it just damages more if you go above that value.

The raspberry pi uses about 400mA, so I expect it to run for about 3 hours on the 1500mAh battery. As you are abusing the battery so much, I expect this to drop below the hour within a day or so.

The "10 hour capacity" is printed on the battery: 1500mAh.