Batteries and New Battery Technologies

[CDK]

Bert, 14.4 is the cut-off voltage of the regulator, measured at the alternator B-terminal when (almost) no current flows. A modern maintenance-free battery will survive that at least during the warranty period.

In my opinion, cars sold anywhere between the tropics should have a temperature compensating circuit just like Siemens solar controllers have for installation in hot climates. Under the bonnet you may expect extreme temperatures, causing the battery to boil and release gas.
The only purpose I can think of to use the absolute maximum charge voltage is in a vehicle equipped with more than one battery, so a diode bridge is used.

I bought a motorhome this spring and had a gel battery installed for the rear compartment. After two months the capacity was reduced to almost nothing and I had to replace it with a wet battery.
The charger and solar regulator supplied 13.8 volts, the limit for gel batteries, but the alternator kept on charging until the battery died.

[BertKu]

Originally Posted by CDK

Bert, 14.4 is the cut-off voltage of the regulator, measured at the alternator B-terminal when (almost) no current flows. A modern maintenance-free battery will survive that at least during the warranty period.

I measured the 14,4Volt over the battery terminal. I thought already that the German, Japanese, French, etc car designers all live in a cold country, never had the pleasure in designing a car in a tropical place.
Thanks, I will make a switch + 2 diodes in series (0,8 Volt) to compensate for the long distances.

Thank you all for the inputs. Back to some modern batteries again.

lets now sort out what the best way is to connect the super capacitors over the LiFePO4 batteries.

[BertKu]

Originally Posted by portacruise

porta

THE BATTERIES WILL NOT REPLACE ENERGY USED BY THE CAP FOR THE MOTOR UNLESS THE MOTOR IS OFF. THE CAP AND BATTERIES WILL SHARE INPUT INTO THE MOTOR AT ANY SET VOLTAGE WHILE IT IS RUNNING. IF THE MOTOR DRAWS MORE CURRENT THAN THE BATTERIES CAN SUPPLY, THE CAP MAKES UP THE DIFFERENCE, UNTIL THE CAP'S RESERVE IS USED UP AT THAT PARTICULAR VOLTAGE.

Assuming I have a 24 Volt motor and a surge of 300 Ampere for 50 milliseconds. The batteries are only 5 Amperhour and 8C = 40 Ampere.

Although I have a number of batteries via diodes parrallel, lets for the sake of theory have only 1 set of 24 Volt batteries.
The battery specs will be exceeded, if on its own. Thus now the cap will supply the 300 Ampere, as the capacitance is 183 Farad.
The resistance is 24 Volt : 300 Ampere = 0,08 Ohm
RC = 0,789 x 183 Farad x 0,08 Ohm = 11,5 seconds. Thus the Voltage does not drop enough over the 50 millisecond to exceed the battery's specs.

The battery will only supply the normal current of lets say 5 Ampere, should that be the normal usage current. If the battery wants to supply more current, the voltage over the diode will increase and compensate, in view that the capacitor is so large, the voltage over the capacitor has maybe dropped by only a few millivolt.

I got it. It is thus worthwile to have some caps added to the battery bank.

But nothing is for nothing, thus the battery will slowly feed some extra current to the cap until voltage equalised.

We all agree???

[sparky_wap]

In 50 mS, you will drop at least 80 mV on you caps. The battery pack internal and wire resistance will be such that most of the surge will come from the caps. I'm not sure of your diode arrangement or its purpose.
What could cause a 300 amp surge on a low powered e-boat voltage bus?
How do you plan on charging 183 Farads of cap safely?

[BertKu]

Originally Posted by sparky_wap

In 50 mS, you will drop at least 80 mV on you caps. The battery pack internal and wire resistance will be such that most of the surge will come from the caps.

I agree with you. The voltage drop of 80 mV will not make the batteries exceed the maximum specifications.

Originally Posted by sparky_wap

I'm not sure of your diode arrangement or its purpose.

Depending the succes of the 40 LiFePO4 batteries and lateron expanded to 1000 batteries. I do not believe in connecting batteries parrallel, and that they are able to feed each other. if they are isolated by a diode or in my case 2 diodes parallel, any bad battery will be picked up by the BMS. It is fine when all batteries are new, but over a periode of time, I will have too many differences, whereby parralel switching of batteries will be a bad decision.

In view that normal silicon diodes have a voltage drop of 0,5 Volt. I will loose at 100 Ampere approx. 50 Watt. By using Shotky diodes, the loss wil only be about 10 Watt.

Originally Posted by sparky_wap

What could cause a 300 amp surge on a low powered e-boat voltage bus?

We took this value as an example to philosofy what would happen and whether it could course poblems for the batteries. Pure theoretical, just for the discussion. I could have said 50 Ampere or 500 Ampere. With 40 LiFePO4 batteries, we will only start with 5 strings parralel. i.e. maximum 125 Ampere or 200 Ampere pulsed for a short time.

Originally Posted by sparky_wap

How do you plan on charging 183 Farads of cap safely?
?

Each string of 8 (24 Volt) or 16 (48 Volt) LiFePO4 batteries in series will be charged seperately. The strings are thus connected parallel via the shotky diodes to the plus bar. The plus bar has the capacitor connected to ground with a special tripswitch. Also each string is charged via a shotky diode.
Our aim is to go for up to 100 strings parallel. i.e. up to 1000 batteries.

Again the strings are charged only with a maximum of 1C for a periode of time. i.e. 5 Ampere. It depends thus whether we have used 600 watthour or in the future with 100 strings some 10 or 20 Kwhour.
Example: at 24 Volt and 5 strings x 5 Ampere (1C), it will take about 2,5 hour. At 0.1 C during the night. The capacitor will be charged via the batteries and strings.

You are indeed correct that it is only a problem to charge the cap, when the cap is connected for the first time. I have to charge the cap separately to approx 20 volt and only then connect it to the plus bar. Thereafter I will bring the voltage up for the LiFePO4 batteries. It totally depends at what voltage level the LiFePO4 batteries will be delivered. 2 Volt? 2,5 Volt? or flat, but then I can throw them away.

That is the reason why I still like NiMH batteries.
There is a new one on the market. 12 Amperhour at 1.2 Volt in a "D" can.

One should not forget, one has not to charge a battery, any battery, fast. There is a trade off between fast charging and developing heat in the batteries or to do it slowly and you need time to do so. I have lots of time.

Sparky it is an experiment. Expensive, yes, but whatever the outcome of the experiment, I will be able to use it either for the boat or for the house or for camping or for my business.

[brian eiland]

Developing an efficient battery management system for the Mercedes-Benz S 400 HYBRID

What goes into developing an efficient battery management system for the just-launched Mercedes-Benz S 400 HYBRID? The ECU’s algorithms were developed in a joint venture between Johnson Controls and SAFT. dSPACE TargetLink is used to generate the ECU software.
http://www.designfax.net/enews/20091201/feature-1.asp

[BertKu]

Very interesting. Some of the features of their BMS are the same as proposed for the "pipe" system. But again, the car-designers are more concerned about the cold countries then the semi/tropical countries. I am wondering how that 25kg pack will perforem at 34 degrees C ambient temperature, like here in the Karoo. Intriging how they will do the balancing with 35 batteries in series, love to know.

I received the 40 LiFePO4 and professor I need your help. The 40 are flat at the top and flat at the bottom, with the wrap around. It does not let me put those batteries easy in series. I solved it temperary by using a South African Rand 0.10 coin between the batteries. But I am concerned they will vibrate away. I could glue them with that special conductive glue. However would you be willing to post me 60 x 10 mm to 12mm in diameter. I could ship you a nice bottle of red South African wine or must find a way to pay you your cost.

The batteries were charged and are 3.27 Volt. Should I be able to get those 2200 Farad Caps, I have to charge the super caps to at least 3.3 Volt before connecting them to the batteries. Otherwise I will probably destroy the batteries.

[CDK]

I have finally downloaded the Chinese alphabet required to read the datasheet of the IFR32900 5Ah LIFE cell from Everwin. Not that the datasheet contains even one Chinese character, but it's the way Adobe Acrobat reader works.

Maybe I didn't read the datasheet correctly, but share my disappointment.

The nominal voltage is 3.2 V., but after consuming the 5 Ah (or 4.5 Ah if the factory had a bad day), there is only 2.0 V left, which is 62.5% of the nominal voltage. That means you have to either accept a steady decrease of available voltage (and current) or must use a PWM controller that simulates a LA cell (62.5% pulse width for a full battery, 100% for an empty one).
In other words: for a 12 V application you need 6 cells, not 4!

To charge the cell, you have to apply 2.5 A @ 3.65 V until the current has dropped to 50 mA, which takes approx. 4 hours, or 5 A under the same condition for approx. 2.5 hours. The latter is the maximum allowed for fast charging.
Assuming that the decay of charging current is linear, you have to put in 18.25 Wh to get 13 Wh out, an efficiency of only 71%.

After carefully charging and discharging with 2.5 A, there will be at least 70% of the initial storage capacity left, 3.5 Ah for a good one, 3.15 Ah for a poor one that still passed the factory exit test and costs the same.

Because you cannot distinguish a good one from a bad one, you should calculate worst case. When building a large pack with one poor cell tucked in somewhere, the whole array must be treated as 4.5 Ah instead of 5 Ah because the poor one will drop below 2.0 V, get a higher internal resistance and may be damaged during charging.

A naked cell weighs 188 grams, say 0.2 kg when provided with terminals and an isolating sleeve, so a worst case 12V/4.5Ah pack puts 1.2 kg on the scale.
A Panasonic 12V/4.5Ah weighs 1.45 kg., so the savings in weight are 17%. Is it worth it?

[erik818]

BertKu,
Why don't you softstart the capacitors the usual way with a current limiting resistor when connecting them to the batteries? After they have reached equilibrium, you shortcircuit the resistor with a switch.
Erik

[BertKu]

Originally Posted by erik818

BertKu,
Why don't you softstart the capacitors the usual way with a current limiting resistor when connecting them to the batteries? After they have reached equilibrium, you shortcircuit the resistor with a switch.
Erik

Hi Erik,

Yes indeed, we could do that. But remember it is only a problem for the first time, when the caps are connected to the plus bar and batteries.
Thereafter the normal charging of the batteries will take care for the charging and voltage of the caps.

Something I still have to figure out is whether I need a 1 MegaOhm resistor or lower resistor over each in serial connected caps to ensure that all voltages are equal per in series switched capacitor.

[BertKu]

Originally Posted by CDK

Maybe I didn't read the datasheet correctly, but share my disappointment.

The nominal voltage is 3.2 V., but after consuming the 5 Ah (or 4.5 Ah if the factory had a bad day), there is only 2.0 V left, which is 62.5% of the nominal voltage. That means you have to either accept a steady decrease of available voltage (and current) or must use a PWM controller that simulates a LA cell (62.5% pulse width for a full battery, 100% for an empty one).
In other words: for a 12 V application you need 6 cells, not 4!

Hi CDK, Yes I share your disappointment. I paid Rand 4313.80 for the consignment, but is an experiment. I hope to come up with a consept whereby I gamble that we will have better and improved batteries in the future.

But CDK, All batteries have an initial Voltage and an end voltage. Your SLAB (Sealed Lead Acid battery) will only provide 4,5Ah at 0,5C but as soon you apply 5C, your battery will only have 2Ah. This is not the case with LiFePO4 batteries. (I hope to prove that to myself, haven't tested that yet)

Originally Posted by CDK

A Panasonic 12V/4.5Ah weighs 1.45 kg., so the savings in weight are 17%. Is it worth it?

No, not at this stage. I pay for a SLAB Rand 88,- versus R 4313.80. But for me it is a learning curve. To experiment with making a different and simpler BMS, making a consept with loose batteries in a pipe system in a boat environment etc. I agree with you, I may have wasted all that money. But somebody elso may smoke, or drink a lot. I calculated that a smoker waste a Mercedes Benz every so many years. I don't smoke.

[CDK]

Bert, this wasn't meant as an attack on your project.
I just heard and read lots of boasting about the rapid developments, exceptional performance, blablabla, but never scrutinized a real data-sheet.

Of course all electrochemical processes have an initial and an end voltage, but we've gotten used to batteries which have these values fairly close together.
A NiCd or NiMh cell shows 1.35 V after charging, but the actual discharge bandwidth is from 1.25 to 1.20. I thought that newer technologies would also be that stable.
One could argue that the drop from 3.2 to 2.0 V offers the opportunity to precisely indicate the remaining charge, while in fact it is an enormous disadvantage.

[BertKu]

Originally Posted by CDK

Bert, this wasn't meant as an attack on your project.
I just heard and read lots of boasting about the rapid developments, exceptional performance, blablabla, but never scrutinized a real data-sheet.

No, I did understand that it was not an attack. I am also a little disappointed.

Originally Posted by CDK

One could argue that the drop from 3.2 to 2.0 V offers the opportunity to precisely indicate the remaining charge, while in fact it is an enormous disadvantage.

The 2.0 Volt is like the 10,5 Volt for a Lead Acid battery. If this is exeeded, the battery will be damaged for life. I don't think it is the linear end voltage, that maybe 3.1 Volt. I will have to test that. But somebody like Rick Willoughby could put more light on this. He has been experimenting with a LiFePO4 battery pack.

[BertKu]

Originally Posted by CDK

In other words: for a 12 V application you need 6 cells, not 4!

The curve is reasonable linear until you hit the last little bit of power in the battery, therefore 4 batteries are good enough for 12 Volt.

Originally Posted by CDK

To charge the cell, you have to apply 2.5 A @ 3.65 V until the current has dropped to 50 mA, which takes approx. 4 hours, or 5 A under the same condition for approx. 2.5 hours. The latter is the maximum allowed for fast charging.

CDK, I am now confused, I am making a charger which is 1C constant current charging. The current should not drop to 50 mA, but the Voltage will rise and the temperature. If the Voltage exceed 3.50V I will have to switch over to constant Voltage maximum 0.01C i.e. maximum 50 mA until Voltage is risen to 3.65 and then I have to switch the charger off. If I leave the constant current at 1C till 3.65 Volt, I will probably damage the battery for life.

To stay at the safe side I will probably start charging at 2,5 or even 3 Volt and stop charging at 3,55 to 3,6 Volt. Like I have said before I don't believe in pushing everything to the limits.

Rick, please, are you willing to explain to me what it actual means : balancing of batteries. If I charge 8 batteries of 3,2 Volt nominal in series. And I charge and discharge them every day (highly theoritical) with a constant current and make sure that the specs are not exceeded. Will after some months the 8 batteries have different voltages?? Or does that only happen when batteries are placed parallel and charging take place with 8 batteries placed parallel and each battery is measured by means of temperature and each battery is switched off seperately when temperature is reached.??

[portacruise]

Originally Posted by BertKu

The curve is reasonable linear until you hit the last little bit of power in the battery, therefore 4 batteries are good enough for 12 Volt.




CDK, I am now confused, I am making a charger which is 1C constant current charging. The current should not drop to 50 mA, but the Voltage will rise and the temperature. If the Voltage exceed 3.50V I will have to switch over to constant Voltage maximum 0.01C i.e. maximum 50 mA until Voltage is risen to 3.65 and then I have to switch the charger off. If I leave the constant current at 1C till 3.65 Volt, I will probably damage the battery for life.

To stay at the safe side I will probably start charging at 2,5 or even 3 Volt and stop charging at 3,55 to 3,6 Volt. Like I have said before I don't believe in pushing everything to the limits.

BERT, SOME CHEMISTRY BATTERY TYPES WILL LOSE CAPACITY IF NOT CHARGED TO FULL CAPACITY EACH TIME- LEAD ESPECIALLY. MIGHT BE TRUE FOR LITHIUM, HAVEN'T SEEN ANYTHING EITHER WAY. NICADS NOT AFFECTED BY PARTIAL CHARGING.

Rick, please, are you willing to explain to me what it actual means : balancing of batteries. If I charge 8 batteries of 3,2 Volt nominal in series. And I charge and discharge them every day (highly theoritical) with a constant current and make sure that the specs are not exceeded. Will after some months the 8 batteries have different voltages??

I'M NOT RICK, BUT THE ANSWER IS YES. THIS IS BECAUSE ARE INDIVIDUAL CELLS ARE NOT CLONES AND WILL HAVE DIFFERENT VOLTAGES WHEN MANUFACTURED AND ALSO BECAUSE ALL CELLS WILL NOT BE AT IDENTICAL TEMPERATURES WHEN GOING THRU THE CYCLES. YOU CAN PURCHASE "MATCHED" CELLS THAT ARE COMPUTER MATCHED TO WITHIN MAYBE A FEW MILLIVOLTS OF EACH OTHER, BUT THE TEMPERATURE WILL STILL MAKE THEM UNEVEN VOLTAGES AFTER A WHILE.


Or does that only happen when batteries are placed parallel and charging take place with 8 batteries placed parallel and each battery is measured by means of temperature and each battery is switched off seperately when temperature is reached.?? DIDN'T QUITE UNDERSTAND THAT ONE.

Hope above helps.

Porta