Charging from a 12v battery + general charging issues

Do not discharge any NiMHs you will ruin them.

Thanks Yith.

Luckily, you won't need to worry as you now have an 'intelligent' charger. Set the left dial to 7 cells, the right dial to 600mA (or 300ma for the non-1400 batteries), switch on and plug your battery in. When the light turns green you are done! Don't worry about over charging either as it has now dropped to a really low trickle charge and safe to turn off when you are ready (heh, I don't mean after a four week cruise!).

Thanks again CW.

My new OMNI intelligent charger now received and set to 7 cells and 300 mAh, 8.4V 1000mAh NiMh battery connected, charger plugged into mains and switched on.

According to the charger's box info, as my power source is 230VAC and I have set to 7 cells my output voltage is 9.8V

Where C = Battery Capacity, the box info suggests a charging current of between 0.25C - 1C

As I understand it, NiMh and NiCd trickle charging is normally at 10% of battery capacity, ie 0.1C (but I am not trickle charging, I am fast charging until the battery reaches its delta peak whereupon the charger switches to a trickle, which explains the suggested rate of 0.25C - 1C).

Again referring to the box info, a 1200mAH battery at 0.4C requires a rate of 480mA (ie, 1200 x 0.4 = 480) so for my 1000mAh battery, selecting the charge rate of 300mA gives me 0.3C (ie, 300 / 1000) which is within the recommended range of 0.25C - 1C.

Hopefully all this is right so far but this is where I start to get a bit unsure.

According to the box info, a 2 cell battery pack of 2.4V is given a charge voltage of 2.8V. Thus when on mains input, with the dial at the 2 cell setting the charger is shown as producing an output of 2.8V, but more importantly for my intended use, it's outputting 9.8V at the 7 cell setting.

So, for my 7 x 1.2V cell battery pack totalling 8.4V, the charger is pushing 9.8V through, which is 116.67% of the battery packs voltage (as 2.8V is 116.67% of 2.4V).

CW has already written about the ineffciencies of charging and I read somewhere that for fast charging NiCd are only about 80% efficient meaning that to charge a 1000mAh battery in 1 hour, 1200mA need to be pushed through (which perhaps coincidentally equates to the factor of 1.2 which CW quotes in a post above, from which I take the equivalent ratio for NiMh to be 1.4).

Watts = Volts x Amps

So, using the box info example as basis to work from, to achieve a charged state in a 2.4V 1200mAh battery pack of Watts = 2.4V x 12A = 28.8W, then the charger needs to output Watts = 2.8V x 12A = 33.6W, ie 33.6W into the battery to achieve a charged state of 28.8W, ie an input ratio of 116.67%.

As the process is inefficent some of those watts are lost to heat etc. So, presuming that the input and ouput watts are the same, then the inefficiency loss will be in the Amps, with the input voltage having been increased to compensate for this.

In other words:

Fully charged battery state = 2.4V x 12A = 28.8W

Output from charger to battery = 2.8V x 12A = 33.6W

The OMNI's charging process is geared to compensate by a percentage ratio of 116.67%; in other words it is notionally 85.7% efficient (ie 100 / 116.67 x 100).

After efficency losses, input charge through battery = 2.8V x 10.29A = 28.8W

Note that 10.29A is 85.7% of 12A.
And note that the ratio of 1.17 (or 117%) is very close to the NiCd constant of 1.2.

As the thread originally suggested, I wanted a 12V charging capability. From a 12V power supply (input) the OMNI is labelled as having a maximum output of 8.4V, which it says is good for 6 (ie not 7) cell battery packs. Presumably though, putting through a lower voltage simply means that the battary will charge but not to its optimum.....such that, simplistically, 9.8V into a 8.4 pack gives 100% charge, so 8.4V into an 8.4V pack will only ever give about 85.7% charge, which would still be useful?

And in all of this I don't need to worry about time as the charger will sort that bit for itself.

Right???????

Sorry for the length of this - I completely understand if no one fancies a go at an answer!

cor dear... thats complicated... with a propeak, you just plug it in and press a button!

Yes, the charger at 600Ah (or 300 for a low capacity battery) is fast-ish charge but not a flat out blat charge many chaps do (and you frequently see done at site safezones to save the day and where having a battery that can be used takes priority over a well-looked-after battery).

After that you might be over-thinking the issue. Put at volt meter on a freshly charged '8.4v' battery (7 cells @ 1.2v per cell) will actually show anything up to 10.5v actual. Don't worry, in fact ignore it!

So, for safe and gentle intelligent charging of low capacity 8.4v batteries (say 1400Ah) use 300 setting, beefier batteries (say 3300Ah) use 600 setting. Plug in and remove when light turns green, couldn't be easier.

Err, beyond that, what exactly is the question?

Blimey - that was quick. Thanks.

you might be over-thinking the issue

Wouldn't be the first time.

what exactly is the question?

Fully accept that all I need to do is as you've previously said (which is damn easy, set dials, plug in, switch on).

For reasons of personal education I just wanted to understand what was actually happening. So I guess the question was..... is my understanding right?

Cheers for your patience!

Ooh-err, delta peak charge sensing isn't my area of speciality (that's why I like a charger you just plug in and forget - I understand red, amber and green lights ) but I rather suspect McV will be happy to deal with the maths...

No worries. Thanks for help so far tho.

Sorry! Been on holiday, am back now....

Well, for your own educational indulgence, woodlander (and anyone else interested), here's a brief explanation of that mysterious phrase "Delta Peak".

Delta, in maths terms means a change, or transition in state or value. so "Delta Peak" means the change in the charge voltage at it's peak (peak as in maxima, 'highest point') As a battery charges, the potential (voltage) stored within - due to the charging cycle - increases. This increase is a positive change (a +ve Voltage delta).

When a battery is fully charged it goes into a state of overcharge (which isn't a good thing). This is due basically, to the internal components of the battery changing state (usually producing gas) and it is this charge in state that leads to battery to "become hot" when nearing or passed the end of its full charge cycle.

Due to the charge in the property of the batteries, and the change in power transfer (heat, instead of stored potential) the change in charge voltage (voltage delta) stabilises at full charge, and then starts to drop off in overcharge - a decrease in voltage potential (a -ve Voltage delta).

Now, these clever 'Delta Peak' detection chargers sense this decrease in stored voltage/potential (while proving a charge current out of the two terminal wires, the charger is always monitoring the potential available on the battery across these two terminals too) from a positive (+ve) to a negative (-ve) change (delta) at this maxima (peak), and turns off the fast charge current and associated circuitry, and - with some chargers - switches back to trickle charge.

Hope that's made sense!
Haven't read it back to myself...

As I suspected - to summarise, plug battery in and switch off when the light turns green...

Haha, nice one McV, knew you wouldn't disappoint!

Thanks for indulging me McV- most kind and useful!

Would it be shameful for me to ask another question?

Have put my 1000 mAh and 1050 mAh NiMh batteries through the new 'intelligent' charger with much better results than in using the old dumb one (and no, I'm, not referring to me!). Both have significantly higher starting voltages than I have achieved before......though this does drop away (I think I read somewhere that NiMh batteries lose 25% of charge in the first 24 hours).

So, to my question. As it nears the end of the charge the 1050 mAh battery gets significantly warmer ... in comparison to the 1000 mAh it gets hot and when tested with a multimeter it shows a lower 'charged' voltage than the 1000..... why is this? A previously damaged cell, maybe?

PS: Welcome back!

The difference between the 1000mAh and 1050mAh should be very insignificant (a difference of 50mAh is effective nothing and most of it can be considered swallowed up due to tolerance issues when comparing two batteries).

I would just put the warmer feeling of the 1050mAh batt' down to manufacture quality and/or less significant negative delta/change allowing the battery to overcharge a little longer before the chargers delta paek detection circuitry kicks in.

Just as an aside, here's a cr*ppy little pic' I've made to show a charge curve, and the effects of the delta peak detection as it kicks in. Please note, the charge voltage shows itself staying the same, but before (to the left of) the dotted line [indicating the delta peak detected] it's a high current (fast) charge, and after (to the right of) the dotted line it's a low current (trickle) charge:

The axis of the graph (which I forgot to stupidly label) should be: X = Time, and Y = Voltage.

Thanks McV. Much appreciated.

This is actually quite interesting.

McV - your graph would seem to indicate that if you cut off the charge say half way through, half way along the x axis, the potential available in the battery would only be about half. Does that mean that if you only half charge a 10V battery you only get 5V? Which would then mean, since resistance in the gun is constant, you only get half the current. But you can run a motor on a (very small) part charge. Surely the voltage stays relatively constant until it collapses at the end of its life in a NIMh?

Or am I reading the graph wrong?

Or more likely displaying remarkable ignorance of all things electrical....

EDIT - I get that the graph is right, what I don't understand is how come a battery can work the same at half, and lower, voltage.

Sorry, yeah, I see that I've correctly labeled the red line as a Potential Energy, and the blue line and the axis incorrectly as Voltage. they should be all energy potential, in this case Watts, which is 'VoltAmps', not just volts.

EDIT - I get that the graph is right, what I don't understand is how come a battery can work the same at half, and lower, voltage.

Sorry, so with this in mind, the voltage remains roughly/theoretically the same, it's just there isn't enough current to drive. The Delta Peak detection can tell it's the measured voltage potential caused by the increasing energy held by the battery as the current supplied from the charger is constant.

Voltamps or power makes more sense to me and would explain/answer my questions above, but having done a quick google everybody else in the world labels the y axis as cell voltage too.

So I think you've labelled it right McV.

But that just doesn't tally with being able to part charge a battery and get the same performance as a full charge, just for less time.

I now officially understand even less about electricity than I thought I did.