Is a B&D Simple Start safe to use?

[MSS]

Originally Posted by SD1too

I think the confusion arising from some recent posts arises from the incorrect use of terminology.

"Charge" is a term derived from static electricity. As Harry pointed out, a capacitor can be charged. When we use the same word in connection with a lead acid battery it is a different principle entirely. Such a battery is a collection of acid and metals which, when an electrical load is applied, undergoes a chemical reaction to produce an electric current. When we "recharge" it we are causing this chemical reaction basically to reverse. The battery does not hold "charge" in the same way as a capacitor does. This is why the above quote is fundamentally incorrect and the analogy of buckets of water does not work.


Here we have another extreme generalisation: "energy", which only serves to confuse. Electricity cannot be stored, and a battery does not store it. The process is chemical, more akin to "generation" I would say, as described above (documentary source: Lucas). Use of the term "energy" is too wide-ranging and vague to be useful in this discussion about how the Black & Decker 'Simple Start' works. I think we are making progress with this. It appears to have more sophisticated circuitry than first thought but, whilst we await Vinnie's measurements, it would be helpful if we restricted our attention to the appropriate fundamentals of voltage and current, avoiding "charge" and "energy" which only confuse and complicate unnecessarily.

Simon

Simon,

I think the "problem" here is that there are two parallel exchanges taking place in this thread. Firstly, your attempt at getting to the facts with Vinnie. For that, I agree that voltage and current are the only variables that should be used as they represent the correct terminology.

The other exchange is more entertainment than fact based and centres around the volume of the entity that is transferred between the two batteries when connected together. For this current and voltage will not suffice and thinking of the battery as a charge reservoir with the input/output characteristics of a lead acid battery (as opposed to a capacitor or indeed a bucket) is reasonable and helpful.

It's for the above reason that I actually posted two successive entries - one aimed at the "enterainment" and the other at the fact finding exchanges.

Maninder.

[MrDoodles]

So is it safe to use or not then?

[SD1too]

We're waiting to hear from Supervinnie40 with the device's output voltage.

Simon

[HarryM1BYT]

No matter how many times some of you guys repeat it, it still will not make it so....

A flat battery cannot and will not ever be charged, or even partially charged, from a fully charged one of the same voltage simply by connecting the two together.

All you will get, is the voltage of the flat battery raised to match that of the charged battery temporarily. No actual energy will be transferred. Why do you think the B&D device includes three batteries, in its attempts to make a device which would be able to charge a car battery?

If you doubt any of the above, then all you need do is try it.

[kaiser]

Originally Posted by HarryM1BYT

No matter how many times some of you guys repeat it, it still will not make it so....

A flat battery cannot and will not ever be charged, or even partially charged, from a fully charged one of the same voltage simply by connecting the two together.

All you will get, is the voltage of the flat battery raised to match that of the charged battery temporarily. No actual energy will be transferred. Why do you think the B&D device includes three batteries, in its attempts to make a device which would be able to charge a car battery?

If you doubt any of the above, then all you need do is try it.

I have given you good reason why it should, and I have asked two simple questions.
You are talking against physics here, and I cannot accept that, unless you can substantiate it.

So, the questions are quite simple an answer to that will suffice, continuing to repeat some unsubstantiated dogma, will not.

[COLVERT]

Originally Posted by SD1too

That's an interesting interpretation Colvert. If true, it's the first specification I've seen of a piece of electrical equipment which describes not the product, but the apparatus to which the customer connects it. But I have a question related to your theory. Here is part of the Black & Decker spec:

12 volt DC outlet connects the booster from inside your vehicle without any need for jump leads
12 Volt DC accessory outlet ideal for charging mobile phones, cameras, mp3 players and all other 12 Volt electronics

Now why would the first 12v DC outlet described be not on the B&D equipment, yet the second 12v DC outlet is on the B&D equipment?
If you're right, that's extraordinarily bad technical writing. But I'm afraid that I don't think you are right. As far as the English language is concerned, a 12v DC outlet to connect the booster is describing a socket on the B&D booster, because it's an output, i.e. the current flows out of the B&D product. In this situation the part in the car (the cigar lighter or 12v accessory socket) is the input.

Simon

Agreed that the car socket is being used as an INPUT in this case but it still remains a car OUTPUT socket.

Chuck away the booster and what do YOU call the lighter socket fitted to the car ??? ( Input or output ? )


Colv.



PS. Could very well be as you suggested-----Badly written suppliers info. Especially if its Chinese made. Lol. )

PPS. Going into detail a little bit more.

The confusion here is that we all know inputting 12 volts into a 12 volt system will not charge a battery ( In this case. )

It is sensible to accept the fact that the booster is outputting 18 volts and is doing so into a 12 volt socket on the car. The fact that you have changed the normal function of part of the car does not mean you also have to change its name. When you have finished playing with the socket it is still what it was---an OUTPUT socket.



PPS. If you think I'm repeating myself here it's only cos I'm drunk. ( Hic. )

[COLVERT]

Originally Posted by kaiser

Thermodynamics, pressure, heat, charge, levels, ANYTHING!! in nature, connected in a fluid form, will equalize if given time. I am not able to think of ONE, not one! single exception.

It is true that a voltage difference needs to exist, to drive a charge, but when a voltage difference does exist, a charge will always flow.

To say that the one battery is able to charge another is thus perfectly true, but it will of course never be able to charge it fully!

A graph of voltage as a function of charge has already been shown, and it shows that the battery voltage will be increasing directly, although not linearly, with charge. The line is at no point level or linear!
That means that there WILL be a difference in voltage between a charged and a discharged battery, and it will persist, until the charges and the voltage are equal!.
The voltage difference will be gradually reduced, as the voltages (and thus charges) of the two batteries approach, but I can see no reason whatsoever for it to cease at all, before the voltages are equal.

I don't know where and how any other stop would come into the process, (apart from natural discharge exceeding the differential in residual charge), and that is the flaw in Harry's argument.

If, Harry, you would educate us on that point, I (for one) would be most interested.

But let me formulate the question very simply thus:

1. Are there, always and at any time, an increasing voltage with an increasing charge of a battery ?
2. Will current always flow from higher to lower voltage, in a connected system.?

the argument can be settled from the two questions above, and if the answers are yes to both questions, it requires the interference of an unknown factor into the equation, one that I am not aware of, and one that I would be most interested in getting to know, if the process were to stop while any potential difference exists.

We can quickly agree that a fully charged battery will never charge another battery fully, that is impossible. But a fully charged battery will charge a flat battery, until they both share the same voltage and charge, given enough time, (and that will be increasingly long towards the end).

The sum of the two charges will be equal to the original charge of the fully charged battery, minus what has been lost in transmission.

The unknown factor.

Which Harry has mentioned is ------

There is a very fast build up of voltage on the discharged batteries plates. This stops the transfer of energy even though the CAPACITY of the charged battery is far far greater than that of the discharged battery.

Please don't theorise on it--try it for yourself. You HAVE to take account of the CHEMICAL changes going on inside the battery. This you are not doing.

------------------------------------

The natural state of a battery, by the way is discharged !!!

By charging it we make a chemical change take place inside the battery.
When we place a load across the terminals we give the battery the opportunity to return to its NATURAL state. ( Just thought I'd add this little snippet. Lol. )


Colvert.

[HarryM1BYT]

Originally Posted by COLVERT

The unknown factor.

Which Harry has mentioned is ------

There is a very fast build up of voltage on the discharged batteries plates. This stops the transfer of energy even though the CAPACITY of the charged battery is far far greater than that of the discharged battery.

Please don't theorise on it--try it for yourself. You HAVE to take account of the CHEMICAL changes going on inside the battery. This you are not doing.

------------------------------------

The natural state of a battery, by the way is discharged !!!

By charging it we make a chemical change take place inside the battery.
When we place a load across the terminals we give the battery the opportunity to return to its NATURAL state. ( Just thought I'd add this little snippet. Lol. )


Colvert.

I do believe we are both wasting our energy trying to get the 'flat earth society' members to understand even the most basic of principles of batteries, so for this reason alone, I will unsubscribe rather than continue.

I am especially concerned about a certain member who keeps trying to suggest he may have a degree, to help bolster his argument, yet seems not to possess even schoolboy knowledge of the subject he professes to understand.

[Supervinnie40]

Can we (again...) refrain from making personal comments. Saying things like 'flat earth society' and/or 'he may have a degree' is pointless and only creates a hostile atmosphere. You don't have to agree with somebody else, but that doesn't mean you need to make it personal.

You might consider yourself to be right Harry, but you are not the one with the best manners. Others have criticized you, but rarely has anybody tried to attack you personally, yet you attack others on a personal level. It would be great if you could just keep it ontopic and talk about the discussion and not fall into a personal war.
I'm sure other won't do it if you won't do it.

Back ontopic:
I haven't had a chance (yet) to repeat the test with a lightbulb. I'll try to do it as fast as I can. Besides that test I'm actually trying to work something out that will mimic the situation in which a car battery is drained (not empty).

[Supervinnie40]

The test with the lightbulb didn't work.
So here are the next best things:

If the device contains 3 battery's, and it should give out 18v, surely each battery should be 6v and the total should be 18 volt. (this doesn't mean that the output of the device is 18v, it could still have some circuitry that lowers the output to 12v)

These are the results:


Now this is weird... shouldn't it be 18v?

Let's see what each battery gives.




Seems one of my batteries isn't performing the way it should.
What does the label say?


Okay, one of my batteries isn't in good shape.

4.30+6.52+6.54=17.36. (everything checks out then...)

So what is the other 'next best thing'?
I took out the 2 batteries that are almost equal (6.52v and 6.54v) and put a lightbulb on one. The other nothing.
In an hour orso, the one with the lightbulb will be a bit drained. Mimicing a drained car battery.