Is a B&D Simple Start safe to use?

[HarryM1BYT]

Originally Posted by VMax1000

Oops.......no need to get personal here.

Stored energy will equalise as will 2 buckets of water.
If you have proof of the opposite, please share.

Don't be so silly, these are batteries. I'll ask you again - take one flat battery, one fully charged one, connect them + to + - to -. then come back and tell us whether your 'buckets' equalise. They will not, but enjoy the waste of time.

But as you said before, I don't know about qualifications of people, nor their background, but they haven't given any proof either.

Only you have at least twice in this forum alluded to having a degree. I don't need to, I am very well known and some are already aware of my qualifications.

So if you would care to explain why it wouldn't work as 2 buckets of water
(and please include the formulas, as I can understand those) feel free.

There are no formulae, why would there need to be, it is basic common knowledge, first year Electrical Engineering and common sense. It was something I discovered long before my teens and was bright enough to question why.

A charge, even a partial charge, cannot be transferred from one secondary battery to another of a similar voltage. It needs a substantial difference in voltage to enable the current and the energy to flow.

You can however connect two capacitors together and achieve the charge equalisation you dream of.

[SD1too]

Originally Posted by COLVERT

The 12 volt DC outlet is a description of the part that is IN YOUR car.

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

[SD1too]

Originally Posted by VMax1000

... could you post a link to the site containing the formulas rather than the formulas themselves?
Allows for context you see.

But Herman, you asked me to do this, but then you said that you didn't accept the relevance of the link. So what's the point?

Simon

[MSS]

Originally Posted by mss

Here you go.








=


>> stored energy equalisation only applies to linear capacitive networks.

>> Buckets no equal battery.

QED

Originally Posted by VMax1000

Interesting, but could you post a link to the site containing the formulas rather than the formulas themselves?

Allows for context you see................

That was intended to be a bit of light hearted input given your repeated ask for formulas.

The formulae above are just random extracts from maths associated with Schrodinger's equation about the probability of a partcile such as an electron being in a certain place at a certain time.

As for providing mathematical proof of what a number of us are saying, I dare say that such mathetical modelling could be attempted by first modelling the physical processes in the battery that lead to the possibility of the movemnt of electrons and then modelling the probability of an ensemble of electrons being in battery B rather than A. I might have had a go 25 years ago but now have other things to enjoy in life, such as driving my 75's.

The problem with what you ask for is that for a person to understand what would be highly complex mathetical modelling, the person would have to have such knowledge that he would not have to ask for it in the first place. He/she would also easily understand how lead acid batteries and their charging processes worked.

Why not do the following simple test?

Directly connect a fully charged 12V battery to an identical one that is fully discharged. After say 24 hours, place two 55W bulbs across each battery and measure the current flow as well as time for which the bulbs remain glowing.

As we know, Q=IT Coulombs, so multiply the measured current by the time for which the bulbs remained glowing and you will have the charge that existed in each battery as a result of the process of one charging the other. I can guarantee that the answer will not be the same for the two batteries.

[MSS]

Originally Posted by Supervinnie40

Lol, you're free to criticize me if you wanted.

But the 0v was measured with my multimeter.
I will make a video later to try and capture what happens when I turn it on. I'm just a bit busy (2 cars that need to get done with a very tight deadline...).

It may be that the SMPS is driving the output to the battery and it's not "kicking-in" due to there being no external load.

Vinnie - could you repeat the test suggested by Simon but with a low load on the output that would be conneted to the battery to be boosted. The best load to use would be a 5W interior light bulb.

[SD1too]

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

Originally Posted by kaiser

Two batteries .... will perfectly equalize their charges given time ... Current will flow until the charge density is equal ...

"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.

Originally Posted by VMax1000

... it has nothing to do with capacity or pressure, but with stored energy.

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

[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.

[Supervinnie40]

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:

Simon

You're right. There is a socket on the top of the device. You can use it to power accessories like a phone charger, a flashlight etc. When this socket is in use, the battery pack can't be used as a start-aid.
There is also (obviously) the 12v plug with a bit of wire. This can be used to either charge the device or to use it as a 'booster'.
You can only use 1 function at a time.

Originally Posted by HarryM1BYT

There are no formulae, why would there need to be, it is basic common knowledge, first year Electrical Engineering and common sense. It was something I discovered long before my teens and was bright enough to question why.

I don't wanna get caught in the discussion to much, but I don't believe that. In science pretty much EVERYTHING can be explained with equations. Without equations, you can't predict what some is going to do, and the big power of science is that it can predict exactly what is happening.

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.......

Simon

I've already said this several pages back. In post #30 (page 3) http://www.the75andztclub.co.uk/foru...1&postcount=30.

Originally Posted by mss

Vinnie - could you repeat the test suggested by Simon but with a low load on the output that would be conneted to the battery to be boosted. The best load to use would be a 5W interior light bulb.

I could try. If it works, do you want me to do something? Like see how long the bulb burns?

[SD1too]

Originally Posted by Supervinnie40

There is also (obviously) the 12v plug with a bit of wire. This can be used to either charge the device or to use it as a 'booster'.

Thank you for clarifying. This is the "12v outlet" mentioned in the spec. in my view, not anything attached to the car as Colvert suggested.

I've already said this several pages back. In post #30 (page 3) http://www.the75andztclub.co.uk/foru...1&postcount=30.

You did indeed. I felt that, judging by some recent posts, your request needed repeating.

I could try. If it works, do you want me to do something?

Yes please; whilst the bulb is burning measure the voltage across it and tell us what it is.

Thank you, by the way, for confirming the reading of 0v when connected to a charged vehicle battery. That's unheard of with mains powered chargers, but it's possible that it was chosen for this device due to its limited battery capacity.
Did you manage to recharge the unit's internal battery pack successfully? What's the voltage reading now (you previously posted a shot of your multimeter reading something like 4v I think)?

Simon

[Supervinnie40]

Originally Posted by kaiser

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.

Actually, one of the things I remembered best at science-class is that energy is attracted by a lower point of energy. Just like how water flows towards the lowest point. Energy is not flowing towards the lowest point, but is attracted by the lowest point.

Originally Posted by kaiser

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!

I concur good sir... (puts pipe in his mouth, tries to look smart and soffisticated )

Originally Posted by kaiser

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.

Yes, but I would also like to point out that the different in power between both batteries could have an affect on how long and how well the charge gets shared between batteries.

The way I was taught about this type of stuff:
Battery 1 is 12v.
Battery 2 is drained: 9v.
Apart from batt 2 being drained, they are exactly the same in any way.

Because there is a difference, the higher voltage of batt 1 is pulled towards batt 2. In the beginning, the difference is big and the 'flow' will be fast. After a little while, the difference becomes smaller, and the 'flow' will be much slower.
When you put a multimeter at the poles of the drained battery, it will show the highest voltage in the circuit; i.e. it will show 12v, but will go down with time. If you disconnect batt 1 from batt 2, and you put your multimeter on the poles of batt 2 (the drained one), it will be slightly more than 9v (depending on how long batt 1 was connected).
The difference will be big in the beginning, but the 'flow' will slow down when they are getting close to equal.

What Harry is saying (I think...)
Eventually, batt 1 will have dropped from 12v to (approx) 10v (or something like it). Batt 2 will have risen to about the same and will show 10v (or something).
Batt 1 has dropped 2v, but batt 2 has only gained 1v.

What the other are saying:
Eventually, batt 1 will have dropped from 12v to 10.5v. Having dropped 1.5v.
Batt 2 has risen to 10.5v, having gained 1.5v.
They both have lost/gained the same, and are now equal.

In any case, both battery will never reach full capacity.