Golf buggy charges

[studentspark]

Charging a 48v golf buggy.
I've been trying to find out the charging requirements.
Basically how many amps a charger will require to charge up the batteries.
And how that current draw will change depending on the charge state of the batteries.

Any help appreciated

 

[EFLImpudence]

Do you just mean - What do you need?

or

Do you want to know how they work?

 

[ericmark]

In the main I work on 12 volt, so times 4 to 12 volt rating, so float at between 13.2 and 13.8 volt. If using stage charging then drops to 13.2 but if simple voltage control then 13.8 volt.

There are a number of ways to charge lead acid batteries, to fast charge normal is a series fixed currents, these change as key voltages are hit.
For example
3.8A until 14 volt
3A until 14.2 volt
0.8A until 14.4 volt
0.2A until volts drop to 12.8.
As the AH rate of battery change so do the currents and triggure voltages, if a battery is charged with a lot of current from 10% to 80% charged it does not upset it too much, but that last 20% has to be slow, so the better chargers bang it in to start with, then easy off at the end.

 

[davelx]

You might find this useful:

https://www.victronenergy.com/upload/documents/Datasheet-Skylla-Charger-EN.pdf

 

[ericmark]

The old idea was a latched relay, at around 56 volt it would reduce the charge rate, So a fork lift may start charging at 100A as the battery becomes part charged the current starts to drop, but at 56 volt the relay will reduce current to maybe 10A and start a timer, after 1/2 hour it then switches off.

The 1/2 hour is called an equalizing charge and it lets low cells catch up to high cells. This means the battery needs topping up once charged. The use of valve regulated lead acid (VRLA) means you can't top up batteries so the so called 3 stage charger has been replaced with 7 stage chargers an pulse chargers.

In the old days we had problems with golf trollies, people would fit car batteries instead of traction batteries, and the user would find he soon got down to 14 holes rather than 18 before batteries failed, and would try to return the batteries. However using a car battery instead of traction invalidated the warranty which upset them some what.

So what type of lead acid are you using? What AH are you using, using 54 volt fixed voltage would in fullness of time recharge the batteries, but how long are they on charge, be it fork lift, or milk float the aim was to recharge within 10 hours, if you have 24 hours it becomes easier.

 

[studentspark]

Im looking at a golf cart charging station.
The whole install is quite poor.
There are charging points for 12 carts, (Outdoor 13amp sockets, flex on a 13amp plug to blue commando type socket. Protected by a 100ma RCD!) but physically you can only fit 6 carts in.
They charge overnight, but quite often the 20 A MCB trips.
Basic maths looking at the charger inputs suggest overload, but Im not sure how much they do actually draw.
The chargers say 12-6 A input.
Its been in place for some time, so an overload issue would be apparent straight away,I would presume.
So thats the reason why I am wondering about the charging cycle amp draw.

Many thanks for the replies. Battery charging seems to be a bit of an art.

 

[JohnW2]

They charge overnight, but quite often the 20 A MCB trips. Basic maths looking at the charger inputs suggest overload, but Im not sure how much they do actually draw. The chargers say 12-6 A input.

Even if the current is sustained continuously, it should take at least 22.6A to trip a 20A MCB. That current at, say, 240V (easy maths!) would correspond (with 100% efficiency) to around 113A at 48A48V. The efficiency will obviously be well below 100% but, even accepting that, do you think that charging 6 batteries (well, 6 sets of batteries) would be using that much current? (it's a real question, not a rhetorical one).

Kind Regards, John
Edit: typo corrected

 

[EFLImpudence]

"113A at 48A"

Typo ?

 

[JohnW2]

"113A at 48A" Typo ?

Quite so - now corrected. Thanks.

I actually noticed that one just before I posted it, and resolved to correct it, but somehow forgot!

Kind Regards, John

 

[studentspark]

No i don't think it will.
Its obvious, I have a big hole in my understanding, which is embarrassing and frustrating me.

I see the relationship (x5), but if you could expand further, I would really appreciate it

 

[JohnW2]

No i don't think it will. Its obvious, I have a big hole in my understanding, which is embarrassing and frustrating me. ... I see the relationship (x5), but if you could expand further, I would really appreciate it

I don't honestly think that a lot more 'expansion' is really necessary.

Since power cannot disappear, the input power has to be equal to the output power plus 'losses' (inefficiencies in the transformation process, usually manifested by the 'lost' power turning into heat).

If one considers the simple hypothetical scenario with zero losses (100% efficiency), then input power will be equal to output power. Hence (assuming power factor = 1):

Input voltage x Input current = Output voltage x Output current

... or, re-arranged ...

Output current = Input current x Input voltage ÷ Output voltage

which you can think of as

Output current = input current x (Input voltage ÷ Output voltage)

... the bit in brackets being the ratio of the voltages - as you say, 5 in the case of 240V being 'transformed' to 48V

Does that help?

Kind Regards, John

 

[EFLImpudence]

Few queries:

Student says charger states 12-6A input.
It's on 13A plug
Do we know the output voltage? The batteries are 48V; isn't it likely to be a bit more?

 

[studentspark]

Thank you for taking the time. Very helpful
I have always struggled with transformer ratios, I see the principle, but it won't click.

 

[JohnW2]

Student says charger states 12-6A input.

He does. However, whatever it means (12.6A?) I poresume that, as with any power supply, it will be the maximum input current (at maximum output current)

Do we know the output voltage? The batteries are 48V; isn't it likely to be a bit more?

Sure - I was just presenting simplistic rough figures. Call it 60V is you want - hence a 4:1, rather than 5:1 voltage (and current) ratio.

Kind Regards, John

 

[JohnW2]

Thank you for taking the time. Very helpful I have always struggled with transformer ratios, I see the principle, but it won't click.

Does the 'clicking' work any better with levers/pulleys or hydraulics - since the principle is essentially the same?

For example, with a pulley system, distance moved ('current') multiplied by force ('voltage') (i.e. when multiplied = 'power') will be equal at both input and output sides of the system.

Kind Regards, John