# How to effectively make use of a 12V 1.2Ah battery?

Discussion in 'Homework Help' started by Devika B S, Mar 8, 2017.

1. ### Devika B S Thread Starter Member

Mar 8, 2017
144
1
I have a 12V 1.2Ah rechargeable lead acid battery. I understand that by 1Ah, I can use the battery to supply 1A current in one hour time. So using this logic, I can use my 12 V 1.2Ah to draw 2.4A from the battery to last half an hour. But how exactly can I do this practically? How do a make my circuit (a boost converter) to draw 2.4A from this supply?

2. ### ErnieM AAC Fanatic!

Apr 24, 2011
7,957
1,827
Practically you do not want to do that. True the AH rating suggests you can but if you have the full specs on that battery it will inform you that you want to draw that current at a lesser value over a longer time, typically say 20 hours for a desired current of 60 mA.

Additionally, you never want to fully draing a battery. 50% is best but will work with reduced life up around 75%.

If you still continue to persist in drawing 2.4 amps set your load up to be around 28.8 watts. Less actually as that's for 100% conversion efficiency. (Power out = Power in, or 12V x 2.4A = 28.8 Watts)

3. ### ScottWang Moderator

Aug 23, 2012
6,458
1,000
In parallel is easier, otherwise you can't reach it.
Wi = Vi * Ii
= 12V * 1.2A
=14.4W

Wo = Vo * Io
= 12V * 2.4A
= 28.8W

Normally the Wo should be considering the efficiency of the converter as 90%~98% of the input voltage.
So how can this Wi < Wo happens?

4. ### Alec_t AAC Fanatic!

Sep 17, 2013
8,994
2,122
Welcome to AAC.
Your understanding is correct for a hypothetical perfect battery. In practice there are limitations. For example, there are internal losses in the battery and you shouldn't discharge the battery below ~11V if you don't want to harm it.
The converter efficiency and the load placed on it would determine the current drawn from the battery.

5. ### Devika B S Thread Starter Member

Mar 8, 2017
144
1
Ok.I understand that its bad. I just need to draw 2.4A current just for 5 minutes. Its for a project. But how do I do that? How can I control the amount of amps drawn by the circuit?

6. ### Devika B S Thread Starter Member

Mar 8, 2017
144
1
I need 25.5 watts exact for my circuit to function as per MATLAB simulation. All I did was to connect a voltmeter and an ammeter on the input side, multiply V and I and found that I need 25.5 watts for the converter to function. I have a 12V, 1.2 Ah battery. Is it possible? What else can I do regarding this? Please give suggestions. This is my first project.

7. ### ian field AAC Fanatic!

Oct 27, 2012
6,306
1,147
It doesn't work like that - the Ah rating is a convenient theoretical standard of how much energy a battery stores.

Can't remember which - but its either a 10h or 20h discharge rate at 1/10 or 1/20 respectively of the Ah rating.

If you draw 1A from a 1Ah battery - you'll get somewhat less than 1h run time.

8. ### Devika B S Thread Starter Member

Mar 8, 2017
144
1
OK, how do I select a battery that gives me the required 25.5 or 30 watt power?

9. ### Alec_t AAC Fanatic!

Sep 17, 2013
8,994
2,122
The simplest way is to use a suitably sized resistor (but a 50W resistor, allowing a safety margin, is going to be expensive). An alternative is to construct a constant current ciruit as the load, dissipating the power in, say, a big MOSFET on a large heatsink.

10. ### Devika B S Thread Starter Member

Mar 8, 2017
144
1
Yes thats for decreasing the amount of amps right? Here I need to increase it. Like, I want it to draw 2.4A when it usually supplies 1A.

11. ### Dodgydave AAC Fanatic!

Jun 22, 2012
7,651
1,271
Use a Constant current source, like an LM338k or two transistor mosfet, current is set by Rs resistor..

Last edited: Mar 8, 2017
12. ### Alec_t AAC Fanatic!

Sep 17, 2013
8,994
2,122
Wrong. I think you are confusing Ah (Ampere-hours) with A (Amps). The current drawn is what the load demands, which in your 2.4A case is much less than the battery can probably provide. Your battery could likely provide >50A through a short circuit (NOT recommended, sparks would fly), albeit for a short time.

13. ### ian field AAC Fanatic!

Oct 27, 2012
6,306
1,147
Unless you have a bias source at least 8V higher than the main rail; you'd be better off redesigning the circuit to fit an IRF9530 P-channel device - or put the N-channel one in the negative feed.

14. ### Devika B S Thread Starter Member

Mar 8, 2017
144
1
Yes, my load requires 2.4 A current from a 12 volt supply. My battery is 12V, 1.2Ah spec. How much current can this battery supply? (My simulation battery in MATLAB is a simple 12V battery with no Ah spec in it which is why I am so confused when I ordered the battery). I have done quite a number of MATLAB simulations. But this is the first time I am doing hardware.

This 2.4A value is the reading from an ammeter which I connected on the input side (i.e. immediately after the battery in my simulink diagram)

Regarding the circuit, I have a 100uH inductor, two 30uF capacitors, a diode, a logic level MOSFET powered by Arduino and a 100 ohm resistive load such as some 7 led lamps in series. My circuit is a boost converter that converts 12V to 48V.

Last edited: Mar 8, 2017
15. ### Dodgydave AAC Fanatic!

Jun 22, 2012
7,651
1,271
Why don't you use a better battery like a 2.6 or 7aH, that way it will last longer and will take the current longer.?

16. ### djsfantasi AAC Fanatic!

Apr 11, 2010
3,621
1,312
I found this article. I am not an expert, but following the article, I get the following.

You need 28.8 Watts of power. That is 12V times 2.4 A. Your battery is rated for 14.4 amps-hour. This suggest you could run your circuit for 30 minutes. But, all isn't 100%. Depending on the type of battery you have, you can't run it below a certain voltage. You need to specify battery type and calculate the time it will take to draw it down to a safe voltage b

17. ### ErnieM AAC Fanatic!

Apr 24, 2011
7,957
1,827
I see alot of noise here but very little signal.

What is this load you need to power for five minutes?

RichardO likes this.
18. ### tcmtech Distinguished Member

Nov 4, 2013
2,809
2,703
The simplest way t look at it by how many amps you need for a given run time then work that out to the equivalent of Amp per hour.

So if you need 2.4 amps for 5 minutes that's 12 amp minutes (12/60) = .2 Amp hours (200 milliamp hours) ~15% total usage of the 1.2 Ah available.

Also the general rule I use for figuring out what a typical good condtion lead acid battery of any design can handle for peak amps without excessive voltage drop and self heating effects to just take the Ah rating and multiply that by 10 which your case I would suspect that any current daw under 12 amps for a few minutes or less at a time wont bother it one bit.

Lastly batteries are a consumable since they do wear out so don't worry too much about how well or badly you treat one. In normal non critical applications consider if it works for the job it's obviously still good enough.

Dec 4, 2009
308
46
Your load determines the amount of current drawn from the battery. 1.2Ah spec. means that the battery can supply
- 2.4A for 1/2 hour.
- 1.2A for 1 hour.
- 0.6A for 2 hours.
Time = Ah capacity / Current in Amps.
But the Ah capacity is rated at 20 hours discharge, that is 60mA for 20 hours. If you discharge higher than that, the discharge time will be less than the computed time. Older the battery, the Ah capacity itself comes down.
Fully charge the battery, and connect your load that draws 2.4A. You could use it for about 20mins.

20. ### Devika B S Thread Starter Member

Mar 8, 2017
144
1
Its a high gain converter that converts 12V to 48V. I need to demonstrate this.