questions on Sealed lead acid battery

Hello Friends

i am confusing about using Sealed acid battery and i have some questions, but before i ask i would to put what i am understanding to find out if i am right or wrong

i have lead acid battery 12 v/ 7.5 AH i want to use it for lighting 12 leds Red color
red color led VF = 2 V and IF = 25 mA
if i connect them in parallel , then the total current consumed by them = 25 X 12 = 300 mA
to calculate resistor in series it should be = (12-2)/0.3
R = 33 Ohm
power dissipation by resistor = IV = 10*0.3 = 3 W , so i should use 5 watt right ???
since the battery 7500 mA and the led consume 300 mA , so the battery can power leds for about 7500/300 = 25 Hours
my first question
1- Is the current consumption by leds i.e. 300 mA for one hour?
2- If i use 3 batteries with the same specification, so it will serve for 3*25 = 75 Hours right??
that was the first section of my question
second section

i want to build simple charging Sealed lead acid battery for the same one that i have mentioned
i want to use DC power supply for example 18 DC, if i am using charging current for example 500 mA from supply so the resistance between supply and battery is
R = 18-12/0.5 = 12 Ohm
power dissipated in resistor = 12 * .5 = 6 W
so i should use resistor 8 watt right ??

3- If i use 500 mA charging current, then it will take
7500 / 500 = 15 Hours for complete charging right??
4- That was simple and direct charging , but some sites mentioned the term overcharging what is that ?? is it over current or over voltage ???

5- i want to use that type of batteries to power a device and that device consumes 200 Watt-Hour
my calculations is
power delivered by battery = 7.5 * 12 = 90 watt-Hour
no. of batteries = 200/90 = 2.222 then i use 3 batteries
so 3 batteries can be used for one hour right ??
6- can i use that battery to power up light bulb(60 W) directly with no need inverter ?
Best regards

It might be better to have a resistor for each LED. The leds wont all have the same forward voltage and current. A led failing might take out the other leds.

(12-2)/0.025 = 400, so a 1/2 Watt 390R resistor for each led would be ok.

A lead acid battery should be charged with a constant voltage supply. Normally ~13.2 Volts.
The battery will draw less current as it charges. You should not have to worry about the charging current, although the power supply could have a current limit circuit.

Having a 18 Volt supply with a series resistor will not work, as the battery will draw less current as it gets charged and the charging voltage will rise.

andro said:

Hello Friends

i am confusing about using Sealed acid battery and i have some questions, but before i ask i would to put what i am understanding to find out if i am right or wrong

i have lead acid battery 12 v/ 7.5 AH i want to use it for lighting 12 leds Red color
red color led VF = 2 V and IF = 25 mA
if i connect them in parallel , then the total current consumed by them = 25 X 12 = 300 mA
to calculate resistor in series it should be = (12-2)/0.3
R = 33 Ohm
power dissipation by resistor = IV = 10*0.3 = 3 W , so i should use 5 watt right ???

Click to expand...

No... not at all... if you are using LEDs in parallel then you have to calculate the resistor for each LED with its current 25mA so that will be (12-2)/0.025 = 400 ohms with each LED ; 33 ohms will blow out your all LEDs by passing 300mA through each.

the power for resistor is 25m^2 * 400 = 0.25W so 0.5W resistor is Enough for that.

andro said:

since the battery 7500 mA and the led consume 300 mA , so the battery can power leds for about 7500/300 = 25 Hours
my first question
1- Is the current consumption by leds i.e. 300 mA for one hour?


2- If i use 3 batteries with the same specification, so it will serve for 3*25 = 75 Hours right??
that was the first section of my question

Click to expand...

1)No it is the current that all LEDs are consuming continuously.
2)yes

andro said:

second section

i want to build simple charging Sealed lead acid battery for the same one that i have mentioned
i want to use DC power supply for example 18V DC, if i am using charging current for example 500 mA from supply so the resistance between supply and battery is
R = 18-12/0.5 = 12 Ohm
power dissipated in resistor = 12 * .5 = 6 W
so i should use resistor 8 watt right ??

3- If i use 500 mA charging current, then it will take
7500 / 500 = 15 Hours for complete charging right??
4- That was simple and direct charging , but some sites mentioned the term overcharging what is that ?? is it over current or over voltage ???

5- i want to use that type of batteries to power a device and that device consumes 200 Watt-Hour
my calculations is
power delivered by battery = 7.5 * 12 = 90 watt-Hour
no. of batteries = 200/90 = 2.222 then i use 3 batteries
so 3 batteries can be used for one hour right ??
6- can i use that battery to power up light bulb(60 W) directly with no need inverter ?
Best regards

Click to expand...

1)yes but why you want to use the resistor between charger and battery? using power supply for charging a SLA battery is dangrous. you have to design a 3 stage charger for that purpose.

2) for charging 12V SLA battery ,,, 14VDC is enough... no need for 18V.. a sla battery got fully charged on 13.5V while fully discharged on 10.5V

3) yes right.

4) over charging refers to the supply of current from the charger after the battery is fully charged. This thing may damage SLA Battery while Li-ion batteries in Laptops and mobiles may explode.

5) yes right.. you have to connect 3 batteries in parrall for that.

6) in case of AC operated bulb the answer is NO.... for 12V DC operated 60W bulb,,, you can use it.

Regards

A series string of LEDs would minimize dropped-voltage losses, so you may connect up to 4 series LEDs for a total of 8-9V. For 12 LEDs we obtain 3 strings of 4 LEDs, each string using a series resistor = 120 – 150 ohms.

Power dissipation by each resistor = 3V - 4V * 0.025A= 0.075W up to 0.1W, so use normal 0.25W resistors.
Total power consumption from the battery 0.075A / hour. In theory full charged battery may last up to 100h ~ 4 days. :-D

Hi-brite RED LEDs are normally drawing 15mA max. with 1.4VDCto 1.6VDC as the forward drops .Recommended that you go for five LEDs to drop 8.0VDC to 8.5VDC max..
As such you can go for three strings with five in each and have 15LEDs lit up.
To minimize power dissipation use a PNP transistor 2N2905 with collector driving the string.Ie =1amp.

vimalkhanna said:

...To minimize power dissipation use a PNP transistor 2N2905 with collector driving the string.Ie =1amp.

Click to expand...

Can you elaborate? :roll:

Thank you all for reply

btbass said:

Having a 18 Volt supply with a series resistor will not work, as the battery will draw less current as it gets charged and the charging voltage will rise.

Click to expand...

do you mean if i use 18 volt maybe the charging volt arise and exceeding more than 12 ??
so if i use 13 volt and diode in series the voltage drop will be
13 -0.7 = 12.3 v to insure that when volt starts to rise up it the diode will open , so using diode as switch. Is it right?

A lead acid battery should be charged with a constant voltage supply.
You could use a diode, then increase the supply voltage so you have between 13.2 - 13.8 volts on the battery terminals when charging.

The voltage wont rise if you don't have a resistor in series.

Usually settings of the charge voltage is in ranges from 13.8V to 14.7V for a ''12V''(6 cell batteries). Once fully charged the battery should not dwell at the topping voltage for more than 24-48 hours and the applied voltage must be reduced to the float voltage level (13.5-13.6V). If you are using a charger that stays on topping charge and does not drop below 13.8V, you need to remove the charge after maximum 48 hours of charge.

Simple charger diagram here:

https://www.edaboard.com/threads/140976/

If possible ,why not go for 2x 12vdc/7ah sla..The voltage is 24VDC.
There is a single string an currents remain at 20mA.
The NPN/PNP bipolar device like 337 /338 can regulate the currents
as series regulator keeping the load currents constant .

andro said:

..... if i use 18 volt maybe the charging volt arise and exceeding more than 12 ??

Click to expand...

If you use 18V you will permanently damage the battery. btbass and Qaisar Azeemi have both given you more suitable values.

If you adhere to the previous replies you are ok, but be careful when dealing with power supplies like the lead acid accumulator. Do not short circuit the terminal and connecting the LEDs each with a seperate resistor of 0.5 watts will be better.

maybe you are right in the experience ...However , could you elaborate the statement re: not coupling with leadacid accumulators ...