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