Hello basically i have a 12v 1.3ah battery (you can see it here: http://www.maplin.co.uk/sealed-lead-acid…
and have two devices which together draw 2.5A per hour...am i right in thinking that this battery would die in less than an hour or is the 1.3Ah something totally different.
I have basic understanding, so please keep answers basic and to the point - all i want to know is how long would the battery last with a 2.5A draw
thankyou
and have two devices which together draw 2.5A per hour...am i right in thinking that this battery would die in less than an hour or is the 1.3Ah something totally different.
I have basic understanding, so please keep answers basic and to the point - all i want to know is how long would the battery last with a 2.5A draw
thankyou
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Yes it will last longer than 30 minutes. How long the 12 Volt,1.3Ah battery will furnish 2.5 Amps at 5 Volts depends on the efficiency of 12V to 5V converter.
Power required at 5 Volts = (5V)*(2.5A) = 12,5 Watts
Amps required at 12 Volts to furnish 12.5 Watts = 12.5W/12V = 1.042 Amps
Maximum time the 12V,1.3Ah battery can last furnishing 1.042 Amps of current without recharging = 1.3 Ah/1.042A = 1.248 Hours.
Actual time = (efficiency factor of 12Vdc to 5Vdc converter)*(1.248 Hours)
Note: The efficiency factor of the 12V to 5V converter = Converter output power / converter input power = [(5V)*(converter output Amps)] / [(12V)*(converter input Amps)]
Power required at 5 Volts = (5V)*(2.5A) = 12,5 Watts
Amps required at 12 Volts to furnish 12.5 Watts = 12.5W/12V = 1.042 Amps
Maximum time the 12V,1.3Ah battery can last furnishing 1.042 Amps of current without recharging = 1.3 Ah/1.042A = 1.248 Hours.
Actual time = (efficiency factor of 12Vdc to 5Vdc converter)*(1.248 Hours)
Note: The efficiency factor of the 12V to 5V converter = Converter output power / converter input power = [(5V)*(converter output Amps)] / [(12V)*(converter input Amps)]
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Constantly the battery can supply 1.3A at 12V for 1 hour.
If you take out more energy with more current then the stored energy in the battery runs out sooner - and in ratio to the currents used and quoted battery capacity
So it will last
1.3 / 2.5 of an hour (about 32 minutes but you can do the accurate sums)
If you take out more energy with more current then the stored energy in the battery runs out sooner - and in ratio to the currents used and quoted battery capacity
So it will last
1.3 / 2.5 of an hour (about 32 minutes but you can do the accurate sums)
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12v×1.3a=15.6watts
5v×2.5a=12.5watts
15.6÷12.5=1.2
Ideally (good battery condition, full charge, no power use by converter) an hour and 12 minutes.
Realistically, 30 minutes.
5v×2.5a=12.5watts
15.6÷12.5=1.2
Ideally (good battery condition, full charge, no power use by converter) an hour and 12 minutes.
Realistically, 30 minutes.
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... and don't forget the current draw of the dc/dc converter!