Choosing right battery for circuit

Thread Starter

Track99

Joined Jun 30, 2022
18
Hello my friends. I have a rechargeable battery that is running a small circuit. I just want you to check the mathematics.

The entire circuit is drawing 1.2 Amps at 5 Volts. ( 6 Wh )
So I calculated that the circuit consumes 6 Watts of power in 1 hour.

I found a battery that is a 2S Lipo Battery 7.4V 6.5Ah. ( 48.1 Wh)

I have found that this 2S Lipo battery can provide me with 8.01 hours of uninterrupted power ( 48.1 / 6 = 8.01).
Can you tell me if the mathematics is correct?
Thank you
 

Papabravo

Joined Feb 24, 2006
18,978
Your units are incorrect.
\( \text{Volts}\times\text{Amperes}=\text{Watts} \)
not Watt-hours. Watt-hours are used to measure the capacity of a battery. That is how long can a battery deliver a specific amount of current for what period of time. This information is normally presented as a discharge curve which shows the evolution over time of voltage and current delivered to a fixed load.

As the battery runs down the voltage will drop and the current will increase in order for the power delivered to the load to remain constant. At some point the chemical reaction will be unable to maintain the power to the load and the voltage will begin to drop faster until the battery is naerly dead.
 

Ian0

Joined Aug 7, 2020
5,839
The figures are correct, but your units are in a mess.
Don't confuse units of power (Watts) with units of Energy (Watt-hours).
Your circuit consumes 6 Watts
6 Watts x 8 hours is 48 Watt-hours.

If you keep confusing power with energy, you'll probably become a science journalist for a daily newspaper and before you know it, you'll be working in units like elephants, double-decker-buses and the size of Wales.
 

Thread Starter

Track99

Joined Jun 30, 2022
18
Your units are incorrect.
\( \text{Volts}\times\text{Amperes}=\text{Watts} \)
not Watt-hours. Watt-hours are used to measure the capacity of a battery. that is how long can a battery deliver a specific amount of current for what period of time. This information is normally presented as a discharge curve which shows the evolution over time of voltage and current delivered to a fixed load.
Ok my friend. I fixed the problem.

The entire circuit is drawing 1.2 Amps at 5 Volts. ( 6 W )
So I calculated that the circuit consumes 6 Watts of power in 1 hour.

I found a battery that is a 2S Lipo Battery 7.4V 6.5Ah. ( 48.1 Wh)

I have found that this 2S Lipo battery can provide me with 8.01 hours of uninterrupted power ( 48.1 / 6 = 8.01).

Can you tell me if the 8.01 hours answer is correct?
 

Thread Starter

Track99

Joined Jun 30, 2022
18
The figures are correct, but your units are in a mess.
Don't confuse units of power (Watts) with units of Energy (Watt-hours).
Your circuit consumes 6 Watts
6 Watts x 8 hours is 48 Watt-hours.
Ok my friend. Thank you. Tell me please, in this case, if I have a battery that is a 1S Lipo Battery 3.7V 6.5Ah. ( 24.05 Wh) will the circuit still work?

I am asking because the circuit needs 5 volts.

What will happen if I give it only 3.7V?

Will the circuit work because at the end of the day, it needs 6W per hour to work and we are providing it with 24.05Wh?

OR

Will it NOT work because, even though the source is ready to give 24.05 Wh for 4.05 hours, the circuit needs a minimum voltage of 5V?

Thank you
 

Papabravo

Joined Feb 24, 2006
18,978
Ok my friend. I fixed the problem.

The entire circuit is drawing 1.2 Amps at 5 Volts. ( 6 W )
So I calculated that the circuit consumes 6 Watts of power in 1 hour.

I found a battery that is a 2S Lipo Battery 7.4V 6.5Ah. ( 48.1 Wh)

I have found that this 2S Lipo battery can provide me with 8.01 hours of uninterrupted power ( 48.1 / 6 = 8.01).

Can you tell me if the 8.01 hours answer is correct?
OK, my casual acquaintance. If the 2S LiPO battery has a capacity of 6.5 Ampere-hours, it will deliver 7.4 volts at that current level which is more voltage than your circuit requires. Assuming you can reduce the battery voltage to 5 Volts, the total regulator power including losses, plus the load power will give you something less than 8 hours. This is because no conversion scheme will be 100% efficient.

The estimated loss will be:
\( (7.4-5.0)1.2= 2.88\;\text{Watts} \)
 
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