# Math Regarding Battery Discharge

Good Day everyone :)

I'm new to electronics and it'll be my 2 week practicing it as my hobby though i've done a lot of DIYs and stuffs just by following tutorials but when my friend ask me regarding the maths of my projects, i can't retaliate at all since even i researched for 2 days, i haven't gone far to understanding the math specially on batteries.

I have made a DIY regarding USB charger and its block sequence is like this:

12v Sealed Lead-Acid Battery -> 5v Voltage Regulator Circuit -> 2x (Li-Ion TP4056 Circuit Module) -> 2x Smartphone Battery

I want to know how would i compute this entire project with regards how much Ah, Volts, power, battery consumption, etc. if there will be 1 smartphone that will be charged and if it will be 2 smartphone also.

Some specs of this entire project are as follows:

a) 12v SLA Battery 10Ah

b) 5v Voltage Regulator 1A output

(i don't know the efficiency of voltage regulator, i just used LM7805 https://www.sparkfun.com/datasheets/Components/LM7805.pdf)

(my 5v voltage regulator circuit is like this: https://i.stack.imgur.com/beL4b.png)

• 100uF 25v cap before 0.33uF cap
• 10uF 10v cap before 0.1uF cap
• some 1n4007 diodes from its ground and output
• feedback 1n4007 diode from its output to its input as protection they said

c) Li-Ion TP4056 charger 1A output

d) Smartphone batteries 4.2v 3000mAh

It's ok for me if only formulas will be given with respect to discharge, power, etc for me alone to do the math, i just need some guidelines because to be honest, i don't know how would i explain if my projects are efficient or later will have a drawback, doesn't last long, not practical, etc. though i really wanted to know how would i compute the battery consumption/discharge on this project.

I hope you can help me because I'm overwhelmed right now, i have a lot of tabs on my browser to search that don't have guide where to start. There's always a miss or a variable or process i don't know that applies on the answers i see on web :(

• You've missed a critical piece of information. What is the efficiency of the 5V regulator? Jul 14, 2017 at 7:54
• Good Day Andrew :) i don't exactly know what you mean by efficiency of 5v regulator, maybe your referring to its percentage efficiency on its output? like 95%? i don't know if what i made had that much efficiency. i had updated my question for details. Jul 15, 2017 at 4:31
• For an ideal switching regulator power out = power in. A good design can hit 95% of that, only 5% of the input power is wasted as heat. A poor design can be in the 70s. You used a linear regulator where current out = current in, any voltage difference represents power wasted as heat. Jul 16, 2017 at 7:53

First off you can use Wh to give an ideal world, everything 100% efficient calculation.
12V, 10Ah = 120 Wh source.
3.7V, 3Ah = 11.1 Wh destination.

So if we could do things perfectly we could charge 10.8 batteries. The real world number will obviously be lower.

Charge rate will only have a minimal impact on the final system power usage, the power supplies will vary in efficiency a little with different current draws but not enough to make a huge difference in the answer.

The battery charger is a linear regulator so the current input is independent of output voltage and only depends on the output current.
So to put 1 Ah into the battery we need to take 1 Ah from the 5 V rail. This makes this stage ~74% efficient. (3.7/5 = 0.74)

Assuming a switching power supply is used to step down from 12 V to 5 V then 1 A on the 5 volt rail will require 5/(12*e) amps from the 12 V supply where e is the efficiency of the power supply.
Since you didn't give a number for this let's go with 85% for now, the real number could be higher or lower.
So 1 A at 5 V requires 5/(12*0.85) = 0.49 A from the 12 V supply.

Which means that to put 1 Ah into your Li-Ion battery you will need to take 490 mAh from your 12 V battery.

If your 12 V battery is 10 Ah and your Li-Ion battery is 3 Ah then you should be able to charge it roughly 6.5 times.

Exact numbers will be a little bit lower due to other losses in the battery charger and the batteries themselves but these will generally be small in comparison to the impact of the factors above.

• Thanks for the great reply Mr. Andrew :D "Assuming a switching power supply is used to step down from 12 V to 5 V then 1 A on the 5 volt rail" - Andrew you mean if i don't use my own 5v voltage regulator circuit and i used a module for stepping down 12v to 5v? " Since you didn't give a number for this let's go with 85% for now, the real number could be higher or lower. " - Andrew if your asking about the efficiency of the 5v regulator circuit, i don't know how to get its efficiency. How would i do that? What is the formula for the 'e' of my 5v rail? Thanks for the help :) Jul 15, 2017 at 4:41
• Looks like you used a linear regulator not a switching one. They are simpler but less efficient. Going from 12 to 5 it will give you an efficiency of 5/12 or 41%. Since your whole system is linear the numbers become really easy, 1 Ah into the battery requires 1 Ah from the source and you can charge 3.3 batteries. Over 2/3 of your power is converted into heat by the regulators. And at a 1A charge current you will be putting 7 W into that regulator, put a heatsink on it, it's going to get hot. Jul 16, 2017 at 7:48