Using a 4.4V lipo battery converts to 9v~12V?

Killerbee65

Joined May 15, 2017
214
Hello all! I know the topic is very vague but for some context I wanted to use a 4.4v lipo battery with a capacity of 10000mAh and transform it into 9v-12v using an adjustable step up module: https://www.amazon.com/gp/product/B07QGGSZF8/ref=ox_sc_act_title_3?smid=A28BKVPRLI4BWN&psc=1

the problem is I don't know how much current will go down as voltage goes up and if the battery I want to use will damage the circuit or vice versa. I tried to do some research on the topic but I know am missing something. The point is to have a lipo battery that can supply 9-12v at 2000mAh or more for a robot I am making with motors.

bertus

Joined Apr 5, 2008
20,465
Hello,

The current at the battery side will be higher than on the output side.
If you want 12 Volts out when the battery voltage is 4 Volts, the input current will be at least 3 times higher.

Bertus

joeyd999

Joined Jun 6, 2011
4,377
Hello,

The current at the battery side will be higher than on the output side.
If you want 12 Volts out when the battery voltage is 4 Volts, the input current will be at least 3 times higher.

Bertus
It'll be exactly Iin = Iout * (Vout/Vin) / efficiency

Of course, efficiency is variable and dependent upon topology, component selection, Vin, and Iout.

Killerbee65

Joined May 15, 2017
214
Hello,

The current at the battery side will be higher than on the output side.
If you want 12 Volts out when the battery voltage is 4 Volts, the input current will be at least 3 times higher.

Bertus
Thank you for the reply! I just need to know if it would be enough (more than 2000mAh) if there exist a mathematical equation then please dont hesitate to fill me in.

Papabravo

Joined Feb 24, 2006
13,559
The basic relationship you have to worry about is the power balance. In simple terms the power out of any conversion process will always be less than the power in. Under some circumstances it will be a great deal less. Do not conflate the battery capacity, measured in Ampere-hours with the maximum continuous current draw. For example are we to conclude that your battery pack can safely provide 10 Amperes for 1 Hour. Maybe it can but, but should it? I don't know. Will it provide 1 Ampere for 10 hours? That might be reasonable. What will the average output voltage be over the 10 hours -- 3.6V perhaps. Multiply 3.6 Volts times 1 Ampere and you get 3.6 watts. Assume for the sake of argument that the converter is 80% efficient, that means after conversion you have 2.88 watts of available output power. At 12 V this will be 240 mA, but at 9V it will be 320 mA. So long as you do the calculations with reasonable assumptions and data you should be able to figure out how to operate without difficulty.

Edit: If the battery will last for 10 hours, then 10 hours @ 240 ma and 12V gets you 2400 mAh, and 320 mA and 9V will get you to 3200 mAh
So yeah you might have a shot. You should run some tests to get a better idea of the module's efficiency.

Last edited:

bertus

Joined Apr 5, 2008
20,465
Hello,

@joeyd999 , You are right about the efficiency.
@Killerbee65 , The input cuttent will depend on the efficiency of the DC-DC converter.
joeyd gave the excaxt formula.
Unfortunately the given link does not mention the needed efficiency for the calculations.

Bertus

Killerbee65

Joined May 15, 2017
214
Hello,

@joeyd999 , You are right about the efficiency.
@Killerbee65 , The input cuttent will depend on the efficiency of the DC-DC converter.
joeyd gave the excaxt formula.
Unfortunately the given link does not mention the needed efficiency for the calculations.

Bertus
I know I am asking for a lot here but is there any way to visualize the equation? Maybe a video link explaining it? Thank you for any reply!

Papabravo

Joined Feb 24, 2006
13,559
Sure!
You live in a neighborhood run by a criminal kingpin. He demands a 15% tribute on every dollar the you earn. There are three immutable rules:
1. You can't win
2. You can't break even
3. You can't get out of the game
Oh wait -- those are the three laws of thermodynamics! Quelle surpise! Mon Dieu!

Killerbee65

Joined May 15, 2017
214
Sure!
You live in a neighborhood run by a criminal kingpin. He demands a 15% tribute on every dollar the you earn. There are three immutable rules:
1. You can't win
2. You can't break even
3. You can't get out of the game
Oh wait -- those are the three laws of thermodynamics! Quelle surpise! Mon Dieu!
I see, I meant an actual equation (1+2=3) but I appreciate the scene

Killerbee65

Joined May 15, 2017
214
I see, I meant an actual equation (1+2=3) but I appreciate the scene
*laughing emoji**thumbs up*

Papabravo

Joined Feb 24, 2006
13,559
Check Post#3 for an equation

Killerbee65

Joined May 15, 2017
214
It'll be exactly Iin = Iout * (Vout/Vin) / efficiency

Of course, efficiency is variable and dependent upon topology, component selection, Vin, and Iout.
Oh thank you so much sorry I didn't see this post and thank you @Papabravo for pointing it out!

Audioguru again

Joined Oct 21, 2019
1,304
Why do you think the Amazon module will produce an output of 9V-12V at 2A when it has No Detailed Spec's?
It says 18650 which is a little 3.7V cell that might be able to produce 2A. The module seems to have a mini 5V USB connector.
The ad says Charging and Discharge.

The USB might be an 5V input that feeds a single cell battery charger circuit. The ad says 9V and 5V but does not say 12V or the maximum output current which might be only (3.7V/9V) x 2A= 0.82A x 80%= 0.66A.

Amazon and their sellers know NOTHING about electronics.

joeyd999

Joined Jun 6, 2011
4,377
Oh thank you so much sorry I didn't see this post and thank you @Papabravo for pointing it out!
'sok. Most everyone ignores me here, anyway.

I'm gonna change my avatar to a wallflower.

Papabravo

Joined Feb 24, 2006
13,559
'sok. Most everyone ignores me here, anyway.

I'm gonna change my avatar to a wallflower.
I guess a little rain must fall on us all.

Killerbee65

Joined May 15, 2017
214
Why do you think the Amazon module will produce an output of 9V-12V at 2A when it has No Detailed Spec's?
It says 18650 which is a little 3.7V cell that might be able to produce 2A. The module seems to have a mini 5V USB connector.
The ad says Charging and Discharge.

The USB might be an 5V input that feeds a single cell battery charger circuit. The ad says 9V and 5V but does not say 12V or the maximum output current which might be only (3.7V/9V) x 2A= 0.82A x 80%= 0.66A.

Amazon and their sellers know NOTHING about electronics.
I second that! I used that device as mainly a reference so I dont sound like an idiot trying to describe what it was that I was looking for but seeing how that specific device was the only one i could really find (in my understanding) it was the one I chose to show off. If there is a better or alternate device/ method I could use please let me know!

Audioguru again

Joined Oct 21, 2019
1,304
If you need 12V then why not use a 3 cells Li-PO battery and charger for it from a hobby store that sells RC cars, airplanes and helicopters? It will be 12.6V at full charge and little ones or huge ones are available.

Killerbee65

Joined May 15, 2017
214
Ok so I did the math, thank you again for supplying the equation, and found that if I want 9v from 4.4v with 10000mAh then theoretically I should have about 4500mAh at 9v as my output using a 94% efficiency rating. Now the question is what device can do this for me? How can I make it myself? Can a device even do this for me in the first place?

Killerbee65

Joined May 15, 2017
214
If you need 12V then why not use a 3 cells Li-PO battery and charger for it from a hobby store that sells RC cars, airplanes and helicopters? It will be 12.6V at full charge and little ones or huge ones are available.
I realistically only really need 9v, that being said I have looked at those kind of batteries but the problem arises when I need to charge the battery. It has to be charged through a micro usb (part of the challenge) and the batteries I have come across use a special kind of charger, using a device like the dc to dc converter with the micro usb saves me time, money, and patience.

Audioguru again

Joined Oct 21, 2019
1,304
An ordinary "3.7V" Li-PO cell is 4.2V when fully charged, never 4.4V. It is destroyed if it is discharged below about 3.0V so a circuit must sense the voltage and disconnect it when the voltage gets low.
As the battery voltage runs down its current rises and its charge is drained faster.
The huge battery cell will take a long time to charge.