# 3.7V 18650 battery

Discussion in 'Power Electronics' started by turbogt16v, Apr 14, 2015.

1. ### turbogt16v

176
4
Mar 27, 2015
Hy,i bought 2, 3.7V 18650 batteries (6000mAh).
i need them to power mp3 and small amp about 5v 1.5A or less.

I have tested battery with multimeter and sparks fly(why),it shows about 20A cant measure voltage(why).
Can i lower amps to about 4 or less,for Step-up booster to handle.

Last edited by a moderator: Apr 14, 2015
2. ### Harald KappModeratorModerator

11,655
2,696
Nov 17, 2011
You cannot measure current from a battery without a load. The battery will drive as much current as is possible due to the resistance of the wires and the internal resistance of the battery. Your multimeter in current range is almost a short circuit!
To measure current, you need to apply a load (e.g. resistor) to limit the current.

To measure voltage you need to set up your meter in the volts range.

The battery will deliver ~3.7V or less (depending on the state of discharge) as long as the current is limited below the max. curent allowed for the battery (see datasheet of your battery). When you connect a booster to the battery, it will draw only the current required to power the load plus a few mA for its own use.

Example:
Load=5V, 1.5A = 7.5W output power
efficiency = 0.75 (efficiency describes how much energy is lost within the boost converter)
Input power = output power / efficiency = 7.5W / 0.75 = 10W
Input voltage = 3.7V -> input current = input power / input voltage = 10 W / 3.7 V = 2.7A (or a bit more when battery voltage drops)

3. ### turbogt16v

176
4
Mar 27, 2015
ok i understand a bit,i need to connect resistor to battery positive and then measure current,what is the value of resistor.
The battery writes 3.7v 6000mAh ,i read from google that max amp is 4.3 so resistor is 0.86046511627ohm
is that 860 ohm resistor,and would it be to high amp for boost up becouse he is rated at 4a max

and one more thing to ask ,i only have 1/4W 860 resistor is that good

Last edited by a moderator: Apr 14, 2015
4. ### Harald KappModeratorModerator

11,655
2,696
Nov 17, 2011
Your asking the wrong question. As per Ohms law, I=V/R. Therefore, given V=3.7V, you can set the currrent by selecting the appropriate resistor. The power in the resistor is P=V*I=I²*R=V²/R = (3.6V)²/860Ω=16mW, therefore a 1/4W (250mW) resistor will be fine.

6000ah? I doubt it. More like 6000mAh (=6Ah), right? This number is the capacity of the battery. It doesn't tell you anything about the max. current you can draw from the battery. The capacity tells you how long the battery will last under a given load current.
If you draw 10A, it will last 36 minutes (10A*0.6h=6Ah).
If you draw 6A, it will last 1hour (6A*1h=6Ah).
If you draw 1A, it will last 6hours (1A*6h=6Ah).
etc.

But what's the use of this measurement? Your load will draw as much current as is required to operate the load.

5. ### Colin Mitchell

1,416
313
Aug 31, 2014
A little knowledge is a DANGEROUS thing.

davenn likes this.
6. ### davennModerator

13,866
1,958
Sep 5, 2009
I have been editing his post to show mAh

7. ### turbogt16v

176
4
Mar 27, 2015
Yes i meant mAh,i did not know about the load determining the power of current,so i wanted to measure exactly.
I am building solar radio with battery bank.I will need more help as i progress in making.

I dont undestand ,the calculator shows 0.86047 ohms not 860 ohms,can someone clarify

8. ### Harald KappModeratorModerator

11,655
2,696
Nov 17, 2011
Which calculator? What do you want to calculate?

9. ### turbogt16v

176
4
Mar 27, 2015
well i already measure it and with 860ohm it rates 10A.
Is it wise to put 5Ohm resistor before boost up to lower amperage (780mAh)and loss from boost up

10. ### Harald KappModeratorModerator

11,655
2,696
Nov 17, 2011
You cannot draw 10A from a 3.7V battery (not even 2 of them in series) through a 860Ω resistor.
3.7V/860Ω=4,3mA

If you were to drive 10A through an 860Ω resistor, this amounts to a power dissipation of I²*R=86kW!

No, this is nonsense. Your misconception is that the battery drives the current into the boost converter. If that were so, it would also drive the same current through the series resistor.
The battery "offers" a voltage (3.7V) to the load circuit, that is the boost regulator. The regulator steps-up this voltage to 5V. The load (amplifier) will drwa as much currrent from the 5V as is required for operation. The boost converter in turn will draw as much current from the battery as is required to generate the 5V from 3.7V. Not more, not less.
A resistor in series with the boost regulator would only dissipate power as heat, it will not reduce the current. Therefore it would increase the loss, not reduce it.

11. ### turbogt16v

176
4
Mar 27, 2015
OK.thank you ,you were very helpful.will come later for more questions

12. ### HellasTechn

1,556
215
Apr 14, 2013

Be very carefull. As far as i know there are no 18650 rated 6000 mah. Only cheap chineese ones that indicate false mah rating. most 18650 are (800mah to 2500mah if i am not mistaking)

The fact that you shorted the battery out with your multimeter to measure the amps output and that the battery still works tells me that the cell is "unprotected" meaning that it does not have an overcurrent protection circuit built in.

This is very important because if you dont use proper charger you risk exploding the battery. youse only chargers that automatically terminate charge whene battery is full.

Harald Kapp likes this.
13. ### HellasTechn

1,556
215
Apr 14, 2013
Theese batteries offer 4.1 to 4.2 volts at full charge and there maximum amps output is usually 2 to 3 Amps.

I do not recommend you useing a regulator because during use it may get hot due to power dissipation.

I think with such a battery you should be fine to use it to power your MP3 straight without any other circuit in the way. Two of theese batteries in paralel should be more than enough to power an MP3 and a small amplifier.

14. ### Harald KappModeratorModerator

11,655
2,696
Nov 17, 2011
I disagree: how else wil you get stable 5V from this battery? Of course some power will be lost within a regulator. This is where a boost converter comes into play which, being a switch mode regulator, will keep losses at a minimum.
Unless, of course, the load (amplifier etc.) accepts the lower and varying voltage directly from a battery.

15. ### HellasTechn

1,556
215
Apr 14, 2013
Your idea is not bad but why not keep it simple enough if it can work like that ?

I am certain that it will work on 4.1 volts. This voltage is well within their tolerance most of the times.

16. ### turbogt16v

176
4
Mar 27, 2015
Well,i am building a solar radio so i need regulator how ever,and it even works best without diode at the end of the solar panel(more power),
the battery by itself can handle amp and mp3 module (tested).I am planing on putting relay to switch power to battery when needed and charge it
with lipo mini charger when not needed.

I would like to build in usb charger do i need to regulate amperage,or does the solar acts same as battery

Last edited: Apr 16, 2015
17. ### HellasTechn

1,556
215
Apr 14, 2013
I dont think you will need a regulator since most solar panels only deliver a couple of hundred miliamps. But that again depends on your panel.

176
4
Mar 27, 2015
3v 2A

19. ### HellasTechn

1,556
215
Apr 14, 2013
3 volts are definetly not enough to charge the 18650. Now you do need to make a charge circuit.

If you use a 5 volts regulator to boost voltage that would be danferous for the battery even if you regulate it to provide 500ma or less.

I think that things get a little more complicated here. i dont know what to suggest. Why dont you just charge the batteries with a 18650 charger ?

Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Continue to site