Making a lamp portable with 18650

Thread Starter

vmazsit

Joined Apr 15, 2020
9
Hey!
I have a lamp which is powered by a 5V power supply and I want to upgrade it by powering it with a battery pack.
The issue is I want to use the lamp while charging its batteries but as I know charging and draining at one time is impossible so I searched for simple switching circuit but I couldn't find one and im not so experienced with electronics so I can't design it myself.
The battery pack is 1S4P 18650 lithium ion battery (is it good or should I change it to 2S2P ???) I charge it with TP4056 and then convert it to 5V with MT3608, and also my lamp needs 1-2A .
My question is how should I do it or is there any products that do that ?
Thanks.
 

KeithWalker

Joined Jul 10, 2017
1,019
Use a charging source that can supply enough current to charge the cells and supply the lamp. Use a manual switch that will disconnect the lamp supply from the converter and connect it directly to the charging supply when you want to charge the lamp and use it at the same time.
Keith
 

Tonyr1084

Joined Sep 24, 2015
4,687
I know charging and draining at one time is impossible
No - that's false. You CAN charge a battery while using the system.

Years ago I got a 12V 20AH battery and connected it to a charger AND a car radio. The charger was capable of playing the radio all on its own. I don't remember the numbers so I'll make something up: The charger could push 4 amps but the radio during normal operation would draw 3 amps. That left 1 amp to charge the battery. When the charger was shut off (the whole thing was self contained) the radio would continue to play. I could shut the radio off and leave it on the charger and restore full charge to the battery. So your assumption that it's impossible to charge and drain at the same time is false. It all depends on the size of the charger.

What you said WOULD be true IF the charger in my example could only produce half an amp. Then the charger would supply SOME current for the radio while the battery would take the lion share of the load. Eventually the battery would be depleted. But that's because the charger would be insufficient for charging the battery AND playing the radio.
 

Irving

Joined Jan 30, 2016
763
That's true for lead-acid batteries, where charging is often more of a brute force approach, but there are some complications with charging Li-ion cells while in use.

For example, when the cell is near fully discharged the charger chip will pulse a small current into the cell to see if the voltage rises, before starting the charge cycle. If the equipment is on the volts won't rise so charging doesn't start.

The next issue is when the charge gets close to completion the charger is looking for the battery open circuit volts to reach 4.2v. But if it's in use it will never do so. For safety most charger chips terminate charging after, typically, 5 or 10 hours if the battery volts haven't reached the expected level, and the charging doesn't progress to the saturation phase. As a result the battery is only 75 -80% full, but the user won't be aware of that.

The solution is to use a charging chip that separates the equipment side from the battery side and a supply that has sufficient capacity to supply the equipment and charge the battery simultaneously. Or build a circuit that achieves the same result.

I have an example somewhere here, I'll try and find it.
 
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Thread Starter

vmazsit

Joined Apr 15, 2020
9
That's true for lead-acid batteries, where charging is often more of a brute force approach, but there are some complications with charging Li-ion cells while in use.

For example, when the cell is near fully discharged the charger chip will pulse a small current into the cell to see if the voltage rises, before starting the charge cycle. If the equipment is on the volts won't rise so charging doesn't start.

The next issue is when the charge gets close to completion the charger is looking for the battery open circuit volts to reach 4.2v. But if it's in use it will never do so. For safety most charger chips terminate charging after, typically, 5 or 10 hours if the battery volts haven't reached the expected level, and the charging doesn't progress to the saturation phase. As a result the battery is only 75 -80% full, but the user won't be aware of that.

The solution is to use a charging chip that separates the equipment side from the battery side and a supply that has sufficient capacity to supply the equipment and charge the battery simultaneously. Or build a circuit that achieves the same result.

I have an example somewhere here, I'll try and find it.
Thanks I really appreciate it :D
 

Tonyr1084

Joined Sep 24, 2015
4,687
No - that's false. You CAN charge a battery while using the system.
That's true for lead-acid batteries, where charging is often more of a brute force approach, but there are some complications with charging Li-ion cells while in use.

For example, when the cell is near fully discharged the charger chip will pulse a small current into the cell to see if the voltage rises, before starting the charge cycle. If the equipment is on the volts won't rise so charging doesn't start.
I amend my comment from post #3. In light of what Irving said - I have to retract my answer because, as he pointed out, it was no referring to Li-ion batteries. Though I said "no" I must now say "Yes, it's true you can not charge Li-ion batteries at the same time as using them".

I'm thinking as Irving said about a charger that can power both your device AND charge the batteries at the same time. The batteries will need a Battery Charge Controller (that name doesn't sound correct - early yet - no coffee - brain not farting on all cylinders). It's obvious I don't have sufficient experience with 18650's. I'm only starting to learn of them in detail.

Thank you @Irving for clearing that up.
 

Thread Starter

vmazsit

Joined Apr 15, 2020
9
Does your lamp really take that much current, or is that what the power pack says its rated at? What bulb is fitted?
Well it's not just a regular lamp with a light bulb.
I made it with arduino and with 70 WS2812B LEDS.
I thought it is not important so I just wrote that it needs a lot of power.
 

Thread Starter

vmazsit

Joined Apr 15, 2020
9
ok so an LED strip. How long?

Powered from? How is the strip connected to the Arduino?
Its divided to 7x10 (7 strip and one strip has 10 LEDs on it).
The LED is connected directly to the main power supply which is 5V 3A and the Atmega328P (arduino) chip is connected to the power supply in series with a 10 Ohm resistor. That's because the high power would fry the chip because it doesn't have any protection like a full arduino nano,uno etc. .Then the LED's data connected to the Atmega chip and also they share GND

Btw if anyone has any question about the lamp or something else that could help my problem I'll reply.
 

Thread Starter

vmazsit

Joined Apr 15, 2020
9
so the 7 strips are in parallel? and how are they connected to the Arduino...?
No they are in series but I didn't wanted to just wound it up on a PVC pipe so I cut each one on every 10th LED and just connected the DOUT and DIN just like nothing happened.
 

Irving

Joined Jan 30, 2016
763
OK, found the datasheet..

70LED @ 60mA each is 4.2A so you aren't running at full brightness on all LED. We'll take 3A as your input current.
How many mAh are your 18650s?
 

Thread Starter

vmazsit

Joined Apr 15, 2020
9
70LED @ 60mA each is 4.2A so you aren't running at full brightness on all LED. We'll take 3A as your input current.
Well yes, but not quite... I am running it at full brightness and it draws like 1,2A and not continuously because it's an animation so it's always changes between 500 mA and 1200 mA.


How many mAh are your 18650s?
My batteries are 3200 mAh and i have 4 of them in parallel so overall i have 12800 mAh
 
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Irving

Joined Jan 30, 2016
763
so you need 5v x 3A = 15W and you have 3.6V x 4 x 3.2Ah = 46Wh, so at 90% efficiency thats 41.5W which gives you 41.5/15 = 2.75h run time

So next question is it step-up from 1S or step-down from 2S. The conversion efficiency is about the same either way.

What are you going to charge them from?
 

Thread Starter

vmazsit

Joined Apr 15, 2020
9
so you need 5v x 3A = 15W and you have 3.6V x 4 x 3.2Ah = 46Wh, so at 90% efficiency thats 41.5W which gives you 41.5/15 = 2.75h run time
Im not sure if I calculated it right but if I have a 12800 mAh battery pack and I connected it to the lamp and even if it's draws continuously 1200 mA's it would last 10 hours ( 12800 mAh/1200 mA ) but maybe Im wrong cause I only know electricity
as a hobby...

What are you going to charge them from?
Not sure that I understand the question.
Like with what ? (the charger is a TP4056 module)
 
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