TP4056 v2 module reverse current protection?

Does this TP4056 v2 module have any sort of reverse current/voltage protection for the out terminals? |500x500

I am planning to connect this to a device from the out terminals while charging the battery using the module. Also I want the device to be powered from another source while the battery is being charged. I want the battery to act like a Uninterrupted Power Supply so that when the other source is no longer powering, then i want the battery to kick in.

I wanted to know whether if I connect the out terminals to the device from this module while I'm powering that device from another source, will the current from that source flow into the battery and damage it?

I don't think it as protection for that.

But why connect another source? If you connect this to a power supply and a battery it already act as UPS...

septillion: I don't think it as protection for that.

But why connect another source? If you connect this to a power supply and a battery it already act as UPS...

wont it draw power from the battery at the same time?

As long as the charge current is more then the current you need you're fine.

Otherwise you can connect the power rail of your project via a (schottky) diode to the battery and via a (schottky) diode to a power supply. That way the project will use the highest supply (so be sure the external power supply has a higher voltage then a charged battery).

septillion: As long as the charge current is more then the current you need you're fine.

Otherwise you can connect the power rail of your project via a (schottky) diode to the battery and via a (schottky) diode to a power supply. That way the project will use the highest supply (so be sure the external power supply has a higher voltage then a charged battery).

Nope, the charging current is less than the current my project needs.

Interesting, can you explain a bit more? Where all should I connect schottky diodes? I have a 5V 2A wall adapter as main power supply which will power both battery charger and my device.

So you would connect them like this. The higher voltage of the wallwart or the battery supplies the load., The diodes prevent the higher voltage source from reverse driving the lower source. The diode between charger and battery prevents the battery from reverse driving the charger when the charger is unpowered.
You might be able to find slightly lower Vf Schottky diodes, but I don't think they support very much current.

CrossRoads:
So you would connect them like this. The higher voltage of the wallwart or the battery supplies the load., The diodes prevent the higher voltage source from reverse driving the lower source. The diode between charger and battery prevents the battery from reverse driving the charger when the charger is unpowered.
You might be able to find slightly lower Vf Schottky diodes, but I don’t think they support very much current.

Thanks for the Diagram, helps a lot.
No problem with Voltage drop caused by diodes, I will use a Boost converter to maintain voltage.

I have a few doubts.

-Will a IN5822 Axial Leaded Schottky Rectifier Diode do the job? My device needs 6W of power.

-Why a diode between battery and charger - Isn’t the charger designed to prevent reverse power from the battery when the charger is not powered?

-If only reverse current protection is needed, won’t ordinary diodes do the job? ( I mean they have less reverse voltage leakage right?)

-Should the diode between the battery and the load be connected directly to battery terminals or should it be connected to ‘OUT’ terminal of the charger?

Update-
I have uploaded 2 modified diagrams of yours, pls tell if it has any problems.

Only the diagram is slightly wrong. See the correct one attached.

Back to the questions

  1. 6W @ 5V => 1,2A, 6W @ 2,3V (li-ion minimum of the module) 2,6A. So a 3A diode should hold. But it can get quite hot because it will dissipate 1,3W…

  2. Because diagram was wrong :wink:

  3. They will, but the voltage drop over them is a lot bigger. And when dealing with batteries that’s a bad thing most of the time. The reverse leakage current is a bit higher but that’s not a problem.

Ow, and be aware your power supply needs to be able to deliver the 6W for the circuit AND the 1A of the charge at the same time.

[Edit] Now with a image…

septillion: Only the diagram is slightly wrong. See the correct one attached.

Back to the questions

1) 6W @ 5V => 1,2A, 6W @ 2,3V (li-ion minimum of the module) 2,6A. So a 3A diode should hold. But it can get quite hot because it will dissipate 1,3W...

2) Because diagram was wrong ;)

3) They will, but the voltage drop over them is a lot bigger. And when dealing with batteries that's a bad thing most of the time. The reverse leakage current is a bit higher but that's not a problem.

Ow, and be aware your power supply needs to be able to deliver the 6W for the circuit AND the 1A of the charge at the same time.

[Edit] Now with a image...

Thanks for the info,

I have a few more doubts.

1)Heat is not a problem for my project, but will mentioned diode ever get so hot that it starts smoking and gets damaged if I use 24x7(since its UPS)? Do you have any other diode model in mind?

2)Are you sure reverse leakage from Schottky wont cause any harm to batteries or other components?

3)My power supply is a 5V 2A USB Wall adapter (10W). Can I make it fully power my device giving it more priority and give only whatever remaining power available to battery charger?

4)If above mentioned method is not possible then can I limit the current going into the TP4056 battery charger to 200mA or so?., No problem if battery takes longer time to recharge but max power going for the Load is a priority when main power is available.

4)Is there a max current limit for the OUT terminals of the TP4056 module? If it has like 1A or below then should I bypass the OUT terminal and connect directly to battery to get more current?

  1. The first diode that meets the spec (3A schottky) will do the trick. If I would have to grab one I’ll get a SS16 but that’s because I have them.

  2. Yep

  3. It charges @ 1A. So 1A x 5V = 5W. Plus the 6W of the circuit makes 11W. So no, not sufficient.

  4. That would be an option. But you have to swap a SMD resistor on the board. It’s the 1k2 near to the IN- terminal. Swap it with 2k2 will set the charging current to 545mA.

  5. (aka 4.2 :stuck_out_tongue: ) I quickly can’t find the mosfet they use to switch the thing. But in the listing of the modules where I bought it it states: “Battery over-discharge protection current:3A”. So when the battery is full that’s all fine. But when it discharges and the current goes up it will shut itself off. But that’s a controlled shut off when it’s almost empty so it’s okay (not neat but okay).

Don’t bypass the charger unless you use li-ion cells with build in overcharge and discharge protection. Otherwise you can drain them to much and kill them.

septillion: Don't bypass the charger unless you use li-ion cells with build in overcharge and discharge protection. Otherwise you can drain them to much and kill them.

Thanks for the info,

Isn't SS16 an SMD? my soldering skills are not very good, I don't think i can handle an SMD.

Is there any option for me to reduce current flow into the charger before it goes into the module?

I'm using samsung 26F cells, I guess I saw over discharge protection for those, but anyways I'm not going to use directly because I'm not sure they are genuine samsung cells or not, I never weighed them(don't have the equipment).

That way the project will use the highest supply (so be sure the external power supply has a higher voltage then a charged battery).

dumb question from me are you sure that the battery wont try to power the device along with the wall adapter when main power is available? I never understood the concept of schottky diode will select highest voltage?

Noobian:
Isn’t SS16 an SMD? my soldering skills are not very good, I don’t think i can handle an SMD.

Yep, it’s SMD. But it’s the only part I have that meets the specs. I don’t say you should use it. It’s just the first part I can get my hands on that meet the spec. You should just use the first part you van get your hands on that meet the spec. And if you want through hole than that’s a spec for you :wink:

Noobian:
Is there any option for me to reduce current flow into the charger before it goes into the module?

No

Noobian:
dumb question from me are you sure that the battery wont try to power the device along with the wall adapter when main power is available? I never understood the concept of schottky diode will select highest voltage?

No it will not. Current can only flow through a diode when the voltage on the cathode side is less then on the anode side and only from anode to cathode. So when the supply and the battery are connected the power supply diode will conduct. So now the supply voltage (minus diode drop) is present on the power line of the circuit and also on the cathode side of the battery diode. This diode now has Vbat on it’s anode and Vsupply on it’s cathode. Vsupply > Vbat so current cannot flow.

And now you thing you’re smart and start with the battery :smiley: But if the battery diode starts to conduct you have Vbat at the circuit and at the cathode of the power supply diode. So this diode now sees Vbat on it’s cathode and Vsupply on it’s anode. So this will start to conduct. But this brings the voltage up so once again the battery diode will see Vbat on it’s anode and Vsupply on it’s cathode. So it stops conducting.

[/quote]

septillion: But this brings the voltage up so once again the battery diode will see Vbat on it's anode and Vsupply on it's cathode.

What brings the voltage up?

The voltage from the power supply. Because when there is Vsupply on the anode and Vbat on the cathode with Vsupply > Vbat so the diode will conduct. And it will conduct as long as Vanode > Vcathode, bringing the voltage up to Vsupply thus the battery diode stops conducting.

septillion: The voltage from the power supply. Because when there is Vsupply on the anode and Vbat on the cathode with Vsupply > Vbat so the diode will conduct. And it will conduct as long as Vanode > Vcathode, bringing the voltage up to Vsupply thus the battery diode stops conducting.

Ok, everything you mentioned so far is very clear.

Now the problem is the power supply - mine is 10w (5v 2a). device needs 6W, battery charger needs 5W.

I want to provide less current for the charger, like 2W or so, how can I achieve that without changing the SM resistor inside the charging module? How can I divide the power into a 3:1 ratio? I dont want to buy a new one if i can manage to to do this.

What if I change the onbaord 1.2K ohm SMD resistor to a 2K ohm one?