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Li-Po charging circuit design

Sashvat

Member
Hello everyone, I am designing a product which needs a Li-Po battery and I need a circuit to charge it. I am confused as to how do I charge the battery as well as use it as a power source within the same circuit. What I mean to say is, how is it that a phone, or any rechargeable device know that the battery is being charged and knows when the current needs to be sent into the battery?

Can someone share how a circuit looks like this? Will the Battery charging IC I choose automatically have this feature? Or do I need a special component that does it for me? Please do let me know. Thank you
 

Musicmanager

Active Member
What I mean to say is, how is it that a phone, or any rechargeable device know that the battery is being charged and knows when the current needs to be sent into the battery?
It doesn't .. .. .. the charger connects to the battery .. .. the battery to the load.

You might like to look at a TP 4056 module .. .. .. .

1622637758793.png

. .. .. .. there are supposedly 3 slightly different versions, although I've only found two, so far. I'd choose the one with the battery protection chip as it also has LED indicators for 'charging' & 'charged'

One thing to watch for, if you're using a larger capacity battery and USB power for the TP 4056, you need to check that the current limit of the TP 4056 is not overloading the USB.

MM
 

Nigel Goodwin

Super Moderator
Most Helpful Member
One thing to watch for, if you're using a larger capacity battery and USB power for the TP 4056, you need to check that the current limit of the TP 4056 is not overloading the USB.
It shouldn't be an issue, as the TP4056 is a constant current charger, with the current fixed by R3 (on most boards).
 

Sashvat

Member
It doesn't .. .. .. the charger connects to the battery .. .. the battery to the load.

You might like to look at a TP 4056 module .. .. .. .

View attachment 131706

. .. .. .. there are supposedly 3 slightly different versions, although I've only found two, so far. I'd choose the one with the battery protection chip as it also has LED indicators for 'charging' & 'charged'

One thing to watch for, if you're using a larger capacity battery and USB power for the TP 4056, you need to check that the current limit of the TP 4056 is not overloading the USB.

MM
That's it? there's nothing more to it than that? Than does that mean I just copy that design, attach my battery to it?
 

Musicmanager

Active Member
Yep, that's it !

Although you might want to show your schematic and let the Guru's on here cast an eye before discovering you have a stick of gelignite masquerading as a capacitor or something :)

Nigel .. I've been looking for the note of what occurred but I can't find it; I remember that the 2600 mAH battery I was using was drawing 582mA through the R400 CL resistor and the USB is only capable of 500mA ; I changed it to a lesser beast but I can't remember what.

MM
 

Nigel Goodwin

Super Moderator
Most Helpful Member
Nigel .. I've been looking for the note of what occurred but I can't find it; I remember that the 2600 mAH battery I was using was drawing 582mA through the R400 CL resistor and the USB is only capable of 500mA ; I changed it to a lesser beast but I can't remember what.
The battery doesn't 'draw' current from the charger, the charger is constant current, so will only supply what it's set to. If your USB socket isn't capable of decent current, then change the value of R3 to the current you want.

There's a nice table of values here:

 

Musicmanager

Active Member
The battery doesn't 'draw' current from the charger, the charger is constant current, so will only supply what it's set to.
Well, I'm sorry but I think that's nonsense !
However, I'm too much the novice to argue .. .. .. .. .

MM
 

JimB

Super Moderator
Most Helpful Member
However, I'm too much the novice to argue
Since when has that been a barrier to a good old argument about terminology!

Is the battery drawing, or is the charger pushing?

JimB
 

Musicmanager

Active Member
Since when has that been a barrier to a good old argument about terminology!
OH, don't you start !!
I thought long and hard before I made post #2 ( even told myself, I'd regret it ! )

I never learn :(

MM
 

Nigel Goodwin

Super Moderator
Most Helpful Member
Well, I'm sorry but I think that's nonsense !
However, I'm too much the novice to argue .. .. .. .. .
I think it's because I worded it very poorly :D

My point was you said:

One thing to watch for, if you're using a larger capacity battery
And this makes no difference at all - the larger battery can't draw any more current because the charger is constant current, so small battery or large battery, the current will be the same, and is set by the charger (and specifically R3).
 

Visitor

Well-Known Member
Well, I'm sorry but I think that's nonsense !
However, I'm too much the novice to argue .. .. .. .. .

MM
....or to use Google it would seem....

A constant current power supply (the TP4056 while charging) will vary the voltage to maintain a set current through a load. Suppose the set current is 1000mA, and the load is 1 ohm. A constant current supply will output 1 volt to achieve a current of 1000mA. If the resistance is increased to 5 ohms, the voltage will be increased to 5 volts to maintain the 1000mA set current. If the resistance is set to, say, 50 ohms, a constant current supply will try its hardest to keep the current at 1000mA, but it is limited to its compliance voltage (the supply voltage minus some overhead).

Constant current supplies are used to power some types of transducers like accelerometers. The DC supply current is put on the same conductor as the AC signal. These supplies typically supply 4mA, at up to 18v (2×9v batteries) or 24 volts. The electronics in the transducer biases this current to around 12 volts, and the AC output signal swings around this 12 volts. On the instrumentation end, a DC blocking capacitor isolates the signal from the DC power supply voltage.

It does take a different mindset to understand this arrangement, but it allows power and signal over a single shielded wire. The bias voltage is also a troubleshooting tool:

~12v – transducer is connected

18v+ – cable is open, transducer is not connected or is open

~0v – cable or transducer is shorted.
 

rjenkinsgb

Well-Known Member
Most Helpful Member
Hello everyone, I am designing a product which needs a Li-Po battery and I need a circuit to charge it. I am confused as to how do I charge the battery as well as use it as a power source within the same circuit. What I mean to say is, how is it that a phone, or any rechargeable device know that the battery is being charged and knows when the current needs to be sent into the battery?

Can someone share how a circuit looks like this? Will the Battery charging IC I choose automatically have this feature? Or do I need a special component that does it for me? Please do let me know. Thank you
In everything we have made using internally charged batteries and significant load current, external power bypasses the battery and runs the load directly, while it is connected.

That means the battery charge is fed to an unloaded battery.

In the simplest possible form & not allowing for charge control, imagine three diodes:
One from charger to battery, one from charger to load (whatever regulator/PSU is used) and a third from battery to load regulator.

When the charger is connected, that feeds the load at a slightly higher voltage than the battery so no current is taken through the battery to load diode.

In practice in such as laptops and presumably phones, low on resistance power FETs are used for power "routing" to minimise losses, with those controlled by either the battery management system or device main CPU.
 

Musicmanager

Active Member
And this makes no difference at all - the larger battery can't draw any more current because the charger is constant current, so small battery or large battery, the current will be the same, and is set by the charger (and specifically R3).
And that's not right either !
Specifically R3 is purposed to limit the current available to the battery, not dictate how much current will flow regardless of demand.

As I said yesterday, I'm too much the novice to argue with your theory but I have found the notes I made of the problem I faced .. .. .. .. .

My wife, amongst other problems suffers Dementia which includes a significant short term memory loss so I built these .. .. .. .

Memory Aid.jpgMemory Aid Remote.jpg


.. .. .. .. . the list on the left are her daily tasks and correspond with the buttons on the second remote unit. They are linked via RF so that pressing a button on the remote unit highlights the corresponding number on the main unit. When the question ' Have I had my lunch ? ' arises for the 4th or 5th time, a look at a highlighted number 4 will answer the question.
At the outset, both were powered with 18650 Lipo batteries connected to TP 4056's and whilst prototyping I was using a couple of cheap chinese 750mAh's which when 'on charge' I measured the current at R3 around 320mA.
However, when I finished work, I inserted a good quality 2600mAh Golishi battery and measured the current again to find 852mA. Note - both are within the limit set by TP4056 R3.

My concern .. .. where is that 852mA's coming from - the USB connection ?
Research suggested that most PC mounted USB outlets (USB 1 & 2) are limited to 500mA; some USB 3 are capable of 900mA and some Superspeed are capable of much more, but my luck always dictates working to the worse case scenario, so I changed R3 to limit current to 450mA.

That was what I was alluding to yesterday !
I don't understand much of technical stuff you guys expound at the drop of a hat, but I'm do know when I'm up the cut in a punt !

MM

Edit: Too many 0's .. 750mAh, not 7500mAh
 
Last edited:

Visitor

Well-Known Member
Constant CURRENT means exactly that, within the supply's ability to do that, it was force that much current to flow.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
And that's not right either !
Specifically R3 is purposed to limit the current available to the battery, not dictate how much current will flow regardless of demand.
Obviously constant current only works within certain limits - you can't force a battery to take more current than it wants, ohms law still applies. But, if the battery tries and draws more than the constant current then the charger will stop it going above that.

Your two examples didn't help, as both were under 1A, and the default constant current is 1A (with 1.2K for R3). This is within the capability of most (all?) USB chargers, which is what it's intended to be used with. Your readings on the two batteries were different, because they were at different points on their charge cycle.

But you've already (correctly) changed your charging current by changing R3.

To explain a little - Li-Ion charging is pretty simple:

1) You need constant voltage AND constant current (I know you can't have both at the same time).

2) Set voltage at 4.2V (per cell) and current at what you want (I use 520mA for 2600mA cells).

3) Connect the battery - which if it's discharged will try and take too much current, the charger will current limit to 520mA, dropping the voltage accordingly. This is now charging from a constant current.

4) The voltage will slowly rise, until it reaches it's maximum 4.2V per cell, the charger will now switch to constant voltage and the charging current will gradually fall.

5) Once the current has fallen to 25% or so (so 130mA) the battery is considered fully charged, and charging is ceased.

6) Charging current, and cut off points, makes a HUGE difference to how well the battery charges - with 20% (as I use) a good choice for maximum battery life, at the cost of slower charging. Waiting for the current to fall to 20/25% means the batteries are more charged than using a higher current - at the cost of taking longer.
 

Sashvat

Member
Yep, that's it !

Although you might want to show your schematic and let the Guru's on here cast an eye before discovering you have a stick of gelignite masquerading as a capacitor or something :)

Nigel .. I've been looking for the note of what occurred but I can't find it; I remember that the 2600 mAH battery I was using was drawing 582mA through the R400 CL resistor and the USB is only capable of 500mA ; I changed it to a lesser beast but I can't remember what.

MM
Sure I am designing my schematic I will definitely share it over here
 

Sashvat

Member
Hello everyone, I just wanted to ask you about these 3 screenshots I took from a schematic, I am just not able to understand them, the battery is not directly connected to the charger and also to the load, I am very confused can someone please explain these schematic diagrams? Thank you
 

Attachments

Diver300

Well-Known Member
Most Helpful Member
Vbat is the +ve end of the battery.

The screenshot with the TP4054 shows the battery being charged from Vbus. The TP4054 is an IC that controls the charging. It is supplied from Vbus and the output is connected to Vbat.

The charging is turned on and off by IO32, but I don't know what drives that.

D2 is a diode that lets Vbus power the 5 V rail. Q1 allows the battery to power the 5 V rail. Q1 is turned on when Vbus is low, and Q1 will basically have no voltage drop. A MOSFET (Q1) is used to get the most life from the battery by avoiding the voltage drop that would come from a diode.

The 5 V rail is regulated to 3.3 V by U2.

I suspect that the 5 V rail won't actually be at 5 V. The TP4054 charges a battery to 4.2 V, so it's likely that the battery will be about 4.2 V fully charged and 3.5 V when discharged. At 3.5 V, a diode drop would be a problem because the diode drop would reduce the voltage below 3.3 V.

So when running from the battery, the "5 V" connection will be between 3.5 V and 4.2 V

Vbus is probably the USB voltage of 5 V. Charging ICs like the TP4054 are designed to run from that. When Vbus is on, there will be a voltage drop of about 0.4 V through D2, so the "5V" connection will be about 4.6 V
 

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