Li-Ion battery discharger

Thread Starter

Chochy

Joined Feb 20, 2021
14
Hi, i wanted to desing a battery discharger. It should discharge the cell to a storrage voltage (3,7V - 3,85V). I wanted to use comparator so I could learn to work with it. The comparator compares the reference voltage with the voltage of the cell and when the cell goes down below the CCA 3.8V, it should open the mosfet. I also wanted to have a bacup plan, in case of something happens with the comparator, so I came up with the zener diode wich should be normally opened and pull up the open collector output of the comparator. Also when the voltage of the battery goes under the 3,3V the gate of the mosfet has to be pulled low. I had issues with the R34 "pulldown" rezistor when it was just 4K7. The the mosfet wont close until the voltage was higher than 4,2V. So I experimented and decided on the 100K rezistor. The issue is, the zener needs 1mA to work. I used 3.6V zener on my breadboard and it did open when the "cell aka power supply" was above the zener voltage, but the zener didnt drop the 3,6V but around 1,4V wich isnt a problem, I dont need to stabilize the voltage, but I dont know if it is reliable to use the zener like that. Also the zener closes the mosfet at around 3ish volts, wich is Ok, but I was using a 3,6V zener diode, so maybe I should change to 3,6V zener in the design. I also added a fuse in case of a short circuit. I would like to ask the the plan B solution with the zener diode is reasonable or I was just lucky and it works. I also think the R3 is uselles at this point. Note that I used diferent components because i was asembling it on a breadboard and just wanted to test it. I was using just and LED as a load with a diferent mosfet. Also the final product should be mostly SMD. This is my first proper desing so I would like to know what I did bad and what should be improved and if the idea with the zener diode is bat, what else should I use. I just want to add that I want to keep things simple and that I know I could use and MCU but my goal was not to use and MCU and learn with comparator and things around that. Thx for the feed back.

PS: the screenshot is of what part of the circuit I built on the breadboard.
 

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dl324

Joined Mar 30, 2015
12,929
Welcome to AAC!

You don't have a pull up resistor on the output of the comparator so you'll never be able to turn the MOSFET on. Unless that's what the zener diode is doing. Can't say if that'll work without knowing what VCC and the battery voltage are.

D1 seems odd. What is the purpose of VCC?
clipimage.jpg

I couldn't force myself to read your entire paragraph. Try using paragraphs to organize your thoughts so your objective and questions will stand out.

That's probably why your post has almost 3 dozen views and I'm the first to try to respond.
 

Thread Starter

Chochy

Joined Feb 20, 2021
14
Welcome to AAC!

You don't have a pull up resistor on the output of the comparator so you'll never be able to turn the MOSFET on. Unless that's what the zener diode is doing. Can't say if that'll work without knowing what VCC and the battery voltage are.

D1 seems odd. What is the purpose of VCC?
View attachment 231067

I couldn't force myself to read your entire paragraph. Try using paragraphs to organize your thoughts so your objective and questions will stand out.

That's probably why your post has almost 3 dozen views and I'm the first to try to respond.
The zener diod should be on until the voltage is high enough or until the comparator grounds the get.
The diode is use to prevent batteries charging each other, in the end 4 of these circuit's are joined in parallel. Vcc is powering the comparator and the tlv voltage reference.
Battery voltage is 4,2v when fully charged. I want to cut of the load at 3.8V idealy but when the comparator fails, the battery should go below 3V
 

dl324

Joined Mar 30, 2015
12,929
The zener diod should be on until the voltage is high enough or until the comparator grounds the get.
I assume get means gate.

You still haven't provided sufficient information.

If I assume that the battery voltage is 4.2V, you'll have about 0.9V on the gate. The MOSFET might be on, or might not because the threshold voltage is 0.65-1.45V. Assuming the zener is even conducting; the knee current is likely around 0.25mA and you'll have something like 10uA.

R10 is unnecessary. What do you think it's doing?

What is the nominal voltage for Vref?

EDIT: Made an edit and decided it wasn't relevant.
 
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Thread Starter

Chochy

Joined Feb 20, 2021
14
I assume get means gate.

You still haven't provided sufficient information.

If I assume that the battery voltage is 4.2V, you'll have about 0.9V on the gate. The MOSFET might be on, or might not because the threshold voltage is 0.65-1.45V. Assuming the zener is even conducting; the knee current is likely around 0.25mA and you'll have something like 10uA.

R10 is unnecessary. What do you think it's doing?

What is the nominal voltage for Vref?
Vref is 1.23V

R10 should limit the current when the gate of the fet is charging, but as u said it's unnecessary, I realized that now.

The diode is conductive, but the voltage drop over her is much less than the zener voltage due to the lowe current and that's why the circuit is working.

So I think this solutions isn't correct and I have to find another solution.

I thought of a second N-chanel mosfet wich would pull the output of the comparator high. With another voltage decided in the base. What do you think?

I just need another low precision cut of circuit if something happens with the comparator and is should be as simple as possible.
 

dl324

Joined Mar 30, 2015
12,929
why do you assume he wants to charge to 4.2v when he said he wants to charge them to a storage voltage of 3.6v?
I assumed a nominal voltage for a charged battery to see if the MOSFET would even turn on.

EDIT: Made an edit and decided it wasn't relevant.
 
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wayneh

Joined Sep 9, 2010
17,153
Personally, I'd leave out the zener diode portion. You've got a fuse, and with a pulldown on the gate, you've really got enough built-in safety. Any of the ICs can fail but that's rare and with a fuse as a catch-all last resort, I think you're good.

Using a quad comparator is handy for this kind of application. You can light an LED when a battery is detected, or for indicating various currents levels, or all sorts of things.
 

Thread Starter

Chochy

Joined Feb 20, 2021
14
Personally, I'd leave out the zener diode portion. You've got a fuse, and with a pulldown on the gate, you've really got enough built-in safety. Any of the ICs can fail but that's rare and with a fuse as a catch-all last resort, I think you're good.

Using a quad comparator is handy for this kind of application. You can light an LED when a battery is detected, or for indicating various currents levels, or all sorts of things.
Yeah but Iam not sure if it would also work with the other smd led, but thanks.
 

Thread Starter

Chochy

Joined Feb 20, 2021
14
Hi,
i wanted to make a battery discharger based on a comparator so I can learn how to use it and I came up with this design.

It discharges the cells to 3,8V (optimal storage voltage) with 2power resistors in paralel. Also a led diode indicates if the batery is discharging or not.

The comparator is watching the voltage of the cell and compares it with a reference voltage (1,23V) and if it goes below that it connects the gate of the fet to the ground and opens it. I also added a fuse if a short circuit occurs. The circuit should do just nothing if the cell polarity is reversed.

I would like to add a second undervoltage protection in case of something happening with the comparator so the cell isnt drained and damaged. Can you suggest some realy simple undervoltage protection so the battery cant go under 3V?

I also should lower the Amp rate of the Fuse, the cells are dischargint at about 0.8A so maybe 2amps should be a better choice?

The VCC is powering the comparator itself and the TLV reference, so you can discharge how many cells you want at onece.

Please let me know if you can think of some improvements. This is kinda my first design, so I would like to hear some feedback before I build it.
1613900201129.png
 

wayneh

Joined Sep 9, 2010
17,153
I still don't understand the purpose of D1 to D4, and the connection of Vcc there. It should work fine without that?

Oh maybe I get it - Vcc will be the highest voltage coming from one of the batteries being discharged? I think that voltage will be too low and the reference and comparators may not function properly.

And yes, your fuse could be much smaller. Are you protecting the battery or the load resistors? Choose the fuse rating based on what you are protecting.

Are you sure about reverse polarity hookup? That'd put a voltage on the comparator input far below its negative power rail. That's usually bad news for an IC. Inputs should be within the power rails unless you're sure it's OK (it is OK for some ICs).
 

Thread Starter

Chochy

Joined Feb 20, 2021
14
The VCC powers up the comparator and the reference, then function just fine, the comparator operates from 2V and the TVL from 1.23V so thats fine. The diodes prevent the bateries from charging one another. So you can discharge bateries of diferent capacities and so on.

There are 4comparators in one package so all outputs of the diodes are connected to the VCC of the ICs and also the diodes protect the ICs form reverse polarity, at least i think.
 

wayneh

Joined Sep 9, 2010
17,153
.... and also the diodes protect the ICs form reverse polarity, at least i think.
They do not protect the comparator from the problem I mentioned of an input well below the negative power rail. Check the datasheet - that's a No-No for the LM339. Your 180K resistor may limit the potential damage caused but I'm not sure about that.

Also read about the common mode voltage range. For the LM339, that's from the negative power rail to Vcc - 1.5V. If your battery is at 3.8V, the diode drops Vcc to 3.1V. Knock off 1.5V to get into common-mode range and you're looking at a sensing range of just 0-1.6V. I guess your reference voltage is in that range? Maybe you're OK on this.
 

wayneh

Joined Sep 9, 2010
17,153
Agree on C2. Good catch on R3.

But did you read what he said about Vcc coming from one of potentially 4 cells, ORd together by diodes?
 

dl324

Joined Mar 30, 2015
12,929
Whatever he's doing, he needs the pull up voltage to be high enough to turn the MOSFET on. 1.25V isn't high enough to guarantee the device will turn on. The threshold voltage can be as high as 1.45V. 4V will give a typical on resistance of 20-30 milliohms.

EDIT: The information about the threshold voltage was irrelevant. I kept thinking the voltage divider voltage would be applied to the gate. Regardless of that, it's still better to strive for having a lower on resistance.
 
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Thread Starter

Chochy

Joined Feb 20, 2021
14
They do not protect the comparator from the problem I mentioned of an input well below the negative power rail. Check the datasheet - that's a No-No for the LM339. Your 180K resistor may limit the potential damage caused but I'm not sure about that.

Also read about the common mode voltage range. For the LM339, that's from the negative power rail to Vcc - 1.5V. If your battery is at 3.8V, the diode drops Vcc to 3.1V. Knock off 1.5V to get into common-mode range and you're looking at a sensing range of just 0-1.6V. I guess your reference voltage is in that range? Maybe you're OK on this.
The reference voltage is 1.23V so it should be okay. I Know about the common mode voltage and I kept that in mind.

so waht do you suggest as a reverse polarity protection?
 
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