How to build a simple volatge buffer

[Adhith]

Hello there friends...
I had made a thread related to the same topic in here 2 days ago but I think I wasn't able to clearly express my question there so made this thread. Sorry for the inconvenience.

I have a 4S battery pack which is connected to a BMS module. To get a stable 12V output from it, a buck/boost module is connected to its output, but the high frequency switching signal of the buck/boost module is causing the BMS to shut off after 3 mins . Its confirmed that the problem is with the BMS in my previous thread. I' now trying to solve this problem by using a voltage buffer between the BMS and the buck/boost module so that it could smoothen the output from the BMS. So does this idea work and if yes then anyone could  help me with a simple circuit.

[tszaboo]

Connect the output of an opamp to the negative input. 

[Adhith]

Connect the output of an opamp to the negative input. 

Sir could you please elaborate. I'm a beginner in electronics. The current draw would be around 1 to 1.5A so could an opamp deal with it??

[capt bullshot]

An OpAmp buffer won't help you here (I've read your other thread now).

From your description, I see three ways that could cause the BMS to turn off:

1) EMC issues:
I'd suppose you to try a Pi type L/C filter between the BMS output and the converter input - A capacitor (ceramic 1uF ... 10uF, parallel with 220uF ... 1000uF electrolytic) at each side, and a 100uH ... 470uH inductor between.

2) low load current induced auto-turn off (IMO the least probable, or you've got a really strange BMS)
Test your system (battery, BMS and converter) with a constant load (e.g. a power resistor - 22 Ohm for about 0.5A) applied to the output of the converter, and no other load and no filter connected.

3) weak battery
Watch the total battery voltage (input of the BMS), and watch each cell voltage while running your system until it turns off.

Test each one for itself (not the combination of both) to find out the root cause.

[tszaboo]

Connect the output of an opamp to the negative input. 

Sir could you please elaborate. I'm a beginner at electronics. The current draw would be around 1 to 1.5A so could an opamp deal with it??

What? Anything with 1A is not a "buffer".

[Adhith]

An OpAmp buffer won't help you here (I've read your other thread now).

From your description, I see three ways that could cause the BMS to turn off:

1) EMC issues:
I'd suppose you to try a Pi type L/C filter between the BMS output and the converter input - A capacitor (ceramic 1uF ... 10uF, parallel with 220uF ... 1000uF electrolytic) at each side, and a 100uH ... 470uH inductor between.

2) low load current induced auto-turn off (IMO the least probable, or you've got a really strange BMS)
Test your system (battery, BMS and converter) with a constant load (e.g. a power resistor - 22 Ohm for about 0.5A) applied to the output of the converter, and no other load and no filter connected.

3) weak battery
Watch the total battery voltage (input of the BMS), and watch each cell voltage while running your system until it turns off.

Test each one for itself (not the combination of both) to find out the root cause.

Thank you very much Sir for reading both the threads and helping me out.
I believe low load current turn off wont be the case since the BMS has no problem in directly driving a small digital voltmeter display attached to its output.

The batteries are brand new and every thing is in the range of 3.5 volts. It was fully charged at first time and have been used to power an amplifier board through the BMS to reach the current 3.5V. So the cells are pretty much in good condition also.

So I think I'll try the filter that you have mentioned and will let you know about the outcome. Since my BMS has been damaged I'm currently waiting for the new one that i have been ordered online.

I have attached a circuit diagram of it, could you please check it and see its correct or not
You said "at each sides" so by that you meant that the filter should be connected both at the input and output sideof the buck/boost module right??

[Adhith]

Connect the output of an opamp to the negative input. 

Sir could you please elaborate. I'm a beginner at electronics. The current draw would be around 1 to 1.5A so could an opamp deal with it??

What? Anything with 1A is not a "buffer".

OK i understood. sorry about that, my mistake

[capt bullshot]

I have attached a circuit diagram of it, could you please check it and see its correct or not
You said "at each sides" so by that you meant that the filter should be connected both at the input and output sideof the buck/boost module right??

No, I've meant to put the filter between the BMS and the converter. At each side (of the inductor) shall be two capacitors, the ceramic in parallel with the electrolytic.

BTW 3.5V is rather at the midrange to discharged end for a Lithion-Ion cell (in case you have Li-Ion batteries, LiFePo have different levels that I'd have to look up). I'd recommend to recharge the battery and try again with the cells at about 4.0V (again, in case they are Li-Ion).

[Kleinstein]

With a battery pack, it is probably OK to use just one inductor, I don't think there is common mode filtering required.

The slightly odd point is having a separate buck/boost stage and a LED driver. If the led driver is also switched mode, they might interact and a good led driver should work with the voltage range from the Li cells. So the best way would be to make the LED driver to work directly from the cells.

[Adhith]

No, I've meant to put the filter between the BMS and the converter. At each side (of the inductor) shall be two capacitors, the ceramic in parallel with the electrolytic.

BTW 3.5V is rather at the midrange to discharged end for a Lithion-Ion cell (in case you have Li-Ion batteries, LiFePo have different levels that I'd have to look up). I'd recommend to recharge the battery and try again with the cells at about 4.0V (again, in case they are Li-Ion).

Thank you again for your reply. Yeah I have understood. updated the circuit diagram and has been attached with this post.
Since the BMS is now damaged and I'm waiting for the new one that have been purchsed, I cant currently charge the pack. I'll let you know about the progress as soon I get the new one.

Also one thing that I would like to ask is that, does adding this filter have any trouble in driving the buck/boost when powered from a DC adapter??

[Adhith]

With a battery pack, it is probably OK to use just one inductor, I don't think there is common mode filtering required.

The slightly odd point is having a separate buck/boost stage and a LED driver. If the led driver is also switched mode, they might interact and a good led driver should work with the voltage range from the Li cells. So the best way would be to make the LED driver to work directly from the cells.

Yes I have thought about directly connecting the LED driver to the BMS, but the problem is that the LED driver is capable of input from 5v to 28v so I guess it could be used to drive any LED strip rated form 5V to 28V and thus no restriction or regulation at 12V. So I guess the 16.8V battery pack could damage the 12V strips.

[Kleinstein]

The easy type of LED driver would need an input voltage that is higher than the LED strip nominal voltage. So for a 12 V LED strip the driver would need something like 12.5 V - 28 V.  It might still work with 12 V. So one should check the details on the LED driver. A 12 V LED stip might be 3 LEDs of maximum 4 V in series and thus more like 11 V actual voltage.

So the 4S  (thus 12-16.x V) pack should be likely OK and no need for an extra voltage converter. One might loose a little of capacitance at the low voltage end, but not really much and it comes with less stress to the cells.

[Adhith]

The easy type of LED driver would need an input voltage that is higher than the LED strip nominal voltage. So for a 12 V LED strip the driver would need something like 12.5 V - 28 V.  It might still work with 12 V. So one should check the details on the LED driver. A 12 V LED stip might be 3 LEDs of maximum 4 V in series and thus more like 11 V actual voltage.

So the 4S  (thus 12-16.x V) pack should be likely OK and no need for an extra voltage converter. One might loose a little of capacitance at the low voltage end, but not really much and it comes with less stress to the cells.

OK I understood that Sir. but I have a doubt. What about the additional voltage above the 12V?? does it get dissipated in the LED strip resistors and heats than the normal limit??

[Kleinstein]

It depends on the LED strip and the driver. The usual LED drivers give out an defined current (e.g. 350 mA). So the voltage over the LED strip will be whatever it takes to get that current, but not more than available.

For the LED strips, there are 2 types: some with a series resistor that can be operated at constant voltage and those made for constant current drive. If there is a series resistor in the LED chain, it will get hot if the LEDs are lit.

So you have to get and understand the manuals / data-sheets  for both the driver and the LED strip, or at least have a look at the circuit / description. Depending on the LED chain the best way is either a constant voltage regulator or a constant current regulator (e.g. the LED driver), but not both.

[capt bullshot]

Thank you again for your reply. Yeah I have understood. updated the circuit diagram and has been attached with this post.
Since the BMS is now damaged and I'm waiting for the new one that have been purchsed, I cant currently charge the pack. I'll let you know about the progress as soon I get the new one.

Also one thing that I would like to ask is that, does adding this filter have any trouble in driving the buck/boost when powered from a DC adapter??

I've mentioned to put two capacitors on each side of the inductor, four in total. Each one ceramic and one electrolytic in parallel.

No, there shouldn't be any problems running this filter from an DC adaptor.

I'm still believing that a discharged battery caused your root problem, so I'd recommend to try again with a fully charged battery and without the filter when you get your new BMS. Anyway, if you want to ensure your LED strip supply doesn't exceed 12V, a step-down regulator will do the job, no buck-boost required. This is because you shouldn't discharge your cells below 3.3V in normal (low current) use, this leaves 13.2V at the battery when it is assumed empty.

[Adhith]

It depends on the LED strip and the driver. The usual LED drivers give out an defined current (e.g. 350 mA). So the voltage over the LED strip will be whatever it takes to get that current, but not more than available.

For the LED strips, there are 2 types: some with a series resistor that can be operated at constant voltage and those made for constant current drive. If there is a series resistor in the LED chain, it will get hot if the LEDs are lit.

So you have to get and understand the manuals / data-sheets  for both the driver and the LED strip, or at least have a look at the circuit / description. Depending on the LED chain the best way is either a constant voltage regulator or a constant current regulator (e.g. the LED driver), but not both.

Ok understood Sir. I'll check the specifications and figure out the things

[Adhith]

I've mentioned to put two capacitors on each side of the inductor, four in total. Each one ceramic and one electrolytic in parallel.

OK now I get it. thank you very much Sir.

This is because you shouldn't discharge your cells below 3.3V in normal (low current) use, this leaves 13.2V at the battery when it is assumed empty.

I though 2.5V was the terminal voltage of a battery up to which safe discharge is possible. or I'm wrong about it and its actually 3.3V??

[capt bullshot]

Yes, these cells can be discharged down to 2.5V, if the cells rest for longer periods at this voltage, you'll damage them. Discharging to 2.5V may happen if you have large discharge currents (> 1C ... 2C), but the cells recover pretty fast to voltages abobe 3V when discharging stops.

For normal use, the cells shouldn't be discharged that deep at lower (< 1C)  currents. Most of the batteries capacitance gets discharged at voltages above 3.0 ... 3.3V, so if you go lower, there will be very little additional capacity. A Li-Ion cell at 3.3V and no discharge current, after some resting time, is considered empty, though you'll see lower voltage while discharging the cell, depending on the current.

You may want to look up some datasheets of these cells to get an idea of the discharge voltage over discharged capacity. At the end of discharging, the voltage falls quite rapidly. From my experience, when discharging these cells at rather low currents (less than 0.5C ... 1C), 3.3V is a good point to stop discharging.

[Adhith]

Thank you very much for your detailed explanation. Didn't know about these facts and I'll make a habit that not to discharge lower than 3V from now onwards. I'll let you know about the progress about the filter when the new BMS arrives. Like you said I'll fully charge the pack first and connecte the BMS alone to see whether the capacity of the battery was the problem.

[capt bullshot]

You're welcome!

[Adhith]

Sorry for the delay in updating the result, I was busy with my college thesis presentation and all. The pi filter is doing the job. I used a 470uH toroidal inductor with two 1000uf electrolytic caps and two 10uf ceramic. Thank you again for your sincere help.