Low Voltage Cutoff

[LukeKnepp]

I found the attached circuit on this website, and that is all I need, but I have 2 questions.
I have a 20V Input Li-Ion battery, and I need this to cut off at 16.1V.

How do I set the cutoff voltage?
Also, how much current can this handle?

Thank You all in advance for the help!!!!

 

[Nigel Goodwin]

Stick a protection board on your battery pack, this will deal with the over discharging issue, as well as a number of others - a battery pack without a protection board is a serious danger risk.

 

[crutschow]

How do I set the cutoff voltage?

Adjust R2 to give 2.5V at the R input (about 8.6k ohm).
You could use a 10k ohm pot for R2 to make it adjustable.

how much current can this handle?

That's determined by the choice of MOSFET for M1.
For the fds4685 it's about 6A with no added heatsink area on the pcb.

But I would follow Nigel's advice and get a protection board designed for the battery.

 

[LukeKnepp]

Thank you so much!!!
The problem with buying a production board designed for these batteries is that I have not found one that will work, and I want to put it all in 1 board along with a speed control for a fan....

Thanks Again!!!!

 

[Nigel Goodwin]

The problem with buying a production board designed for these batteries is that I have not found one that will work,

Why not?, there are numerous protection boards available, for very little money, and 5S are quite common.

Without a protection board how are you monitoring individual cell voltages during charging?, and correcting any problems?.

 

[LukeKnepp]

Sorry, I meant that I have a protection board that works, but I want to incorporate this board into the same one as the speed control so that it fits into my small space I have.

 

[Nigel Goodwin]

The protection board is normally built-in the battery pack, which is why they are commonly available shaped to fit in such a configuration.

 

[LukeKnepp]

I am using a portable tool battery, and they have the shutoff built into all of the tools, to make the batteries cheaper.

 

[Nigel Goodwin]

I am using a portable tool battery, and they have the shutoff built into all of the tools, to make the batteries cheaper.

Are you sure there's no protection board inside the battery packs?, you need individual connections to each cell in order to provide a decent level of protection. There could be a low voltage shut-off in the tools, and individual cell protection in the battery packs.

 

[LukeKnepp]

I don't know, but I work for a company that just puts a low voltage shutoff in all of our products, so I figured I better do the same.

 

[Nigel Goodwin]

I don't know, but I work for a company that just puts a low voltage shutoff in all of our products, so I figured I better do the same.

The protection board should do that anyway.

 

[LukeKnepp]

What does this mean?
The Green line is the Input from the source and the blue line is the output.
They cross at my desired shutoff voltage???
What should I do to fix it??

 

[test4]

Set the correct signal V1. Add R1 for hysteresis.

 

[LukeKnepp]

Which shunt diode are you using, and can I find it in the spice libraries?

THANK YOU SO MUCH!!!

 

[test4]

Enjoy!

 

[LukeKnepp]

I am trying to get a complete low voltage cutoff using this schematic...
Can anyone tell me what is wrong with this one?
I am using the voltage controlled switch as a relay...

 

[crutschow]

The way you have S1 connected, there never is a voltage from V1 applied to the circuit.
S1 should be moved to just in front of Vout.
Why did you move the switch's position from that shown in post #13?

Also the polarity of S1's control input is reversed.
As connected, S1 will be off when U1 is on (high battery voltage condition)..
A P-MOSFET, connected as shown in post #13, is ON when its gate is pulled to near ground, and OFF when the Gate is near its source voltage.

 

[LukeKnepp]

The reason I moved the switch is because I need as little battery drain as possible after cutoff...
Is their a way to calculate the drainage current after cutoff?

 

[crutschow]

The reason I moved the switch is because I need as little battery drain as possible after cutoff...
Is their a way to calculate the drainage current after cutoff?

It is the current through R2, R3, and R5 (assume the voltage at U1's cathode is about 1V when it is on.
For the resistor values shown in post #16, the current draw at 16.1V would be about 1.7mA.

If that's too much, you can use a TLV431, which can operate with higher resistor values.
With that you could reduce the current by at least a factor of 10.
The main difference is that it has a 1.24V trip point at its control input, rather than the 2.5V for the TL431, so the values of R2 and R3 will need to be changed to get the same trip point.

 

[LukeKnepp]

Refer to #13

According to my calculations, R6 would have to be a 40W resistor...
Voltage Across it - 20V
Resistance - 10 ohms
20/10=2A at 20V DC...

Amps x Volts = Watts, so this will take a 40W resistor...

I am using this exact setup, except with a 10k resistor, since my 10 ohm burnt to a crisp...
It does not shut off until about 12V...

These are the links to my parts I bought...
https://www.digikey.com/product-detail/en/texas-instruments/TL431CLPR/296-1290-1-ND/404938
https://www.digikey.com/product-detail/en/on-semiconductor/FDS4685/FDS4685CT-ND/2094417
https://www.digikey.com/product-det...cs-inc/RNF14FTD16K9/RNF14FTD16K9CT-ND/1975110

and then I just used my own resistors...

Can anyone tell me why it doesn't shut off at 16.1?

THANK YOU!!!!