2868 Cut-Off Tunning Help (S9V11MACMA)

Hello ,
I recently bought one 2868 Step up/down and it works fine.

But just because i am very new generally on electronics
i can not well understand the tunning process for the cutt of potensiometer.

(Admins…Feel free to combine/move this post according best forum readability)

So…i have a battery pack 3AA rechargables that gives apporximatelly 4.35 Volt on start up
and i ve set the primary/basic potensiometer to give me 4.0 Volts.

And i want to cut of at 3.3 Volt meaning no less than 1.1 for each rechargable AA

According the formula
EN / 0.7 = VIN / VIN cutoff

i must have(with the secondary potensiometer)
EN adjusted @
= (4.35 / 3.3 ) * 0.7 = 0.92

Now the problem is that ok…i am doing like this and lets say working perfect.

But what will happen if sometime that i will go to replace the batteries
…the starting voltage (for some reason) is not 4.35 but lets say 4.2

Then finding cutt off with formula
EN / 0.7 = VIN / VIN cutoff

VIN cutoff = (VIN * 0.7) / EN
= (4.2 * 0.7 ) / 0.92
= 3.20

If all those i wrote untill now are ok,then
Can you please clarify if
this is “as it is”?

and …is not possible to have a
STABLE Cutt off voltage
(as i was imagine (as newby))

I mean …i though i could exactly set up the cutt off coltage.

Ah.!..and not only this …but the question
(sorry if i am confused and confusing you)

Lets say…that ok have it started at VIN=4.35
and EN=0.92 meaning cutt off at 3.3

But if i switch off my application
when the batteries are 4.0 volts

and after 2 days switch on (without recharge batteries)

then the cuttof will be less that 3.2 ??? !!!?

(Is there somekind of internal
"hold in 2868 memory the start voltage)?

VIN cutoff = (VIN * 0.7) / EN
= (4.0 * 0.7 ) / 0.92
= 3.04 ???

Is this 3.04 / 3AA =1 Volt will “destroy” the efficiency of my batteries?

Thank you in advance and sorry if i am not very clear…


The low-voltage cutoff stays stable at the same voltage once it is set unless the potentiometer is turned. The enable pin on the regulator IC will disable the regulator when it is pulled below 0.7V and enable the regulator for voltages above about 0.8V. The cutoff feature uses a voltage divider from VIN to EN to set what VIN voltage corresponds to an EN voltage of 0.7V. Therefore, as the VIN changes over time the voltage at EN will change proportionally and the cutoff voltage will remain the same. If you want to prove it to yourself, you could measure VIN and EN a few times over the course of one battery discharge cycle.


Firstly thank you for your imidiate responce
and secondly…

…even if i didnt well understand give me sometime to finish
my “how long my batteries will keep run the project” experiment
(that my 2868 is buzzy)
And i will come back after doing experiments with my
2868 and 2869.

And again,Thank you.

OK…in 2 days i will make my question again if still not understand.

Just a secondary question ,a little out of topic,
(i am placing it here because i am a beginner
and most propably all know this and didnt want to create a new topic)

When the specs are saying that the regulator
has not reverse voltage protection
then this simply mean that if i have it configured
to step up/down Vout= 4 volts
when Vin can be from 3.3 up to 4.5(3 AA batteries)

then …if i disconnect the batteries and power the arduino from
usb that has 5 volts …
then?..those 5 volts that will touch VOUT of regulator
will damage something?

This is the concept of reverse voltage protection in regulators?
That i must have a switch(at least) in such a case?

Or no problem for this case?

(Sorry again …that i am using this post for different question
and also for general electronics …lessons…)
Thank you in advance.

“Reverse-voltage protection” for our regulators is referring to a feature that protects the board in the event that the input power is connected with reversed polarity (i.e. connecting the negative lead on your power source to VIN and the positive lead to ground). The #2868 regulator does not have reverse-voltage protection, so connecting the power with the wrong polarity will damage it.

None of our regulators have any specific circuitry for protecting them if their outputs are connected to the voltages from another source, and we do not generally characterize them for that. You can find some more information on the topic in this post by Jan. Connecting multiple power sources together can potentially damage all of the sources, so you should make sure you either know the inner workings of each source (usually not practical) or connect them with extra circuitry specifically made for connecting multiple power sources together. In some cases, you can get by with just some diodes, and there are also specialized products like this power multiplexer.


Thank you very much.
As i understood from the link-topic

“We do not generally characterize our regulators for reverse current from output to input, though I expect it to usually be okay if the current does not get big…”

I most propably will not damage the regulator if i have
a)disconnect the batteries from its VIN
b)5Volt of usb(powering arduino) touch also Vout of 2868.

So may no need a switch for this.
But ok…in my design i will be able to
c)put out of the circuit the 2868(when powering from usb)

If you have anything else to advise (and if have time) feel free.!
In either case Happy New Year.!!!

I will come back for the main question about Cut-Off Tuning.
Be well.

Ok…finally i have free my
a)2868(with cut off pot)
b)2869 (without cut off pot)

and after a little experiements i think i am understanding the process
of adjusting the cut-off voltage of EN pin.

As i understood …it does not matter how much voltage will start to operate but JUST how much voltage is applied(as VIN)
THE TIME that we are measuring-adjusting
the EN voltage.

In my first experiment i had measured the following
Pololu 2868 Vout = 4.39 (from 4 AA)
Starting Voltage 5.82 (of the 4 AA)
Because i wanted to cut-off at 4.4 (1.1 minimum per cell)
EN=0.7 *( 5.8 / 4.4 ) = 0.9227
and doing this also for 4.5 gives
EN=0.7 *( 5.8 / 4.5 ) = 0.902222

So my thinking is to put somewhere below 0.92
and i adjusted the potensiometer at EN=0.91

And the last data (from arduino via 2868-4AA batteries)
01/01 10:07 = 4.97
01/01 17:07 = 4.96
01/01 21:07 = 4.96
02/01 06:07 = 4.95
02/01 12:27 = 4.95
02/01 20:27 = 4.94
03/01 00:27 = 4.94
03/01 06:27 = 4.93
03/01 15:27 = 4.93
03/01 19:07 = 4.92

03/01 22:00 FINISH
Cut off tooked place somewhere there.
…at 4.9 Voltage and NOT 4.4 .

Now i ve made some experiments and saw
that if i feeding 2868
via 2869 (simulating the batteries)
then the 2868 is Cutting Off at about 0.73 and not 0.7(of EN)
and also is turning back on at about 0.78 and not 0.8.

And because my post most propably is boring
(and not well understandable)

**A simple question. **(for if and when someone find a little free time)

Is 0.7 stable ?
**… or is it possible to be different (0.73) **
in my specific 2868 module ???

(that in this case i must use 0.73 in my formula
EN=0.73 *( 5.8 / 4.4 ) = 0.962 (vs 0.92 with 0.7)
and doing this also for 4.5 gives
EN=0.73 ( 5.8 / 4.5 ) = 0.94 (vs 0.90 with 0.7)
that in my case that i adjusted at 0.91
(but in actual off ~0.73 of my module)
computing Vin (cutt off) = 0.73 * (VIN /EN)
gives Vin=0.73
(5.8/0.91)=4.65 Voltage will shutt down.

And this is more closer to what happened
(stopped at about 4.9 and not 4.4 that i have computed).

PS.My Voltage-Ampere meter is the chiepest in the market (10 euros)…meaning i dont know if my measurements are good.

Your calculations look correct. The 0.7V falling and 0.8V rising enable thresholds are typical, so your unit’s actual thresholds might vary a little.


Great !!!
Thank you all again and happy new year.

can someone help me i am feeling expecially dumbfounded. I have everything set up correctly, i assume…

I have am currently setting the cutoff voltage and seem to be complete misunderstanding on how to set it.

The initial battery voltage i am using is 7.4 (3.81v per cell for storage in between use)

with the formula:
EN= (7.4/3.4)*0.7 im getting 1.52v for EN…

My problem is that when reading the voltage on EN im getting 3.23 v and i cannot get the POT any lower. how am i suppose to read the voltage on EN.

I am currently using GND / EN to get voltage on EN.

If i do the math differently
i get 0.76V… but again I have a voltage reading of 3.23 v on EN…

I am not in support team but untile they are arrive…

Ok… 1.52 must be the EN if now you are 7.4 and want to stop at 3.4

Did you try turn the EN potensiometer the other way around?
(Is reverse turning)

Hi, Lordprimate.

It seems like something is wrong with your board or setup if you cannot get the EN voltage to read lower than 3V. Could you post pictures of your setup that show all connections and solder joints? Does the voltage setting potentiometer on your board work as expected?

By the way, from your calculations it seems you are trying to set the undervoltage lockout to 3.4V. Is that correct? Assuming you are using a 2-cell LiPo battery (7.4V), that seems very low. A standard cutoff would be 3V per cell.


ok , thanks for the replys, Jim and Claire:
I can say that i believe i solved my problem with EN…

You are correct Claire. I was not thinking when i typed , and did maths. I ment 6.8 (3.4 per cell)

So I was just testing with a 7.4 lipo. Ultimately I want the cut off to be around 3.4v per cell on my 2s lipo. In the OP its obvious that it doesnt matter what the starting voltage is. I did my maths again using 8.4 v (a full 2s lipo at full cap) so when i do my maths again i get that EN should be 0.855 if i want my cut off to be 6.87.

am i correct. I am getting the voltage i want to see on EN. I was turning the wrong way… a complete failure on my part on several fronts, however, i was able to figure it out, after some more trial and error.

You can also fix a simple excel sheet to do the hard work!!


PS.advise admins can you fill some jscript at
the EN adjustments (for non excell masters like me !! heheh)

That math looks correct, though I recommend measuring the actual VIN voltage when you go to adjust the cutoff in your setup and recalculating.