voltage drop after short period of time !

hello.
since i am newbie and i am going to ask simple questions any help will be appreciated.
i have a project where i have to create maze solving robot using "Left Hand Rule" , i bought the component and finished the assembling ,
( 3 ultrasonic sensors HC-SR04 , 4 wheel robot )
i have a serious problem is the robot after a complete 5 ends ( solve the maze 5 times ) it kind of start moving slowly
my connection is like that
3 battery of ( 3.7V , 3800MAH) connected Sequentially forming ( 11.1 V ) connected directly to L298N dc drive and the 5+V I/O from L298N goes to the arduino Vin
my question :

  1. is the slow of the robot movement after X no. of times normal ? if yes , how could i prevent that ?
  2. does the ultrasonic take a lot of power ?

JumperIQ:
3 battery of ( 3.7V , 3800MAH) connected Sequentially forming ( 11.1 V ) connected directly to L298N dc drive and the 5+V I/O from L298N goes to the arduino Vin

Hello there!

Quick question. Why is the 5V from the L298N going to the Vin port? 5V at the Vin port will not put 5V on the 5V rail on the Arduino because of the dropout voltage of the regulator.

bos1714:
Hello there!

Quick question. Why is the 5V from the L298N going to the Vin port? 5V at the Vin port will not put 5V on the 5V rail on the Arduino because of the dropout voltage of the regulator.

well, i am using only one source of power formed in a holder of 3 battery space ( btw i am using 3.7 V as mentioned before)
do you have any suggestions better than my power source or how to provide better power to arduino ?
thank you in advance

I see. Basically, the type of regulator that is on your Arduino board has a characteristic known as dropout voltage. At its simplest form, the dropout is the minimum voltage difference between the output and input voltages to the regulator. For an example, a 5V regulator with a 1V dropout voltage will need a minimum of 6V to correctly output 5V. Putting 5V into the Vin port would not be enough to get a solid 5V at the output of the regulator, which is what powers the board.

I am not familiar with the L298N, but with the similar L293D device, the 5V pin is used to power the logic systems, and I imagine it is the same case for the L298N. I don't believe the pin is a power output.

What you could do is apply the 11.1V from the batteries to the Vin port of the Arduino instead of the 5V pin, and connect the 5V pin on the Arduino to the 5V pin on the L298N. Make sure you also connect the ground connections from the batteries, Arduino, and L298N together somehow.

Common L298N boards with heatsink do have an onboard 5volt regulator that can be used to power the Arduino.
As bos1714 explained, 5volt from the motor driver should be connected to the 5volt pin of the Arduino.

Current draw of the Arduino and ultrasonic sensor is insignificant compared to what motors use.
The batteries or the L298N might have trouble powering/driving the motors.
You should measure battery voltage under load (when the bot slows down).
Leo..

The ancient L298N driver wastes a great deal of battery power as heat, in your case up to 1/3 of the available power.

Pololu has an excellent selection of much more efficient, modern motor drivers.

bos1714:
I see. Basically, the type of regulator that is on your Arduino board has a characteristic known as dropout voltage. At its simplest form, the dropout is the minimum voltage difference between the output and input voltages to the regulator. For an example, a 5V regulator with a 1V dropout voltage will need a minimum of 6V to correctly output 5V. Putting 5V into the Vin port would not be enough to get a solid 5V at the output of the regulator, which is what powers the board.

I am not familiar with the L298N, but with the similar L293D device, the 5V pin is used to power the logic systems, and I imagine it is the same case for the L298N. I don't believe the pin is a power output.

What you could do is apply the 11.1V from the batteries to the Vin port of the Arduino instead of the 5V pin, and connect the 5V pin on the Arduino to the 5V pin on the L298N. Make sure you also connect the ground connections from the batteries, Arduino, and L298N together somehow.

well,i read once that the arduino will always drop any input power to be 5V on it's board !.
and i tried your advice but the motor was really really slow the car couldn't move on the ground !, i guess the L298N need more voltage than the 5V output from the arduino

I see. What kind of motors are you using? I noticed your battery setup is 11.1V and 3800mAh. Is that enough current?

bos1714:
I see. What kind of motors are you using? I noticed your battery setup is 11.1V and 3800mAh. Is that enough current?

sorry each battery is 3800MAH , 3.7V .
well, i am using a normal dc motor that came with 4 wheel kit "Geared Motor "

JumperIQ:
sorry each battery is 3800MAH , 3.7V .
well, i am using a normal dc motor that came with 4 wheel kit "Geared Motor "

Did you buy the batteries on Ebay and did they come from China?

Paul

Since you measured (I at least hope you measured) the voltage of all three batteries you are using at 11.1V, that means you connected them in series, which makes the voltages of each one add up to one voltage. However, that means the current does not change, and stays at 3800mAh.

Speaking of which, there is a big difference between MAH and mAh (and it isn't the 'h'). Do you know which you have? I would be shocked :slight_smile: at a battery that has 3800MAh.

I recently read a long article in a national magazine pretty much proving anything above 3500 MAh is bogus!

Paul

Paul_KD7HB:
I recently read a long article in a national magazine pretty much proving anything above 3500 MAh is bogus!

Paul

They clearly don't understand the point being made. M is mega, m is milli. 9 orders of magnitude different...

Anyway for batteries you can tell from the size if the capacity figure is crazy.
The energy density of LiPo is typically 250 to 730 Wh/l (for a single cell this translates to 70 to 200Ah/l)

So 10Ah cell should be about 50 to 140 cm^3, exclusing case and wiring/protection circuitry.

bos1714:
Since you measured (I at least hope you measured) the voltage of all three batteries you are using at 11.1V, that means you connected them in series, which makes the voltages of each one add up to one voltage. However, that means the current does not change, and stays at 3800mAh.

Speaking of which, there is a big difference between MAH and mAh (and it isn’t the ‘h’). Do you know which you have? I would be shocked :slight_smile: at a battery that has 3800MAh.

sorry for misleading you about the battery specifications , i will copy paste the on battery info. again
LC 18650 3800MAH , i think there is a mistake in “MAH” is should be “mAh” as you said ,for the size of the battery it is in "hand finger size "
as we as i am really really new to the powering issues .
i just figured out the drop in voltage after the sensor works , firstly i test the movement of the car without any ultrasonic in and the car was really fast and took to much time to recharge it but after i implemented the ultrasonic in the movement that car starts to move slowly !
can you guide me to better way in powering i just think that i will buy two jacks one for the arduino and one for the dc motors ? is that ok or is it going to solve my problem ?
thank you .

Paul_KD7HB:
Did you buy the batteries on Ebay and did they come from China?

Paul

i bought them from a market in my country but of course they are not original !
and if i want to buy something from ebay or amazon it will take 30-40 day to arrive to my place !
,btw same market was working on a project " drone " and they advice me to buy a battery same as the drone has !

thank you