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Author Topic: Power problem with 40$ line follower  (Read 1114 times)

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Offline ashourTopic starter

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Power problem with 40$ line follower
« on: August 24, 2010, 06:26:21 PM »
Hi

i've successfully made the the 40$ line follower and it works well expect that the motors have very low current that they barely rotate and friction force of surfaces is higher than their torque, also the leds glows less whenever a motor works and when both motors are on , the circuit dies but for the Green power indicator red,

I think that the problem is that i can't get my hands on a quality 9V batteries , and i won't be able to do so because it something related to importing economical goods into my country , so i was thinking of increasing the current using more batteries which would in turn increase the torque , but how would i do so?

Also why not use a cell phone battery i've read on one of them that it can give out current up to 800 mA while my battery can be short circuited to give at max 100 mA :(

Someone have an idea about this?

P.S. i've checked that there are no voltage drops in the circuit it gives 6V at the battery cap and 6V at the output of the motors (i know i use 9V battery but i told u before its of very low quality :( )

Thanks


Offline Razor Concepts

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Re: Power problem with 40$ line follower
« Reply #1 on: August 24, 2010, 06:34:31 PM »
For a 9v battery, 6v means its completely dead and will not supply current.

A cell phone battery could do it, the problem is they are only about 3.3 to 3.7 volts. If you have multiple batteries, you could string three or two of those batteries in series, and that should work in place of a 9v battery.

Offline Soeren

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Re: Power problem with 40$ line follower
« Reply #2 on: August 24, 2010, 07:08:39 PM »
Hi,

I think that the problem is that i can't get my hands on a quality 9V batteries , and i won't be able to do so because it something related to importing economical goods into my country
Even the best 9V (PP3) will have way to high internal resistance anyway.
Which country do you reside in?


Also why not use a cell phone battery i've read on one of them that it can give out current up to 800 mA while my battery can be short circuited to give at max 100 mA :(
Why not use AA cells?
Use 6 cells if using Alkaline cells and 7 cells if using NiMH cells for a 9V replacement.

A cell phone battery will be marked in mAh (mA-hours), not mA and an 800mAh battery can give out a couple of amperes at the minimum - And it has to, since cell phones transmits in bursts drawing that much current in the pulses.
Don't ever short a lithium battery to test it - the safety circuit inside it could fail and you'd have a nice little fireball removing the skin of your pinkies or worse.


P.S. i've checked that there are no voltage drops in the circuit it gives 6V at the battery cap and 6V at the output of the motors (i know i use 9V battery but i told u before its of very low quality :( )
The voltage drop is inside the battery. A battery could be viewed as an ideal voltage source in series with a resistor. As long as you don't draw current from it, it will be 9V (for a 9V battery), but the harder you load it, the more voltage is dropped across the internal resistance.

If the short circuit current is 100mA and the open loop voltage is 9.0V, you can draw an X-Y map with voltage 0 to 9 [V] on the Y-axis and current 0 to 100 [mA] on the X-axis which will be a straight line from (X,Y) 0,9 to 100,0 - you can then read directly that if you load it with eg. 50mA, you will drop half the available voltage over the internal resistance, getting just 4.5V at the terminals.
In your case, around 3V is dropped internally, which means you load it with around 33mA.

Shifting to AAs will cure that problem.

What does the battery measure when you remove the load?
Regards,
Søren

A rather fast and fairly heavy robot with quite large wheels needs what? A lot of power?
Please remember...
Engineering is based on numbers - not adjectives

Offline ashourTopic starter

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Re: Power problem with 40$ line follower
« Reply #3 on: August 25, 2010, 03:23:49 AM »
Quote
Which country do you reside in?
I live in Egypt , unfortunately project components are not widely spread here , just electronic stuff for replacing parts of devices and stuff like that.

Quote
What does the battery measure when you remove the load?
When i measured the volt i didn't measure it while having the circuit running i just put the multimeter terminals on the battery wires and it it gave 6V reading

also when i created a short circuit i did put the battery and a 1k Resistor , and the multimeter in series with each other and it gave that 100 mA reading.

Thanks

Offline Soeren

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Re: Power problem with 40$ line follower
« Reply #4 on: August 26, 2010, 07:00:30 AM »
Hi,

I live in Egypt , unfortunately project components are not widely spread here , just electronic stuff for replacing parts of devices and stuff like that.
You could just order something online - DealExtreme.com (a Chinese business) have very low prices and the shipping is free - and they have lots of different Lithium cells.


Quote
What does the battery measure when you remove the load?
When i measured the volt i didn't measure it while having the circuit running i just put the multimeter terminals on the battery wires and it it gave 6V reading
Oh, if 6V unloaded, that 9V battery is very very dead.


also when i created a short circuit i did put the battery and a 1k Resistor , and the multimeter in series with each other and it gave that 100 mA reading.
That's impossible!
9V/1kOhm = 9mA
Perhaps check your meter, what range it was in and so on (including the real value of the resistor), because you certainly cannot get 100mA that way.
Regards,
Søren

A rather fast and fairly heavy robot with quite large wheels needs what? A lot of power?
Please remember...
Engineering is based on numbers - not adjectives

Offline Libraoct7

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Re: Power problem with 40$ line follower
« Reply #5 on: November 08, 2010, 10:58:16 AM »
Quote
Which country do you reside in?
I live in Egypt , unfortunately project components are not widely spread here , just electronic stuff for replacing parts of devices and stuff like that.

Quote
What does the battery measure when you remove the load?
When i measured the volt i didn't measure it while having the circuit running i just put the multimeter terminals on the battery wires and it it gave 6V reading

also when i created a short circuit i did put the battery and a 1k Resistor , and the multimeter in series with each other and it gave that 100 mA reading.

Thanks
Hi I am doing this project for my school please help. I did the robot but the motors are not running. The LED both red and green are on. But motors are not. I check the voltage coming out to the motors and they are 0. I don't understand why I check my circuit again but nothing wrong with the circuit. Could someone please tell what could be wrong!!!

Offline Soeren

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Re: Power problem with 40$ line follower
« Reply #6 on: November 08, 2010, 11:24:03 AM »
Hi,

Could someone please tell what could be wrong!!!
The first thing that's wrong is that you double post in old treads with different subjects - please start a new one with a descriptive subject line.
And if you want help, please state what equipment you have for troubleshooting, what you have done so far and how it did or didn't react.
Regards,
Søren

A rather fast and fairly heavy robot with quite large wheels needs what? A lot of power?
Please remember...
Engineering is based on numbers - not adjectives

Offline Libraoct7

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Re: Power problem with 40$ line follower
« Reply #7 on: November 08, 2010, 03:02:25 PM »
Hi,

Could someone please tell what could be wrong!!!

The first thing that's wrong is that you double post in old treads with different subjects - please start a new one with a descriptive subject line.
And if you want help, please state what equipment you have for troubleshooting, what you have done so far and how it did or didn't react.

Thank you very much for your fast reply. I am sorry. I am new to this Forum. I haven't gotten the hand of it yet. I have posted a new thread, but waited for days and there wasn't any response, so I had to get into a thread and ask the question. I am sorry again for my inexperience action in the forum. I will try to learn more from others.

I build the $40 Line follower exactly as it is described in this site http://www.societyofrobots.com/member_tutorials/node/63
1- I used 9v battery to power the robot.
2- I used L293NE, but it says to use L293D. Does that matter? Could this be the problem?
3- I used 22k preset, but it says to use 20k preset. I could not find 20k preset. Is that OK to use 22k instead?
4- I used BC548 PNP
5- I used 100rpm motors
6- I used a multimeter to check for any bad wiring and for the amount of voltage that is going into the motors.
7- I plugged the battery in, and the two Red LED and the green LED lit up. That's all it did. The motors are not running.
8- I used the multimeter to check if there is any current going into the motor, but there is zero current going into the motor. I don't know why.
Please help me in solving the problem. It really matters to me a lot to finish this for my school. Thanks again.

Offline Soeren

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Re: Power problem with 40$ line follower
« Reply #8 on: November 08, 2010, 04:38:17 PM »
Please start a NEW tread!


(And if you had read this tread, you'd have noticed that a 9V PP3 battery won't work).
« Last Edit: November 08, 2010, 04:41:21 PM by Soeren »
Regards,
Søren

A rather fast and fairly heavy robot with quite large wheels needs what? A lot of power?
Please remember...
Engineering is based on numbers - not adjectives

 


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