Author Topic: problem with my $40 Line follower project!!!  (Read 4592 times)

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Offline Soeren

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Re: problem with my $40 Line follower project!!!
« Reply #30 on: December 11, 2010, 08:00:07 AM »
Hi,

Quote
Your LDR's are of a higher resistance when in light, so the circuit have to be adapted for this.
Either you swap the trimmer potentiometers for some 200k to 500k, or you keep the ones you have and add a resistor to each trimmer where it connects to the +9V line. Break the connection and insert the resistor in between.
The trimmer will have much less influence if you just add a resistor, but it will be cheaper and easier.

Start with say a 100k resistor and test if the trimmer can now get it in range.
Measure the voltage over the LDR with the trimmer in each outer position and report back your findings
I have changed all my trimmer potentiometers to another ones with 250K.
I measured the voltage over the LDRs. Here are the readings.
right LDRs : 7.38   and   5.84
left LDRs   : 5.87   and   7.29
I measured the voltage of the battery and it was at that time  9.09
The resistance needs to be much higher then. The circuit is a bit strangely designed, with a NPN transistor in that position, but as far as I can see, the voltage over the LDR's needs to get down to around 3V.
If 250k gets you down to ~5.8V (I assume the readings are with the trimmers in each outer position and over the same color paper?), adding a 330k resistor in series with the 250k trimmer should get you down to around 3V with the trimmer approximately midway - try this and it should work.
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 AberTopic starter

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Re: problem with my $40 Line follower project!!!
« Reply #31 on: December 11, 2010, 02:26:00 PM »
Hi,
Quote
The datasheet for your LEDs will list their operating voltage.
I bought the LEDs from the shop near by. There wasn't any datasheet because I bought them individually. I need to know what kind of blue LEDs I should buy. Can you help?
Aber Aljadi

Offline AberTopic starter

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Re: problem with my $40 Line follower project!!!
« Reply #32 on: December 11, 2010, 02:40:07 PM »
Hi,

Quote
Use the LED's in parallel with a resistor for each...
That was my solution to. I need to buy me more 1k resistors.
Aber Aljadi

Offline AberTopic starter

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Re: problem with my $40 Line follower project!!!
« Reply #33 on: December 11, 2010, 02:45:55 PM »
Quote
(I assume the readings are with the trimmers in each outer position)

I don't think I understand what you mean by that!!!
« Last Edit: December 11, 2010, 03:18:48 PM by Aber »
Aber Aljadi

Offline AberTopic starter

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Re: problem with my $40 Line follower project!!!
« Reply #34 on: December 11, 2010, 02:51:29 PM »
while we are at it, I have another question.
I have asked this question before in this thread, but no one has answered it yet.
When I turn the resistance of the trimmers down to 0, aren't my Red LEDs suppose to go off?
Well, they are not. They are not even going dimmer or brighter. Any suggestions on that?!!!
Thanks for you time and help.
Aber Aljadi

Offline Soeren

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Re: problem with my $40 Line follower project!!!
« Reply #35 on: December 11, 2010, 04:17:31 PM »
Hi,

I need to buy me more 1k resistors.

Not quite... They shouldn't be more than 270 Ohm maximum.

With a full 9V battery and 20mA through the LED's, you need:
  (9V-3.6V)/0.02A =  270 Ohm
When the battery is down to 7V (and thus still have power left) you need:
 (7V-3.6V)/0.02A = 170 Ohm
At the end-of-life, the battery supplies 5.4V and then you need:
 (5.4V-3.6V)/0.02A = 90 Ohm

Clearly, a compromise must be found - selecting 180 Ohm, the current will be:
From (9.6V-3.6)/180 Ohm = 33mA
 to (5.4V-3.6)/180 Ohm = 10mA
And the medium will be (7.5V-3.6V)/180 Ohm = 21.7mA

The battery falls fairly rapid to a bit under 9V when loaded and so, the 33mA on a freshly charged battery shouldn't be a problem, but the 10mA at the end of battery life is much weaker.

A much better way is to make a 21mA constant current circuit like this, to replace the resistors. That way the current will stay the same over the battery life.



BTW. You need to use 6 AA-cells or similar, as a PP3 (box shaped) 9V battery just won't have the power to drive it.
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 Soeren

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Re: problem with my $40 Line follower project!!!
« Reply #36 on: December 11, 2010, 04:21:15 PM »
Quote
(I assume the readings are with the trimmers in each outer position)

I don't think I understand what you mean by that!!!

These measurments:
Quote
I measured the voltage over the LDRs. Here are the readings.
right LDRs : 7.38   and   5.84
left LDRs   : 5.87   and   7.29
I measured the voltage of the battery and it was at that time  9.09
7.38V and 5.84V represents what then?
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 Soeren

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Re: problem with my $40 Line follower project!!!
« Reply #37 on: December 11, 2010, 04:26:04 PM »
while we are at it, I have another question.
I have asked this question before in this thread, but no one has answered it yet.
When I turn the resistance of the trimmers down to 0, aren't my Red LEDs suppose to go off?
Well, they are not. They are not even going dimmer or brighter. Any suggestions on that?!!!
What do you mean by turning the trimmers to 0 (you cannot get them to 0V).
But the idea is to adjust them to a point where the red LED extinguish when over white and turns on when over black.

That is exactly what you're gonna solve by adding the 330k resistors I mentioned.
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 AberTopic starter

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Re: problem with my $40 Line follower project!!!
« Reply #38 on: December 18, 2010, 12:17:13 PM »

Hi

Quote
(I assume the readings are with the trimmers in each outer position)

I don't think I understand what you mean by that!!!

These measurments:
Quote
I measured the voltage over the LDRs. Here are the readings.
right LDRs : 7.38   and   5.84
left LDRs   : 5.87   and   7.29
I measured the voltage of the battery and it was at that time  9.09
7.38V and 5.84V represents what then?


Sorry, I must have gotten myself confused. When I measured the voltage over the LDRs, I put the black probe on positive lead and the red probe on the negative lead of the LDR. I thought that how you taught me how to measure the voltage over the LDR.
But I am not sure what the trimmers have to do with measuring the voltage over the LDR. That's why I got confused when you said "I assume the readings are with the trimmers in each outer position"
Aber Aljadi

Offline AberTopic starter

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Re: problem with my $40 Line follower project!!!
« Reply #39 on: December 18, 2010, 03:09:12 PM »
Hi,

WOOW, ok now this is too advanced for me. I thought this project is supposed to be easy and not complicated for beginners. From what I have been reading through threads, I believe some beginners have built the circuit with the exact parts mentioned in the tutorial, and it worked for them.

Quote
I need to buy me more 1k resistors.

Not quite... They shouldn't be more than 270 Ohm maximum.


I thought you said "Use the LED's in parallel with a resistor for each..."

Quote
With a full 9V battery and 20mA through the LED's, you need:
  (9V-3.6V)/0.02A =  270 Ohm
When the battery is down to 7V (and thus still have power left) you need:
 (7V-3.6V)/0.02A = 170 Ohm
At the end-of-life, the battery supplies 5.4V and then you need:
 (5.4V-3.6V)/0.02A = 90 Ohm
.
.
.

Now this is advanced and new to me.

Quote
A much better way is to make a 21mA constant current circuit like this, to replace the resistors. That way the current will stay the same over the battery life.



This is way advanced for me to understand. First, it looks like I need to make a big change in my circuit. Second, I don't quit understand this schematic. I have spent hours searching the net to find out what of these schematics mean. Now I understand that Q1 and Q2 refer to the collector lead of a transistor. But there are things I don't understand about it:
-Is B+ means the positive side of the battery? If so, I don't see the B- . What is 0V
-You are using BC547 and BC337, different than what the tutorial is suggesting which is BC538 for all four transistors.
-R1 and R2, are these resistors? If so, what is 3k3 and 33R?
-LED- and LED+ are for Red LED?

Quote
BTW. You need to use 6 AA-cells or similar, as a PP3 (box shaped) 9V battery just won't have the power to drive it.

I am using 9V Ni-MH 250mAh Rechargeable battery. Doesn't this battery have the power to drive it?
« Last Edit: December 18, 2010, 03:16:35 PM by Aber »
Aber Aljadi

Offline Soeren

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Re: problem with my $40 Line follower project!!!
« Reply #40 on: December 21, 2010, 04:30:05 AM »
Hi,

WOOW, ok now this is too advanced for me. I thought this project is supposed to be easy and not complicated for beginners. From what I have been reading through threads, I believe some beginners have built the circuit with the exact parts mentioned in the tutorial, and it worked for them.
It is easy, but you didn't get the exact parts... The LDR's you got are of a much higher resistance.


Quote
With a full 9V battery and 20mA through the LED's, you need:
  (9V-3.6V)/0.02A =  270 Ohm
Now this is advanced and new to me.
Let's break it down then:
You have 9V on the battery (when fresh)
Blue LED's needs to drop around 3.6V
That leaves 9V-3.6V = 5.4V to drop over the resistor

To get 20mA (0.02A) for the LED, the resistor should thus be: 5.4V / 0.02A = 270 Ohm


This is way advanced for me to understand. First, it looks like I need to make a big change in my circuit. Second, I don't quit understand this schematic. I have spent hours searching the net to find out what of these schematics mean. Now I understand that Q1 and Q2 refer to the collector lead of a transistor.
Q1 and Q2 are just the component designations in the schematic and it's a coincidence that they're both close to the collectors.
However, you should probably just stay with single resistors for now.


But there are things I don't understand about it:
-Is B+ means the positive side of the battery? If so, I don't see the B- . What is 0V
-You are using BC547 and BC337, different than what the tutorial is suggesting which is BC538 for all four transistors.
-R1 and R2, are these resistors? If so, what is 3k3 and 33R?
-LED- and LED+ are for Red LED?
B+ is the positive side of the battery. You may be used to thinking that the other end is -9V, but that would make it an 18V battery. You use one side of the battery as 0V and then you get +9V from the other side (or -9V if you F'd up ;D) - hence the 0V. Calling it B- would indicate a second battery with its positive side connected to 0V.

The transistors in the schematic is BC548 (not BC538). The reason I used other types is that I reused an old schematic made for a slightly higher current draw, hence the BC337 (which handles more current than the BC547.
The only difference between BC546 (not used here), BC547 and BC548 is the max. voltage they can handle. BC546 can handle 65V, BC547 can handle 45V and for the BC548 it's 30V - all other parameters are equal.
I always keep a stock of BC547, as I sometimes use them above 30V, but rarely above 45V.
You can use BC548 for both of them, in case you'd like to try out the circuit - perhaps for some different purpose.


I am using 9V Ni-MH 250mAh Rechargeable battery. Doesn't this battery have the power to drive it?
It's a bit low on capacity, but might work due to the lower internal resistance compared to an alkaline - however, you'd have a low runtime. I imagine that your motors use around 1A (1000mA) and in that case, the apparent capacity of your battery is even lower, perhaps around 150 to 200mAh.
Assuming 200mAh freshly charged, the runtime would be 200mAh/1000mA=0.2h (12 minutes).

A set of 6 AA cells of say 1500mAh would give 1.5 hours of runtime (all this based on a 1000mAh current draw from the motor).
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 AberTopic starter

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Re: problem with my $40 Line follower project!!!
« Reply #41 on: April 15, 2011, 04:00:40 PM »
Hi,
I am very sorry I haven't been in contact lately about the progress of my project. I have been busy doing some other work, now, two weeks ago, I am back to my project.
I thank everyone who has helped me in doing this project, specially Soeren who was an excellent helper and who has helped me all the way to the end without getting tired of me asking many questions, but hey the more we ask the more we learn.

My robot worked very good. The Science Fair project is due April 21. There is only one catch about this project I just need to point out or rather to ask about.

First,
In the "Electronics (Circuit Explanation)" Pratheek explained the following "for following white line and for using the robot as a photophobe, there is a slight change in the circuit. The left sensor output is connected to Pin 7 and right sensor output to Pin 10. Pins 2 and 15 remain unconnected."
However, in the "Putting everything together" Pratheek has showed the procedure differently. He only let pin 7 and 10 unconnected.

Second,
in the "Putting everything together" He has mentioned that connecting pin 7 and 10 to Vcc would make the robot "To use the robot as a black line follower, obstacle avoider and photophobe, make the following connections in the main circuit board." I think connecting pin 7 and 10 to Vcc would make the robot photovore not photophope.

Lastly,
I am not sure how this robot would be an obstacle avoider!!!
Aber Aljadi

 


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