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Problem with MUX

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Simply GUESS at the max allowed current for your LEDs then use Ohm's Law to calculate the value of a current-limiting resistor.
Buy an assortment of resistor values.
 
ruzfactor said:
It's awesome! LEDs are very bright!!! I've used 100ohms as resistance to connect the 3 LEDs in series. I have a dumb question to ask: Can I connect the LEDs to +12V instead of +9V? Will it burn out the LEDs?

To audioguru: I have 100 ohms and after that I have 220 ohm resistors. Don't have any 150 ohm. :(

hi,
Yes you will burn out the LED's.

Recalc shows that for 9V use a 150R to reduce the LED current to a safe level.

If you use 12V then you must increase the resistance value.

Use Ohms law.... [12volts - Total Voltage drop across the LEDs]/0.02 = resistor value.
 
calculation showed that I should use 345 ohm for +12V. So Can i use 320 ohms?
 
ruzfactor said:
calculation showed that I should use 345 ohm for +12V. So Can i use 320 ohms?

hi,
I would suggest 390R

I thought that the LDR's were going to be used in a line follower project.

If the LED's are VERY bright, is it possible that LDR's could be saturated due to excessive light reflections.

I would go for a LED intensity that gives the best contrast line/no line at the LDR's...:)
 
hi

Yes LEDs are to be used for a Line follower project. I have these values of resistors: 100ohm, 220ohm, 270ohm, 470ohm(found it in my old box) and 560ohm. I can use 100ohm and 270ohm as 370 ohm. Is it fine?
 
At 440ohm and 12V , my LFR worked well. I think too much illumination makes it hard to tune the comparator with POT.
 
ruzfactor said:
At 440ohm and 12V , my LFR worked well. I think too much illumination makes it hard to tune the comparator with POT.

hi ruzfactor,
This is the point I was trying to explain, adjust the intensity of the light from the LED's to give the best response from the LDR's

Also by lowering the light intensity by reducing the LED current, they will draw less current from the battery and it will last longer..
More is not always better..:)

The other power saving method would be to have all SIX LED's in series from a 12V supply and use a 120R series resistor.
 
hi ericgibbs,

I need to clear another thing. I'm a bit confused. Two resistors are used as voltage divider in the comparator at the inverting I/P. Now I my question is that do I need to use the same value of resistors for both Comparators I'm using?? I have used two 57K resistor for the first comparator and two 27K for the second comparator. Plz clear me out.
 

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ruzfactor said:
hi ericgibbs,

I need to clear another thing. I'm a bit confused. Two resistors are used as voltage divider in the comparator at the inverting I/P. Now I my question is that do I need to use the same value of resistors for both Comparators I'm using?? I have used two 57K resistor for the first comparator and two 27K for the second comparator. Plz clear me out.

hi,
Looking at your photograph of the LED's and LDR's I would say the two comparators should be the same.?
The use of the 27K and 57K in the comparators will set the switching voltage levels for the comparators.

What switching levels have you designed the circuit for.?

EDIT:
Added drawing, with approximate voltages for the two resistor values.
 
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Actually I have connected like this schematic. Please check it. For Both inverting I/Ps i get about 2.9~3.0 Volts (I'm using +6V instead of +5V). Is it necessary to use same values of resistors for both comparators?
 

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ruzfactor said:
Actually I have connected like this schematic. Please check it. For Both inverting I/Ps i get about 2.9~3.0 Volts (I'm using +6V instead of +5V). Is it necessary to use same values of resistors for both comparators?

hi,
If you use two equal values for both comparators you will get the same voltage at the Vref input.
The variable resistor in series with the LDR's will set the comparator switch point voltage [light/dark]

Are the two LDR's going to be located either side of the line they have to follow.?

Do you have details of the Line Follower project you are building, not just the circuit.?
 
I don't have any details on this LFR project except the circuit diagram. But the LDRs should be positioned like this pic. Assume the red dots as LDRs. Also the I/P voltage should be tuned to the voltage close to the Ref. voltage at start.
 

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ruzfactor said:
I don't have any details on this LFR project except the circuit diagram. But the LDRs should be positioned like this pic. Assume the red dots as LDRs. Also the I/P voltage should be tuned to the voltage close to the Ref. voltage at start.

From the picture, I would say the two LDR detectors are identical.

Whats the voltage at the LDR's when dark and light.?
 
hi,
Track background is black and two 470ohm used for the two LED series.
At first I tune the pot to set the Input voltage(2.9~3.05V) close to Reference voltage (which is around 3V). When No light is detected, I/P voltage becomes about 3.4~3.5V. And when the LDR directly comes below the white tape, the I/P voltage becomes very low(around 1.17V) and slight displacement would make it vary between 1.85~2.4V.

Plz correct me if I'm wrong: when V I/P is less than V ref then the O/P is LOW

When V I/P is greater than V ref then the O/P is HIGH.
 
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ruzfactor said:
hi,
Track background is black and two 470ohm used for the two LED series.
At first I tune the pot to set the Input voltage(2.9~3.05V) close to Reference voltage (which is around 3V). When No light is detected, I/P voltage becomes about 3.4~3.5V. And when the LDR directly comes below the white tape, the I/P voltage becomes very low(around 1.17V) and slight displacement would make it vary between 1.85~2.4V.

Plz correct me if I'm wrong: when V I/P is less than V ref then the O/P is LOW

When V I/P is greater than V ref then the O/P is HIGH.

The LDR signal is connected to the Non Inv input of the Comp, so when the LDR signal is less than the Vref the Comp output is Low and vice-versa.

It sounds OK to me.
 
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