[SOLVED] Circuit flow of infrared .

[WhyWhy]

Hi guys,

Actually i wants to know my concept is correct or not.

1) Starting the 5V will supply to the IR Transmitter because of the resistance is the lowest among R1,R2,VR1 and VR2. 2) Then the IR receiver will get the voltage and pass it to the inverter.
3) The Output of the LM324 is depend on the comparison between Voltage of VR1(10kohm) and IR Receiver.

Condition:
If Voltage VR1 is lower than IR Receiver then the Output of LM324 will equal to low,whereas Voltage VR1 is higher than IR Receiver then the Output of LM324 will equal to high.

Question:
1) Did the R1,R2,VR1 and VR2 just limit the current only?
2) what is the voltage of VR1 (10Kohm)?
3) Why i need the VR2(4.7Kohm)?
4) Can i just connect the Comparator Pin 7 to microcontroller?

Thanks for help :razz:

 

[adi0803199]

The value of the resistors R1,R2 are dependent on the operating point of the IR sensor.When R1=100 Ω,the current is limited to 50 mA.The output power must be made sure that it is sufficient for the IR detector to detect the radiations.The value of R2 is set to a specific value as it is the value that determines how much this sensor is sensitive to the next stage(i.e) comparator stage with a resistor of 10kΩ.
Answer:
1)It acts as a Current Limiting Circuit only as far as I can see but these designs determine the sensitivity of the sensors present.
2)The voltage from VR2 is dependent upon where the slider of the Potentiometer is present and it determines the crossing point(threshold) of the power from sensor to determine high output or low output value.
3)4.7 kΩ is required to limit the amount of current that flows through the LED branch in case the ouput is low from LM... comparator.This value is obtained by considering the above mentioned constraint.
4)It depends upon how much voltage the pin 7 of the comparator generate and remember to add an appropriate resistance so as to avoid the pin of the microcontroller from getting damaged.

 

[WhyWhy]

Hi adi,

The IR Sensor detector actually get the output power from the IR transmitter by using using the P=IV ( I x 2 x R) method right?

Question:
1)For the output power of R2 and VR1 is 2.5mA which is same (without adding the power from IR detector) because the Resistor is same 10k.
When the IR Detector get the power from the Transmitter.,The output power R2 will add with the power IR Receiver,hence the power of inverting input is higher than the non-inverting which output goes low right?

2) Since starting output power of R2 and VR1 is 2.5mA which is same,then the output will be high or low ?

Thanks for help!:lol:

 

[adi0803199]

1) is wrong as even though the resistor R2 and Vr1 are the same.The transducer that recieves the signal must have one or the parameter changed(resistance I suppose) so as to reflect that change in the output voltage measured that is then fed to the terminal of the comparator.
You need to understand that the currents are limited so that the sensors work properly,but we are interested only in the voltage output that is given to the comparator.The presence of POT ( Vr1) will determine the crossing point in the minimum strength of the recieved IR signal.

Hope this helps n clarifies your doubt.

 

[WhyWhy]

Hi adi,

Thanks,but for IR Transmitter part,actually the intput for the IR receiver to receive is Current or Voltage?

Regards,
YY

 

[adi0803199]

The input for the IR receiver is voltage as far as the website that I have referred,

http://www.futureelectronics.com/en/optoelectronics/infrared-receivers.aspx

Hope this answers your question.

 

[WhyWhy]

Thanks adi..