| T O P I C R E V I E W |
| kenocs09 |
Posted - Jun 06 2010 : 08:39:44 AM
Hello to all members of aaroncake forum.
i'm a newbie here.
just need some help with INFRARED LED.
We have an existing thu-beam type circuit using IR LED as a transmitter and IR photodiode as the receiver.
The sensing range of the circuitry is about 25cm.
Supply voltage is 12VDC(using AN7812 voltage regulator).
The circuit works just fine but there is a slight problem with the resistor in series with IR LED.
once the circuit is activated,the resistor gets too hot after a few seconds you can't even touch it with your finger.
The value of the resistor is 100ohms, 1/4watts.
Any help with the correct/appropriate value of the resistor to be used in series with the IR LED considering that the supply voltage is 12Volts.
Any response will be very much appreciated.
Thanks in advance.
Ken |
| 6 L A T E S T R E P L I E S (Newest First) |
| kenocs09 |
Posted - Jun 08 2010 : 09:34:57 AM
Hi Mr Audioguru.
I have already tried what you have suggested.
R2 was already removed from the circuit.
I have also tried using higher values of resistor for the R3 and after several trial, i have come up with a value of around 20k to suit my requirement. i also replaced the 100Ohms resistor with 560ohms to limit the value to around 20mA and the circuit works just fine with the said combination.
the coil resistance of my relay based on multitester reading is around 120ohms, so to limit the voltage across it to 6V, i will put a resistor with 120 ohms resistance (please correct me if i'm wrong).
What's left for me to do is to put an input and output capacitor for my regulator since i haven't purchase a capacitor yet.
Regarding, the diodes, the connection on the actual circuit were correct. its only on the diagram where the pin of the diode was reversed.
Mr Audioguru, A BIG THANKS to you for the suggestions. It really helped me a lot in making my circuit work and I also Learned a lot from you.
Although i still have to finalize the complete circuit, i am confident that this circuit will already work.
Again, Thank You Very Much for your help.
I'll let you know when the circuit is already 100% functional. ;)
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| audioguru |
Posted - Jun 08 2010 : 08:47:28 AM
With the IR diode connected backwards then it conducts all the time. With nothing limiting the current then the IR diode and/or the base-emitter junction of Q1 will blow up. |
| kenocs09 |
Posted - Jun 07 2010 : 6:39:50 PM
Thanks mr. audioguru.
1. I'll review my 7812 datasheet to for the input and output capacitor that you have mentioned.
2. Regarding the IR Diode, will the circuit works even if the IR Diode is connected backwards? Anyway, I'll try reversing my IR diode to see what will be its effect on my circuit.
3. I'll try to get different values of resistor between 330K and 1M to test which one is suitable for the circuit.
4. I'll also try to check my D1 on the actual circuit.
5. I"ll try to look for a datasheet of the relay that we are currently using to see the coil resistance. perhaps, i'll just try to put a resistor in series with this to decrease the voltage to 6v.
Again, thank you very much for the suggestions. I'll give you an update of whatever the result of my testing will be.
Have a Good Day!
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| audioguru |
Posted - Jun 07 2010 : 10:22:13 AM
1) Your voltage regulator IC is missing the important input and output capacitors shown on its datasheet.
2) Your IR receiving diode is shown backwards so its current and the current in the base-emitter of Q1 are extremely high all the time.
3) The value of R3 is much too low so it is shorting the very low output current of the IR receiving diode. Try 330k ohms to 1M ohms.
4) D1 is shown backwards.
5) R2 does nothing and can be removed.
6) A 6V relay will get hot when fed 12V. |
| kenocs09 |
Posted - Jun 06 2010 : 8:41:55 PM
Thanks audioguru for the reply.
i have also computed the circuit using the same formula. however, the value which i have used for the IR LED was 2V giving me 100mA current and 1Watt for the power.
I have also tried replacing the resistor with 1 watt and even tried 2watts resistor. the circuit lasts longer with lesser heat on the 100ohms resistor, but with the 2 watt resistor, other components gets hot also thats why i decided to join the forum and seek advice from the experts.
I have tried googling for circuit similar to mine but i can't seem to find one. So i need your help on this.
I have also tried decreasing the current to about 20mA by using a 470ohms but the sensing distance is decreased.
i have also attached the circuit that i am using right now for your reference. http://aaroncake.net/forum/uploaded/kenocs09/201066201313_IR%20SENSOR2.JPG
I would really appreciate if some one can give me suggestions on how to improve the circuit.
I think i need to focus more on modifying the receiver part, as what audioguru have suggested, but i don't know how that's why i attached the circuit.
Thanks.
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| audioguru |
Posted - Jun 06 2010 : 09:41:59 AM
An IR LED is about 1.3V. The 100 ohm resistor feeding it supplies a current of (12V - 1.3V)/100= 107mA which is probably much too high for an ordinary little LED. The max allowed continuous current in the LED might be only 30mA.
The heat dissipated by the resistor is 107mA squared x 100= 1.15W. So your little 1/4W resistor is very overloaded.
An IR photo-diode needs an amplifier. If you use a photo-transistor instead then the transistor part amplifies the signal so the IR LED does not need such a high current. |
