Unable to find out problem

[Nathaniyelu]

Thank you for your kind information..
What about receiver side..?
I.e demodulation is needed..
But in the circuit it is an amplifier..
How the photo diode will demodulate the received signal..?

 

[Les Jones]

Think of Rich D's example of a light being switched on and off at a frequency of 1 Hz. During the 0.5 seconds that the light is off the photo diode will give no output. During the 0.5 seconds when the light is on the photo diode will give some output depending on the amount of light reaching it. Looking at the output of the photodiode with an oscilloscope you will see a 1 Hz square wave. This is what you have in the unit that you are trying to repair. The only difference is that the light (IR source) is flashing at 6.7Khz (Rather than 1 Hz) The AC signal is easier to amplify than a DC signal. (It could be improved by inserting a filter with a centre frequency of 6.7 Khz. This would reduce the effect of ambient light from fluorescent lights etc.) In the case of remote controls the infra red is modulated twice. First with a frequency of about 38 Khz. The data steam being sent then modulates the 38 Khz

Re post #21 The photo diode does the demodulation. It gives no output when it receives no infra red. When it does receive an infra red signal it gives an output depending on the strength of the infra red signal. As the infra red source is being switched on and off at 6.7 Khz you will see a 6.7 Khz square wave on its output.
Les.