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I have simple IR receiver without any modulators, which reacts for IR wave (940 nm) link , I connected it as on this schema enter image description here

I beam on this receiver with laser diode with 650 nm lenght wave light. It reacts, but I would like to improve this reaction.

Right now, when I don't blink there is 1.6 V bettwen IR receiver (LED is OFF), and when I blink with laser diode (650nm) there is 1.8 V (LED is ON)

The question is How can I improve the difference to be not only 0.2 V but much higher ? at least 1 or 2 Volts.

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  • \$\begingroup\$ When the LED is on you have 1.8 volts and if you had 1 or 2 volts higher across the LED you'd have a dead LED. Look at your LED data sheet and realize that it's current you need to put thru the LED. Then rearrange your question along the lines around amplifying the photodiode's output so it is more sensitive to your laser diode and can therefore activate your LED from a much greater distance. \$\endgroup\$ – Andy aka Nov 7 '15 at 0:52
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Without changing your general idea, you need to be aware that the voltage across an LED changes very little with current, once there is any current at all - such as room light will provide. The LED than acts as a clamp at about 1.6 to 1.8 volts. To get around this, try

schematic

simulate this circuit – Schematic created using CircuitLab

You will note that I have not included a resistor value. You will have to determine this yourself. In principle, you ought to be able to get a 2 volt swing easily, but you will not be able to get more than about 3 volts. Higher resistance will give greater sensitivity.

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Why not try this simple circuit:

simple photo gate

The idea come from this website.

You will get a nice 0 and 5 V output.

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The best/easiest way to increase an optical TRF CW receiver's sensitivity is to choose, for the receiver, a photodiode sensor with an inherent spectral response which more closely matches that of the transmitter, and to isolate the sensor from the load with some gain, which will keep its bandwidth low, and its Q high.

The next best way might be to use a sensor with a receiver touting flat response over the band of wavelengths to which it'll be exposed and to use an optical filter to limit the wavelength of the light into the receiver's sensor to be close to that of the emitter's output.

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  • \$\begingroup\$ I know but laser diode 650nm costs 1$, and 940nm oround 30$, so having cheap laser diode i should use photodiode for 650nm , but that would catch me daily light, wo I use 940nm \$\endgroup\$ – kosnkov Nov 7 '15 at 8:25

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