Hi,
I am applying square wave to the LED, and i want to detect it at the other end using photodiode.
I measured the output of Photodiode using DMM it shows +0.23V when light falls on it and 0.01 otherwise.
(PIN 5D |1544-1 is written on the diode casing)
so, i implemented the given circuit but it is not working at all (may be 0.23v is too small {correct me if I am wrong} ).
please provide any suggestions.

 

The (-) input of the op amp should be connected to the op amp's output, not to ground. That way the op amp will operate as a unity-gain buffer (i.e., a "voltage follower"). The way it's connected now, it won't do much of anything.

 

thanks for the help.
unity gain means input(+) will reflect at output.
but the input is very small i think so if light falls on diode, will the output become +5V ?

 

thanks for the help.
unity gain means input(+) will reflect at output.
but the input is very small i think so if light falls on diode, will the output become +5V ?

 

unity gain means input(+) will reflect at output.
but the input is very small i think so if light falls on diode, will the output become +5V ?

No, the output will only rise as much as the (+) input. If you want more output than that, use the op amp in a non-inverting amplifier configuration.

 

Hello,

@ian field , Why do you post such a dark picture?
Using the GIMP it is easy to make it more bright using the colors>levels function.



I also did cut out the schematic part of the picture.

Bertus

 

OK,
I will try that on hardware. one doubt-> do i need to change the resistor of 10K given in my schematic,or the way in which i connected it?
thanks.

 

I will try that on hardware. one doubt-> do i need to change the resistor of 10K given in my schematic,or the way in which i connected it?

The way you have it connected is fine. If you decrease the value of that resistor, you will get less output; if you increase it, you will get more, although increasing the resistance will also increase the response time of the circuit. Depending on what you're doing, this may or may not be a problem.