Compare the output of PIN Photodiode

Thread Starter

Chetan_Jadhav

Joined Mar 2, 2017
52
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.
 

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OBW0549

Joined Mar 2, 2015
3,566
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.
 

Thread Starter

Chetan_Jadhav

Joined Mar 2, 2017
52
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 ?
 

bertus

Joined Apr 5, 2008
22,121
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.

Ian_Dusklight-Block_corr.jpg

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

Bertus
 

Thread Starter

Chetan_Jadhav

Joined Mar 2, 2017
52
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.
 

OBW0549

Joined Mar 2, 2015
3,566
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.
 
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