how to avoid effect of ambient light on photo diode

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k_ishan

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hi friends, I am using LED - photo diode pair for line tracing........
but due ambient light from the sources like tube lights ,halogen lights,sunlight...etc output of diode is getting saturated to the supplied voltage.

So, can u please suggest something to avoid this saturation of diode due to ambient light?????
 
hi,
Whats the spectral response range of the emitter and detector, is it IR.?
 
To stop saturation, increase the load (decrease the load resistor) on the photodiode.

If you want the photodiode to work over a large range of lighting levels, you can use a string of (silicon) diodes as the load - this will give a logarithmic response to the photocurrent rather than linear.

Another method is to modulate the emitter & use an AC amplifier on the photodiode output - the ambient signal is consequently removed.
 
A couple things you can do:

1. Turn the LED on. Record the value. Turn the LED off. Record the value. Subtract and use the difference.
2. Shield it. I've competed in outdoor line following contests and almost everyone uses a shield around the sensor array to block ambient light (Sun).
 

hey i got the first point of using series of diodes..........
Can u please elaborate the second point.................i am not getting what is "modulate the emitter"?
 
k_ishan said:
hey i got the first point of using series of diodes..........
Can u please elaborate the second point.................i am not getting what is "modulate the emitter"?
Click to expand...
Turn it on and off quickly, e.g. >20kHz
 
actually...i have applied square wave of 1khz to led...... so a wave which is somewhat like a square wave appears across the photo diode.......
and when ambient light falls on the photo diode this wave gets DC shifted .......and I thought that i can get this square wave by some AC coupling or clamping.....................
but the problem is that the DC Shift is too high that the square wave gets totally clipped............
 
Reducing the resistance will stop the diode 'saturating' as it can't pass sufficient current.

Can you post a picture of what you are seeing?

What is the part number of the photodiode?
 
dougy83 said:
Reducing the resistance will stop the diode 'saturating' as it can't pass sufficient current.

Can you post a picture of what you are seeing?

What is the part number of the photodiode?
Click to expand...
i have attached the image of my circuit ...............
in normal conditions it gives proper output as expected ...........
but when i take this circuit near the lamp output gets DC shifted and the output square wave is totally clipped ...................

so i want to reduce this DC shift and want to get the square wave
 

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You forgot to draw the power supply; I'm assuming it's 5V. The 33K resistor is a little on the large side. Try 4k7 or 1k (this will reduce the output swing however). what values have you tried already?
 
dougy83 said:
You forgot to draw the power supply; I'm assuming it's 5V. The 33K resistor is a little on the large side. Try 4k7 or 1k (this will reduce the output swing however). what values have you tried already?
Click to expand...

the photo diode i am using is like a LDR (light dependent resistor) .
Its resistance decreases with the increase in light following on it.
so when i take the circuit near the lamp, resistance gets nearly zero so output voltage which is across 33k resistor becomes 5v.
so i think there is no need to change the value of resistance.........
 

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Right......
 
Hi,

I have the same situation. When I place the PD directly to the sunlight the PD seems to be saturated. But when It's not directly facing the sun my circuit can operate perfectly.
Any solution how to avoid this scenario?
 
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