Light switch with photodiodes instead of resistors

Thread Starter

karamba

Joined Dec 29, 2016
2
I am trying to come with the simplest solution to produce a signal depending whether one photosensor gets more light than the other. The one on the picture uses photo resistors and LM324 comparator that would work. Unfortunately photo resistors are too slow for what I intend to use it. How would I change the schematic to use photodiodes with minimal additions such as amplifiers etc ?
Scan1.JPG
 

dannyf

Joined Sep 13, 2015
2,197
conceptually what you have will work. in reality, the resistors should be increased substantially, which can create a speed issue.

I would configure the amplifier differential and put the diodes back-to-back across the inputs. this requires a dual rail supply.

for single rail, use two such amplifiers, 1 for each diode.
 

GopherT

Joined Nov 23, 2012
8,012
I am trying to come with the simplest solution to produce a signal depending whether one photosensor gets more light than the other. The one on the picture uses photo resistors and LM324 comparator that would work. Unfortunately photo resistors are too slow for what I intend to use it. How would I change the schematic to use photodiodes with minimal additions such as amplifiers etc ?
View attachment 117692
Get rid of the 10k resistor connecting output to ground. That will slow your response.

Also, what kind of speed and what wavelength are you looking for? Infrared or something in the visible range?

Also, LM 324 is not s comparitors but it can make a close substitute. If a photoresist or is not good enough, then you might also want to upgrade to a real comparitor.
 

crutschow

Joined Mar 14, 2008
27,424
You might use an LM339 comparator for more speed, but you will need to add a small amount of positive feedback (hysteresis) to avoid oscillations around the trigger point.

How much difference in light do you want to detect?
 

Bordodynov

Joined May 20, 2015
2,890
dannyf,
I also like to increase at the beginning of the resistor values (in 100). But then he changed his mind. Necessary ratings are determined by the size of lighting.The photodiode BPW 34 S (Osram) has a sensitivity of 75nA / Lx. Let illumination is 1000 lux. Then on 10k resistor will decrease the voltage equal to 75nA / lx * 1000lx * 10kOhm = 750mV. The hysteresis value is not more than 3.6V / 3Meg * 10k = 12mV. The sensitivity is equal to 12mV / (75nA / lx * 10kOhm) = 16lx. If you want to compare the smaller illumination, it is possible to increase the quantities of resistors. When comparing the pulsating illumination required filter capacitors.
 
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