# Calculation help, light level measuring with CdS photo cell

Discussion in 'Electronic Basics' started by ITSME.ULTIMATE, Feb 28, 2006.

1. ### ITSME.ULTIMATEGuest

Ok, I put together a simple sensor circuit consisting of a resistive
divider and a CdS. Since CdS have a negative resistance characteristic
with input and I want positive output with input, I put the resistive
divider between the ground and sensor so that output increase with
light.

http://img518.imageshack.us/my.php?image=cdscalculation9bq.png
This is what my circuit looks like and I measured the Vo four times a
minute for a few hours and I have the data in form of voltage as a
function of time.

The chart is a CdS light input vs resistance chart. I'm not sure what
kind of chart it is from, but it is logarithmic. Assuming my CdS
follows a similar characteristic, how should I process my output voltage
so I can get a graph of relative light input that is roughly linear to
level of light?

2. ### BanGuest

Hello you,
if you want a precise measurement, do not use a photoresistor.
Apart from the not very straight logarithmic curve, there are more
disadvantages. For low light levels the temp dependency gets awful, at 0.1lx
the resistance doubles from 20 to 50C, but not so at 1000lx. And worst:
there is a memory effect, after long exposition to higher light levels the
resistance increases, it needs days in the dark to come back to the original
values.
A photodiode or -transistor or even a solarcell or an LED will produce a
*current* proportional to the light level. This current flows opposite the
normal forward direction, so you have to connect the +marked side to +V in
the same way as the CDS resistor. The bottom resistor you choose to get the
desired output swing with the max. light intensity. With +5V supply you can
reach a full output range of 5V, since the current flows even at 0V reverse
bias in a photo diode, LED or solar cell, a phototransistor will need 1-2V
across it (C to +V, E to resistor/ADC, B open) but is much more sensitive,
especially the darlington ones.

3. ### redbellyGuest

I agree that a photodiode is a much better way to do this. Here is a
simple circuit:

..
.. ,-------|<|-----,------o
.. | ph.d. |
.. + | >
.. ------- < R
.. Vb --- >
.. | <
.. | |
.. '---------------'------o

Your signal is the voltage across R. Choose R so that the maximum
light level in your application, whatever it is, produces an output
voltabe less than the bias voltage Vb. If the output signal ever gets
higher than Vb + 0.1 or 0.2 volts, then the signal will saturate.

You can increase the range by replacing R with a switch that connects
to one of two resistors, one of which is say 100 x the other:

.. |
.. \
.. \
.. . .
.. | |
.. [R1] [R2]
.. | |

Regards,

Mark

4. ### ChrisGuest

Hi. Everything Ban says about CdS photoresistors is true. They're
crude and sloppy, and they have a lot of hysteresis. And yet, they're
still used in many outdoor light sensors. First, they're responsive to
a wide range of light frequencies all across the visible spectrum,
where photodiodes and phototransistors are usually very
frequency-specific. Also, your eye's perception of light is also
logarithmic. Compared with your perception of light intensity, your
output voltage will seem more linear.

Possibly you could explain more about where that voltage output goes,
and what you're trying to accomplish? Are you trying to make a light
intensity meter, or are you trying to do something else?

Good luck
Chris

5. ### redbellyGuest

Hi Chris,

It turns out silicon photodiodes are also responsive across the visible
spectrum, plus the near IR and near UV. (If some are not this
responsive, it is because an optical filter has been added to the
sensor).

Photoresistors are okay for a simple "on/off" application, but it
sounded like the OP wants more than that, since he was asking about
getting a linear response.

Regards,

Mark