I am rusty at circuit design and have not done much with this. I would really appreciate any advice/input. I have an IR LED that I know is working and I want to use a photodiode. The photodiode has a photocurrent up to 90 uA, which I know is too small for arduino. So I did a lot of research on transimpedance amplifiers and decided this would be a good option to relate that small photocurrent to a voltage between 0 V and 3.3 V. I completed the calculations and designed the attached circuit. The problem that I have is based on the output values, there is no clear response of the photodiode and it seems to be steady at a number, even when I cover the photodiode all together. I am not sure if I miscalculated or missed something in the research but I would appreciate an advice.
Components
IR533C LED
SFA-213-FA Photodiode
TLV2462 Op amp
Arduinio Due
Code consists of analogRead() and Serial.print() in the main loop just to see values.
rutkeyh:
The photodiode has a photocurrent up to 90 uA, which I know is too small for arduino.
Ahh, I didn't know that.
I thought that Arduino input pins take NO current at all.
What are you trying to do.
A transimpedance amplifier is to get a linear voltage from a linear light level.
Not needed if you just want to detect light/no light.
For a transimpedance amp you connect photo diode between -input and ground (not 3.3volt).
With anode to ground.
Leo..
Correct, my apologies.
Ideally, I am trying to get a linear voltage from a linear light level.
I made that change to my schematic, does this change the mode the photodiode is in?
Wawa:
For a transimpedance amp you connect photo diode between -input and ground (not 3.3volt).
With anode to ground.
jremington:
because the op amp output has to go negative.
No, the photo diode generates a negative voltage/current, and the opamp fights that with a positive voltage/current. You measure that (always positive) correction voltage of the opamp.
Because the opamp cancels any voltage buildup across the photo diode, the diode is only delivering current.
And that is linear with light.
Leo..
No, the photo diode generates a negative voltage/current
A better way to think about it is that the photocurrent in the posted circuit (OP), which flows in the opposite direction of the diode symbol, has to flow from 3.3V through the feedback resistor, keeping the op amp (-) input at ground potential, which requires the op amp output to go negative.
Wawa:
No, the photo diode generates a negative voltage/current, and the opamp fights that with a positive voltage/current. You measure that (always positive) correction voltage of the opamp.
Because the opamp cancels any voltage buildup across the photo diode, the diode is only delivering current.
And that is linear with light.
Leo..
If the diode is biased at +3.3V with anode to the opamp input, the current flows the wrong way, the opamp
needs dual supply and will drive negative.
Cathode to the opamp with zero bias (ie PV mode), the current goes the right way and opamp output goes
positive. This mode is much slower though, as the diode's capacitance is much higher at zero bias compared
to reverse bias. Its also less linear.
Thank you, jremington, for the "overview" of photo diodes. It
explains much more than we need to know here. We are exploiting
the reverse leakage current property of the diode not the normal
diode function. So it must be reversed biased. That current is
sensitive to light intensity and is very small. Thus, it needs a high
amplifier gain.
Herb
@herbschwarz: Sorry the brief overview material proved to be too deep for you, but perhaps the OP will find it enlightening.
The circuit posted in reply #5 is using the photodiode to measure light intensity in standard photovoltaic mode and works with zero reverse bias, as designed and intended.
We are exploiting the reverse leakage current property of the diode not the normal diode function. So it must be reversed biased
. "We" are not so doing. Please review the posted links to improve your understanding.
