I'm trying to create a photodiode interface for my Arduino Uno board that can measure very high lux values. This is to be part of my lightcontroller project for monitoring high wattage lamps (~1000w). I've made the following design to simulate this in Proteus, but since I don't own a lux-meter, I don't have a faintest clue how much current will go through the photodiode when exposed to a 1000w lamp.
Could I go with this approach in real-life?
Any suggestions/comments would be most appreciated!
Best Answer
A transimpedance amplifier is the circuit of choice for combining high sensitivity and high speed. Since neither seems necessary (EDIT - and your C1 guarantees no high speed), your approach seems adequate. A few suggestions:
1) There is no obvious reason to include your bias resistor. On the one hand, you're running your photodiode in photovoltaic mode, so bias should not be important. On the other hand, you're running at high levels (~1 mA in your schematic) so dinking around with a few microamps doesn't seem worth the effort.
2) There is no need for your voltage follower, since the output simply feeds the + input of the gain amp, and there is no need to make the impedance change. That is, the gain amp has exactly the same input impedance as the follower, so you don't gain anything.
3) Frankly, I'd also get rid of the zener as unnecessary. Where, exactly, do you think high voltage is going to come from? The photodiode? You should read the data sheet (even better, provide us a link). For most signal photodiodes, as opposed to solar cells, a saturation current of about 10 mA applies. But you can leave it in if you like.
4) And most importantly, you'll need to do some experimenting to determine just exactly what sort of light intensities you'll be dealing with. Nominally, noon sunlight has a brightness of about 100,000 lux, and a total optical power of about 1000W/m^2. However, this is about 8% UV, 40% visible, and 50% IR. An incandescent lamp, of course, will have a somewhat different overall output curve (no UV to speak of, different peak wavelength, no atmospheric absorption of IR), and at this point I have no information about the response curve of your photodiode, its area, or the distance it will be from the lamps being measured. As a result, it's impossible to predict what the out of the photodiode will be. And that's why you need to do some measuring.