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I need to detect room luminosity. I was thinking to use normal LED and op amp to bring the signal to PIC adc.
Will the selected op amp in the below circuit be OK?
it is an photodiode, not a led. The Photodiode need to be biased in reverse operation mode. So the dc voltage at the non inverting input should be higher than 0V.
Hey,
so you use a diode (in your case the led) also in the reverse biasing mode. Due to this biasing the minority carrier density is reduced by the boltzman factor so only a few carrieres are left which are available for a current. So you see, led emitts in the forward biasing mode photons while it is in reverse mode nothing else than a diode. But be careful. Normally leds are made of III/V semiconductor material which has a bandgap typically about 2eV. So only wavelength shorter than 615nm can be detected.
The circuit itself must be reconsiderd. The output will response depending on the time constant in the feedback. The output swing is depending on R and the photo current which seems to be very low. But you are right. 320M makes no sense.
Not true. Reverse-biasing photodiodes improves their high-frequency response by decreasing their junction capacitance, but if speed is not a concern then this is not necessary; they will still produce a current (in effect, acting like a tiny little solar panel). For DC purposes, reverse-biasing is undesirable because it introduces a leakage current sometimes called "dark current" -- this translates to a DC current offset. In OP's circuit, it's obvious that he doesn't care much about high-frequency operation, so it's definitely not needed and is instead undesirable.
For PIN photodiodes then reverse-biasing does improve sensitivity (since the captured photons are photo-multipled through the avalanche effect), but using a PIN photodiode is not indicated here.
nikens,
If you intend to use an LED it will have reduced sensitivity compared to an actual photodiode which has been engineered for the purpose. Like analog-designer said, its sensitivity will only be to light that has higher energy than the color the LED emits, so you should choose a red or IR LED for greatest sensitivity to the visible spectrum.
Your circuit should work, but the LM741 doesn't operate if its inputs are at its negative rail. This is referred to as a limitation in its input common mode voltage range. You may consider using an LM358 instead, since its input common mode range includes its negative rail.
A point of consideration with the LM358 (and any bipolar op-amp) is that its input bias current (for the LM358, typically 45nA at room temperature) will translate into a DC offset, and therefore your microcontroller will read a different light level. You can compensate for it in code, but the bias current will change with temperature so it gets painful. You may also consider instead using a CMOS-input op-amp, such as the MCP601 or the LMC662. The typical input bias currents for these op-amps are 1pA and 2fA, respectively.
Another point of concern is the 320Meg resistor. You may have trouble finding a resistor of such a high value. If you are attempting to measure very low light levels, you may consider using multiple photodiodes or a solar panel, so that you can use a lower resistor value. As a point of reference, a single LED configured as a light detector under room lighting typically produces somewhere between 100nA and 1uA of current. A resistor value of 10Meg or less is generally more reasonable.
"Another point of concern is the 320Meg resistor."
With those resistor values, the current leakage thru a PWB surface could be significantly higher, and it will vary with humidity and cleanliness of the board surface.
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