I am a newbie in electronics and try to design a smoke detector incorporated with Arduino Uno.
I am trying the circuit given below. I used oscilloscope to measure the voltage across the resistor. The resistor chosen here is 1 Mohm.
When there exists light, the voltage is approximate 300 mV. However, when the diode isolated from the light, the voltage has strong resonance with frequency of 100 Hz, an amplitude of 200 mV and mean of 6 mV.
May I know why the photodiode has such performance? Is it due to the dark current and the practical photodiode with the equipment circuit of a current source, a shunt resistor Rj, a shunt capacitor Cj, and a series resistor Rs?
heng:
When there exists light, the voltage is approximate 300 mV. However, when the diode isolated from the light, the voltage has strong resonance with frequency of 100 Hz, an amplitude of 200 mV and mean of 6 mV.
It is not "resonance".
You are working in an artificially lit room with either fluorescent or LED lighting, which is flashing with every half cycle of the power mains, so it is pulsating at 100 Hz and your photodiode is picking that up because you are not isolating it from the room light.
heng:
So the photodiode will not generate negative voltage if it really isolated from the light？
The voltage is actually positive - the anode becomes positive with respect to the cathode (as your diagram indicates).
The voltage generated is always less than the forward conduction threshold of the photo-diode. Most certainly, no light, no voltage generated.
For maximum detection sensitivity, photo-diodes are generally reverse-biased whereby you detect current, not voltage. This also widens the depletion zone and thus reduces the capacitance so improving high frequency response.
heng:
When there exists light, the voltage is approximate 300 mV. However, when the diode isolated from the light, the voltage has strong resonance with frequency of 100 Hz, an amplitude of 200 mV and mean of 6 mV.
With light carrier pairs (electrons and holes) are created as photons are absorbed and these drift
towards the ends of the device due to the built-in electric field of the junction, causing an external
voltage and current through the resistor. The diode's dynamic resistance will be much lower than
1M ohm in this regime.
With no light there are only thermally generated carriers (much much fewer) and the diode is
almost perfectly insulating. The signal you see is mains pick-up from nearby conductors or your
body. With 1M of resistance the circuit is very sensitive to capacitive pickup like this. Place the
entire circuit over a ground plane and avoid touching it or the 'scope probe while observing it and
the signal should reduce significantly - and exclude all light of course.
