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Photo Diode as current source




lab05:fig-120_inverting_op-amp_photo_diode_housing.svg

Fig. 1: Inverting Op-Amp: Photo Diode BPW 34 S


lab05:fig-110_inverting_op-amp_photo_diode_diagramms.svg

Fig. 2: Inverting Op-Amp: Diagramms of BPW 34 S


lab05:fig-100_inverting_op-amp_photo_diode.svg

Fig. 3: Inverting Op-Amp: Photo Diode as current source

$U_{\rm DD}{\rm~=10~V},~U_{\rm SS}{\rm~=-10~V}$


We assume a good illuminated room of 300 lx, illuminated by a white LED. White light is a mixture of many wavelengths across the visible spectrum, roughly 380 to 780 nm.
For a typical white LED, the spectrum usually comes from a blue LED chip with a peak around 450 nm, plus a broader phosphor emission that spreads across green, yellow, and red wavelengths.
For an easier calculation, we take a mean value of 500 nm which is close to the peak value of the blue LED (in reality a greenish light) and 300 lx for the illumination.
The graph in figure 2 shows that the photodiode sensitivity at 500 nm is only 30%. The maximim current (100%) at 300 lx is 30 ${\rm\mu}$A.
We can now estimate the current we would expect from the diode at 300 lx:

$I_{\rm 1} = 30 {\rm~\mu A} * 0.3 = 9 {\rm~\mu A}$
$I_{\rm 1} \approx 10 {\rm~\mu A}$

30% of 30 ${\rm\mu A}$ is roughly 10 ${\rm\mu A}$.

Complete the arrows in the scematic of the circuit.
Calculate ${\rm R_2}$ so that $U_{\rm OUT}$ = 5 V at 300 lx. Take a resistor from the E6 series that is as close as possible to the calculated value.


$I_{\rm 1}{\rm~=}$


$I_{\rm 2}{\rm~=}$


$U_{\rm OUT}{\rm~=}$


$U_{\rm 2}{\rm~=}$




















$R_{\rm 2}{\rm~=}$



What value would you expect for $U_{\rm D}$ and why?


$U_{\rm D}{\rm~=}$


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What value would you expect for $U_{\rm D}$ at 300 lx when it is not connected to the Op-Amp or any other electronic component (open-circuit voltage)?


$U_{\rm D}{\rm~=}$


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