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Photodiode as current source

A photodiode is a special type of diode which, in the absence of light, exhibits a current-voltage relationship very similar to that of a standard diode (see the dark current characteristic in the \(I-V\) diagram).

When illuminated, it generates additional electron-hole pairs within the crystal.

Photodiodes are often operated in reverse bias, where the charge carriers (electrons and holes) generated by the incident light cause an increased reverse current flow (third quadrant of the I-V diagram). The higher the light intensity, the greater the reverse current. Forward bias operation is also possible, where the photodiode behaves like a solar cell (first quadrant of the I-V diagram).

Applications include remote controls (IR range), galvanic isolation (optocouplers), light measurement, positioning, and light barriers.



Fig. 1: Inverting Op-Amp: Operating principle of a photodiode







Fig. 2: Inverting Op-Amp: Photodiode BPW 34 S




Fig. 3: Inverting Op-Amp: Diagramms of BPW 34 S














Fig. 4: Inverting Op-Amp: Photo Diode as current source

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


We are assuming a well-lit room with an illuminance of 300 lx, lit 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 and 300 lx for the illumination. (500 nm is in reality a greenish light and not blue)
The graph in figure 3 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 photodiode 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 ${\rm\mu A}$.
We will assume a current of 10 ${\rm\mu A}$ at 300 lx for our calculations.

Complete the arrows in the circuit diagram in figure 4.

Calculate $R_{\rm 2}$ 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.
Also enter the values for $I_{\rm 1}$, $I_{\rm 2}$, $U_{\rm 2}$ and $U_{\rm OUT}$.




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


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


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


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




















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



What value would you expect for $U_{\rm D}$ in figure 4 and why?


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


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


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


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Measure or calculate the values given in the table below.