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lab05_en:inverting_op-amp_photo_diode_as_current_source [2026/04/02 13:01] – mexleadminlab05_en:inverting_op-amp_photo_diode_as_current_source [2026/05/07 12:37] (current) – mexleadmin
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-=== Photo Diode as current source ===+=== Photodiode as current source ===
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-{{drawio>lab05:Fig-100_inverting_op-amp_photo_diode.svg}}+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.\\ 
 +{{drawio>lab05:Fig-150_inverting_op-amp_photo_diode_function_principle.svg}} 
 + 
 +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.\\
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 +{{drawio>lab05:Fig-160_inverting_op-amp_photo_diode_i_v_diagramm.svg}}\\
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-<imgcaption Fig-100_inverting_op-amp_photo_diode | Inverting Op-Amp: Photo Diode as current source> </imgcaption>+<imgcaption Fig-160_inverting_op-amp_photo_diode_i_v_diagramm | Inverting Op-Amp: Operating principle of a photodiode> </imgcaption>\\
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-$U_{\rm DD}{\rm~=10~V},~U_{\rm SS}{\rm~=-10~V},~R_{\rm 1}{\rm~=10~k\Omega}$\\ \\ 
-Use the values from <imgref Fig-20_inverting_op-amp> for $U_{\rm IN},~U_{\rm OUT},~R_{\rm 2}$. 
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-Complete the arrows in the scematic of the circuit.\\ 
-Take the values for $U_{\rm 1},~U_{\rm 2},~U_{\rm OUT}$ from <imgref Fig-30_inverting_op-amp_inv_input>.\\ 
-Use these values to calculate the sum of the voltages at node ${\rm N_{12}}$.\\ 
-Compare your result by measuerement. 
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 +{{drawio>lab05:Fig-120_inverting_op-amp_photo_diode_housing.svg}}\\
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 +<imgcaption Fig-120_inverting_op-amp_photo_diode_housing | Inverting Op-Amp: Photodiode BPW 34 S> </imgcaption>\\
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-$U_{\rm 1}{\rm~=}$ 
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 +{{drawio>lab05:Fig-110_inverting_op-amp_photo_diode_diagramms.svg}}\\
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 +<imgcaption Fig-110_inverting_op-amp_photo_diode_diagramms | Inverting Op-Amp: Diagramms of BPW 34 S> </imgcaption>\\
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-$U_{\rm 2}{\rm~=}$ 
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-$U_{\rm OUT}{\rm~=}$ 
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-Calculated $U_{\rm 12}{\rm~=}$ 
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-Measured $U_{\rm 12}{\rm~=}$+\\ 
 +\\ 
 +{{drawio>lab05:Fig-100_inverting_op-amp_photo_diode.svg}}\\ 
 +\\ 
 +<imgcaption Fig-100_inverting_op-amp_photo_diode | Inverting Op-Amp: Photo Diode as current source> </imgcaption>\\ 
 +\\ 
 +$U_{\rm DD}{\rm~=10~V},~U_{\rm SS}{\rm~=-10~V}$\\ 
 +\\ 
 +\\ 
 +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 <imgref Fig-110_inverting_op-amp_photo_diode_diagramms> 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 <imgref Fig-100_inverting_op-amp_photo_diode>.\\ 
 +\\ 
 +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 $U_{\rm OUT}$.\\ 
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 +$I_{\rm 1}{\rm~=}$\\ 
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 +$I_{\rm 2}{\rm~=}$\\ 
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 +$U_{\rm 2}{\rm~=}$\\ 
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 +$U_{\rm OUT}{\rm~=}$\\ 
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 +$R_{\rm 2}{\rm~=}$\\ 
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 +What value would you expect for $U_{\rm D}$ in <imgref Fig-100_inverting_op-amp_photo_diode> and why?\\ 
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 +$U_{\rm D}{\rm~=}$\\ 
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 +${\rm ................................................................................................}$\\ 
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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?\\ 
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 +$U_{\rm D}{\rm~=}$\\ 
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 +${\rm ................................................................................................}$\\ 
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 +Measure or calculate the values given in the table below. 
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 +\\ 
 +{{drawio>lab05:Table-1_inverting_op-amp_photo_diode.svg}}\\ 
 +<tabcaption lab05:Table-1_inverting_op-amp_photo_diode | Photodiode measured and calculated values> </tabcaption>\\ 
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-What are your results?\\ 
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-What will happen if you short-circuit $R_{\rm 2}$?\\ 
-Try it and explain your results.\\ 
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