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2

What I meant was to use a transistor with a current source in the base, which is meant to represent the luminosity. Something like this: This is not meant to represent the CNY70, rather to show how it could be done. You will need much more than a wet finger in the wind (my approach) to model the diode and the transistor (I used some round numbers; the β is ...


3

Imagine a series resistance Rs in S1 normally closed contact. The current If flowing through the feedback resistor Rf1 will result in the voltage at the op-amp output being higher than the ideal voltage by Rs*If. But the voltage at the right hand side of Rf1 will be unaffected (because it's inside the feedback loop). If you add the second switch you can pick ...


2

According to this diagram the current at zero bias is down->reverse current.How is that even possible? First you must understand that this graph shows what happens when an external voltage is applied across the photodiode. By convention, when voltage is applied across a component any current flowing into it is considered positive, while current coming ...


1

As the light intensity increases there is increased photocurrent. As can be seen from the graph this causes a voltage to exist across the diode at zero current coming from the pins of the device. If the diode is biased in reverse the photo-current acts as increased leakage current, ie it tends to bring the voltage towards zero. (and ultimately to a positive ...


9

simulate this circuit – Schematic created using CircuitLab Andy's anti-parallel diode idea works nicely in a transimpedance amplifier. The dynamic range is maximized so long as the opamp doesn't run up against its DC supply rail with too much gain. Choose a feedback resistor value (R1, or R4) to stay within the opamp linear range. The photodiodes and ...


9

For my application, the only relevant data is the difference in luminosity between these photodiodes. If both photodiodes are the same, wire them antiparallel to a single TIA. Wiring them antiparallel means that the only current feeding the TIA will be the difference current. Now, the problem of noise is halved. Use a balanced transmission system like this: ...


0

They are independent and you would need a current limiting resistor with applied voltage. Include link to datasheet next time pls.


2

Yet, the minimum optical power source range that this photodiode is recommended to be used at is 1nW/cm^2 which is way, way below the 10uW/cm^2 that was being used in the estimate above. Note that the photodiode is advertised for use in photovoltaic mode, but the dark current is speced with a reverse bias (so in photoconductive mode where dark current is a ...


2

The concern when your dark current approaches your photocurrent is SNR. The three big noise components you are likely to encounter are dark current shot noise, photocurrent shot noise, and noise from your readout circuit. The first two can be easily calculated as: \$S=2qI\$ Where \$S\$ is the input referred noise power spectral density in A2/Hz. The third is ...


1

Typically for precision work, you have to calibrate out the dark current (along with other offsets). This may happen implicitly if you're measuring the AC component of the optical signal. If you need to work with DC, then you're in the world of periodic recalibration, temperature control to reduce drift, etc.


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