Consider the following simple photodiode circuit:

enter image description here

I've created a real-life version of this circuit, and I've detected that $I$ is oscillating: $$I(t)=I(t+\tau)\tag{$\tau\approx 0.016~\text{secs}$}$$ Since I'm certain that the incident light upon the diode is constant, I'm suspecting that the cause is an oscillating voltage $V_S$: $$\text{Hypothesis}:~~V_S(t)=V(t+\tau_V)$$ However, based on the Shockley diode equation as given below, $$\text{Shockley diode equation}:~~~~I=I_S(e^{V_D/(nV_T)}-1$$

$I$ is $\textbf{ independent}$ of $V_S$. However, the Shockley diode equation may just be a quasi-static approximation of the behavior of a pn-junction.

Based on your knowledge of the physics of a pn-junction, could the Shockley diode equation simply not hold if $V_S$ is oscillating sufficiently rapidly? If so, under what regimes?

  • 2
    $\begingroup$ Are you saying that the period of the observed oscillation is 0.016 seconds? This corresponds to 60 Hz and may be an artifact (pickup) in your as-built circuit... (e.g. the light source has a 60 Hz component and is not "constant", or you have electrical pickup in your test equipment) $\endgroup$ – user3814483 Sep 15 '14 at 19:38
  • $\begingroup$ Yes - 63 Hz exactly, but 60 Hz is still plausible because of sources of error. What is electrical pickup? $\endgroup$ – Dave Sep 15 '14 at 19:44
  • $\begingroup$ Typically capacitive coupling between your circuit and a 60Hz source (e.g., the very power supply, DMM or oscilloscope you use to power/measure the circuit if said device is not filtered properly). In your case, the photodiode might directly "see" 60Hz or in some cases 120Hz if you expose it to indoor/fluorescent lighting. Turn the lights off, and use a battery-powered flashlight - do you still see oscillations? $\endgroup$ – user3814483 Sep 15 '14 at 19:49
  • 1
    $\begingroup$ Shockley's equation is good at least to mm-waves. $\endgroup$ – hyportnex Sep 15 '14 at 19:58
  • $\begingroup$ One would, normally, not operate a photodiode in conduction, to begin with. Photodiodes are usually operated in reverse at the highest voltage, at which the leakage current is not increasing, yet. Operation under a reverse voltage reduces the capacitance of the diode, which is the actual performance limiting parameter for detection of signals. $\endgroup$ – CuriousOne Sep 15 '14 at 20:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.