Photodiode bandwidth: why does power decrease with frequency

[Jalo]

Hi,

I'm studying p-i-n photodiode (PD) at the moment and understand that the photodiode's response will depend on the frequency of the light signal going into it. I am struggling however to understand the concept of bandwidth, and why is it that the photocurrent at the PD decreases with higher frequencies.

1. What exactly is bandwidth? My understanding it that when we launch light into the PD, it will be absorbed in the PD's depletion region, originating an electron-hole pair that will be swept across the depletion range (drift) and afterwards diffuse through the bulk semiconductor region (diffusion) until it reaches the metal contacts. If we have an oscillating light signal going into the PD, how exactly does the signal's frequency change this process, or make it less effective? Is the collection of electrons/holes less efficient at high frequencies somehow? If so, why?

2. When we talk about 3dB bandwidth, am I correct in that this is the frequency at which the photocurrent measured in the PD is half of that measured when the signal is not oscillating (0 Hz)? If so, is the process of measuring bandwidth a matter of sending first a constant signal into the PD and then simply increase the signal's frequency, keeping it's amplitude, and waiting until the photocurrent collected at the PD halves?

Apologies if the questions are a bit daft, but I've just started studying this and am struggling to get a good intuition of the PD's dependency on the light signal's frequency. Thank you very much.

 

[Charles Link]

For a photodiode in the linear region, you will get a total photocurrent generated that is proportional to the average power of the incident signal, (the total charge that results is proportional to the incident energy), but if the incident signal is modulated at some frequency, in general you can not expect the instantaneous photocurrent of the photodiode to respond in such away that it reproduces the waveform. The linear response of the photodiode to a delta function incident power input, in the case of a perfectly linear response, will experience some delays. The result is linear response theory works very well, but do expect a frequency response with a characteristic high frequency roll-off. $\\$ To express this mathematically , photocurrent $I(t)=\int\limits_{- \infty}^{t} P(t')m(t-t') \, dt'$, where $m(t)$ is the linear response function, and $P(t)$ is the incident power. $\\$ The convolution theorem says $\tilde{I}(\omega)=\tilde{P}(\omega) \tilde{m}(\omega)$, where the Fourier transform $\tilde{F}(\omega)=\int\limits_{-\infty}^{+\infty} F(t)e^{-i \omega t} \, dt$. $\\$ (So that $\tilde{m}(\omega)=\int\limits_{-\infty}^{+\infty} m(t)e^{-i \omega t} \, dt$, and similarly for $\tilde{I}(\omega)$, and $\tilde{P}(\omega)$ ). $\\$ (The photocurrent response at any frequency will be proportional to the incident power at that frequency). $\\$ If $m(t)$ is not instantaneous (i.e. if it is not a delta function), then there will be a (finite) frequency response associated with $\tilde{m}(\omega)$ with a high frequency roll-off in the response.$\\$ If $m(t )$ is instantaneous=a delta function, in that case the bandwidth of the response is infinite, and $\tilde{m}(\omega)$ is a constant, independent of frequency. $\\$ ( For the sake of completeness, the inverse Fourier transform $F(t)=\frac{1}{2 \pi} \int\limits_{-\infty}^{+\infty} \tilde{F}(\omega) e^{+i \omega t} \, d \omega$ ). $\\$ And note: This is the same kind of linear response that is used in ac circuit theory. The response of the photodiode is assumed to be perfectly linear. Any delay in the response that is spread out over time results in a frequency dependent response. $\\$ For a quick review: In the ac circuit linear response theory, we write $V_{out}(t)=\int\limits_{-\infty}^{t} V_{in}(t') m(t-t') \, dt'$, and $\tilde{V}_{out}(\omega)=\tilde{V}_{in}(\omega) \tilde{m}(\omega)$. $\\$ e.g. For a simple RC circuit, $\tilde{m}(\omega)= \frac{-i/(\omega C)}{R-i/(\omega C ) }$ by simply using known impedances, etc., where $V_{out}$ is measured across the capacitor. Alternatively, it could be computed from $m(t)=(\frac{1}{RC})e^{-t/(RC)}$ for $t>0$. $\\$ Also note of $M(t)$ is the response to a unit step input for $V_{in}(t)$, then $m(t)=\frac{dM(t)}{dt}$, and $m(t)$ is the response when $V_{in}(t)$ is a delta function. $\\$ For the above photodiode, the empirical function $m(t)=\frac{A}{\tau} e^{-t/\tau}$, ($t>0$), for constants $A$ and $\tau$ can be a good starting point for simple modeling of the response.