Hello Arpit, how are you? A photodiode is different from a RF antenna, since the photodiode is a square-law detector. What does it mean?

A photodiode is a device that produces a current which is linear with the input light power. Light at 1550 nm has a frequency of 200 THz (more or less). So the electric field of the signal is E(z,t) = E0 cos(wt+kz), with w = 2pi200 THz. The photodiode will provide a photocurrent that is proportional to |E(z,t) |^2. Therefore, |E(z,t) |^2 = E0^2/2 {1+cos[2(wt+kz)]}. Since the photodiode has a slower time response compared with the period of the input light, the photocurrent will be proportional to E0^2/2 . Therefore, if the power amplitude is modulated, then the amplitude of the photocurrent will be modulated as well.

Best,

Lucas.

Hello Arpit, how are you? A photodiode is different from a RF antenna, since the photodiode is a square-law detector. What does it mean?

A photodiode is a device that produces a current which is linear with the input light power. Light at 1550 nm has a frequency of 200 THz (more or less). So the electric field of the signal is E(z,t) = E0 cos(wt+kz), with w = 2pi200 THz. The photodiode will provide a photocurrent that is proportional to |E(z,t) |^2. Therefore, |E(z,t) |^2 = E0^2/2 {1+cos[2(wt+kz)]}. Since the photodiode has a slower time response compared with the period of the input light, the photocurrent will be proportional to E0^2/2 . Therefore, if the power amplitude is modulated, then the amplitude of the photocurrent will be modulated as well.

Best,

Lucas.

The response time of photodiode is slow in comparison to the frequency of light electric field. So it can not show variation in electric field intensity. Its signal will be proportional to the light intensity. If the light source is pulsed then photodiode signal will be pulsed. If it is chopped light source, then signal will vary with the chopping frequency.

Hi Arpit,

Some details answers have been given, but just to be clear in answer to your specific question, the signal from a photodiode would be DC as the direction of current would not change. The magnitude of the current and voltage measured across the device would vary according to the incident light intensity as described but the direction would be fixed.

I think the Photocurrent current output from the Photodiode will be DC.

It can give out both AC and DC currents.If constant light is made to incident on PD,a constant dc current proportional to light power is obtained.If light is ON-OFF in nature,current is also AC like ON-OFF proportional to the input....

HI every one

Up to my knowledge, such current has theoritically an exponential component

Ashok Panchapakesan , a photodiode, by definition of being a diode, only lets current flow in one direction, therefore the signal is DC. The magnitude of the current changing does not make it AC. This would just be a varying analogue signal, but still DC.

Ok clearly you have gotten a lot of answers, and they are all correct, but lets quickly ask our selves WHY its DC rather than just saying 'its cos its' a diode' or something of that effect.

The way that a diode works, one can simplify , to be a bunch of electrons stranded on one side of a gap, and due to a voltage difference, which arise from different inherit charge distributions (doping), they all want to go to the other side. Now, the gap in this case (the depletion zone) acts as a energy barrier, with the effect that electrons need a lot of energy to cross the gap. However, when we apply light to this set up, what happens is a photon will come by and hit an electron within the gap, causing it to gain enough energy to actually escape from it atom. once this electron is free however, it sees a whole lot of free electrons to one side, and a lack of electrons to the other (the inherit voltage difference). Thus, by the laws of electrodynamics, it is pulled/pushed to the positive terminal. This then goes down a wire to be detected. And what happens on the other side of the terminal, is that an electron will then see this newly available spot in the gap, and it will have enough energy to migrate over, and so it will. Now of course, this process happens in the same manner every single time a photon strikes the junction, as it is purely a matter of voltage distribution. Thus current will only flow in one direction.

Please note, there may be some other aspects that I did not cover, and that this was all just simplified down to pass on the logic of what is happening. I advise if you are going to use this material that you confirm it with the literature.

Thank you all for the answers. These were immensely helpful. I would like to summarize my gain out of these answers. Please correct me if I am wrong.

1. For a single CW optical signal with constant intensity - output of photodiode is DC.

2. For a single pulsed (on/off) optical signal - output of photodiode will be a similarly pulsed signal provided that its response time is less than pulse repetition rate.

3. For 2 CW optical signals at different frequencies - output of photodiode will be an AC signal with frequency equal to the difference of both optical signals.

Arpit -- the last item (3) in your summary is an interesting one. The frequency that you talk about could be an amplitude modulation frequency, say in the RF range. In that case, the combination of two amplitude modulated light signals with slightly different modulation frequencies would include a DC component and tones at each frequency but NOT a difference tone because a photodiode is a linear device. However, if the frequencies that you mention are optical frequencies (mentioned by VNR above, e.g., frequency = c/lambda) and the two optical frequencies differ by a small amount so that the difference is within the bandwidth of the photodiode, then the two electric fields would interfere within the diode producing a field that would then be squared (as mentioned by LR above). To be clear, these two optical signals are not amplitude modulated in the RF range. The result of the non-linear squaring process inside the photodiode DOES produce a difference tone. This optical beating case can be exploited to lock the optical frequencies of two optical sources together or to a fixed difference. The literature is full of clever example of this method. The key to understanding the difference between the RF modulation case and the optical frequency case is notion of interference of E-fields within the active portion of the photodiode. (Note that you have to assume that the two RF modulated signals, mentioned first, are either optically broad or different enough in optical frequency that their beating is out of the diode bandwidth.)

As a last remark, this same phenomenon occurs in your ears which are remarkably linear like photodetectors. Two different audio frequencies, say from two different guitar strings, don't produce a difference tone. Therefore, the sound of a whole orchestra is clean rather than filled with interharmonic distortion which would make it sound terrible (like through very bad speakers). However, if the two guitar strings are close enough in vibrational frequency to be within the bandwidth of a single sound reciever in your ear (in analogy to the photodiode), then you can hear a difference or beat tone. This is how some guitarists tune their instruments. But, you can only hear beats up to a few tens of Hz because each separate sound receiver in your ear has a narrow bandwidth.

Output current is DC

Dear Arpit

The statement #3 of your summary is not right. The photodiode output current will still be named DC since it will not change its sign. Remember that DC simply means no change of polarity. So, even a time-varying electric signal is called DC whenever it has constant polarity. Constant-amplitude signals are only a special case of DC signals.

Best wishes!

Hmmm. Seems like there are varying opinions on the definition of DC. For example, see https://engineering.mit.edu/engage/ask-an-engineer/whats-the-difference-between-ac-and-dc/ which suggests that DC is a constant in time and not varying as suggested by CM above. On the other hand, Wikipedia agrees with CM here https://en.wikipedia.org/wiki/Direct_current.

The same is not the case for AC. Everyone seems to agree that AC means that the current changes direction.

But rather than getting hung up on definitions, let's agree that a photodiode produces a current that is proportional to the light hitting it and in a single direction. No amount of light will create a negative current. So yes, it is therefore DC by CM's and Wikipedia's definition.

As for CM's comment about Arpit's #3, correct -- the output of a photodiode that is time-varying due to a time varying input light signal will not reverse direction. Rather, it will have a constant (DC, by some definition) component and a time-varying but always unipolar component.

But sometimes, for example in the case of Fourier analysis, the time varying part of a signal is referred to as the AC part because the direction varies relative to the DC component. Example: DC component = +1A; AC component = 0.5A x sin(wt) so that minimum current = 0.5A and maximum current = 1.5A, all positive and never crossing 0A becoming negative. Definitions get twisted to fit the needs of each context.

Thanks for the clarification Cesar Moreno . By AC, I did not mean a reversal in polarity of the signal but rather an alternating signal superimposed on a DC value, as explained in the last paragraph by David A. Ackerman .

It's depends of the illumination and the device cut-off frequency but for classical commercial PD the time response of the device is too small to follow the variations of the optical power. You can have a look to photomixing of an optical beatnote in photodiode to better understand the concepts related to cut-off frequency of photodiodes (transit time of free carriers...). It also exists "high speed" photodiodes as for example uni-traveling-carrier photodiodes where the design of the device improves the cut-off frequency.

While reading the answers, another doubt came to my mind.

Bandwidth of photodiode is always stated in terms of highest frequency of intensity variations that it can detect. Does a photodiode has lower limit on the frequency of input intensity variations that it can efficiently convert into electrical signals?

No lower limit. A photodiode will respond to a DC light signal with a DC response. And here, the meaning of DC is a signal that is unchanging in time (practically speaking, one that changes to a steady value and then stays at that value for a long time.)

But won't the noise sources such as Intensity Noise or 1/f noise have some limiting effect on the photodiode response. Suppose, the input signal to photodiode has intensity variations at 1 mHz, will the noise not come into effect?

Yes, noise will affect measurements, including the 1/f noise that you mention. Lots of literature to learn from. See, for example, https://www.st-andrews.ac.uk/~ctab/MSc_Oft/Lecture_Notes/lecture_10.pdf and lots of others. But the point is that the photodiode has no intrinsic low frequency cutoff. (Unless your detection circuitry itself has a low frequency cutoff -- but that is not fundamental to the diode.)

Any photodiaode will give signal depending on the nature of light source. If light source is continuous then signal will show a DC shift. If source is pulsed then signal will be pulsed. If the light is chopped at certain frequency then photodiode signal will vary at that frequency. However noise frequencies will be observed superimposed on the signal produced due to light.

The photo diode is operated normally in a reverse biased mode so that the dark current is very small and is in the order of a reverse saturation current. On illuminating the diode with a photon flux phi(t) the diode will generate a photo current proportional Iph to the photon flux such that

Ip(t)= consatnt Phi(t). As a consequence the photocurrent reflects the time variation of the incident photon flux. For more information please follow the paper in the link: Conference Paper A comprehensive study of an optical transceiver

Best wishes

A photo diode can give dc current proportional to light intensity . First we have to calibrate the system with the used load line. Second we start to take voltage measurement. The photo diode current can be in to BJT. The BJT is used as a current to voltage converter. This way we can relate intensity to voltage reading.