I Photodiode: pulsed vs. continous signal

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Summary
I can't explain, why maximum signal coming out of photodiode is different for two laser beams of same intensity. First beam is coming from cw laser, second beam is a chopped from cw using Pockels cell.
Hello,
I have done an experiment with cutting out a pulse from continous laser using Pockels cell. The setup is cw laser -> Pockels cell with crossed polariazers -> photodiode.

The photodiode is connected in reverse bias, a capacitor is connected parallel to photodiode.

If I put a half wave plate between the polarizers, continous light goes on the photodiode. The photodiode measures voltage U.
Then I remove half wave plate and I cut out a 10 ns pulse from the same continous light using Pockels cell. Now the photodiode measures a pulse, but its maximum is approximately 1/10*U.

I would expect the maximum of pulse signal to be the same as the signal from continous light, because both pulses have the same intensity and photodiodes measure measure the intensity of light.

Can you explain, why the signals don't have the same maximum voltage output from photodiode, even though they have the same intensity?
 

scottdave

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Have you thought about how the capacitor behaves in both situations?
 

scottdave

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Also you may want to look at the spec sheet for the photodiode, to see what response characteristics it has.
 
6
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Thank you for your response.

I attached the spec sheet for the photodiode (APY 13 II) as well as electrical cicuit of the diode. Instead of voltmeter, there is an oscilloscope with 50 Ω termination.
I have also measured the response time of the photodiode. Its rise time is 0.5 ns and the fall time is 1 ns.

circuit.png


I have thought about the capacitor, but I am not very familiar with electrical circuits.
The capacitor is fully charged to voltage 25 V, when there is no light. Then I think the capacitor behaves the same for both cases. Incident photons decrease the resistance of photodiode and current flows from the capacitor through the diode to earth. For the pulse beam, the current flows for a shorter time, but that shouldn't change the value of the current.

If the continous light was going on photodiode long enough, the capacitor would loose its charge/voltage and the signal would decrease. I am pretty sure that didn't happen, and even if it did, it wouldn't explain, why the signal from pulse beam is lower.
 

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I have not used one for quite some time but isn't 10ns very fast for a Pockels cell?
 
6
2
I have not used one for quite some time but isn't 10ns very fast for a Pockels cell?
In my opinion. If you make voltage signal that has the right maximum value and length of 10 ns, you should be able to open the Pockels cell for 10 ns. I have yet worked only with this Pockels cell and it was already set up.

However I measured that the (FWHM) length of pulse cut out from continous signal using Pockels cell was 10 ns, so it is probably possible to open the cell for 10 ns.
 

Paul Colby

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Even if the diode became conductive instantaneously, the RC of that circuit is 0.5 micro seconds. It simply can't switch that fast, right?

[edit] Suggestion, I would dump the 500 ohm resistor and put the photodiode in it's place.
 
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The 500Ω resistor is the current sensing device in the circuit. Please reconsider your analysis.....you certainly want to measure the photocurrent.
 

Paul Colby

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The 500Ω resistor is the current sensing device in the circuit. Please reconsider your analysis.....you certainly want to measure the photocurrent.
A resistor doesn't sense anything. The sensor is a scope measuring the voltage drop across the 500 ohm resistor. How does the circuit shown get around the 0.5 microsecond time constant shown? Inquiring minds would like to know.
 
6
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Even if the diode became conductive instantaneously, the RC of that circuit is 0.5 micro seconds. It simply can't switch that fast, right?

[edit] Suggestion, I would dump the 500 ohm resistor and put the photodiode in it's place.
Well, I cannot explain the fast switch, but next week I can look at the electrical circuit, if I read the parameters correctly. The photodiode was built by someone else. On the capacitor was written 100k x 40 V, which should mean, that the capacitance is 100 nF and maximum voltage is 40 V.

Can you please explain, how would dumping the resistor help? I think that I would measure the voltage on the capacitor, if I dumped the resistor. And that's not any good, is it?
 

Paul Colby

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I was suggesting replacing the 500 ohm resistor with the photodiode and the diode with a wire. With 50 ohm termination on the scope, I think this isn't a great suggestion as the bulk of the photo current flows into the scope. Assuming all the photo current flows through the 50 ohms of the scope, then the time constant is 50 ns rather than the 500 ns I was complaining about prior. This is still 5 times the 10 ns pulse with you are attempting to measure.
 
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A resistor doesn't sense anything. The sensor is a scope measuring the voltage drop across the 500 ohm resistor. How does the circuit shown get around the 0.5 microsecond time constant shown? Inquiring minds would like to know.
I will elucidate. The photocurrent is controlled only by the illumination of the photodiode (and very linearly). The measured voltage is simply the photocurrent x(500Ω) independent of C. The effect of C is to keep the backbias on the photodiode more stable.

Inquiring minds now know.
 
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I was suggesting replacing the 500 ohm resistor with the photodiode and the diode with a wire. With 50 ohm termination on the scope, I think this isn't a great suggestion as the bulk of the photo current flows into the scope. Assuming all the photo current flows through the 50 ohms of the scope, then the time constant is 50 ns rather than the 500 ns I was complaining about prior. This is still 5 times the 10 ns pulse with you are attempting to measure.
An oscilloscope measures voltage, not current. This is just a very bad idea, please analyze the circuit again..
 

Paul Colby

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I agree. Missed a few zeros and missed the real problem. The rise time should be dominated by the the diode capacitance and the input capacitance of the scope. I stand corrected.
 

Paul Colby

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An oscilloscope measures voltage, not current. This is just a very bad idea, please analyze the circuit again..
So, the photo current at DC will flow through the 500 ohm resistor assuming the scope is capacitively coupled. For a short pulse, the majority of the current will flow through the 50 ohm impedance of the scope, 50 being 10 times smaller than 500. For roughly the same current, ohms law would suggest 1/10 the voltage would be developed across the scope for the pulse than for the DC case.
 
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The intrinsic input impedance of most scopes is quite high (MΩ) and that is how they are normally used. The 50Ω input is a purely resistive load used for very high frequency where the reflection inside the cables (50Ω impedance) could be an issue (1 nanosecond per foot!!). It will affect the signal as you say but the capacitive coupling part is not correct. It will affect both the steady state and pulse signal equally (I assume DC coupling for the steady state and therefore also for the transient).
So unless the input parameters for the scope are switched I don't think that is the root of the discrepancy.
 

Paul Colby

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Ah, this is interesting, thanks. I think I see now. Even with a high impedance scope the current is established as soon as charge carriers are present, this raising the voltage drop across the 500 ohm resistor to it's static level for a 10 ns period (thanks to the cap). nice. I think I see what you are getting at. Perhaps the OP should consider a different scope impedance termination?

So, if the scope input impedance is 50 ohms as stated, this is problematic as I suggest.
 
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Well if he just wants to compare the two it should be consistent at 50 ohm DC coupled. The actual size of the photocurrent will need use the 50 ohms in parallel with the 500 ohms to get the actual number from the measured voltage. I still don't really see why the peak should differ in size unless Pockels cell is not getting full rotation of polarization. Maybe we're 'missing something...I've modulated LEDs at nearly 100MHz but not lasers with a shutter.
 

tech99

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Thank you for your response.

I attached the spec sheet for the photodiode (APY 13 II) as well as electrical cicuit of the diode. Instead of voltmeter, there is an oscilloscope with 50 Ω termination.
I have also measured the response time of the photodiode. Its rise time is 0.5 ns and the fall time is 1 ns.

View attachment 249488

I have thought about the capacitor, but I am not very familiar with electrical circuits.
The capacitor is fully charged to voltage 25 V, when there is no light. Then I think the capacitor behaves the same for both cases. Incident photons decrease the resistance of photodiode and current flows from the capacitor through the diode to earth. For the pulse beam, the current flows for a shorter time, but that shouldn't change the value of the current.

If the continous light was going on photodiode long enough, the capacitor would loose its charge/voltage and the signal would decrease. I am pretty sure that didn't happen, and even if it did, it wouldn't explain, why the signal from pulse beam is lower.
The 15k and 11n are causing problems with pulsed/continuous signals. I should get rid of these. Then the average current in the 500 Ohm resistor will depend only on mark/space ratio (the on/off ratio). The diode always conducts the same when the light is on, so its linearity is not an issue. A voltage indicator with an averaging characteristic is needed, such as an analogue microammeter. A traditional CRO will not perform averaging and is peak-reading. A high value resistor in series with the CRO, followed by a shunt capacitor, such that the pulse is completely smoothed out, will make it average-reading.
 
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The OP is not interested in a rms reading. As I understand the question he is interested in peak so I don't understand @tech99 comments.

However I just rechecked the specification for this photodiode and this is a slow diode.!! In fact with 10kΩ the rise time spec is 20μs. Case closed. We can all say "DUH" together now.....
 

Paul Colby

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Case closed. We can all say "DUH" together now.....
Except the 11kOhm is used to charge the cap. The diode discharges (slightly) the cap through the parallel combination of the 500 ohms and the impedance of the scope. So, I'm confused by your comment.
 

Paul Colby

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Ah, the plot on the lower left of page 45 of the spec show the ##f_g## of 20 kHz for a 0 ohm load. That's slow if ##f_g## is what I think it is.

##f_g## is indeed the cutoff frequency of the diode.
Explanation of ##f_g##
 
Last edited:
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Except the 11kOhm is used to charge the cap. The diode discharges (slightly) the cap through the parallel combination of the 500 ohms and the impedance of the scope. So, I'm confused by your comment.
Yes that is a good point (that I forgot about .....it is now tomorrow)
But they don't even spec the diode that fast and as you note later the extrapolations do not look good. Are germaniun photodiodes slower than silicon?? I should know but i don't.
Anyhow I am no longer surprised by the result.
 

tech99

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The OP is not interested in a rms reading. As I understand the question he is interested in peak so I don't understand @tech99 comments.
My mistake, I should have read the question better.
 
6
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Thank you all for answering the original question. However, I still have a few follow-up comments and questions and I would be glad, if you helped me.

However I just rechecked the specification for this photodiode and this is a slow diode.!! In fact with 10kΩ the rise time spec is 20μs. Case closed. We can all say "DUH" together now.....
Thanks for finding the missing piece.
But in my defence, I measured the rise time of the photodiode before looking for the spec sheet, because my colleagues told me, I wouldn't find it. I shined a laser pulse with rise time and fall time each 200 ps at the photodiode. The measured voltage from the diode had rise time of 500 ps. I think that this is the method to experimentally determine the rise time of a photodiode.

I thought that if rise time is 20 μs and any light pulse shorter than 20 μs (e.g. 200 ps) reaches the photodiode, the output voltage always has the rise time of 20 μs. Is my assumption wrong? Can light pulses shorter than rise time create voltage pulses with shorter rise times?

fg is indeed the cutoff frequency of the diode.
Is the cutoff frequency equal to the bandwidth frequency (in case of photodiodes)?
In other words can the rise time be determined from trise = 0.35/fg ?
(I found the equation on https://www.thorlabs.com/tutorials.cfm?tabID=787382FF-26EB-4A7E-B021-BF65C5BF164B )

Are germaniun photodiodes slower than silicon??
Yes, they are, according to the web page above.
 

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