# Response time of photodiodes

by TChi
Tags: measurements, photodiodes
 P: 4,664 Hi Tran- If the light pulse is much faster than the photodiode response (primarily transit time spread), then the photodiode slows down the signal rise time. You can also stretch out the pulse duration (integrate the charge) by paralleling the photodiode output with a high value discharge resistor in parallel with a capacitor. Most of my experience is with fast vacuum photodiodes, such as the RCA 935. These photodiodes have fractional-nanosecond risetimes. Using such a RC integration circuit permits directly measuring the photodiode charge from the 1 ps pulse on the capacitor. Don't use a slow photodiode. Bob S
 P: 17 Hi Bob S, Actually, the original pulse duration in my case is 30fs, the 1ps value is the stretched one. Even if we use the photodiodes as you suggested, do you think that we can gauge the light intensity by the photodiode current? Please let me know, I am not quite clear here. What decides the response time of one photodiode? And would you tell me why I should not use the slow photodiode. IMO, the slow photodiode is too slow, therefore the light pulse is finished while the electrical signal is rising. When the light stops, will the electrical signal continue to rise (inertial property of electrical circuit) or stop rising (will we lose part of the rising as it is supposed to rise)? Is there any other reason that I should not use the slow photodiode? Please teach me and I am happy to hear from you. Thank you very much.
P: 4,664

## Response time of photodiodes

If the photodiode circuit integrates (on a capacitor) the total charge released in the pulse, the voltage output will be proportional to the charge (number of photons) and not the instantaneous current (number of photons per ps). I believe you want to measure the number of photons and not photons per ps. In the case of the vacuum photodiode, the geometry of the cathode, and the transit time spread of the electrons arriving at the anode determine the risetime. I have seen some parallel-plate vacuum photodiodes with a semitransparent (wire screen) anode, but not in recent years. Remember that the photodiode (especially a vacuum photodiode) is an ideal current source. As long as the discharge (or charge recombination) time of the charge stored in the photodiode circuit is long relative to the digitization time, the information will be accurately recorded.

Bob S
 P: 17 Dear Bob S, Thank you for your messages. You were definitely right. What I want is to measure the number of photons that come to the photodiode. However, I am lack of knowledge in this specific topic. Would you please give a reference or any typical electric circuit to get into reality. I want to make it done. Thanks in advance.
P: 4,664
 Quote by TChi Thank you for your messages. You were definitely right. What I want is to measure the number of photons that come to the photodiode. However, I am lack of knowledge in this specific topic. Would you please give a reference or any typical electric circuit to get into reality. I want to make it done.
The total charge Q out of the photodiode is proportional to the total number of photons hitting the cathode (vacuum photodiode) or Si PIN semiconductor. depleted depth. If the photodiode is output paralleled with a capacitor C, the risetime of the voltage is proportional to the laser pulse area integral to time t, and the voltage V stored on the capacitor is equal to Q/C. This final voltage is proportional to the area of the laser pulse. The decay time of this voltage is exp[-t/RC], where R is the discharge resistor in parallel with C. This discharge time should be long enough to digitize it before the voltage decays more than a few %, or it is stored in an an analog sample-and-hold circuit.

Bob S
 P: 17 Hi Bob S, Thank you so much. What you said is exactly what I am trying to understand. Please do me one more favor: please give a reference list or a name of a book or whatever you think that I can read and learn more about this. I have only the fundamentals of electronics. Please just give me the names, I can do a research about them. Regards.
 P: 4,664 Here is a simple charge integrator for very fast pulses from a current source, like a vacuum photodiode (see attachment). The plot shows the input current pulse (red, right vertical scale), and the output voltage pulse (black, left scale). The horizontal scale is 10 microseconds long. The light pulse (and charge pulse) occurs at 1 microsecond. A 1-milliamp, 10 nanosecond current pulse (red trace) produces a 10 picocoulomb pulse that immediately gets stored on the 100 pF capacitor (V = QC = 100 mV). The resistor drains off the charge with a 10 microsecond time constant. Bob S Attached Thumbnails
 P: 17 Dear Bob S, This is very clear and this will be a practical solution for me. However, please tell me again about the references, textbooks or articles, etc that I can gain more knowledge of this field. Since I want to learn and design it myself in case the real problem is different than the one you gave me. For example, this is 1 microsecond pulse, of course I can change the parameters to make it suitable to my experiments but still this is your suggestion. I am quite nervous when I face new and different electrical problems. So please tell me where I can learn about this. Thanks to your help, I think that my topic is fulfilled. Thank you very much! Have a nice day, Regards, Tran Trung Luu
 P: 4,664 This vacuum planar photodiode is the fastest I know of, with risetime measured at ~100 ps or less. It has a semitransparent wire-grid anode biased up to about +5 kV, and a dc-coupled solid cathode plate. Both ITT and Hamamatsu made versions of this. The cathode plate (seen through anode grid in photo) and coaxial ground shield is impedance-matched to 50 ohms, with a UHF connector on the back. I was told once that the ITT version was linear from about 1 nanoamp to about 10 amps for very short pulses. See http://www.sphere.bc.ca/test/phototu...3-u-03-box.jpg Bob S

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