I am trying to accurately capture the video latency and response time curve of an LCD display from a black image to a white image for the purpose of reviewing the accuracy of a couple of video latency measurement tools. I am using a digital storage oscilloscope and a phototransistor. This is the spec sheet for the phototransistor. I would like the values of the curve captured by the oscilloscope to be delayed no more than 100 microseconds or so. What resistance should I use to capture a luminosity curve that has this accuracy of timings?

Here is more background on my progress so far:

I have created a circuit that is similar to the "Testing principle diagram" on page 3: I have a Vcc of 4.5 V (3 AA batteries) and the oscilloscope is wired to measure the voltage across the resistor that connects to the phototransistor and ground. When using the LCD screen that I would like to measure, I get a maximum of about 60 µA through the circuit.

I have noticed that the phototransistor's rise and fall times are affected by the resistor that I use in the circuit. To demonstrate this effect clearly in a controlled manner, I have placed the phototransistor in front of an LED that is switched on using a debounce circuit. Here are some screenshots that show the voltage change across the resistor when different resistors are used:

1.3 kΩ 1.3 kOhm - red LED

7.8 kΩ 7.8 kOhm - red LED

140 kΩ 140 kOhm - red LED

It seems that as resistance increases, the rise time of the phototransistor also increases. Does this mean that a resistance that approaches 0 is the most accurate? I have noticed that the signal to noise ratio decreases as I reduce the resistance. What is a good resistance that will keep the curve accurate to within 100 microseconds while also keeping the voltage across the resistor high enough to not be problematically noisy?


After reading Andy's answer, I took the advice to consider using a photodiode instead of a phototransistor for this project. I actually had a couple on hand, but when I first tried them out I didn't understand anything I was doing and disregarded them after the phototransistor seemed to be "easier" for me to get visible on my oscilloscope. To be honest, I'm not sure why I initially struggled with the photodiodes.

Here is the voltage across a 194 kΩ resistor that is in a short circuit with a BPX 61 photodiode when turning on the same LED that was used in the previous examples:

194 kOhm with photodiode

This shows the LED reaching full luminosity in only 90 microseconds. When I reduced the resistance to 37 kΩ, the time to reach full luminosity decreased to 19 microseconds. That's one fast LED!

I also realized that I have an audio device with S/PDIF coax and optical toslink output that I have previously measured using an optical receiver to find that the coax and optical outputs were aligned to within less than 100 nanoseconds of each other. With this in mind, I thought it would be good to compare the voltage from my photodiode circuit when measuring the optical output of this audio device with the coax electrical output of the audio device. Here are my findings (blue is optical through my photodiode circuit and yellow is electrical S/PDIF output from the audio device):

194 kOhm - toslink and coax SPDIF BPX 61

I can see from this that the upwards and downwards changes of the optical signal are aligned with the upwards and downward changes of the electrical signal. This seems to show that the photodiode circuit's rise/fall time is not fast enough to keep up with the S/PDIF optical signal, but it does respond immediately to a change, which is the most important thing for me.

Here's a 100Ω with the same optical and electrical S/PDIF signal (I switched channels, so this time blue is electrical and yellow is optical through my photodiode):

100 Ohm - toslink and coax SPDIF BPX 61

This "works" for this optical S/PDIF signal, but is way too noisy for my use, so I'll go with a higher resistance.

I suspect I would be able to get an even better result with a transimpedance amplifier, but at this point I might just play around a bit with the resistance I am using with this specific photodiode and that might be good enough for my purposes.

  • 3
    \$\begingroup\$ Some years ago I tried the same in a study where we wanted to measure the exact response time of a human to a very short optical stimulus. A photo diode performs much better if you need a precise timing, but then you may need an amplifier. We decided to use BCS2015G1 and MTD5052 because their spectral sensitivity is close to human eyes. \$\endgroup\$
    – Jens
    Commented Jun 7, 2022 at 20:52
  • 1
    \$\begingroup\$ It seems that for 100 Ohm, the time response should be 2us (datasheet). > The reaction time (rise) tr : for Vdd=10V, Iss=5mA, RL=100 -> 2 μS. The reaction time (down) tf = 2 μS. \$\endgroup\$
    – Antonio51
    Commented Jun 7, 2022 at 21:01
  • 1
    \$\begingroup\$ Beware of capacitance of devices and circuit board. 10:1 scope probes are around 10 to 15pF. If you are using a proto-board (SuperStrip), the electrode capacitance between rows is 2pF minimum, but can go as high as 5pF depending on adjacent wiring. Also consider the capacitance of the sensor. Using a reverse biased photo diode is a better option for fast events. \$\endgroup\$
    – qrk
    Commented Jun 7, 2022 at 22:47
  • \$\begingroup\$ While it's easy enough to build a photodiode circuit for a few tens of dollars, these amplified modules are quite handy and not outrageously overpriced: thorlabs.com/thorproduct.cfm?partnumber=PDA100A2 \$\endgroup\$ Commented Jun 8, 2022 at 15:25
  • 1
    \$\begingroup\$ Do not bother measuring stuff without an active amplifier. It's a good learning experience, but you won't get solid results that way. The best way is probably an optical-to-voltage converter chip. They are fast enough for what you want, and they are very simple to use: just decoupled power, and output straight to a scope probe. \$\endgroup\$ Commented Jun 8, 2022 at 20:02

1 Answer 1


It's the self-capacitance of the phototransistor that is the problem. It's a problem with photodiodes too but to a significantly smaller degree. So, consider using a photodiode instead of the "notably slow" phototransistor.

However, if you want to stay with the phototransistor, you can use a common base NPN BJT circuit where the phototransistor is connected from emitter to 0 volts. This keeps the voltage across the phototransistor at an almost constant level thus, its internal capacitance cannot produce negative feedback inside the device that "hits" rise and fall times so badly.

An alternative circuit is the TIA (transimpedance amplifier). These are often used with photodiodes for very fast response times (sub 10 ns) but, there's no reason why a phototransistor cannot also benefit from the same circuit (with a volt or so of bias): -

enter image description here

The above picture came from Pullup vs. transimpedance amplifier and it explains the mechanisms behind why a phototransistor with pull-up resistor is basically slow.

  • \$\begingroup\$ Thanks, Andy! The reason I was using the phototransistor instead of a photodiode was because it showed promise of not needing any sort of amplifier circuit to easily view on an oscilloscope with minimal noise. If I need to build a transimpedance amplifier, then I might as well just use a photodiode… \$\endgroup\$ Commented Jun 8, 2022 at 1:42
  • \$\begingroup\$ That said, maybe it would be easy to make this common base NPN BJT circuit. I’ll try looking into what that is. I do the tiniest bit of hobby electronics once every three or four years, so my knowledge going into a project usually starts and ends and a very basic understanding of Ohm’s law 😅 \$\endgroup\$ Commented Jun 8, 2022 at 1:51
  • \$\begingroup\$ @AllenPestaluky shout up if you are struggling but, you might as well use the op-amp circuit on the right. \$\endgroup\$
    – Andy aka
    Commented Jun 8, 2022 at 8:46
  • \$\begingroup\$ Thanks, Andy for your great answer and help! I've updated the original question with some progress using a photodiode instead. I think I might have things working well enough for my uses without the amplifier using the photodiode now :) \$\endgroup\$ Commented Jun 8, 2022 at 17:46
  • \$\begingroup\$ @AllenPestaluky good work. \$\endgroup\$
    – Andy aka
    Commented Jun 8, 2022 at 18:09

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