I have a typical NPN phototransistor. I have it working in a common-collector configuration; see figure 2 of this app note.

enter image description here

Increasing Re will increase the sensitivity, but decrease the speed. I have been studying phototransistors for a few days now, and I think that a transimpedance amplifier can give me additional sensitivity without sacrificing speed, since I won't have Re loading the emitter anymore.

However, I can't seem to find straightforward implementations. The vast majority of app notes describe photodiodes. Unlike a photodiode, a phototransistor needs to be biased, and the few application notes that discuss using phototransistors assume the presence of a negative biasing voltage in their transimpedance amplifiers. I need a solution that works with a single-supply op-amp.

Would a virtual ground on the non-inverting input of the transimpedance amplifier correctly bias the phototransistor? Usually virtual ground is half-way between VCC and GND, but I don't think it has to be. My phototransistor saturation voltage is 0.15V; given VCC=3.3V, does that mean my virtual ground could be at ~3V?

Is there a better way to design this circuit? I would like the output to get as close to GND as possible, because there will probably be a second-stage amplifier.


More details on the application. I'm sensing light levels; low, very low, and off. There are no problems with ambient light, so I'd rather not focus on the phototransistor aspect of this question too much. The bandwidth of interest is around 1-10 kHz. The common-collector almost works; I've raised Re as high as it can go while maintaining the bandwidth I want, but I'd still like Re about 2x larger which results in a signal that's too slow.

  • \$\begingroup\$ Check out the receiver designed for the RONJA project: ronja.twibright.com/schematics \$\endgroup\$ Jul 13, 2011 at 6:11
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    \$\begingroup\$ There's no such thing as a "typical NPN phototransistor"? Why don't you just tell us what type it is? Also, like I wrote in my edited answer: what's the application? Light levels, Code reception? Visible light, IR? etc. etc. \$\endgroup\$
    – stevenvh
    Jul 13, 2011 at 6:40
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    \$\begingroup\$ Sure there is such a thing. Search digikey for "npn photo transistor", average all the values in the datasheets together (e.g. Vcesat, Vceo, Ic, etc), and I would call that "typical". Ultimately, it's an indication that the amplifier circuit is more important \$\endgroup\$
    – ajs410
    Jul 13, 2011 at 15:18
  • \$\begingroup\$ Please name the type of the photo transistor, the value range of Re that you have tested ("almost works" = how many Ohms for Re), and describe a bit more what "low, very low, and off" light level means. If you don't know the intensity, describe the light source, distance and the setup or optical path to the photo transistor. \$\endgroup\$
    – datenheim
    Dec 25, 2022 at 10:42

3 Answers 3


I've been trying to do a very low light level project myself the past 2 days with photodiodes and phototransistors. This is for people like myself and the original poster who are pushing light detection without a photomultiplier to the limit (below 0.1 mW/cm^2).

I looked at the first receiver module and its minimum irradiance detection was 0.2 mW/m^2 which is about 10,000 times more (less capable) than what discrete photodiodes and phototransistors can do (maybe they meant cm^2 instead of m^2?). Neither are good for really low light levels according to "Art of Electronics" (1 uA per uW of light page 996), totally not capable of getting near what the human eye can do due to leakage current and noise. He describes using photomultipliers which may be required if your light levels are too low. However, in shining light through my fingers in a well-lit room, I am able to see what my eye can't detect on an oscilliscope (with either PhotoDiode or PhotoTransistor).

Assuming his 1 uA per uW is correct, here is an example: a 5 mm photodiodes and phototransistors have an area of 20 micro m^2. So 1 uW/m^2 (1/1000th of noon sunlight) would generate 20 uA (according to Art of Electr) . [[ 1/1000th of noon sunlight is 1 W/m^2 which is about twice as strong as a 20W incadescent light at 1 meter (6W light output into 12 m^2 surface area of a surrounding sphere). ]]

However, my 880nm phototransistor datasheet indicates 600 uA at 1W/m^2 (0.1 mW/cm^2), wich is 30 times more. This assumes all the light is within the active range of the diode's junction.

Sharp has a much better application note, but it seems to be lacking in explaining which design is best for which situations. Figure 13 is most applicable to what original poster and I need, and figure 10B is very interesting but I don't know what they mean by "improves response". http://physlab.lums.edu.pk/images/1/10/Photodiode_circuit.pdf

When used with an op amp, a phototransistor may not capabale of getting as good of a gain as a photodiode for very low light levels because it to uses a "cheap" method of getting its initial gain (transistor instead of op amp). I suspect a photodiode with a JFET op amp (very low input current) would ultimately provide a higher gain with less noise. In any event, the photodiode or phototransistor with the largest optical receiving area might have the best ability to detect low light levels, but that might also increase noise and leakage by a proportional amount and they are usually the underlying problem. So there is a limit to this type of light detection and ideally efficient phototransistors and photodiodes may ultimately be equally good when used with an op amp, but theoretically I suspect photodiode is a little better. It will be the op amp design that matters and I think figure 13 with a JFET op amp and a well-chosen shunt capacitor across the feedback will be best for phototransistor.

For the dual supply op amp, you can use a "lowish" valued resistor pair (two 1k for 10V Vcc to get 5 mA bias) to split the voltage to create a false ground for the +Vin.

I found R=1M for the feedback resistor much better than R=4.7M. Forrest Mimms in his simple opto book used a 10 M with a parallel 0.002uF and a solar cell instead of a phototransistor or photodiode for "extrememly" low light levels" (maybe a solar cell would be better for your application) It seems all PN junctions seem to operate as a solar cell to some extent, as I've read of using clear-cased small signal diodes to detect light. I'm using a regular 830 nm LED as my "photodiode".

The lens angle of whichever 5mm optical diode you use makes a big difference. +/-10 degrees is roughly 4 times more sensitive than +/-20 degrees....if the light source is coming in from less than +/-10 degrees. If the light source is a big area that is +/-20 degrees in front, then it doesn't matter.

I tested the two circuits below. I could detect 0.3V, 5 ms pulses on the phototransistor's Vo which means 0.3 uA which means 0.05 uW/cm^2 if my reading of the datasheet is correct and if it remained linear (big ifs) all the way down to 0.3uA. Maybe it was 5 uW/cm^2. If 0.05 uW/cm^2 is correct, then the off-the-shelf 830 LED was reading down to 0.5 uW/cm^2. I was shining 10 mW 830 nm light through 1 cm of tissue (my finger). I know that if the light levels I was working with were red, it would have been barely visible. The link below shows using 500 M ohm feedback with a photodiode, indicating much lower light levels. Notice the orientation of their photodiode, which is the same as my LED (backwards from most internet links). I got better results this way.


phototransistor with JFET op amp for low light levels

5mm 830 nm LED in place of a photodiode with JFET op amp

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    \$\begingroup\$ I posted this before editing and immediately got 3 negative votes. Hopefully the edited version will not be disliked so much. \$\endgroup\$ Aug 15, 2012 at 17:56
  • \$\begingroup\$ It's a bit wordy, and you didn't address the whole trans-impedance opamp thing the original poster was asking about. But you're new here, so I'll give you the +1 for effort. \$\endgroup\$
    – user3624
    Aug 15, 2012 at 18:09
  • \$\begingroup\$ He says raising Re decreases bandwidth. I thought it would have no effect. Looking at his subsequent comments, figure 13 of the link I provided is an answer to his request for a specific design. Also figure 10B is an interesting idea that "improves the response" whatever that means (gain, BW, or both?). If he can't get it to work from light levels that are too low, "photomultiplier", "lens", or "increase light source" are possible answers. \$\endgroup\$ Aug 15, 2012 at 18:56
  • \$\begingroup\$ +1 for actually addressing the question, which was about single-supply amplifier circuits for phototransistors. \$\endgroup\$
    – ajs410
    Aug 24, 2012 at 18:48

I was also thinking about an inverting opamp amplifier. The nicest thing would be a dual power supply so that you don't have to bias the inputs to create a virtual ground. The picture shows the schematic. You'll have a positive ground-referenced signal: more light = higher output voltage.

enter image description here


Since you only have a single supply you'll have to use a virtual ground, which I would place at half \$V_{CC}\$. You have to bias both inputs, the output will have this bias as well.

edit dd. 2012-08-15
In this answer Alfred showed that a photodiode will also sink current without a voltage drop across it. That means that we don't need the negative supply, and a single supply is possible:

enter image description here

Make sure that's an RRIO (Rail-to-Rail I/O) opamp.

In the above I presumed that you want to measure light levels, i.e. analog values. But rereading your question it nowhere says you do. The mention of speed suggests pulse code reception. If that's what you want, what does the signal look like? What's the wavelength (IR or visible light?) Can't you use an IR receiver module?

  • \$\begingroup\$ I've updated the question with extra application information. Unfortunately, I need to use a single-supply op-amp. Also, from what I've read, the transimpedance amp will probably need a feedback capacitor in parallel with the resistor to compensate. Fortunately, the bandwidth reduction of this feedback cap probably won't be a problem for me. \$\endgroup\$
    – ajs410
    Jul 13, 2011 at 15:51

If you really need flexibility, consider using a photodiode instead of a phototransistor-- you're already building a transimpedance amp, so why not go all the way?

Also, there's a great book on the subject, with many detailed circuit examples, for low noise and/or high speed.
Building Electro-Optical Systems: Making It All Work, by Hobbs.

  • \$\begingroup\$ +1 for "why not use a photodiode". \$\endgroup\$ Jul 13, 2011 at 6:09
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    \$\begingroup\$ Photodiodes are too fast. Photodarlingtons are too slow. Phototransistors are juuuust right - with a good amplifier. Photodiodes also require more gain in a circuit with an already large gain; they would probably require a transistor somewhere, and at that point a phototransistor has the transistor on-die, rather than at the end of some traces. \$\endgroup\$
    – ajs410
    Jul 13, 2011 at 15:21
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    \$\begingroup\$ "Too fast?" You can always slow things down with cheap, tiny capacitors. Or use a larger, slower photodiode-- not all photodiodes are fast. And, op-amp circuits can give you ludicrously high gain-- in normal cases, you can use an op-amp OR a transistor. \$\endgroup\$ Jul 13, 2011 at 22:51
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    \$\begingroup\$ Why use a discrete transistor on my PCB when the phototransistor has it integrated on-die? And with ludicrously high gain comes a requirement for ludicrously low offset voltage. I'm looking for an amplifier circuit to go with a phototransistor. I appreciate the book suggestion, but it does not appear to contain such examples. \$\endgroup\$
    – ajs410
    Jul 14, 2011 at 16:08
  • \$\begingroup\$ Really, photodiodes can not be too fast for anything. Most of the times they are either to slow or not sensitive enough. This is my penny after working for 22 years an ditching photodiodes to their limits. But let's accept the OP has some untold good reason to stick to phototransistors. \$\endgroup\$
    – datenheim
    Dec 25, 2022 at 10:46

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