I've been messing around with photodiodes do measure relative light intensity from very low light to very bright conditions. My search for information on photodiodes and the circuits they are implemented in led me to various different configurations.

I understand that photodiodes are run in reverse bias and are treated as variable current sources. A seemingly very common measurement method is to use a transimpedance amplifier with the photodiode feeding directly into the op-amp's inverting input. Based on the Wikipedia’s article on photodiodes, it would seem that this is the kind circuit one would want to use for measuring light because it supposedly has less noise. I've tried this configuration and I will touch on that in a bit. Continuing, with the use of the transimpedance amplifier, there are at least two different ways to bias the photodiode. One of the methods involves imposing on the cathode a positive voltage relative to the inverting input with the anode feeding the inverting input. A second method is to impose on the anode a negative voltage relative to the inverting input and tying the cathode to the inverting input.

An alternative to using a transimpedance amplifier would be to follow an approach similar to that suggested in the product datasheets for the photodiodes from Thor Labs (shown below) where there is a load resistor between the cathode and anode, and then the potential across the resistor is measured. This voltage can be measured directly or indirectly.

I am generally aware of the effects of biasing, but I have not been able to see the benefits of doing so in my own prototype circuits.

When I used a transimpedance amplifier (AD8015) as a first stage and an OP196 as a second stage differential amplifier, it seemed that the photodiode had two states, on and off. At certain medium intensity light levels the circuit did pick up a bit of light, and high intensity it went to full on immediately. At really low level light conditions, it couldn't measure anything. My circuit at the time couldn't go lower than 0.5 V bias.

I had a different photodiode circuit that use a load resistor to compare against, and it detected the ramp up of light intensity sooner than the transimpedance circuit. This meant that it could see the transient response and seemingly higher resolution. The loaded circuit could also see low to medium intensity light in situations where the transimpedance circuit could not. However, once the light intensity hit a certain point, the photodiode behaved like an avalanche diode despite being a PIN diode. The transimpedance circuit had a very rapid and high amplitude response (i.e. it clipped on my 0-10 V output). I have no need to have high frequency response (<10 Hz is acceptable). The transimpedance amplifier circuit had a smaller window of detectable light intensity that was significantly smaller than the loaded circuit and had next to no discernment in light intensity.

I have since replaced my two stage transimpedance circuit with a single stage differential amplifier measuring across a loaded photodiode. What I found is that a large load resistor (1.8 kΩ) introduces a rather high noise floor. By reducing the resistor size (100 Ω) it initially reduced the measured potential across the load. Reducing the load further (50 Ω) cause the potential to increase, and then reducing even further (10 Ω) caused a massive decrease in potential. The first step to 100 Ω immediately removed the high noise floor.

So, my questions are as follows. What are the pros and cons of connecting the anode to the transimpedance amplifier versus connecting the cathode to the transimpedance amplifier? The datasheet for the AD8015 speaks a bit to this, but I believe it was truly intended for fiber optic communications. Why does the transimpedance amplifier seem to have no middle ground in terms of detecting light intensity? How does the load impact the photodiode (i.e. how does one determine optimal load, what determines a photodiodes ability to drive a load)? Can I get a better explanation as to when one would use a loaded circuit versus when one would use a transimpedance amplifier? Is there a way to modify a transimpedance amplifier circuit so that it has better discernment in light intensity?

These photos are referenced from the datasheet for the Analog Devices AD8015. Photodiode Referenced to Positive Voltage Photodiode Referenced to Negative Voltage

This photo is the standard circuit used in photodiode datasheets from Thor Labs. This is the circuit I'm using currently except that I have a differential amplifier across the load resistor. My load resistor is 50 Ω. enter image description here

This is the link to the AD8015 datasheet. AD8015 Datasheet

  • \$\begingroup\$ Detecting a wide range of light levels will be difficult without changing the scaling (TIA feedback resistor). Cheaper commercial light meters do this with a manual scaling knob. \$\endgroup\$
    – Mattman944
    Mar 28 at 22:27

3 Answers 3


A DC-coupled TIA connected to a diode configured in photoconductive mode (reverse bias) can measure "any" level of light, relatively linearly within each decade or two (the slope changes slowly across many decades), up to saturation of its output stage. But we perceive light in a logarithmic fashion, and our eyes have incredible dynamic range, so we're least equipped to evaluate "how much light" hits the photodiode in absolute terms. A TIA output from a photodiode "out in an office" won't be very usable for scope viewing unless the optical power on the photodiode is within an order of magnitude or two, and the transimpedance is set to accommodate that.

I have just connected a random LED as a photodiode, reverse-biased from -40V, into a TIA based on TL071, powered from -5,+30V, nulled the offset to zero with the LED shielded from light, and it measures as expected from the noise floor to op-amp output saturation. Takes a digital voltmeter to see anything useful though. On a typical 8-bit scope you can see maybe 2 orders of magnitude worth of dynamic range, so unless the optical power on the photodiode falls exactly in that range, you'll either see the output "off" or "fully on". On a 6.5 digit DVM, I could use up the full dynamic range of the DVM on two adjacent ranges, with long integration times at the lower range. More would be possible with a better op-amp, although last time I was working with photodiodes and precision measurements based on them was 2.5 decades ago. Breadboard photodiode circuits are a no-go if wide dynamic range is needed: takes a ground plane and shielding around the TIA stage and the photodiode to get output that isn't swamped by mains interference (even with no artificial light).

As for the pros and cons: TIAs use the diode in photoconductive mode, voltage amplifiers use the diode in photovoltaic mode. The photoconductive mode is better for linearity, so finds use where that is important. In data transmission - high-bandwidth applications, the dynamic ranges are small, so photovoltaic mode performs better, since the equivalent source impedance is lower IIRC.

Op-amp TIAs take some work to make them fast, but they are easy to apply for frequencies in the audio range up to 10s of MHz. They act like a voltage source from the diode's perspective, so the diode's capacitance is shorted out, and its shunting effect is reduced. But, because op-amps are not ideal, at higher frequencies the AC voltage across the diode may become large enough that shunting the diode with 50Ω at AC will give better performance than terminating it in a virtual ground of an op-amp.

Then it's still a choice whether to use photoconductive or photovoltaic mode.

Since photodiodes are used with signals into the 10s of GHz, there's no op-amp that will do the job at frequencies that high, and voltage- or current-input RF amplifiers are used.

  • \$\begingroup\$ Regarding the use of the reverse bias, I understood the purpose of it, but what will increasing the bias of it do? One of the photodiodes I've tested has a max of 5V, and the one that I'm trying to compare against has a max of 25V. I don't know what its bias is. Will the bias improve the sensitivity at all. I understand the effects of bias on the linearity, but it seems to me that leaving photovoltaic operation results in lower sensitivity. Is that statement accurate? \$\endgroup\$
    – Ender
    Mar 30 at 18:21

There's a huge difference between an amp with 10kΩ transimpedance and a 50Ω load resistor. The 10kΩ will give you 10V per milliamp of photocurrent. The 50Ω load resistor will give you 50 mV per milliamp, 0.5% of the transimpedance amp. So, perhaps your problem with the transimpedance amp is simply oversensitivity.


So, my questions are as follows. What are the pros and cons of connecting the anode to the transimpedance amplifier versus connecting the cathode to the transimpedance amplifier?

It determines the polarity of your output (if DC coupled) and what rails you need to supply. The negative supply is more awkward since it requires you to either generate a negative rail, or else to raise the opamp voltage above zero (potentially well above zero if you diode needs a large bias voltage).

Why does the transimpedance amplifier seem to have no middle ground in terms of detecting light intensity?

You are using an amplifier with too much gain. Either lower the gain or use dimmer light sources.

How does the load impact the photodiode (i.e. how does one determine optimal load, what determines a photodiodes ability to drive a load)?

That thorlabs page you're looking at explains this: https://www.thorlabs.com/newgrouppage9.cfm?objectgroup_id=285

Doubling the resistance doubles the amplitude but halves the bandwidth.

Can I get a better explanation as to when one would use a loaded circuit versus when one would use a transimpedance amplifier?

If you don't need gain, then the shunt resistance is a very simple solution. The downside is that it lacks gain. If you're comparing operating the amplifier in a transimpedance configuration vs. operating it as a voltage amplifier measuring across a shunt, they do the same thing. Usually the transimpedance amplifier configuration will have better performance, but theoretically you might have some situation where the voltage configuration worked better (maybe at very high bandwidth and very low transimpedance gain?).

  • \$\begingroup\$ Regarding having too much gain, I do have another circuit that lacks any op-amp that is already in use to compare against, and it successfully picks up a very low light condition that my attempts whether with a TIA or with a shunt resistor will not detect at all. Unfortunately, it doesn't pick it up in a way that the detection algorithm can reliably identify. In fact, the TIA actually had a thresholded minimum light intensity that the shunt resistor method did not demonstrate. What I also found was that shunt resistor increases the noise floor with increasing resistance which is bad. \$\endgroup\$
    – Ender
    Mar 30 at 14:25
  • \$\begingroup\$ @Ender Its possible your TIA circuit isn't working as well, they're notoriously difficult to design. But no middle ground with a binary output sounds like too much gain. Try lowering it and see if the circuit works. If not, continue troubleshooting. \$\endgroup\$ Mar 30 at 16:57
  • \$\begingroup\$ I will rebuild the circuit to exclude the AD8015 and use a "custom" TIA because my understanding is that the AD8015 has a fixed gain per the datasheet. I'll end up needing an inverting amplifier, but that shouldn't be a problem. I will report on my results which may be a while. System time isn't readily available. \$\endgroup\$
    – Ender
    Mar 30 at 18:31

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