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I have a system which demands optical data transmission since rotating parts are present. A mixed mechanical-electrical engineering project has led us to use photodiodes to receive data from one rotating end to another. The communication system is through a CAN bus. The transmitter and receiver diodes are:

emitter: VSLB3940

receiver: BPV10NF

In a nutshell: most applications I have thought of and tried don't seem to comply with the speed I'm looking for, which is at least 300 kHz.

The circuits I have tried are the following

Circuit 1 - A simple "RC" (the C stems from the photodiode junction capacitance) circuit as a means to take the output of a square wave. The problem arose when I looked at the scope display (image right below the schematic) and the turn-on and off time were way too high for my implementation as can be seen below.

schematic

simulate this circuit – Schematic created using CircuitLab

The scope associated to both input (yellow square-ish wave) and output (green RC-like response) is shown below:

enter image description here

The operating frequency is 206 kHz, and the response barely reaches the 12 and 0 V level. The lower boundary is at 4.8 V and such delayed reponse might disturb the communication system. If R is increased the time delay is of course even greater, however, by reducing it there's an obvious loss of voltage swing for a comparator (e.g. LM311N) to sense.

Other circuits I have tried are the following:

Circuit 2 - Q1 and R1 sources current towards the photodiode in the range of μA, which is pretty much the order of magnitude from the datasheet, so that when the photodiode is subject to the specific light source, the current amount is increased and voltage could be sensed better. Despite the attempt, no clear result came out and the BW was limited to 10 kHz without complete signal distortion.

schematic

simulate this circuit

Circuit 3- The idea has been to push current towards the transistor base and then switch the output so as to find a better result in the end, which nonetheless didn't work quite well, and the bandwidth was again reduced.

schematic

simulate this circuit

All circuits share the same input method.

I was thinking of the basic transimpedance amplifier method, however, I wouldn't expect an op-amp to be faster than a single transistor network, not to mention the additional PCB space it would take compared to a transistor.

What could be done to improve the system and meet the 300 kHz requirement? Any ideas would be really appreciated.

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    \$\begingroup\$ Do note that you don't have to reach the 12V and 0V level to have effective communication. The receiver only needs to see whether the signal is going up or down, basically. It doesn't need to wait for it to go all the way up or down. Try sending some low-amplitude sine waves (e.g. 1V) with DC bias and see how high of a frequency gets through - you might be pleasantly surprised. \$\endgroup\$ Aug 9, 2022 at 17:56
  • \$\begingroup\$ I recently was going through an online book on this particular subject: Photodetection and Measurement - Maximizing Performance in Optical Systems. You can find a free pdf here: pdfcoffee.com/… Equation 2.4 on page 28 might give you some insight on to how to build out a TIA to hit your intended frequency. \$\endgroup\$ Aug 9, 2022 at 17:58
  • \$\begingroup\$ that is also a good point: the resistor bias is not even close to an optimal receiver circuit; using a TIA will keep the voltage across the diode constant, cancelling out the effect of the diode's capacitance (which makes the signal rise and fall slowly) \$\endgroup\$ Aug 9, 2022 at 18:06
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    \$\begingroup\$ @KyleB That receiver suits me very well, however I'm looking to reduce space taken by the PCB even though it might be a very good solution. \$\endgroup\$ Aug 10, 2022 at 2:41
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    \$\begingroup\$ @IronMaiden. That's just one example of an IrDA receiver. A search on Google should get you alot more (smaller) options. IrDA is a couple decade old IR communication standard. \$\endgroup\$
    – Kyle B
    Aug 11, 2022 at 3:43

3 Answers 3

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Use a transimpedance amplifier; you'll get bandwidth into the MHz at pretty decent gains too: -

enter image description here

Image from The Fundamentals of Transimpedance Amplifiers. And, in many cases (yours included I'm sure) the most negative rail on the op-amp can be ground/0 volts. Vref will usually be half the positive supply rail but, check the data sheet for any op-amp you might choose.

In the past I've used the AD8065 for slow-ish speed applications like yours but even it might be overkill. It is a good device though and, the data sheet has a nice section on using it as a TIA.

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  • \$\begingroup\$ I'm afraid you've picked the wrong example. That's being run in photovoltaic mode, and it has its uses, especially in low-level setups, but it is the slowest TIA. You should have shown the second example in the article, which is running in photoconductive mode. \$\endgroup\$ Aug 9, 2022 at 20:12
  • \$\begingroup\$ It's not the wrong example for the sort of speeds the OP is looking at @WhatRoughBeast. If you have a better circuit then please post an answer. \$\endgroup\$
    – Andy aka
    Aug 9, 2022 at 20:20
  • \$\begingroup\$ @Andyaka I'll certainly give it a try. The only reason I chose not to go straight with that is due to space limitations when compared to a resistor-transistor-photodiode circuit. The question I still don't understand is why the simple resistor-photodiode circuit took way longer (2.5μs taken from the scope) than what the datasheet told me (RC =10e3*2e-12= 20ns)? It says R-load = 50ohm for 80ns rise time, however even that figure doesn't match junction capacitance calculation. \$\endgroup\$ Aug 10, 2022 at 2:50
  • \$\begingroup\$ @IronMaiden none of your circuits in the question appear to me to be using a 50 ohm termination so, did you actually test the photodiode connected to a 50 ohm load and get a waveform? \$\endgroup\$
    – Andy aka
    Aug 10, 2022 at 9:03
  • \$\begingroup\$ @jpa a TIA cancels out photodiode capacitance due to the inverting input being held at a virtual AC ground. Therefore, with a DC bias your observations about capacitance being reduced are correct but, the effect or importance of this advantage is significantly null and void when using a TIA. Do you know how a TIA works? Maybe you don't? \$\endgroup\$
    – Andy aka
    Aug 10, 2022 at 9:56
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Looks like resistors across PD are too big in all cases. PD times in datasheet are measured at 10V and 50 Ohm load. It says nothing about PD internal transient parameters like diffusion capacitance which can help to predict its behaviour in your mode, so practical investigation is required.

In the last circuit resistor between base and emitter should be added in order to remove impulse "tails" and collector resistor should be decreased to speed up desaturation, however, this circuit is unreliable in any case.

Take in mind that transistors in both parts (TX and RX) work in saturation. Remember that most manufacturers are cheating and switch times in transistor datasheet are measured under quite artificial conditions and require recheck in particular circuit.

I suggest to add resistor in series with emitter in transmitter part and reduce R2 to turn this circuit to unsaturated current source. This probably will require bigger transistor. Note that you probably overload the LED (100 mA max.)

Receiver circuit may require some experience depending on working contitions, for example, back-light and variation of distance, like AGC and de-biasing. You can draw inspiration from older IR discrete remote receiver circuits.

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As mentioned by Andy aka, a transimpedance amplifier is a good solution for the purpose. The voltage over the photodiode will stay relatively constant, reducing the effect of the photodiode capacitance. I would suggest connecting the photodiode between V+ and the transimpedance amplifier input, because higher external bias voltage will improve response.

Opamps can be had in very small packages, so I don't think your size concern is valid. But if strictly necessary, it is possible to build a transimpedance amplifier out of a single transistor.

Resistor R2 determines the gain. With 1 Mohm value, a 10 µA photocurrent will result in ideally 10V change in output voltage. In practice this will be less due to limited current gain of the transistor. The bias resistor R1 sets the idle operating point and should be approximately R1 = R2 / B where B is the transistor current gain.

The circuit will be faster with lower values of R2, but the output voltage swing will be correspondingly smaller. For best performance a transistor with high current gain and low input/output capacitance should be selected.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ nice circuit. would a FET achieve better gain? (at lower bandwidth) \$\endgroup\$
    – tobalt
    Aug 10, 2022 at 10:42
  • \$\begingroup\$ @tobalt The current-to-voltage gain is mostly determined by the resistors, so even with a single transistor it is possible to get very high gains. As is, the circuit would not work with a FET because there would be no path for the photodiode current to flow. But a different kind of circuit with a FET could work well. \$\endgroup\$
    – jpa
    Aug 10, 2022 at 10:45

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