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I'm working on a photodiode circuit which has been designed by someone else a few years ago.

The basics of the circuit seem simple enough, photodiode to op-amp, op-amp to ADC, ADC to FPGA.

The circuit has worked on and off for a while and everyone has always told me "It's the photodiode, just swap the photodiode and it works usually".

However looking at the actual circuit something doesn't make sense. I've always read/seen that a transimpedance amplifier always has the photodiode as shown in the picture below.

enter image description here

The circuit I have has the photodiode connected in the opposite direction.

schematic

simulate this circuit – Schematic created using CircuitLab

  1. What would connecting the photodiode in this orientation do?
  2. Is there a reason why you would do it this way?
  3. Could I flip the photodiode round or will I need to change the whole circuit?

Edit:

I've added the full section of the concerning circuit incase it's needed at all

Op-amp supply rails:

ISL2812: -38V   +2V
AD8676:  -2V    +10V

enter image description here

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    \$\begingroup\$ What supply rails do the op amps have? \$\endgroup\$ – pjc50 Aug 24 '16 at 13:42
  • \$\begingroup\$ What are you trying to improve? Dark Current offset? Noise ? or bandwidth? or ?? \$\endgroup\$ – Sunnyskyguy EE75 Aug 24 '16 at 13:57
  • \$\begingroup\$ @pjc50 I've updated the question to include the voltages. \$\endgroup\$ – Doodle Aug 24 '16 at 14:38
  • \$\begingroup\$ @TonyStewart Sometimes I'm getting a fault where I'm getting no voltage output from the opamps, I wasn't sure whether this was due to the connection of the photodiode or some other cause \$\endgroup\$ – Doodle Aug 24 '16 at 14:38
  • \$\begingroup\$ Could ESD failures be a risk? \$\endgroup\$ – Sunnyskyguy EE75 Aug 24 '16 at 15:07
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If you understand the op-amp, you will see that the photodiode has zero volts across it. It is the functional equivalent of the circuit of Figure 6A in https://physlab.lums.edu.pk/images/1/10/Photodiode_circuit.pdf, shown below. enter image description here

Since there is zero volts across the diode, it doesn't really make a difference which direction it points -- the only difference is whether the output generated by the op amp will be positive or negative. Also, there are some ramifications of not using the photodiode in a reverse bias -- it will be slower, and it will have less dark current.

enter image description here

Note that with the voltage across the diode fixed, you are moving up and down the negative y-axis in this graph, changing current as the light on the photodiode changes.

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The only difference is the sign of the output voltage... direction of current flow. The way you have shown is typical for PD circuits I've built, mostly because the PD's I use have the cathode connected to the case which gets connected to ground.

BTW why are you throwing away ~80-90% of your signal in the second stage? (I'd also ditch the bias current compensation R's on the non-inverting inputs... not needed for an FET opamp.)

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  • \$\begingroup\$ The ADC that the board was originally designed with is an AD7655. From the first opamp of the one I've just measured I'm getting -22V so I assume it's to take it to a 0-5V level \$\endgroup\$ – Doodle Aug 24 '16 at 14:44
  • \$\begingroup\$ Also I'm looking at the datasheet for the AD8676 and I can't see where it mentions about it being a FET opamp, is there something in the datasheet which gives away if it's a FET opamp or not? \$\endgroup\$ – Doodle Aug 24 '16 at 14:52
  • \$\begingroup\$ @Hayman, my mistake. I just figured a TIA with 2 Meg feedback resistor would use a FET. (You can typically tell it's a FET by the low bias current ~pA are typical.) Re: throwing away signal. It would be better then to reduce the FB resistor on the first stage, this will give more speed and less noise. (You may not care about either of those things.) \$\endgroup\$ – George Herold Aug 25 '16 at 14:56

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