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PCB with photodiode arrayMy colleague and I recently designed and ordered a PCB intended to capture red laser light using a SFH2440 photodiode. We used the analog devices photodiode circuit design wizard to design the photodiode portion of the PCB. We received the boards, however,upon testing, it looks like we made a mistake with the design and the output is saturating to the negative rail. I think this is because we connected the (V-) pin of the op-amp to ground. We are seeking some opinions on whether we can salvage these boards to make them work or if we need to start over.

photodiode circuit schematic

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    \$\begingroup\$ So what was the mistake, how it should have been? \$\endgroup\$
    – Justme
    Commented May 28 at 19:33
  • \$\begingroup\$ I think the negative power rail being connected to ground instead of (-5v) is causing it to go to 0. \$\endgroup\$
    – Jort2750
    Commented May 28 at 19:35
  • \$\begingroup\$ You need to get 1 channel working, and then copy that fix to the other channels. it might be as simple as lifting the neg supply pin on each op amp and connecting it to -5V. \$\endgroup\$
    – Drew
    Commented May 28 at 19:45
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    \$\begingroup\$ Well how could we have known you have -5V available unless you show or tell. Conncting it as intended is a solution. If you do have -5V available then what other things you have not mentioned could be potential solutions? \$\endgroup\$
    – Justme
    Commented May 28 at 20:32

3 Answers 3

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Suffice it to say all channels are identical? So, concentrating on one:

enter image description here

We have an inverting configuration, which takes the photocurrent (drawn from the +5V bias source) into a virtual ground / summing node, and the same current flows through C1 || R2 from the output.

The photodiode sources current into this node, acting to pull it up. The op-amp in turn must act to pull it down, keeping its inputs at equal voltage (since +IN is GND, -IN will also be held to GND, hence "virtual ground"), thus the output voltage must go negative.

But the op-amp is constrained by its supplies, +5/GND, and saturates for any nonzero photocurrent.

You could invert the current, by flipping the photodiodes, cutting the +5V trace, and introducing a -5V supply to it instead (perhaps via charge pump derived from the +5V boost regulator?). The .1 section then provides the correct output value, without needing an inverter stage.

The same problem applies to the .2 section, which acts as a current balance between R1 and C2 || R3. DC gain is -1, but the output is constrained to above 0V. The input is also constrained to above 0V, so you've made a ground rectifier, so to speak: it outputs either GND or GND depending on whether the input is high or low.

So, you need to bypass this section entirely, perhaps by cutting the SIGn trace, removing R1, and jumpering from the .1 section output to the far SIGn stuff (whatever it's going to, mux? ADC?).

The key insight missed from (or perhaps, missed by) the design wizard tool, is that the op-amps are assumed to operate linearly, or that both supplies are adequate to avoid clipping the output.

The semantically simpler solution would be lifting the +IN pins to a mid-rail voltage reference, thus mimicking e.g. ±2.5V supplies (at some expense to photodiode performance; though I gather you aren't after high bandwidth here, given the huge photodiodes, and modest filtering on the op-amps, so this probably doesn't matter). Or just as well, lifting the VEE pins of the op-amps and wiring to -5V. But cutting those small pins off from the ground plane would be quite a bit more tedious than the above solution, I suspect.

Whether you want to modify the board (cutting traces, lifting pins, adding bodge wires) or order new, is up to you.

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    \$\begingroup\$ If each opamp Vee pin is lifted and tied to a -ve DC bus, it would be advisable to add a surface mount bypass capacitor to GND. I'd add a few along that long bus, perhaps even a capacitor at each opamp's Vee. \$\endgroup\$
    – glen_geek
    Commented May 28 at 20:06
  • \$\begingroup\$ Thank you for such a detailed response. In the end we will be reordering the boards but we're hoping to salvage these boards for initial testing and proof of concept. We are both mechanical engineers, therefore PCB design is not our strong suit. We hired a PCB designer to finalize our design as well as a reviewer prior to procuring these boards, but they did not catch this. Thanks again \$\endgroup\$
    – Jort2750
    Commented May 28 at 20:07
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It looks like you've got an inverting op amp configuration as the first stage, but it's centered at 0V. The op amp can never output less than 0V with a negative supply connected to gnd, so the op amp will never reach equilibrium.

Try lifting the negative supply pin on one of the op amps and connecting it to -5V. If it works, do that on all channels.

Consider testing your op amp circuit on a breadboard next time before manufacturing such a large pcb.

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  • \$\begingroup\$ Thanks for the response. We did some breadboard testing but not this exact configuration. We are both mechanical engineers and hired a PCB designer and a reviewer to help finalize our design, however they did not catch this issue. \$\endgroup\$
    – Jort2750
    Commented May 28 at 20:04
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    \$\begingroup\$ ...and, for a large run of boards look at ordering and testing a few samples first so you can find and fix problems like this before you have hundreds or thousands of bad boards. \$\endgroup\$
    – GodJihyo
    Commented May 28 at 20:06
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    \$\begingroup\$ The other concern is those 8.2pF caps. With small values like this the actual component value you need to use can be quite different due to parasitics and tolerances, so you probably want to check the step response on a real board in case it starts oscillating. \$\endgroup\$ Commented May 28 at 22:21
  • \$\begingroup\$ @user1850479 I believe that's there to reduce the gain at high frequencies. The value probably isn't critical. Regardless, if it were me, I would get one channel working and test it thoroughly, then modify the others to match. \$\endgroup\$
    – Drew
    Commented May 29 at 12:37
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You are right. The negative voltage is required.

You may be able to lift the pin grounded by mistake. Remove the solder with solder wick. Let it cool. Then use a tiny blade, lift the pin while applying heat to melt the solder. “Air wire” the pins together. Use proper routing technique. You can make node points from small single sided squares of pcb material, glued down. Be sure to add decoupling capacitors at each point near the lifted pin.

Lots of creative students have saved their project by being inventive.

Go slow.

Follow good electronic design an layout principles.

Plan your wiring on paper first.

Modify one first then test. Don’t do all at once. Test as you go.

Good luck!

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