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I'm using a Raspberry Pi connected to a custom breakout board. I'm reading in low power signals from an analog opto-isolator and was wondering if there are any specifications for the GPIO pins such as \$V_{IH}, V_{IL}, I_L\$? I'm trying to get the most gain I can from the isolator without adding additional hardware, which means having a large value pull-up resistor.

schematic

simulate this circuit – Schematic created using CircuitLab

I have found for the PI it seems documentation is sparse...

If there is no documentation that anyone knows of, would the illustrated method below be considered reasonable for \$\pm\$5% measurements? I would take the measurement over a few temperature ranges (~0C, ~20C, ~40C).


schematic

simulate this circuit

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  • \$\begingroup\$ I guess \$V_{IL}\$ and \$V_{IH}\$ are most likely \$0.3V_{CC}\$ and \$0.6V_{CC}\$ respectively. Current is mostly irrelevant for an input - it is high impedance. \$\endgroup\$
    – Majenko
    Commented Aug 18, 2016 at 21:32
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    \$\begingroup\$ Yes but I'm feeding the input with a source that could be around 47k to 100k Ohm, so it is no longer irrelevant. I am trying to avoid adding a buffer amplifier to save space... \$\endgroup\$
    – MadHatter
    Commented Aug 18, 2016 at 21:33
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    \$\begingroup\$ Reading between the lines, I suspect that you are trying to make an analog measurement using the Pi's digital GPIO pins. As far as I know, the Pis do not have any analog inputs, so you can't do analog meaurements with them (without adding an external ADC). \$\endgroup\$
    – Peter Bennett
    Commented Aug 19, 2016 at 0:37
  • \$\begingroup\$ Peter Bennett, I am reading a digital signal, but the signal has a very weak drive. I am concerned with input leakage current. \$\endgroup\$
    – MadHatter
    Commented Aug 19, 2016 at 15:28

2 Answers 2

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CMOS digital inputs are normally specified in terms of leakage current, rather than input impedance. I can't find anything specific to the pi, but normal values are around 1uA.

1uA*100kOhm = 0.1 volt, which is a small but potentially significant value. If possible, you'd want to measure the leakage current of the pi and choose a resistor accordingly. Your setup seems like a reasonable idea.

As a side note, the pi's inputs have an input pull up enabled by default on them (last time I was using one, anyways). Make sure you turn it off!

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  • \$\begingroup\$ Thank you for atleast understanding what I was asking. I'm surprised by the number of people that didn't understand, I thought it was clear... I guess I should have specifically said input leakage. \$\endgroup\$
    – MadHatter
    Commented Aug 19, 2016 at 15:30
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As long as you are configuring the GPIO as an INPUT, the current (as Majenko pointed out) should be irrelevant. I found this article that seems to have a good explanation of the RPi GPIOs. Take a look at the schematic for a GPIO. There is no current path when you have the GPIO configured as an input and you turn off the internal pull-up and pull-down.

Your comments suggests that you are worried about the current that the opto-isolator can provide. In that case, you just need to be sure that it can sink 3.3V/47kOhm ~= 70uA of current for the signal you are trying to measure. Otherwise you will have a signal swing of something less than 3.3V.

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    \$\begingroup\$ -1, because with a high impedance input, the leakage/input impedance can start to have a significant effect. \$\endgroup\$
    – mbrig
    Commented Aug 18, 2016 at 23:02
  • \$\begingroup\$ @mbrig I am curious where you got a 1uA leakage number from. When I read the OP question, input leakage wasn't even on my radar because I have never seen such a high number quoted or specified. Certainly no CMOS input will have anywhere close to that much leakage \$\endgroup\$
    – Daniel V
    Commented Aug 21, 2016 at 6:40
  • \$\begingroup\$ 1uA max leakage is the spec'd number for a lot of the atmel chips (like the 328p). A random TI cortex m4 I looked at was max 0.4uA (on fast gpio), and max 4nA on slow gpio. The number is probably lower in practice, but presumably you can reach those in a worst case. \$\endgroup\$
    – mbrig
    Commented Aug 21, 2016 at 6:59

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