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Please bear with me as I'm not too experienced with this.

I'm attempting to use an ADC to get readings from a photodiode. From what I understand, I should likely use an op-amp as the range of the current that will be provided by the photodiode is low (0-150 uA), and I can output a voltage required by my ADC (0-3.3V).

To get an idea of what I should be looking for, I used the site here

enter image description here

The above is what I produced from that site. However, I noticed on the second op-amp's (LT6202) datasheet, is that is has a minimum output current of 30mA. I was unable to find any reference to a maximum analog input current on the ADC's datasheet, and I wonder if this safe as an input, or should I go about finding different op-amps?

The selected photodiode (TEMD6010FX01)

The selected ADC (MAX1249)

The first op-amp (LTC6240HV)

The second op-amp (LT6202)

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  • \$\begingroup\$ How fast\what is the bandwidth the design needs? \$\endgroup\$
    – Voltage Spike
    Commented Jul 22, 2019 at 19:21
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    \$\begingroup\$ Well, you're already way ahead of most people who ask this question merely due to the fact you are using a transimpedance amplifier. But not just a transimpedance amplifier, a transimpedance amplifier that will actually work in the real world because it has a compensation capacitor. \$\endgroup\$
    – DKNguyen
    Commented Jul 22, 2019 at 19:32
  • \$\begingroup\$ The photodiodes are to be used as part of an attitude determination system, so ideally I would like to maximize both bit resolution and sampling rate as much as I can. However, I would say a 10-bit resolution would be acceptable, and as for the sampling rate.. Perhaps 10/s would be the absolute minimum? I don’t have a good idea for this yet, but again ideally I would like to boost it as much as possible. (I hope that answered your question correctly, if not let me know) \$\endgroup\$
    – user226923
    Commented Jul 22, 2019 at 19:42
  • \$\begingroup\$ The "minimum current" mentioned in the datasheet is how much current it can deliver into a short circuit, not how much current it will force through the ADC. \$\endgroup\$
    – JRE
    Commented Jul 22, 2019 at 20:18

1 Answer 1

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The max input current is most likely the mux leakage current.

enter image description here

By inspecting the equivalent input circuit, it is high impedance as there are two capacitors that capcititivly couple the input signals to the ADC core. These currents are likely to be negligible and the current that the LT6202 needs to source is 1uA which is well within its capability.

enter image description here

ADCs generally keep the input impedance high (input current low) because any load from an ADC will contribute to error when sampling\switching.

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  • \$\begingroup\$ I'm assuming you meant the MAX1249 (ADC) instead of the LT6202? This helps put some of my worries to rest and I think I’ll give it a test and see how it fares. I imagine this should probably hold up with no issues as long as the systems operating conditions are within the maximum ratings correct? For reference, it will be operating constantly for a year or so. Regardless, thanks for the help! \$\endgroup\$
    – user226923
    Commented Jul 22, 2019 at 20:34
  • \$\begingroup\$ No, the LTC6202 is sourcing current so it would be the source (almost all of the time) The ADC would be the sink. Currents usually go from source to sink. \$\endgroup\$
    – Voltage Spike
    Commented Jul 22, 2019 at 20:38
  • \$\begingroup\$ Ahh, I see what I got mixed up there now. Thanks again! \$\endgroup\$
    – user226923
    Commented Jul 22, 2019 at 20:42

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