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Simulation of Photodiode connection to ADA4350

Hello Fellows,

I am attaching a LTSpice simulation circuit. Inverting input of ADA4350 is connected to a recommended equivalent circuit of a photodiode. The stimulus is a 2MHz sinusoidal signal of 2.5uAp value.

To calculate the Rfeedback (R1) and Cfeedback (C1), used the literature - snapshot is attached.

/cfs-file/__key/communityserver-discussions-components-files/385/Sinusoidal.zip

My questions are: 

1. When C2 (900pF) is almost nonexistent, input/output waveforms are not distorted. The gain is still not what is expected according to the formulas from the attached sheet. What I might be misinterpreting in setting up the circuit?

2. Datasheet for ADA4350 states that it can be operated from single supply source of 3.3VDC. When I replace VEE with a ground, The output seems to have been disabled. There is hints of sinusoidal signal at output, but no gain. Again, how should I correct this issue.

Note: The photodiode I am looking for will have nominal terminal capacitance of 900pF and will have up to 2000pF.

Thanks for helping.

  • 0
    •  Analog Employees 
    •  Super User 
    in reply to Mohi

    Hi Mohi,

    Thanks for the compliment :-)

    BTW, I've been told that the the AD4008 actually has a lower 1k list price of $4.95 vs the AD7685’s $6.78, and has better performance overall. So the $30 cost difference is questionable.

    Also, some comments below per the Application eng. responsible for these ADC's for your reference:

    "I will make the general comment though that since you're only sampling at 100SPS, that the RC filter recommended by the data sheet is probably much higher bandwidth than it would really need to be to ensure proper settling. The ADC Driver Tool should illustrate just how much easier it is to settle the inputs when sampling at 100 SPS vs. the 250 kSPS upper limit of the AD7685.

    Outside of the improvement in specs (INL, SNR, etc.) one argument I would give to select AD4008 over AD7685 is that the AD4008 inputs being easier to drive means that for a given analog input bandwidth, you’ll be able to sample the AD4008 faster than the AD7685, which would allow for oversampling and averaging to reduce the overall noise of the measurement – but that only matters if you're really trying to implement a very low noise measurement."

    Regards,

    Hooman

  • Excellent suggestions. Thanks Hooman.

  • Hi Hooman,

    Attaching is a mostly completed circuit (yet to add FB components) - but this time it is from the schematic capture.

    https://drive.google.com/file/d/1EV1c2_NZWZRGQg0iJ73Yo8QllCfITLQJ/view?usp=sharing

    Would you please suggest any modification if necessary. 

    One question about the DVDD: should it be connected to 3.3V to work with microcontroller which is a 3.3V?

    Thanks,

  • 0
    •  Analog Employees 
    •  Super User 
    in reply to Mohi

    Hi Mohi,

    I don't see anything obvious, but to be honest I did not spend an inordinate amount of time reviewing your ADA4350 TIA design feeding into the ADC :-).

    I also wanted to point out a TIA device Preliminary datasheet for LTC6563 with ADC driver built-in if you've not seen it before, for your reference, if you wish to go that route:

    https://www.analog.com/media/en/technical-documentation/data-sheets/ltc6563.pdf

    I'm hoping other people on the forum offer suggestions (if any) to your ADA4350 schematic.

    Regards,

    Hooman

  • Hi Hooman,

    I got suggestion to use AD4000 in another thread. I am set for the ADC.

    My primary question is to confirm if any errors in the attached schematic stand out?

    Also confirming that for DVDD my use of 3.3V is correct since uC (DIO and SPI) is based off 3.3V.

    drive.google.com/.../view

    Thanks,

    Mohi