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.

Parents
  • 0
    •  Analog Employees 
    •  Super User 
    on Jan 14, 2021 3:17 AM

    Hi Mohi,

    Several points related to your issues with the ADA4350 LTspice file you attached:

    1. LTspice Oscillation: I can't tell what the screenshot instruction you attached for designing the feedback components (R1, and C1) comes from since it's not from the datasheet. However, if you want a Transimpedance gain of 200kohm, ADA4350 datasheet equation 8 shows you how to compute the feedback across it:

    Plugging in:

    RF= 200k, CS = 900pF, fGBW = 175MHz --> CF = 2pF

    Once I changed CF (your C1) to 2pF (vs. 0.2pF you had used), the output oscillations you were seeing disappear.

    Also, the gain is correct: Vout1 / I(R6) = 0.516V / 2.7uA = 191kohm (vs. 200k).

    I have reduced the stimulus frequency to 100kHz (instead of your 2MHz) because I think at that frequency you're already bandlimited, which might make it look like the transimpedance gain (ohm) is off or lower than expected.

    Here is the simulation file:

    ADA4350_Sinewave_Stimulus modified EZ 1_13_21.asc

    2. Single Supply Operation: For single supply operation, I've modified your circuit and tied IN_P to some voltage (1.3V) to make sure I allow bipolar swing at the FET amplifier using single 3.3V supply. I've also reduced the input stimulus to 1uAp to avoid distortion. Gain expression computes correctly:

    Vout1 / I(R6) = (1.51-1.1)V / 2uApp = 205k (vs. 200k).

    Here is the simulation file:

    ADA4350_Sinewave_Stimulus modified Single Supply EZ 1_13_21.asc

    Hopefully, all this helps.

    Regards,

    Hooman

  • Hi Hooman,

    Thank you for your quick reply and corrections to my schematic. I really appreciate your time and efforts.

    I will be using your first modified schematic for further discussion. Following are more of my questions:

    1. The "design-gain x signal-bandwidth" should not exceed the GBW of the OpAmp. The rule does not seem to be applicable for transimpedance amplifier. The signal I need to process is 2MHz. What gain is achievable for this signal? For the transimpedance gain of 200k and the input signal of 100KHz, the design GBW = 200k x 100kHz = 20Ghz - I must be missing a point here.

    2. What does the "The signal bandwidth" [1/(Rf x Cf)] mean mentioned in the paragraph following equation 4? For this design, it is ~2.5MHz but the input signal is limited to ~100kHz. Output starts to reduce above 100KHz of input signal.

    3. If R4 is reduced, the signal start to become DC biased. Is it because of V3?

  • 0
    •  Analog Employees 
    •  Super User 
    on Feb 3, 2021 12:39 AM in reply to Mohi

    Hi Mohi,

    Thanks for the compliment :-)

    I went through some of the spreadsheet. You've not used the noise_pp cell J6 anywhere but it usually has a 6.6x multiplying factor to RMS noise (not 2*SQRT(2)) as you've used). Also, I could not see why you've calculated Cfb in a separate worksheet and not just computed it interactively for any RF value. At least, that's how I'd have done it. And, you probably won't need to invoke LTspice to compute the output voltage separately because your TIA gain is known based on the feedback resistor you've chosen and you could just use the RF value in your spreadsheet directly. But, overall I think your approach works.

    With regard to your question about selecting an ADC: I'm not really an expert on that, but I'll point you to the parametric selection guide here to do your own evaluation:

    https://www.analog.com/en/parametricsearch/11007#/

    Maybe somebody else on this forum who is more knowledgeable about ADC's can chime in?

    Regards,

    Hooman

  • Hi Hooman,

    I back with another question.

    In another thread related to ADC discussion, Sean suggested to use AD4008 or AD7685.

    One of the advantages of using AD4008 is it has "The analog input incorporates circuitry that reduces the nonlinear charge kickback seen from a typical switched capacitor SAR input." as stated in the datasheet.

    The ADA4350 has built-in driver circuit. Is using AD4008 gives any advantage in our case.

    We will be sampling the signal only few times per second (~100SPS) but continuously. However, input to ADC can be as little as 10mV.

    The reason to asking this question is the price difference between the two ADCs. 08 cost ~$30 while 85 is ~$15.

    Thanks

  • 0
    •  Analog Employees 
    •  Super User 
    on Feb 25, 2021 2:01 AM in reply to Mohi

    Hi Mohi,

    I'm not on expert on the ADC's and the driver you've noted. However, here is my 2-cents:

    As far as driving the input of a SAR ADC like AD7685:

    Usually an RC as shown in Figure 27 of AD7685 datasheet (33ohm series and 2.7nF shunt cap) alleviates most issues with charge kickback from the ADC frontend. You may want to make some measurements with the RC combination to see if you see any issues.

    You may want to consult this excellent article (link below) from Alan Walsh which goes into the design of the front-end RC network for charge kickback mitigation:

    "Front-End Amplifier and RC Filter Design for a Precision SAR Analog-to-Digital Converter."

    I took Alan's idea and simplified it a bit here in this article:

    The Perilous Path from the Transducer to the ADC: What’s an Engineer to Do?

    There is also some help from ADI on the ADC driver Tool in configuring the input network here which does some of the tedious computations automatically for you:

    Precision ADC Driver Tool

    Hopefully this answers your question.

    Regards,

    Hooman

  • Thanks Hooman.

    Once again, your answer are very through. I will go through both articles to help me understand the root of the issue.

    Mohi

  • 0
    •  Analog Employees 
    •  Super User 
    on Feb 26, 2021 9:01 PM 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

Reply
  • 0
    •  Analog Employees 
    •  Super User 
    on Feb 26, 2021 9:01 PM 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

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