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.


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 
    on Jan 14, 2021 3:17 AM 1 month ago

    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.



  • 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 
    on Jan 15, 2021 4:25 AM 1 month ago in reply to Mohi

    Hi Mohi,

    You're welcome!

    To answer your specific questions on the ADA4350 operation:

    1. GBW vs I-to-V Gain: The GBW computation you've done is mixing parameters / units together. The 200kohm is the transimpedance gain in units of ohm. The GBW is in units of V/V*Hz or just Hz. The I to V -3dB BW you can expect will be the same as fN in equation 9:

    Here are some examples:

    RF (ohm) fN (Hz)
    200.0E+3 392480.2
    100.0E+3 554535.6
    50.0E+3 783203.8

    Here is a simulation file that shows the AC small signal response for various values of RF (with the appropriate value of CF computed for each step), as reference:

    ADA4350_Sinewave_Stimulus modified AC EZ 1_14_21.asc

    2. Signal Bandwidth: Figure 58 schematic shows that the photodiode current flows through the parallel combination of RF and CF to arrive at the output. Such an output voltage will roll-off beyond 1/2pi()RFCF (in Hz) thereby setting the bandwidth achievable. "1/RFCF" would be the bandwidth in rad/sec (not Hz). You can follow my description above to see what bandwidth you can expect for a given CS, and CF given that GBW is 175MHz.

    3. Reducing R4 (shunt resistor across photodiode): If you reduce this large (2Gohm) resistor, you are stealing the signal current from the photodiode and diverting it to ground. Not sure why you're doing this. R4 should be what the Photodiode manufacturer specifies. It is generally very large so all photo current flows through the amplifier.



  • Hi Hooman,

    I am back wit a couple of more questions regarding TIA's signal and SNR.

    1. What is the lowest current level that can be amplified and digitized with this TIA.
    Above 100k feedback resistor, I noticed that the value of feedback capacitor start to becomes unrealizable and the signal is unrecognizable in LTSpice.
    I am given the task to detect 300nA current. Is it possible with any of the off-the-shelf TIA without using cryogenics?

    2. I can sum the two noise sources at input of the TIA namely thermal noise from the feedback resistor and TIA's internal current noise. This is for the SNR calculation. How one add the internal voltage noise to the equation? Ultimately how do one calculate the internal resistance of the TIA? Is there a simpler way to calculate SNR?

  • 0
    •  Analog Employees 
    on Feb 1, 2021 3:38 AM 27 days ago in reply to Mohi

    Hi Mohi,

    To answer your specific questions about the ADA4350 interface with your photodiode:

    1. Minimum discernible input current from your photodiode: I'd use the information in Table 17:

    This would be output noise. With a high impedance photodiode, the low frequency noise gain is 1V/V (as Figure 60 bode plot also shows).

    The input noise current contribution will not be dominant, as stated in the paragraph just above Table 17:

    "The effect due to the current noise is negligible in comparison".

    You can then compare your output "signal" [ = i_Photodiode * RF] against the output "noise" [ = RMS sum of the RF noise and the VNOISE from Table 17]. You'll be able to detect the 300nA photodiode current as long this ratio is at least a few dB's.

    2. Individual Noise Terms Effect / Contribution: As the datasheet states and as Table 17 shows, your dominant noise sources are RF thermal and the input voltage noise. Apparently, input current noise is of little significance here.