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OPA838: Achieved SNR is different than the simulation results

Behnam Madadnia
Intellectual 360 points
Part Number: OPA838

Tool/software:

Dear Engineers,

I have designed a transimpedance amplifier as follows: 

Based on the simulation, I should get around 58 dB for SNR @3MHz:

But what I achieved in my experimental results is as follows:

-Supply voltage                                 +/- 1.5 V

-Power consumption                       900 uA (datasheet=950 uA)

Output signal:

Noise:

Calculations:

-〖Signal〗_RMS= 3.8 mV

-〖Noise〗_RMS= 0.25 mv

-SNR (MATLAB)= 23.4 dB  (I used snr() function in MATLAB to calculate the SNR)

-SNR (Calculated)= 20log(〖Signal〗_RMS/ 〖Noise〗_RMS)= 23.6 dB  (calculation by hand)

PCB and schematic details:

So, the achieved SNR is 23.4 dB, but what should I get based on simulation is totally different 58 dB, would you please let me know why there is this much difference in the experimental and simulation results? 

Best regards,

  • 0
    Guru 48195 points

    Hey Benham, 2 things, 

    make sure you know what your noise gain shape over frequency, sim vs board, 

    the thing everyone seems to miss is while the sim will give you the amplifier output noise the source itself is adding shot current noise as well in most cases. That I think has to be added manually to pick up

  • 0 in reply to Michael Steffes
    Intellectual 360 points

    Dear Michael,

    Thank you very much for the good insight.

    Would you please let me know how can I measure noise gain with an oscilloscope? then I can plot it and compare it with simulation.

    I also attached the simulation files for the current design:

    7608.Transimpedance test.TSC

    Best regards,

    Behnam,

  • 0
    TI__Genius 13911 points

    Hello Benham,

    I was able to confirm an SNR of 58.148 dB @ 3MHz for an output signal of 0.05V.  If you had different conditions than I assumed, your calculated value from your circuit setup may vary.  I found the noise gain of the circuit using NG = 1 + 2*pi*Rf*Cin*Freq, which is 19.85 for your circuit.  I then used the frequency range of 3MHz to calculate the amplifier noise with the noise density of 1.8nV/sqrt(Hz) a 1kHz.  The SNR follows as the 20 x log (Vout / (NG * Vnoise) = 58.148 dB.

    I am not sure what is occurring with your experimental data and calculations.  How are you getting your 3.8mV value for signal rms voltage?  Are you certain the TINA simulator is using RMS voltages here, or are we comparing apples to oranges?

    Best,

    Alec

  • 0 in reply to Alec Saebeler
    TI__Genius 13911 points

    Hello Benham,

    For an operating frequency of 2-3MHz, the total bandwidth for which we are examining the noise is over the 1MHz between the 2-3MHz range.  This changes my calculation and results in the first step being to take the sqrt(3MHz - 2MHz) = 1000, multiplied by the noise density to arrive at Vnoise = 1.8uV.  The NG and final calculation are the same process, with a result of SNR = 62.92dB.

    Best,

    Alec

  • 0 in reply to Alec Saebeler
    Guru 48195 points

    For some reason I am not seeing your reply with file, here, but I got it, here is the output spot noise sweep, yes that 1st peak is the noise gain peaking up, and it shoudl be stable, but that 2nd peak suggests instability - might be a Zol interacting with the 33pF feedback C, 

    If I add a 20ohm inside the loop, seems to help - that might be an actual oscillation around 35Mhz folding back into your SNR, 

  • 0 in reply to Alec Saebeler
    Intellectual 360 points

    Hello Alec,

    Thank you for your nice explanation. Would you please let me know how did you considered the output signal 0.05 V?

    If my transducer input capacitance in reality be different than 1nF, for example 700pF, then can we justify the difference between simulation and measuring? (also the exact resonating frequency of the transducer is 2.25 MHz).. In simulation if I change 1nF to 700pF, the result is as follows:

    C=1nF

    C=700pF

    Then there is a bigger peak in output noise, but it seems, the SNR should be even higher?!

    Thanks 

  • 0 in reply to Michael Steffes
    Intellectual 360 points

    Dear Michael, thank you for the insightful idea, I will try this, to see if SNR will be improved or not. Based on the simulation if we add this resistor the SNR value would be : 

    Without he internal R:  SNR=58.1 dB

    With the internal R: SNR=58.08 dB

    It seems the is not that much difference, but you mean it will have and effect in the real set up?

    Besides, I have doubt about the transducers capacitance, recently, I measured it and it seems it is around 700pF, do you think this can the reason of the low SNR? in comparison with my simulation? I did the simulation with 700pF and there is high peak noise:

    Can this be the root cause the problem?

  • 0 in reply to Behnam Madadnia
    TI__Genius 13911 points

    Hello Behnam,

    I selected 0.05V as a baseline for me to investigate the circuit.  I needed an output voltage to proceed with my calculations, and from your signal pulse it appeared to have a peak-to-peak magnitude of 50mV.

    Your selected Cin capacitor will have an effect on the noise gain of the system.  While Rf alone sets the transimpedance gain, the noise gain is set by both Cin and Rf terms.

    Best,

    Alec