This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

THS4541: Question about OPA857+THS4541+LTC2521

Other Parts Discussed in Thread: THS4541, OPA857

none

  • Hi Zhongya Wang,

    Using a TINA simulation to check these things before fabrication is an excellent choice, solving these problems in simulation is always a good move when you can. Now, to make sure I understand your question correctly, you seem to be having unexpected or undesirable linearity within your circuit, as well as concerns with matching your ADC to the rest of your circuit.  I am working on some ideas for your circuit behavior at the moment, but I think I would like to try and simulate the design myself.

    Are there any other parameters, perhaps the exact simulation conditions in TINA (like source settings, etc) I need to be able to replicate your simulation?  The closer I can approximate and see your design behave on my screen, the sooner I can advise you further.  Even parameters such as the values used to choose the ADC could be helpful.  I will link an interesting and useful article for you, in case some context for this type of system is all you need: https://www.planetanalog.com/author.asp?section_id=483&doc_id=565005

    Please let me know if this is not the kind of help you are looking for, or if I can be of any more assistance.

    Best Regards,

    Alec

  • Hi Zhongya,

    Because the OUTN will always be equal to the maximum TIA output voltage, the FDA will only see negative differential input voltage as current increases and TIA output voltage decreases. The FDA gain is positive, so the output is also only negative. In order to use the full bipolar (±) differential output, you must create a bipolar (±) differential input. To do so, the FDA input reference node, now OUTN, should be set to the TIA output voltage midpoint (1.215V). That way, 0 current gives you the maximum, positive FDA output at the ADC input, not 0V. To get the 1.215V, (174Ω+25Ω) reference voltage, you can replace OUTN and R36 with a Thevinin equivalant resistor divider of the FDA's Rg (174Ω+25Ω). Here is my cookbook on this matter. The resistor divider values should be 315Ω and 540Ω in this case (F=1.215V/3.3V, E=174Ω+25Ω=199Ω).

    If you prefer not to isolate the FDA reference from the TIA, you can alternatively put the OUTN through a resistor divider, followed by an increased Rg. These values would be (25Ω), 49Ω to ground at OUTN, and R36=182Ω. Note that it will consume higher power to accommodate the low 25Ω internal resistor.

     Best regards,

    Sean

  • Hi Alec;

    Thank you for your kindly answer. Atteched is my simply simulation file. My question mainly focus on how to match the ADC's full scale. I dont know how to modify the circuit. Thanks again.signal.TSC

  • Hi Sean,

    Thank you very much.Indeed as you said,the output is also only negative. so the input reference node of the FDA THS4541 should be set correctly.

    Your attched file is very helpful,but I dont know why the midpoint here is 1.215V as you said.In addition, the TIA is a differential output, which is not the same as your example. Do you mean that because the TIA OPA857's OUTN is always 1.81V, it  is equivalent to single-ended input to FDA THS4541's INP?

    Look forward to your reply.

  • It is often the case a photodetector is unipolar, 

    To use the full differential input range of the ADC you would want to DC bias the FDA output nearly full scale in the other direction from what the signal will be generating. That needs to be a current source into the summing junction on the non signal side of the FDA. That current source could be as simple as a resistor to a reference voltage (then match that noise gain with a resistor to ground on the signal side) or a real current source constructed in any one of several ways. You eventually care about the noise and Noise gain impact of that offsetting current source. You don't need precision probably, but here is a good one - 

    https://e2e.ti.com/support/amplifiers/f/14/t/796855?tisearch=e2e-quicksearch&keymatch=precision%20current%20source

    Not sure I got the polarity right here, but where I show it here. Actually, I notice now the outputs are crossed, that current source needs to be on the other side.