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Questions about Ti precision lab-SAR-Quiz: Driving a SAR ADC with a Fully Differential Amplifier

Other Parts Discussed in Thread: THS4551

1. Why the linear input range refers to only the input common mode voltage range? Whether the differential signal will also affect the input range? Why we did not consider that into the input range analysis

2. For the "Quiz: Driving a SAR ADC with a Fully Differential Amplifier", question 2 Page 9 of the PPT.W

hen calculate the amplifier output liner range. How was the lower end calcuated as -2.4V, given the power rail -2.5V to 2.5V and open-loop gain 0. <Vo< (V+) -0.1V. Whether it should be 0.1V? or this 0.1 means 0.1V above the lower supply rail? 

3. On the Page 6 of Quiz, there is a type as pointed, I think it should be -3.896V not -3.996V

Thanks,

B.F

  • B.F., Thanks for your questions. The answers are below in RED. Don’t hesitate to ask further questions.

    1. Why the linear input range refers to only the input common mode voltage range? Whether the differential signal will also affect the input range? Why we did not consider that into the input range analysis.

    For the FDA, we are looking at both the output swing limitation and the common mode limitation. If the amplifier is inside of both of these limits, the system differential input signal will also be acceptable.   If you look through the data sheet specification for the THS4551 you will notice that the common mode and output swing are specified limits, but the “differential input” is only specified in the “Absolute Maximum Ratings” (+/-1V). The differential input in the absolute maximum specifications is the differential input directly on the amplifiers inputs. Under normal conditions the differential input at the input pins of the amplifier are the amplifiers offset (in microvolts). I believe that when you are referring to differential input you are referring to the differential input applied to the input of the system (including the resistors), not directly on the pins of the amplifier.

    2. For the "Quiz: Driving a SAR ADC with a Fully Differential Amplifier", question 2 Page 9 of the PPT.W

    When calculate the amplifier output liner range. How was the lower end calculated as -2.4V, given the power rail -2.5V to 2.5V and open-loop gain 0. <Vo< (V+) -0.1V. Whether it should be 0.1V? or this 0.1 means 0.1V above the lower supply rail? 

    This is a very good question.  The data sheet specification means 0.1V above the lower supply rail. The data sheet specification was written with a single supply configuration in mind. It would have been better if the specification was written (V-)+0.1V < Vo < (V+)-0.1V. You will see that the output swing and common mode specifications sometimes use this “single supply” type specification, but it will always be true for common mode and output swing that the voltage is relative to the negative supply. Sorry for the confusion.  I will not edit this question, as this is how the actual data sheet specifies linear range.

    1. On the Page 6 of Quiz, there is a type as pointed, I think it should be -3.896V not -3.996V

    You are correct. Thanks for pointing this out. I will update this document.

  • Hi Art, 

    Thank you very much for making the excellent educational training series and responding to my questions.  I do have questions regarding the input range. 

    1. When we discussing the input linear range of the opamp, we actually only consider the common mode voltage range, such as in first attachment, and also in your opamp video the second attachment.  We defined the opamp input range as the common mode voltage range see attachment 2. All this discussion we did not consider the actual signal applied to this common mode to be amplified. Is this because they are small signals? 

    2. But if we have the situation that the input signal  is large ("the signal " refers to the signal applied on the common mode voltage) such as the case in attachment 3, the FDA input signal is the same order as the common mode 2.5V, if we want to calculate input linear range, whether we should consider the range of the "signal" + common mode voltage? In that case, the input voltage range should not be just the input common mode voltage as shown in the attachment 2?

    PS: The attachment 1 is from  TI precision lab SAR adc  lecture 2-2 determine ADC linear range using opamp P3

    The attachment 2 is from TI precision lab opamp lecture input output limitation -1  P4

    The attachment 3 is from TI precision lab SAR lecture 2-5 determine ADC linear range using FDA P7.

    Thank you very much,

    B. Fan.

  • Thanks for your great questions.  See my comments in red.

    1. When we discussing the input linear range of the opamp, we actually only consider the common mode voltage range, such as in first attachment, and also in your opamp video the second attachment.  We defined the opamp input range as the common mode voltage range see attachment 2. All this discussion we did not consider the actual signal applied to this common mode to be amplified. Is this because they are small signals? 

    The goal here is to find the maximum input range for linear operation. The actual signals can be any voltage equal to or less than this maximum range. In this example the common mode range is -0.1V to 3.4V. That means that the input signal can be anywhere inside of this range, so technicaly, you could apply a peak-to-peak waveform from -0.1V to 3.4V.

    1. But if we have the situation that the input signal is large ("the signal " refers to the signal applied on the common mode voltage) such as the case in attachment 3, the FDA input signal is the same order as the common mode 2.5V, if we want to calculate input linear range, whether we should consider the range of the "signal" + common mode voltage? In that case, the input voltage range should not be just the input common mode voltage as shown in the attachment 2?

    You have to be careful here as this section was written about op amps not FDAs.  The principle is similar, but in the case of FDAs the input signal is applied to a set of input resistors.  The actual common mode voltage on the FDA is the voltage on the input pins of the amplifier.  The feedback network, input signal, and Vocm signal will all impact the common mode on the input pins.  The best way to verify the common mode input signals on an FDA is to do a simulation and probe the common mode pins.  Here is a simulation for an FDA: 5775.look at common mode on FDA.TSC.  Notice in this simulation that the common mode is held at a near constant value of 2.5V.  This is because the FDA acts like two combined inverting op amp circuits. 

    If you look through the op amp material for common mode you will notice that inverting amplifiers do not normally have issues with common mode limitations. This is because the non-inverting pin is held contestant, and by the virtual short the inverting pin is also held constant. A similar phenomena is true with FDA amplifiers as they act like two inverting amplifiers combined.

    .