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AMC7812EVM-PDK: Hi, I have a question about AMC7812 EVM.

Yunsik Chung
Intellectual 1150 points
Part Number: AMC7812EVM-PDK

Hi,

I did experiment to check if the EVM converts data properly by measuring DAC output and ADC input.

When i checked the DAC output after modified program DAC voltage, i could get the voltage what i put.

However, when i checked the program ADC voltage after supplying voltage to ADC input, I could not get the voltage i supplied.

For example, when i supplied 4V, I could only get 3.1V. When i supplied 1V, I could get 1.4V. (Default voltages are about 2V)

Only in ADC_input 2,3, i could get the exact voltage i put.

I think this is because there are buffers in front of ADC_input 2,3.

But, I wonder why the buffer makes this difference.

Also, I wonder how i can get the exact voltage i supplied except ADC_input 2,3.

Also, I wonder why there is default voltage around 2V when there is no voltage supplied?

thanks,

  • 0
    TI__Guru 52663 points

    Hi Yunsik,

    This features a SAR type ADC, which does have a sample cap that needs to be charged.  If your source is high-impedance and has low current output capability, then it is possible that the sample cap is not being charged in the sample window.  A few things can be done to improve this, such as reducing any series resistance, adding a cap to the input pin of the ADC, or buffering the input.  You could also reduce the sample rate, which increases the acquisition time. 

    The default voltage can vary as the input is 'floating' or relatively high impedance, so it usually just floats to approximately mid-supply.

    Thanks,

    Paul

  • 0 in reply to Paul_Frost
    TI__Intellectual 1150 points

    Hi Paul,

    When I changed the sample rate from 500kSPS to 62.5kSPS, I could get the result i want.

    I have one more question about source impedance.

    The reason that high-impedance source limits the current is that high-impedance decreases the current because of ohm's law?

    Also, could you let me know how to calculate proper sample rate when source impedance is around 16kohm in this EVM?

    Thanks, 

    Yun

  • 0 in reply to Yunsik Chung
    TI__Guru 52663 points

    Hi Yun,

    I think it is better to consider the RC filter that is created with the input impedance and the sampling capacitor.  In the datasheet, we specify the ADC input capacitance is 73pF.  The acquisition time depends on the sample rate.

    Recall that τ (tau) is τ= RC.  If the R = 16kΩ and C=73pF, then your τ = 1.168µs.  

    Now consider that the RC charging function is V(t) = V0(1-e^(-t/τ)).  For example, if your input was 4V, then you would have: V(t) = 4V(1-e^(t/1.168µs))

    At the fastest sample rate there is only 0.375µs to acquire the input voltage, so V(0.375µs)= 4V(1-e^(0.375/1.168µs)) = ~1.1V.

    At 250kSPS, the output would be 3.476V.

    At 125kSPS, the output would be 3.983V.

    And at 62.5kSPS, the output would be 3.9998V, so very close.  Reducing the input impedance really helps with SAR topologies :)

    Thanks,

    Paul

  • 0 in reply to Paul_Frost
    TI__Intellectual 1150 points

    Dear Paul,

    Thank you so much!

    you really helped me understanding this topic:)

    I have one more question.

    Here in ADC cookbook when they calculate the acquisition time they multiplied K. They said for 14-bit, K is 11.

    What is this K value? I could not find in Analog engineer's pocket reference.

    Do i also have to multiply k in my previous question?

    Thank you so much for your kind response,

    Yun

  • 0 in reply to Yunsik Chung
    TI__Guru 52663 points

    Hi Yun,

    So K is the number of τ you must wait to get within 1LSB of your signal.  There is a typo in the cookbook as K = 11 for 16-bit devices, not 14 bit.  

    Check out page 151: https://www.ti.com/seclit/ml/slyw038c/slyw038c.pdf

    K = ln(2^N) where N is the number of bits of your ADC.

    If we look at our previous equation V(t) = V0(1-e^(-t/τ)), how does K factor into it?  Well really K is our t/τ value.

    Let's say we have a fullscale input in the device, ideally code 4096 (not possible, but will illustrate this well).  Instead of V0 being in units of volts, let's just use LSB.

    Code(t) = 4096(1-e^(-t/τ)), but recall that t/τ = K = ln(2^N) = ln(2^12)

    so Code(t) = 4096(1-e^(-ln(2^12)))

    We evaluate that as Code(t) = 4095!

    So with K = ln(2^12) ≈ 8.318, we can say that we need 8.318×τ to for our input to settle to 1LSB.   

    Recall that τ (tau) is τ= RC.  If the R = 16kΩ and C=73pF, then your τ = 1.168µs.  

    With your 16kΩ impedance, you need 1.168µs × 8.318 to settle or 9.715µs!

    So TACQ ≥ 9.715µs for 1LSB accuracy.

    Thanks,

    Paul

  • 0 in reply to Paul_Frost
    TI__Intellectual 1150 points

    Dear Paul,

    I could fully understand it!

    Thank you for your support:)

    Thanks,

    Yun