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Hi,

As you know, the maximum throughput rate of AD7091R-8 is 1Msps.In the condition of all 8 channels enabled by internal sequencer, convert from channel 0 to channel 7 sequently, what’t the actual throughput rate? Is it still 1Msps or down to 1Msps/8=125Ksps?

 

Below is different connection of AD7091R-8, are there any differences between their throughput rate?

Thanks!



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[edited by: @skowalik at 11:01 PM (GMT 0) on 26 May 2020]
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    •  Analog Employees 
    •  Super User 
    on May 26, 2020 10:59 PM

    Jerry,

    In the condition of all 8 channels enabled by internal sequencer, convert from channel 0 to channel 7 sequently, what’t the actual throughput rate? Is it still 1Msps or down to 1Msps/8=125Ksps?

    To answer your first question the throughput rate per channel will be reduced to 125 KSPS as the repetition period will now be 8 us.

    With regards to your second question regarding the connection affecting the throughtput rate.  

    The simplest answer is in most cases it will not, however as with anything there are always corner cases that need to be considered.  Obviously one would like to always use the mid-amp topology (Single amplifier) as it saves component count and power.  This is typically fine for frequency signal content where the time constant that is formed by the on-resistance of the mux switch and the parasitic capacitance of the MUX_OUT node (input to amplifier) allows for any signal errors to settle out prior to conversion.  Also you can only use this topology when the signal swing of interest is inside the rails of the data converter otherwise we risk corrupting the sampled data on the effected channel as well as adjacent channels due to ESD diodes turning on and charge loss occuring as a result.

    In cases where the input signal must first be attenuated or filtered it may in fact be advantageous to use the second example with an amplifier per channel to accomplish the signal chain response you want while maximizing throughput.  

    Hope that helps.

    Sean

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  • 0
    •  Analog Employees 
    •  Super User 
    on May 26, 2020 10:59 PM

    Jerry,

    In the condition of all 8 channels enabled by internal sequencer, convert from channel 0 to channel 7 sequently, what’t the actual throughput rate? Is it still 1Msps or down to 1Msps/8=125Ksps?

    To answer your first question the throughput rate per channel will be reduced to 125 KSPS as the repetition period will now be 8 us.

    With regards to your second question regarding the connection affecting the throughtput rate.  

    The simplest answer is in most cases it will not, however as with anything there are always corner cases that need to be considered.  Obviously one would like to always use the mid-amp topology (Single amplifier) as it saves component count and power.  This is typically fine for frequency signal content where the time constant that is formed by the on-resistance of the mux switch and the parasitic capacitance of the MUX_OUT node (input to amplifier) allows for any signal errors to settle out prior to conversion.  Also you can only use this topology when the signal swing of interest is inside the rails of the data converter otherwise we risk corrupting the sampled data on the effected channel as well as adjacent channels due to ESD diodes turning on and charge loss occuring as a result.

    In cases where the input signal must first be attenuated or filtered it may in fact be advantageous to use the second example with an amplifier per channel to accomplish the signal chain response you want while maximizing throughput.  

    Hope that helps.

    Sean

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