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ADA4895-1 output noise

I would like a sanity check on my noise calculation and the actual result and any other comments.

Two ADA4895-1 op-amps in series each with a non-inverting gain of 10x. Rf = 226 ohms. Rg = 25.1 ohms, Rs = 51 ohms.

They are followed by an AD8021 with a non-inverting gain of 1. That is followed by a single pole R-C filter with 249 ohms and 18pF into an AD9240 ADC. The R-C filter frequency should be 35MHz. 55MHz was used due to the 1.57 single pole factor.

     From the ADA4895-1 datasheet:

236 MHz, −3 dB bandwidth (G = +10)

voltage noise is 1nV / SQRT(Hz)

current noise is 1.6pA / SQRT(Hz)

With the input grounded what is the predicted amplifier output noise into the AD9240? For my first approximation I ignored the AD8021 and assumed the first stage ADA4895-1 would dominate.

Scott

  • FormerMember
    0 FormerMember

Scott,

I just noticed you haven't gotten a reply since 5/15, so we owe you something.

I'm lazy, so I  won't calculate the whole thing, but remember Friis' equation:

Friis formulas for noise - Wikipedia, the free encyclopedia

Even though they are talking about microwaves, the same principle applies.

If you have enough gain in the first stage, the other stages matter very little.

So to a first approximation, it's simply 10nV/rt-Hz integrated over the bw.

Also see:

http://www.analog.com/media/en/training-seminars/tutorials/MT-050.pdf

Harry

  • Harry,

    Sorry, I didn't expect someone to do the full calculation. Was just looking for a sanity check.

    The actual results measured using the AD9240 are below. The application is sensitive to peak voltages so that is shown along with the RMS conversion using 6.6, 99.9%. AD9240 is at the left of the list.

    No buffer, AD9240 inputs shorted = 1.53mV pk-pk = 231uV RMS

    AD8021 buffer = 2.14mV pk-pk = 324uV RMS

    AD8021 buffer + AD8021 at x2.5, x5 and x10 all = 2.75mV pk-pk = 416uV RMS

    AD8021 buffer + x5 AD8021 + x10 ADA4895-1 = 4.58mV pk-pk = 694uV RMS

    AD8021 buffer + x10 AD8021 + x10 ADA4895-1 = 9.16mV pk-pk = 1.38mV RMS

    The x10 noise from an AD8021 is the same as the noise from a 10x ADA4895-1 as the only amp.

    The result is that gains over 50x do not help as the noise from 50x to 100x doubled.

    Scott

  • Harry,

    Also, Why 10nV/rt-Hz rather than 11nV/rt-Hz since the noise gain is x11 and the amp noise is 1nV?

    Scott

  • FormerMember
    0 FormerMember

    Sloppy thinking on my part.  Yes, the NG =11 and you are correct.

  • Harry,

    The Friis formula says to use the power gain. When the output of an op-amp is driving the high impedance input of another op-amp the resistance is near infinite. How does power gain relate to voltage gain in that situation?

    Scott

  • Harry,

    I tried to make this clear. Hopefully it sums up why I am baffled.

    All bit noise is peak to peak into the AD9240, +5V full scale. Each histogram has 2^32 samples (>4 billion). The highest and lowest values typically have less than 20 counts (they are very rare) but they matter in our application.

    The board has several op-amps that can be switched into and out of the signal path. An ADA489595-1 configured for +10x produces 9 bits of noise. An AD8021 configured for +10x also produces 9 bits of noise..

    Those same amplifiers cascaded with the ADA4895-1 first produce 28 bits of noise. If the noise from the first stage is supposed to dominate why do the +10x amps cascaded produce so much more noise than either do individually?

    The graph previously posted has three grey dots. They are AD8021s with gains of +2.5x, +5x and +10x. Their noise levels are all the same individually, 9 bits.

    Two AD8021s cascaded, the first set to +5x the second to +2.5x. produce 11 bits of noise.

    Cascaded the AD8021 +10x and the +2.5x produce 12 bits of noise

    Cascaded the AD8021 +10x and +5x produce 19 bits of noise.

    Cascaded the ADA4895-1 +10x with the AD8021 +5x produces 15 bits of noise.

    Scott

  • FormerMember
    0 FormerMember

    Scott,

      Sorry.  I was using it as an example/analogy.    The general concept of the Friis formula is later stages

    appear referenced to input by gains.  For RF, they use power/noise figure/noise factor.

    In the case of a chain of op amps, you use volts.  So if stage 2 has an input referred noise of 10 nV/rt-Hz,

    and the first stage has a gain of 5, then the second stage will contribute 2nV/rt-Hz referred to the input

    of stage one.  If stage one had an input noise of 10 nV/rt-Hz also, you would RSS the two and get

    10^2 + 2^2 = 104 so sq-rt of 104 is 10.2, IOW, the second stage hardly matters.   If you have a chain

    of four op amps, you don't want to use a low noise quad;  too much money and power.  You can use

    noisier op amps for stages 3 and 4.

    Harry

  • FormerMember
    0 FormerMember

    Scott,

    I'll look at it in detail Monday, but a couple of quick comments.

    Can you do a histogram on several thousand samples with the AD9240 inputs shorted?

    Then do another with only the 8021 driving the ADC and the 8021 input AC shorted.

    See fig 13 on d.s.

    Once you get past 12 bits, you really should do differential input drive rather than single input;

    gives better accuracy and rejects common mode noise.

    You talked about switching op amps in and out;  how are you doing it?   You could have some

    undesired coupling if you have traces running everywhere......

    Op amps over 50-100 MHz gbw can have peaking at higher frequencies and layout is very important;

    You could have something that is trying to oscillate or is oscillating at 100-500 MHz, so that would

    appear as noise.

    Harry

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