From: Win Hill
Subject: Re: High impedance low noise
Date: 10 Sep 2002 15:53:13 -0700
Organization: Rowland Institute
References: <3D760B64.CC303206@scs.uiuc.edu> <3D763949.4B6D795E@scs.uiuc.edu> <email@example.com>
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>Win Hill wrote...
>> Hi Jim. On Wednesday, September 4, 2002, you wrote:
>>> Win wrote,
>>>> In one opamp, yes, but that's not necessary a good approach. When
>>>> measuring voltage, the reason for considering an FET opamp operated
>>>> as unity-gain follower is to drive a second high-bandwidth low-Zin
>>>> amplifier stage that uses conventional components, such as current-
>>>> feedback opamps that have low noise compared to the FET, obviating
>>>> the need for the FET stage to have any gain. Well, a gain of 2x in
>>>> the FET stage would be nice, and has over 1MHz bandwidth with any of
>>>> the parts below. The follow-on G = 50 low-noise 1MHz stage is easy.
>>> Well, this was exactly my thinking too, but I keep hearing from others
>>> that "all (or most of) the gain should be in the first stage, for max
>>> sig/noise. We were pondering the trick of multiple fet op amp inputs
>>> all paralleled, then summed. I read somewhere that this is a trick for
>>> cancelling random noise. I think another engineer here tried that a
>>> while back and it did not work well. We are all having a meeting
>>> tomorrow, and I should know more then. I just don't have enough
>>> details yet.
>> Hi Jim, best wishes in your meeting. To explain the issue of one stage
>> vs. several stages. The situation is accurately understood by evaluating
>> the effect of each stage on the effective input noise. It's useful to
>> use input-noise-density squared, for a simple easy-to-understand formula.
>> The formula below is for a three-stage amplifier. Please rewrite it on
>> paper, to see it easier without the poor ASCII-character limitations.
>> v_n(in)^2 = v_n1^2 + (v_n2/G1)^2 + (v_n3/G1G2)^2
>> We can see that if the gain G1 of the 1st-stage in a 2-stage amplifier
>> is 1, and both stages have the same noise level v_n, the effective input
>> noise will be v_n(in)^2 = v_n1^2 + v_n2^2 = 2 * v_n^2, which means that
>> V_n(in) is sqrt 2 = 1.4 times the noise of the 1st stage alone. That's
>> where the common wisdom comes from, assuming that the 1st stage uses the
>> lowest-noise amplifier, and saying that G1 = 1 is a bad thing.
>> However, we know that high-Z JFET amplifiers are much noisier than good
>> low-Z wideband amplifiers. For example, an AD8011 has 2nV of noise, as
>> compared to a JFET opamp with say 6nV of noise. Using the formula, we
>> see that the overall noise is sqrt(6^2 + 2^2) = sqrt 40 = 6.32nV. This
>> is so close to 6nV we can ignore the extra noise from the 2nd stage.
>> If the 1st stage has a gain of 2x, the effect is more striking, 6nV vs
>> 6.08nV. Clearly the old rubric is wrong under these conditions.
>> . f_T e_n Cin max Vcc
>> . (MHz) (nV) (pF) (+/-V)
>> . ----- ----- ---- ------
>> AD743 4.5 2.9 18 18
>> LT1792 5.6 4.2 14 20
>> AD711 4 16 5.5 18
>> LT1057 5 13 4 20
>> OPA655 240 6 1.0 5.5
>> AD8011 * 2 low-Z 6.3 current-feedback, 57MHz at G = 10.
>> - Win
> Also don't forget the TI winners
> f_T e_n Cin max Vcc
> (MHz) (nV) (pF) (+/-V)
> ----- ----- ---- ------
> THS4021 350 1.5 1.5 +/-16 stable G>10
> THS4031 100 1.6 1.5 +/-16 stable G>2
> THS4011 290 7.5 1.2 +/-16
Indeed those are nice amplifers, but they aren't JFET input
oamps, and they have an input current of up to 6uA, which
would be a killer for Jim, I suspect!!