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Martin Shepherd wrote:<br>
<blockquote type="cite"
cite="midPine.LNX.4.58.0401131408520.3750@goblin.caltech.edu">
<pre wrap="">On Tue, 13 Jan 2004, Brian Mason wrote:
</pre>
<blockquote type="cite">
<pre wrap="">...
ps, It would be useful (particularly for the detector circuit
discussion) to review Randy's latest design...:
<a class="moz-txt-link-freetext"
href="http://wwwlocal.gb.nrao.edu/%7Ebmason/gbt-dev/KaRxDetAmpRevH.bmp">http://wwwlocal.gb.nrao.edu/~bmason/gbt-dev/KaRxDetAmpRevH.bmp</a>
</pre>
</blockquote>
<pre wrap=""><!---->
I have updated my one-page calculation of the noise contribution of
the first stage amplifier in Randy's latest design. This can be found
at:
<a
class="moz-txt-link-freetext"
href="http://www.astro.caltech.edu/%7Emcs/GBT/preamp_noise.ps">http://www.astro.caltech.edu/~mcs/GBT/preamp_noise.ps</a>
As explained before, this calculation is based on the fact that in a
differential amplifier with high open-loop gain, negative feedback
ensures that the inverting and non-inverting inputs of the op-amp are
always at the same potential (unless the amplifier saturates), so all
noise voltages appearing at either input can be summed to a single
noise voltage that is common to both inputs. This sum has to be
performed in quadrature to get the RMS noise voltage.
For a signal applied to the non-inverting input of this amplifier,
this circuit has the standard non-inverting configuration, so the
noise voltage accumulated there is amplified by the non-inverting gain
of this amplifier, which is (1+Rf/Rm). Note that if you look at the
amplifier from the point of view of a signal appearing at the
inverting input of the amplifier (ie. not the input of the whole
amplifier circuit), the gain is still that of a non-inverting
amplifier, since the two inputs have the same potential. The noise
voltage would have to appear at the other side of the series resistor
to be amplified by the inverting gain of the amplifier, and this isn't
the case.
Martin
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href="http://listmgr.cv.nrao.edu/mailman/listinfo/gb-ccb">http://listmgr.cv.nrao.edu/mailman/listinfo/gb-ccb</a>
</pre>
</blockquote>
Martin,<br>
<br>
I'm afraid that your line of reasoning here is flawed...<br>
<br>
In an inverting amplifier configuration employing negative feedback,
the potential<br>
realized at the non-inverting input is brought about <u>strictly</u>
by those sources which <br>
appear in the non-inverting input branch. Please note that the
amplifier's <u>negative</u>
<br>
feedback path has <u>no</u> means by which to influence what happens
at the
OP AMP's<br>
non-inverting input. Therefore, any potential which appears within the
inverting<br>
input branch will have <u>no effect whatsoever</u> on the potential at
the non-inverting<br>
input.<br>
<br>
As it turns out, those potentials which appear anywhere within the
inverting input <br>
branch <u>actually don't influence the potential at </u><u>either
of the OP AMP's inputs</u>!<br>
Why? Because the OP AMP's high-gain output works back through the <u>negative</u>
<br>
feedback path, forcing the potential at the inverting input to always
be equal to that <br>
seen at the non-inverting input (as established <u>strictly</u> by
those potentials appearing<br>
within the non-inverting input branch). <br>
<br>
As a result, those potentials which appear <u>anywhere</u> within the
inverting input branch <br>
only serve to bring about a direct
change in the current flowing withing this branch;<br>
and, since the OP AMP's inverting input must exhibit arbitrarily high
impedance,<br>
this change in current is forced to flow through the feedback resistor
(Rf). This,<br>
in turn, results in those potentials appearing <u>anywhere within the
inverting input</u><br>
<u>branch</u> experiencing a gain of <b>Rf / Rin</b>, where Rin is
the <u>total</u>
impedance existing<br>
within the inverting input branch.<br>
<br>
On the other hand, those potentials existing anywhere within the
non-inverting input<br>
branch will appear at <u>both</u> of the OP AMP's inputs. Again,
this is brought about by<br>
the OP AMP's output working back through the <u>negative</u> feedback
path to force<br>
the potential at the inverting input to equal that established at the
non-inverting
input. <br>
And, since the potentials existing anywhere within the non-inverting
input
branch <br>
are "mirrored" at the inverting input, they experience a gain of <b>1
+
(Rf / Rin)</b>,<br>
where Rin is the <u>total</u> impedance existing within the inverting
input branch.<br>
<br>
Now let's examine the effect of the <u>placement</u> of a noise source
within
the inverting<br>
input branch of an inverting amplifier configuration employing negative
feedback...<br>
<br>
First, the potential realized at the non-inverting input is established
<u>strictly</u> by those<br>
potentials appearing within the non-inverting input branch.<br>
<br>
Second, the amplifier's negative feedback path forces the potential at
the inverting<br>
input to equal that established at the non-inverting input (whatever
that
might be).<br>
<br>
Third, any voltage source introduced <u>anywhere</u> within the
inverting input branch<br>
doesn't influence the potential realized at <u>either</u> input.
Instead, it brings about a<br>
change in the current flowing in the inverting input branch which is
then forced to<br>
flow through the feedback impedance (Rf). Thus, the gain experienced
by any<br>
voltage source introduced <u>anywhere</u> within the inverting input
branch
will be<br>
<b>Rf / Rin</b>, where Rin is the <u>total</u> impedance existing in
the
inverting input branch.<br>
<br>
Summarizing here (for our inverting amplifier configuration)...<br>
<br>
1) <u>Any</u> potential appearing <u>anywhere</u> within the
inverting
input branch will<br>
experience a gain of <b>Rf / Rin</b>, where Rin is the <u>total</u>
impedance existing<br>
within the inverting input branch.<br>
<br>
2) <u>Any</u> potential appearing <u>anywhere</u> within the
non-inverting
input branch<br>
will experience a gain of <b>1 + (Rf / Rin)</b>, where Rin is
the <u>total</u> impedance<br>
existing within the inverting input branch.<br>
<br>
Therefore, you should adjust your noise formulations accordingly,
applying<br>
"inverting gain" ( <b>Rf /Rin</b> ) to those potentials appearing
anywhere within the<br>
inverting input branch, and applying "non-inverting gain" ( <b>1 + (Rf
/ Rin)</b> ) to <br>
those potentials appearing anywhere within the non-inverting input
branch.<br>
<br>
Also keep in mind that, since all of the noise sources being
investigated are <br>
uncorrelated, their combined effect would properly be obtained using
an <br>
RSS
formulation.<br>
<br>
<br>
Randy<br>
<br>
<br>
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