A lot of years ago (38 to be exact) a wise old "rule of thumb" engineer
taught me a slick way of getting maximum gain out of an opamp without
resorting to very high or very low values of resistors.
As we all know, for an inverting opamp, the gain is given simply by Rf/Ri,
where Rf is the feedback resistor from output to inverting input and Ri is
the resistor between signal and inverting input. The DC level of the output
may be set anywhere you choose by an appropriate bias level on the
noninverting input. For AC amplifiers from a single supply, this is
generally Vcc/2 with capacitive coupling between Ri and the signal.
However, for very large AC gains, either Rf must be rather large or Ri
rather small. Rf being rather large makes the input voltage/current errors
become significant as regards quiescent DC output point and Ri being rather
small requires large capacitors for coupling and loading errors from the
signal source.
So, sez old wily rule-of-thumb, just break Rf into two reasonable sized
equal value resistors equal to Rf/2 and run them in series from output back
to (-) input. And, from the midpoint tap on these two resistors run a
series RC circuit to ground. Bingo, the AC gain improves greatly.
And guess what, it works. How do I calculate the R in the series RC circuit
I asks old wily. The answer comes back "Tweak it until you get the gain you
want." (Assume that C can be made appropriately large to get the
low-frequency gain you want.)
I haven't used that trick in an awfully long time, but I've got an
application that needs it. And, if I want to use Diddle's constant in a
simulation program I can fool around (ahem, heuristically experiment) to get
the gain I need.
However, I can't convince myself that I can mathematically come up with the
resistor value. I have googled the problem and come up short. Anybody got
a pointer to a URL that goes through the math of how this configuration
works? And what I'm doing to my phase margin?
Jim
(Refer to Jim Thompson's schematic in alt.binaries.schematics.electronic)
In what follows, I'll assume the capacitor is replaced with a short and only first order
effects will be considered.
An easy way to see what is happening in the "T" (multi path) network feedback
arrangement is to transform the "T" to a "Pi" network.
http://encyclopedia.thefreedictionary.com/Y-delta+transform
First, consider what happens in an inverting amplifier of the sort in Jim's schematic
when you connect a resistor from the "-" input of the amp to ground. The "signal" gain
remains unchanged, but the "noise gain" of the circuit changes.
Now apply the well known "Y-Delta" transformation. Jim shows a "T" with (left to right;
input to output side) a series arm of 9000 ohms, a shunt arm of 1000 ohms, and a series
arm of 99200 ohms. This is equivalent to a pi network (left to right) with a shunt arm of
10090.7 ohms, a series arm of 1.001 megohms, and a shunt arm of 111222 ohms. The 10090.7
ohm shunt arm on the left goes from the "-" input to ground and so has no effect on the
signal gain. The 1.001 megohm series arm provides essentially the same feedback as the
1.000 megohm resistor in the two resistor amplifier circuit. The 111222 ohm shunt arm
from output to ground is simply an additional (small) load on the amplifier and has no
effect on circuit performance.
The noise gain in the two-resistor circuit is 1 + R3/R1; in the multi path feedback
circuit, the noise gain is 1 + R3/R1 + R4'/R1, where R4' is the 10090.7 shunt arm in the
equivalent delta network. Since R4' is about .01 times the value of R3, we should expect
no easily observable low-frequency difference in the performance of the two circuits, as
Jim's simulation shows. The noise gain is also the gain applied to the op-amp offset
voltage, and the capacitor in the multi path circuit prevents the increase in this gain at
DC. But in this particular multi path circuit, the increase in noise gain is only around
1%, so we could omit the capacitor.
This method (using the Y-Delta transformation) of analyzing the circuit indicates to me
that Jim's assertion:
"...the 1:1 case has no useful value toward improving (lowering)
the total impedance in the feedback loop."
is incorrect. Consider the following:
In Jim's multi path circuit, make R4 9803.92 ohms, make R6 98.0392 ohms, and make R5
9803.92 ohms. Now the Pi equivalent network has the two shunt resistors equal to 10000
ohms, and a series arm equal to 1.000 megohms, which gives a (low-frequency) circuit
performance identical to the two-resistor circuit, except for about 1% higher noise gain.
The total impedance in the feedback loop is much lower than the two-resistor circuit.
