Op Amp Calculations

I don't seem to have access to alt.binaries.schematics.electronic - perhaps someone can tell me how.

I do take your point Jim, that the T circuit is not "bad". Engineering is about making "how much" judgements. I don't think I wrote off the circuit, in fact I suggested using unequal resistances each side of the tap to minimise the penalty.

My point is - use the ratio of resistances at the opamp minus node to ESTIMATE what you are doing to noise, offset, drift and bandwidth. Applies for any opamp circuit, not just the T. Most of the time you have extra performance to burn.

The T feedback can be evil. Easy as pie to peel 6 or 10db off your equipment spec and make it worse than the competition.

The original post disturbed me because it advertised the "T" as a free lunch and seemingly equivalent to a feedback resistor.

Roger Lascelles

(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.

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

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

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

first order

schematic

"signal" gain

(left to right;

a series

shunt arm of

The 10090.7

effect on the

feedback as the

shunt arm

has no

feedback

arm in the

I made a transcription error here. The expression for noise gain of the multi path network should be 1 + R6'/R1 + R6'/R4', where R6' is the series arm in the equivalent delta network, and R4' is the 10090.7 ohm shunt arm in the equivalent delta network.

circuits, as

This sentence should read "Since R4' is about 10 times R1, we should expect..."

this gain at

only around

This should be "around 10%".

indicates to me

make R5

to 10000

circuit

This also should be "about 10%".

circuit.

I don't know what that means.

Yes, it assumes that the op-amp has infinite open-loop gain. Not a bad approximation at frequencies very much smaller than the unity-gain bandwidth.

You can 'prove' all sorts of things by changing a circuit. If you measure the signal voltage on the inverting input, you find, unless you are using the op-amp at an unwisely-high frequency, that the voltage there is TINY compared with the input voltage. Not 'most of the input signal'.

But considering it so low an impedance, that you call it just a divider, is a degenerate case... and certainly NOT the most useful one.

...Jim Thompson

In the "interesting" cases the "divider" impedance is within the same order of magnitude as the first feedback resistor.

...Jim Thompson

Hi John,

obviously your "no signal" is quite a bit larger than my "no signal"

thats a convenient approximation, but it is most surely an approximation.

to prove it, place a 10 Ohm resistor from the -ve input to 0V, and watch the circuit perform differently.

I blame Jiri Dostal.

Cheers Terry

I read in sci.electronics.design that The Phantom wrote (in ) about 'Op Amp Calculations', on Mon, 26 Sep 2005:

Instead of that, go back up the tread and see my ASCII art re-draw of the feedback as a simple potential divider across the output, with the feedback resistor taken from the tap. It's FAR easier to analyse.

I read in sci.electronics.design that Jim Thompson wrote (in ) about 'Op Amp Calculations', on Mon, 26 Sep 2005:

Would you care to enlarge on that? It's always a divider, for finite resistor values, so I don't see your point.

I guess you and I will have to disagree about what "...FAR easier..." means. You first give an expression which is only approximate and then to make it exact, you say: "replace R4 by R4 + (R2R3/(R2 + R3)". By the time this is done, the amount of algebra is not significantly (if any) less than a Y-Delta method:

Knowing that the shunt arms of the equivalent delta have no effect on the signal gain, we need only compute the series arm and use the old Rf/Ri gain formula for the two-resistor case, with Rf replaced by the expression for the delta series arm.

Thus, the series arm is given by (R3*R4 + R2*R3 + R2*R4)/R3. Since the input resistor is R1, we have the exact gain expression of Rf/Ri = (R3*R4 + R2*R3 + R2*R4)/(R1*R3).

I can't see that, as you said, "Gain = (R4/R1) x {(R2 + R3)}/R3, if R4 is very much larger than R3. If it isn't, replace R4 by R4 + (R2R3/(R2 + R3)."

is "...FAR easier..." than recognizing that the equivalent Rf is (R3*R4 + R2*R3

• R2*R4)/R3, which when divided by R1 gives the signal gain.

I remember Robert Pease in his column in Electronic Design magazine dealt with this topic:

Since he shows a circuit with no input resistor, but rather just an input current (this would be equivalent to an infinite input resistance), his comments on noise gain would have to be modified to apply to the circuit under consideration here.

means. You first

say:

of algebra is

signal gain,

arm.

input resistor

R2*R4)/(R1*R3).

very much

R2*R3 +

That's why in my discourse on the use of the Y-Delta transformation, I began by saying: " In what follows, I'll assume the capacitor is replaced with a short and only first order effects will be considered."

But for those who care, using Woodgate's reference designators, the signal gain with infinite Aol is: (R3*R4 + R2(R3 + R4))/(R1*R3) and if the amplifier gain is finite, the signal gain is:

Aol*(R3*R4 + R2*(R3 + R4)) -------------------------------------------- R1*(R2 + R3 + Aol*R3)+ R3*R4 + R2*(R3 + R4)

As one might expect, when trying to get a gain of 1000 out of the closed loop circuit, an open loop gain of 100 times that, or 100,000 would seem reasonable to reduce errors to the 1% level, and in fact that's what I get when I compute the error in the infinite gain expression compared to the more exact finite gain expression with an Aol of

The T-PI transformation is a natural here because you don't care about the two shunts arms most of the time and the poles/zeroes fall right out: View in a fixed-width font such as Courier.

- - - Z - - -

---[R1]--+--[R2]--- | [R3] | === |C | ---

R1R2+R1(R3+1/SC)+R2(R3+1/SC) Z= --------------------------- R3+1/SC

(R1R2+R1R3+R2R3) ---------------- SC R1 + R2 + 1 Z= (R1 + R2) ------------------------ R3CS + 1

(R12 + R3 )CS + 1 Z= (R1 + R2) ----------------- ;R12=R1||R2 R3CS + 1

(R1 + R2) (R12 + R3 )CS + 1 Vout/Vin=(-) --------- x ----------------- Rin R3CS + 1

Add the input shunt component:

---[R1]--+--[R2]--- | | [R3] Zs | === | |C | ---

(R12 + R3 )CS + 1 Zs= (R1 + R2) ----------------- ;R12=R1||R2 R2CS

R1 Rs= ( 1 + -- ) (R12 + R3 ) R2

C Cs= ---------- R1 ( 1 + -- ) R2

I can't imagine analysing it any other way !

Graham

Just take the source resistance of the divider and add it to the attached feedback R. It resolves very easily.

Graham

I read in sci.electronics.design that Jim Thompson wrote (in ) about 'Op Amp Calculations', on Mon, 26 Sep 2005:

In which case the 'bottom arm' is two resistors in parallel, and the 'first feedback resistor' is one resistor in series with two in parallel. Not rocket science, jut the normal 'loaded potential divider'.

Us old farts only understand loop and nodal analysis... short cuts in the head always wander off ;-)

...Jim Thompson

my "no signal" is zero.

indeed.

at 1% of the UGBW, OL gain is 100, far from infinite

We both disagree with Roger then.

I recently built about 50,000 of this circuit, with a feedback cap too (mathcad rather than mathematica, and a pencil to start with for the analysis), and 15 inputs thru 100k resistors. the effect of the 14 "grounded" resistors shifted the center frequency by about 10% - Aol was about 50. power consumption (and cost) constraints meant I couldnt use a faster opamp, so instead I stopped assuming and started calculating :)

Cheers Terry

What was the transfer function you were shooting for, and which amp?