Op Amp Calculations

[Jim Thompson]

I read in sci.electronics.design that Jim Thompson


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

 

[John Woodgate]

I read in sci.electronics.design that The Phantom <[email protected]>

I would agree, we need to be "far, far below UGBW" to get a closed loop
gain of 1000 with a fairly small error from neglecting the shunt arm on
the minus input.

I would really be reluctant to try to get a gain of 1000 out of one
op-amp at today's prices. I suppose that sometimes there is no choice,
but I think that's very rare. 31.62 is your friend.

 

[Terry Given]

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

a summing band-pass (ish) filter. 40 x TLV274.

I didnt want to AC-couple the inputs (that would have cost me 240
capacitors) so I used the bridged-T feedback network with an RC shunt to
give a DC gain of about 1/16 - any DC is basically common-mode, and the
next stage was AC coupled. 3 Rs and 2 Cs was a whole lot cheaper than an
RLC. But 100k/14 = 7k in parallel with the -ve shunt arm, enough to move
Fc 10% or so.

SPICE clearly showed it, so I went back and re-did my opamp analysis
using Dostals approach (originally I did it using the Woodgate
approximation), and voila - out popped the same answer. Mr HP3577 also
agreed with spice and mathcad. Dostals method also allowed me to
directly calculate the phase margin. Since then, I have analysed all
opamp circuits thusly - but I use the Woodgate approach with pencil &
paper as a bullshit detector

Cheers
Terry

 

[Jim Thompson]

[snip]

Dostals method also allowed me to
directly calculate the phase margin. [snip]
Cheers
Terry

What IS Dostal's method? The Loop Gain & Phase analyser on my website
is based on R.D. Middlebrook's laboratory technique, and is VERY
accurate, since the loop is never actually broken.

...Jim Thompson

 

[Terry Given]

[snip]

Dostals method also allowed me to
directly calculate the phase margin.
[snip]

Cheers
Terry

What IS Dostal's method? The Loop Gain & Phase analyser on my website
is based on R.D. Middlebrook's laboratory technique, and is VERY
accurate, since the loop is never actually broken.

...Jim Thompson

Hi Jim,

its the same one used by Jerald Graeme.

Hideal(s) = -a(s)/B(s)

a(s) = feedforward factor = signal at opamp -ve input when output grounded

B(s) = feedback factor = signal at opamp -ve input when input grounded

Hactual(s) = Hideal(s)
-------------
1+Aol(s)/B(s)


I got Dostals book about 12 years ago, but until last year never used
this method, as the -ve input = 0V method worked well enough. Its only
when I started really pushing an opamp that it became necessary, mostly
because I wasnt happy with simply twiddling component values in SPICE.

I havent yet managed to get your SPICE trick to work in Simetrix

but I did read RDMs paper.

Cheers
Terry

 

[The Phantom]

a summing band-pass (ish) filter. 40 x TLV274.

I didnt want to AC-couple the inputs (that would have cost me 240
capacitors) so I used the bridged-T feedback network with an RC shunt to
give a DC gain of about 1/16 - any DC is basically common-mode, and the
next stage was AC coupled. 3 Rs and 2 Cs was a whole lot cheaper than an
RLC. But 100k/14 = 7k in parallel with the -ve shunt arm, enough to move
Fc 10% or so.

SPICE clearly showed it, so I went back and re-did my opamp analysis
using Dostals approach

On what page(s) in Dostal is this approach described?

 

[The Phantom]

[snip]

Dostals method also allowed me to
directly calculate the phase margin.
[snip]

Cheers
Terry

What IS Dostal's method? The Loop Gain & Phase analyser on my website
is based on R.D. Middlebrook's laboratory technique, and is VERY
accurate, since the loop is never actually broken.

...Jim Thompson

Hi Jim,

its the same one used by Jerald Graeme.

Hideal(s) = -a(s)/B(s)

a(s) = feedforward factor = signal at opamp -ve input when output grounded

B(s) = feedback factor = signal at opamp -ve input when input grounded

Hactual(s) = Hideal(s)
-------------
1+Aol(s)/B(s)


I got Dostals book about 12 years ago, but until last year never used
this method, as the -ve input = 0V method worked well enough. Its only
when I started really pushing an opamp that it became necessary, mostly
because I wasnt happy with simply twiddling component values in SPICE.

I havent yet managed to get your SPICE trick to work in Simetrix

but I did read RDMs paper.

There is a 2001 paper describing extensions of Middlebrook's method. It's somewhat
obscure since it was published in one of the IEEE magazines rather than a regular journal.
I've posted it over on ABSE with the subject "Small Signal Stability".

 

[Terry Given]

[snip]


Dostals method also allowed me to
directly calculate the phase margin.

[snip]


Cheers
Terry


What IS Dostal's method? The Loop Gain & Phase analyser on my website
is based on R.D. Middlebrook's laboratory technique, and is VERY
accurate, since the loop is never actually broken.

...Jim Thompson

Hi Jim,

its the same one used by Jerald Graeme.

Hideal(s) = -a(s)/B(s)

a(s) = feedforward factor = signal at opamp -ve input when output grounded

B(s) = feedback factor = signal at opamp -ve input when input grounded

Hactual(s) = Hideal(s)
-------------
1+Aol(s)/B(s)


I got Dostals book about 12 years ago, but until last year never used
this method, as the -ve input = 0V method worked well enough. Its only
when I started really pushing an opamp that it became necessary, mostly
because I wasnt happy with simply twiddling component values in SPICE.

I havent yet managed to get your SPICE trick to work in Simetrix

but I did read RDMs paper.

There is a 2001 paper describing extensions of Middlebrook's method. It's somewhat
obscure since it was published in one of the IEEE magazines rather than a regular journal.
I've posted it over on ABSE with the subject "Small Signal Stability".

Cheers
Terry

aargh, my new news server (news.slingshot.co.nz) hasnt seen *any* new
posts on abse for 5 days....

can you please email it to me...

cheers
terry

 

[John Woodgate]

I read in sci.electronics.design that Terry Given <[email protected]>

B(s) = feedback factor = signal at opamp -ve input when input grounded

How do you get any signal at the -ve input with the input grounded?

 

[Terry Given]

I read in sci.electronics.design that Terry Given <[email protected]>



How do you get any signal at the -ve input with the input grounded?

in the simple case,

a(s) = Vin(s)*Zf(s)/[Zin(s) + Zf(s)]
B(s) = Vo(s)*Zin(s)/[Zin(s) + Zf(s)]

H(s) = -Zf(s)/Zin(s)

Cheers
Terry

 

[Fred Bloggs]

[snip]

Dostals method also allowed me to
directly calculate the phase margin.
[snip]

Cheers
Terry

What IS Dostal's method? The Loop Gain & Phase analyser on my website
is based on R.D. Middlebrook's laboratory technique, and is VERY
accurate, since the loop is never actually broken.

...Jim Thompson

Middlebrook is still going strong too:
http://ardem.com/index.asp

 

[The Phantom]

On Wed, 21 Sep 2005 21:03:48 +0100, John Woodgate

I read in sci.electronics.design that "RST Engineering (jw)"
about 'Op Amp Calculations', on Wed, 21 Sep 2005:

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?

I believe Win has sort-of proprietary rights on this circuit. But it's
easy if you re-draw it. Why do these things come up in s.e.d. instead of
a.b.s.e?

ASCII art; use Courier font.

_______R4_______
| - |
o----R1--+---|\ |
| \_____ out |
| / | |
o------------|/ R2 |
+ |----'
R3
|
o--------------------+------ Gnd

---

I believe the circuit he described looks more like this:

Vcc------------+
|
[R1]
|
+----------|+\
| | >-----+-->Vout
Vin>------[R3]-------+----|-/ |
| | |
[R2] +-[R4]-+-[R5]-+
| |
GND>-----------+-[C1]--[R6]-+ A


Where, for DC,


R4 + R5 Vcc R2
Vout = -Vin --------- + ---------
R3 R1 + R2

For AC, however, (assuming Vin and Vout are perfect voltage sources)
we wind up with Vout trying to force the - input of the opamp to
stay at whatever voltage the + input is set to by R1 and R2.

For AC, assuming that C1 is very large so that it can be treated as an AC short, and
the op amp gain is infinite, the gain is:

-((r5*r6 + r4*(r5 + r6))/(r3*r6))

If A is the op amp gain and C1 is included in the calculation, the gain is :

-A*(r4 + r5) - A*c1*(r5*r6 + r4*(r5 + r6))*s
----------------------------------------------------------------------- --
r3 + A*r3 + r4 + r5 + c1*(r5*r6 + r4*(r5 + r6) + r3*(r5 + r6 + A*r6))*s

Well, you've forgotten the GBW in all that

No, I didn't forget at all; it's implicit in A.

If we set WT=2.pi.GBW then we have

-A - B p
----------------- with
1 + C p + D p^2

A0(R4+R5)
A = ----------------
R3(1+A0)+R4+R5

A0.C1(R4 R5 + (R4 + R5) R6)
B = ----------------------------
R3(1+A0)+R4+R5

(A0(R3 + R4 + R5 ) + C1.WT((R3 + R4)R5 + R6(R3 + A0.R3 + R4 + R5))
C = --------------------------------------------------------------------
(R3 (1 + A0) + R4 + R5 ) WT


A0.C1 ((R3 + R4) R5 + (R3 + R4 + R5) R6)
D = ------------------------------------------
(R3 + A0 R3 + R4 + R5) WT


Less than 2 min to work this out from scratch, incl. sign error correction.

What sign error is that?

Isn't that Mathematica lovely? ;-)

Unfortunately, the result doesn't seem to be correct. Put the
expressions for A, B, C (the numerator of the "C" expression seems to
be missing a closing parenthesis), and D into the first expression,
namely:

-A - B p
-----------------
1 + C p + D p^2

Then when you have it written out in all it's glory, find the limit
as A0 --> infinity, and then the limit as C1 --> infinity. You
*should* get: -((r5*r6 + r4*(r5 + r6))/(r3*r6))

But, alas, you don't.

 

[The Phantom]

(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.

Why hasn't someone pointed out that by exchanging the left and right sides of the T
network, we could have a shunt arm on the minus input of 111222 ohms, instead of 10090.7
ohms, a factor of 10 better. The 10090.7 ohm arm will then be a load on the amplifier
output, still without effect. The series arm is unchanged of course, so the net effect
may be an improvement. A possible undesirable consequence might be that the input
capacitance of the op amp might be greater than the stray capacitance at the junction of
the 3 T-network resistors, so with the reversed arrangement the op amp capacitance is fed
with 99200 ohms, rather than 9000 ohms. One would have to check this possibility.

Why didn't *I* point this out before now?

 

[Fred Bloggs]

a summing band-pass (ish) filter. 40 x TLV274.

I didnt want to AC-couple the inputs (that would have cost me 240
capacitors) so I used the bridged-T feedback network with an RC shunt to
give a DC gain of about 1/16 - any DC is basically common-mode, and the
next stage was AC coupled. 3 Rs and 2 Cs was a whole lot cheaper than an
RLC. But 100k/14 = 7k in parallel with the -ve shunt arm, enough to move
Fc 10% or so.

SPICE clearly showed it, so I went back and re-did my opamp analysis
using Dostals approach (originally I did it using the Woodgate
approximation), and voila - out popped the same answer. Mr HP3577 also
agreed with spice and mathcad. Dostals method also allowed me to
directly calculate the phase margin. Since then, I have analysed all
opamp circuits thusly - but I use the Woodgate approach with pencil &
paper as a bullshit detector

Cheers
Terry

You can achieve a wild increase in effective GBW by going to current
mode feedback. The peaking and rapid rolloff due to that low impedance
-ve shunt is eliminated from any frequency bands usable with the voltage
feedback circuit. The output gain of 2x deals with the CMR input range
of the TLV274- requires about a volt of headroom to V+ - facilitation
odds and ends not shown...
View in a fixed-width font such as Courier.

..
..
.. >--[Ri]-+-------+------[R1]--+--[R2]------------+
.. | | | |
.. o | | [R3] +--[R]---+-->Vout
.. | | +5V | | |
.. | | | C | | |\ |
.. o | | +-----||-+ +--|-\ |
.. | | | | | >--+
.. | | +-----[Rc]-----+------|+/
.. o | | | | | |/
.. | | | | |
.. >--[Ri]-+ | +-----------+ | [R]
.. | | | | | |
.. >--[Ri]-+ +------|>|---+ | | |
.. | | | | | | |
.. >--[Ri]-+ +--------------------------+
.. | | | | | |
.. | | | | | |
.. | | |\| c | |
.. | +-|+\ |/ | | OA TLV274
.. | | >-+---| c |
.. 2.5V>-+-----|-/ | |\ |/ |
.. |/| | e+ c
.. | | |\ |/
.. | [Rb] e+
.. | | |\
.. | | e
.. | | |
.. GND---------+--+-----------+------------
..
..
..
..
..
..
..
..
..
..
..
..

 

[Terry Given]

a summing band-pass (ish) filter. 40 x TLV274.

I didnt want to AC-couple the inputs (that would have cost me 240
capacitors) so I used the bridged-T feedback network with an RC shunt
to give a DC gain of about 1/16 - any DC is basically common-mode, and
the next stage was AC coupled. 3 Rs and 2 Cs was a whole lot cheaper
than an RLC. But 100k/14 = 7k in parallel with the -ve shunt arm,
enough to move Fc 10% or so.

SPICE clearly showed it, so I went back and re-did my opamp analysis
using Dostals approach (originally I did it using the Woodgate
approximation), and voila - out popped the same answer. Mr HP3577 also
agreed with spice and mathcad. Dostals method also allowed me to
directly calculate the phase margin. Since then, I have analysed all
opamp circuits thusly - but I use the Woodgate approach with pencil &
paper as a bullshit detector

Cheers
Terry

You can achieve a wild increase in effective GBW by going to current
mode feedback. The peaking and rapid rolloff due to that low impedance
-ve shunt is eliminated from any frequency bands usable with the voltage
feedback circuit. The output gain of 2x deals with the CMR input range
of the TLV274- requires about a volt of headroom to V+ - facilitation
odds and ends not shown...
View in a fixed-width font such as Courier.

.
.
. >--[Ri]-+-------+------[R1]--+--[R2]------------+
. | | | |
. o | | [R3] +--[R]---+-->Vout
. | | +5V | | |
. | | | C | | |\ |
. o | | +-----||-+ +--|-\ |
. | | | | | >--+
. | | +-----[Rc]-----+------|+/
. o | | | | | |/
. | | | | |
. >--[Ri]-+ | +-----------+ | [R]
. | | | | | |
. >--[Ri]-+ +------|>|---+ | | |
. | | | | | | |
. >--[Ri]-+ +--------------------------+
. | | | | | |
. | | | | | |
. | | |\| c | |
. | +-|+\ |/ | | OA TLV274
. | | >-+---| c |
. 2.5V>-+-----|-/ | |\ |/ |
. |/| | e+ c
. | | |\ |/
. | [Rb] e+
. | | |\
. | | e
. | | |
. GND---------+--+-----------+------------

I'll study that a bit later. unfortunately it also achieves a wild
increase in parts count and cost - this circuit is replicated many, many
times

Plus of course the original works just fine, and is in production.

Cheers
Terry

 

[Fred Bloggs]

Fred Bloggs wrote:



Terry Given wrote:


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


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


a summing band-pass (ish) filter. 40 x TLV274.

I didnt want to AC-couple the inputs (that would have cost me 240
capacitors) so I used the bridged-T feedback network with an RC shunt
to give a DC gain of about 1/16 - any DC is basically common-mode,
and the next stage was AC coupled. 3 Rs and 2 Cs was a whole lot
cheaper than an RLC. But 100k/14 = 7k in parallel with the -ve shunt
arm, enough to move Fc 10% or so.

SPICE clearly showed it, so I went back and re-did my opamp analysis
using Dostals approach (originally I did it using the Woodgate
approximation), and voila - out popped the same answer. Mr HP3577
also agreed with spice and mathcad. Dostals method also allowed me to
directly calculate the phase margin. Since then, I have analysed all
opamp circuits thusly - but I use the Woodgate approach with pencil &
paper as a bullshit detector

Cheers
Terry

You can achieve a wild increase in effective GBW by going to current
mode feedback. The peaking and rapid rolloff due to that low impedance
-ve shunt is eliminated from any frequency bands usable with the
voltage feedback circuit. The output gain of 2x deals with the CMR
input range of the TLV274- requires about a volt of headroom to V+ -
facilitation odds and ends not shown...
View in a fixed-width font such as Courier.

.
.
. >--[Ri]-+-------+------[R1]--+--[R2]------------+
. | | | |
. o | | [R3] +--[R]---+-->Vout
. | | +5V | | |
. | | | C | | |\ |
. o | | +-----||-+ +--|-\ |
. | | | | | >--+
. | | +-----[Rc]-----+------|+/
. o | | | | | |/
. | | | | |
. >--[Ri]-+ | +-----------+ | [R]
. | | | | | |
. >--[Ri]-+ +------|>|---+ | | |
. | | | | | | |
. >--[Ri]-+ +--------------------------+
. | | | | | |
. | | | | | |
. | | |\| c | |
. | +-|+\ |/ | | OA TLV274
. | | >-+---| c |
. 2.5V>-+-----|-/ | |\ |/ |
. |/| | e+ c
. | | |\ |/
. | [Rb] e+
. | | |\
. | | e
. | | |
. GND---------+--+-----------+------------

I'll study that a bit later. unfortunately it also achieves a wild
increase in parts count and cost - this circuit is replicated many, many
times

I can't comment on your application since only you know what that is.
You may find this strange, but the idea is to overcome the limitations
of low inverting input shunt impedance and not to improve your product.

 

[Fred Bartoli]

Sorry for the late reply.

No, I didn't forget at all; it's implicit in A.

Well, yes, I somewhat figured that.
But not mentionning A(s) when you explicitly use s somewhere else might lead
the not so careful to some error, i.e. forgetting its phase which is -90d
over almost the bandwidth.

correction.

What sign error is that?

That was in the starting equations (wrong opamp gain sign).

Unfortunately, the result doesn't seem to be correct. Put the
expressions for A, B, C (the numerator of the "C" expression seems to
be missing a closing parenthesis), and D into the first expression,
namely:

-A - B p
-----------------
1 + C p + D p^2

Then when you have it written out in all it's glory, find the limit
as A0 --> infinity, and then the limit as C1 --> infinity. You
*should* get: -((r5*r6 + r4*(r5 + r6))/(r3*r6))

But, alas, you don't.

But then I do
I've doubled checked the results the brute force way, taking the A,B,C,D
directly from here back into mathematica and all the limits are OK.

 

[Terry Given]

Terry Given wrote:

Fred Bloggs wrote:



Terry Given wrote:


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


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


a summing band-pass (ish) filter. 40 x TLV274.

I didnt want to AC-couple the inputs (that would have cost me 240
capacitors) so I used the bridged-T feedback network with an RC
shunt to give a DC gain of about 1/16 - any DC is basically
common-mode, and the next stage was AC coupled. 3 Rs and 2 Cs was a
whole lot cheaper than an RLC. But 100k/14 = 7k in parallel with the
-ve shunt arm, enough to move Fc 10% or so.

SPICE clearly showed it, so I went back and re-did my opamp analysis
using Dostals approach (originally I did it using the Woodgate
approximation), and voila - out popped the same answer. Mr HP3577
also agreed with spice and mathcad. Dostals method also allowed me
to directly calculate the phase margin. Since then, I have analysed
all opamp circuits thusly - but I use the Woodgate approach with
pencil & paper as a bullshit detector

Cheers
Terry




You can achieve a wild increase in effective GBW by going to current
mode feedback. The peaking and rapid rolloff due to that low
impedance -ve shunt is eliminated from any frequency bands usable
with the voltage feedback circuit. The output gain of 2x deals with
the CMR input range of the TLV274- requires about a volt of headroom
to V+ - facilitation odds and ends not shown...
View in a fixed-width font such as Courier.

.
.
. >--[Ri]-+-------+------[R1]--+--[R2]------------+
. | | | |
. o | | [R3] +--[R]---+-->Vout
. | | +5V | | |
. | | | C | | |\ |
. o | | +-----||-+ +--|-\ |
. | | | | | >--+
. | | +-----[Rc]-----+------|+/
. o | | | | | |/
. | | | | |
. >--[Ri]-+ | +-----------+ | [R]
. | | | | | |
. >--[Ri]-+ +------|>|---+ | | |
. | | | | | | |
. >--[Ri]-+ +--------------------------+
. | | | | | |
. | | | | | |
. | | |\| c | |
. | +-|+\ |/ | | OA TLV274
. | | >-+---| c |
. 2.5V>-+-----|-/ | |\ |/ |
. |/| | e+ c
. | | |\ |/
. | [Rb] e+
. | | |\
. | | e
. | | |
. GND---------+--+-----------+------------

I'll study that a bit later. unfortunately it also achieves a wild
increase in parts count and cost - this circuit is replicated many,
many times

I can't comment on your application since only you know what that is.
You may find this strange, but the idea is to overcome the limitations
of low inverting input shunt impedance and not to improve your product.

What about the typo? the 2nd transistor shorts out the +5V supply....

Cheers
Terry

 

[The Phantom]

Sorry for the late reply.



Well, yes, I somewhat figured that.
But not mentionning A(s) when you explicitly use s somewhere else might lead
the not so careful to some error, i.e. forgetting its phase which is -90d
over almost the bandwidth.

I think that anyone sufficiently well versed in complex arithmetic
as used nowadays to write transfer functions will know that if they
want the DC gain, they can just use the TF I gave with A constant, and
if they want AC results, then of course they will know that A must be
a function of frequency..

That was in the starting equations (wrong opamp gain sign).

Did you notice that three different people contributed to the
posting that you originally replied to? John Woodgate did the first
ASCII schematic, John Fields did the second and provided an equation,
viz:

R4 + R5 Vcc R2
Vout = -Vin --------- + ---------
R3 R1 + R2

All I (Phantom) posted was a couple of transfer functions. I didn't
use John's equations, and I don't think there was a sign error in what
I posted. Most of your reply was directed to me, so I thought you
were suggesting I had made a sign error.

It would make it easier for others to comment without ambiguity and
misunderstanding on your posting if you would cut and paste in the
equation you think is in error, perhaps even indicating what you think
the correct equation should be.

But then I do
I've doubled checked the results the brute force way, taking the A,B,C,D
directly from here back into mathematica and all the limits are OK.

You set WT=2.pi.GBW, but you haven't indicated what kind of op amp
model you're using. Was it the standard one pole model:

A(s) = WT/(s + Wa)

or something else? I can't comment further until you tell me your
model, but I still think there is something out of kilter with your
expression:

-A - B p

 

[Fred Bartoli]

I think that anyone sufficiently well versed in complex arithmetic
as used nowadays to write transfer functions will know that if they
want the DC gain, they can just use the TF I gave with A constant, and
if they want AC results, then of course they will know that A must be
a function of frequency..

OK, fair enough.

Did you notice that three different people contributed to the
posting that you originally replied to? John Woodgate did the first
ASCII schematic, John Fields did the second and provided an equation,
viz:

R4 + R5 Vcc R2
Vout = -Vin --------- + ---------
R3 R1 + R2

All I (Phantom) posted was a couple of transfer functions. I didn't
use John's equations, and I don't think there was a sign error in what
I posted. Most of your reply was directed to me, so I thought you
were suggesting I had made a sign error.

It would make it easier for others to comment without ambiguity and
misunderstanding on your posting if you would cut and paste in the
equation you think is in error, perhaps even indicating what you think
the correct equation should be.

Oh, I see.
Sorry for the misunderstanding, I wasn't implying somebody made a sign
error.
It was just *me* that made the sign error when first writing the problem and
got obviously wrong results.
Error corrected in a snap.

By saying:
I was just emphasing how easy it is to correct such errors vs the
paper/pencil method.

You set WT=2.pi.GBW, but you haven't indicated what kind of op amp
model you're using. Was it the standard one pole model:

A(s) = WT/(s + Wa)

Yes it is.

or something else? I can't comment further until you tell me your
model, but I still think there is something out of kilter with your
expression:

-A - B p

Well, you can't just make A0 and C1 go to the limit and expect the DC
response. You also have to either make WT go to the limit, or make s=0.

Then all goes fine.

But then you can see that it's easy, even for someone "sufficiently well
versed in complex arithmetic
as used nowadays" (to take your words), to forget something in the process.
Hence my preference for writing A(s). It's just one thing less to remember.

From some of your previous posts I believe you have mathematica, so you'll
find the code herewith.


--
Thanks,
Fred.


In[1]:=
\!\(Eqs = {\[IndentingNewLine]Vo == \(-Vn\)\ \[Omega]T\/\(\[Omega]T\/A0 + \
p\), (*\
opamp\ gain\ *) \[IndentingNewLine]\(Vn - VA\)\/R4 + \(Vo - VA\)\/R5 \
+ \(0 - VA\)\/\(R6 + 1\/\(C1\ p\)\) == 0, \ \ \ (*\
node\ A\ = \
T\ common\ node\ *) \[IndentingNewLine]\(Vin - Vn\)\/R3 + \(VA - Vn\
\)\/R4 == 0\ (*\
opamp\ minus\ input\ *) \[IndentingNewLine]}\[IndentingNewLine]\
\[IndentingNewLine]
sol = \(\(Vo\/Vin /. Solve[Eqs, Vo, {Vn, VA}] // Simplify\) //
Collect[#, p] &\) // First\[IndentingNewLine]\[IndentingNewLine]
\(Print["\<\nChecking the gain limits:\>"];\)\[IndentingNewLine]
Limit[sol, A0 -> \[Infinity]]\[IndentingNewLine]
Limit[%, C1 -> \[Infinity]]\[IndentingNewLine]
Limit[%, \[Omega]T -> \[Infinity]] // FullSimplify\[IndentingNewLine]
Limit[%%, p -> 0] // FullSimplify\[IndentingNewLine]\[IndentingNewLine]
\(solnum = Numerator[sol] // Collect[#, p] &;\)\[IndentingNewLine]
\(solden = \(Denominator[sol] // FullSimplify\) //
Collect[#, {p, \[Omega]T}] &;\)\[IndentingNewLine]
\(cl = CoefficientList[solden, p];\)\[IndentingNewLine]\[IndentingNewLine]
\(Print["\<\nTF numerator:\>"];\)\[IndentingNewLine]
solnum = \(solnum\/cl[\([1]\)] // FullSimplify\) //
Collect[#, {\ p}] &\[IndentingNewLine]\[IndentingNewLine]
\(Print["\<\nTF denominator:\>"];\)\[IndentingNewLine]
solden = \((cl[\([3]\)]\/cl[\([1]\)] //
FullSimplify)\)\ p\^2 + \((cl[\([2]\)]\/cl[\([1]\)] //
FullSimplify)\)\ p\ + \ 1\[IndentingNewLine]\[IndentingNewLine]
\(Print["\<\nThe results:\>"];\)\[IndentingNewLine]
\(cln = CoefficientList[solnum, p];\)\[IndentingNewLine]
\(cld = CoefficientList[solden, p];\)\[IndentingNewLine]
Print["\<A = \>", a = \(-cln[\([1]\)]\)\ ]\[IndentingNewLine]
Print["\<B = \>", b = \(-cln[\([2]\)]\)\ ]\[IndentingNewLine]
Print["\<C = \>", c = cld[\([2]\)]\ ]\[IndentingNewLine]
Print["\<D = \>", d = cld[\([3]\)]\ ]\[IndentingNewLine]\[IndentingNewLine]
(*\ Checking\ *) \[IndentingNewLine]
\(Print["\<\nFinal check:\>"];\)\[IndentingNewLine]
\((sol - \(\(-a\) - b\ p\)\/\(1 + c\ p + d\ p\^2\))\) // Simplify\)