Op Amp Calculations

[Fred Bloggs]

Fred Bloggs wrote:



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

It is a common collector pre-drive for the third transistor which is CE.
If extreme gain accuracy is not needed it can be pulled, the CE should
be a high beta type at low Ic.

 

[Terry Given]

Terry Given wrote:

Fred Bloggs wrote:



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

It is a common collector pre-drive for the third transistor which is CE.
If extreme gain accuracy is not needed it can be pulled, the CE should
be a high beta type at low Ic.

shouldnt there be something to limit current though? +5V...Vce...Vbe
with nary a resistor in sight. ultimately the base current could be as
high as the opamp output current (assuming negligible contribution from
the summing junction)....

nice artwork BTW

Cheers
Terry

 

[Jim Thompson]

Fred Bloggs wrote:



Terry Given wrote:

Fred Bloggs wrote:



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

It is a common collector pre-drive for the third transistor which is CE.
If extreme gain accuracy is not needed it can be pulled, the CE should
be a high beta type at low Ic.

shouldnt there be something to limit current though? +5V...Vce...Vbe
with nary a resistor in sight. ultimately the base current could be as
high as the opamp output current (assuming negligible contribution from
the summing junction)....

nice artwork BTW

Cheers
Terry

Triple Darlington? Bah! Humbug! It'll fry!

...Jim Thompson

 

[Fred Bloggs]

Fred Bloggs wrote:



Terry Given wrote:

Fred Bloggs wrote:



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

It is a common collector pre-drive for the third transistor which is
CE. If extreme gain accuracy is not needed it can be pulled, the CE
should be a high beta type at low Ic.

shouldnt there be something to limit current though? +5V...Vce...Vbe
with nary a resistor in sight. ultimately the base current could be as
high as the opamp output current (assuming negligible contribution from
the summing junction)....

nice artwork BTW

Cheers
Terry

You're dwelling on open loop anomalies- tie that collector to the
rightmost transistor collector if it bothers you. There is no condition
under which the TLV274 will put sustained maximum current into the base
circuit- not even close- the output will rail at uA drive.

 

[The Phantom]

OK, fair enough.


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.

I understand now.

By saying:

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

You're absolutely correct. It is really nice, having spent hours on
a really complicated problem, discovering a stupid mistake that would
entail more hours of computation after having discovered the mistake.
to be able to just key in a small change, hit enter, and !bang!, the
final result, a result derived from a large amount of algebraic
drudgery!

Yes it is.

Except that you didn't use that particular formulation. You used a
non-standard (in my experience) expression, namely;

A(s) = WT/((WT/A0 + s)) and this caused your results to contain both
A0 and WT. There's no need for this since WT incudes A0.

(I'm going to use Wp for the pole frequency in the single pole model
for an op amp in what follows.)

A standard expression for op amp gain, (which is exactly equivalent
to yours, but not involving *both* A0 and WT explicitly) such as found
on page 19 of the classic Schaumann and Van Valkenburg, is:

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

another alternative would be:

A(s) = (A0 Wp)/(s + Wp)

Solving the circuit with either of these expressions gives results
that explicitly display the effect of Wp, and only one variable that
involves the DC gain of the op amp.

(I personally like to use the expression:

A(s) = A0/(1 + sT), where T is the time constant of the pole.)

Well, you can't just make A0 and C1 go to the limit and expect the DC
response.

Yes, you can, if you use the second version of the op amp gain I
gave above. If you use the first, standard, expression for A(s), then
you have to make WT go to infinity (but not A0 since it's no longer in
the result expressions) and then C1.

You also have to either make WT go to the limit,

Only if you have used an expression for A(s) which involves *both*
A0 *and* WT, which is not necessary.

or make s=0.

In your Mathematica program, you have these four lines (change mode
temporarily from using s to using p ):

Limit[sol, A0 -> infinity]
Limit[%, C1 -> infinity]
Limit[%, WT -> infinity]
Limit[%%, p -> infinity]

The fourth limit operates on the result of the second limit, which
expression already lacks the variable p, so of course the result of
this limit is the same as the result of the second limit and therefore
accidentally gives the correct DC gain; but, in general, just letting
p -> 0 won't give the DC gain if capacitors are involved, as I
explain.

It is the admittance of the capacitor(s) we want to become infinite
to get DC gain; that is, the product p*C1 must go to infinity. I
notice that if you let p -> 0 first (with C1 still finite), and then
let A0 go to infinity, the result is -(R4 + R5)/R3, because the
product p*C1 has gone to zero. Letting both p -> 0 and C1 -> infinity
would only work if p and C1 vary in such a way that the product p*C1
-> infinity. Thus, letting p -> 0 won't give the DC gain if C1
remains finite. But this is not a problem, since letting C1 ->
infinity does the job.

When I originally took your full expression involving A, B, C, and D
and took the limit as A0 -> infinity and C1 -> infinity, I got the
same result you got after the second limit (above) and I couldn't
understand why WT was still in the result; that's why I initially
thought your result was incorrect. Other than that unnecessary
complication, your result is, of course, correct.

The basic three equations are easy to write down by inspection as
you did, and Mathematica did just what you told it to, and yes,
Mathematica is indeed lovely.

 

[Fred Bartoli]

Fred Bloggs wrote:



Terry Given wrote:

Fred Bloggs wrote:



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

It is a common collector pre-drive for the third transistor which is CE.
If extreme gain accuracy is not needed it can be pulled, the CE should
be a high beta type at low Ic.

shouldnt there be something to limit current though? +5V...Vce...Vbe
with nary a resistor in sight. ultimately the base current could be as
high as the opamp output current (assuming negligible contribution from
the summing junction)....

nice artwork BTW

That, (which can be corrected by moving Q2's collector) plus Q1/Q2 operate
at vanishingly low IC so that the input impedance has nothing to do with
that of a CFB, even at moderate audio frequencies.

Take IC3=1mA, HFE=100 for all the Qs. That makes IC1=25nA and gmQ1=1uA/V and
a total transconductance one third of it or gm = 0.33uA/V.

The TLV GBW product is 3MHz, so at 10kHz the input impedance is 3M/300=10K
and inductive (a 0.16H inductance).
With a 3pF input it'll resonate at a low 230kHz and give a 2nd order
roll-off above this frequency.

The 3rd stage miller capacitance will also roll-off in that frequency range.

Hardly a usable CFB amp

 

[Fred Bloggs]

Terry Given wrote:


Fred Bloggs wrote:



Terry Given wrote:


Fred Bloggs wrote:



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


It is a common collector pre-drive for the third transistor which is CE.
If extreme gain accuracy is not needed it can be pulled, the CE should
be a high beta type at low Ic.

shouldnt there be something to limit current though? +5V...Vce...Vbe
with nary a resistor in sight. ultimately the base current could be as
high as the opamp output current (assuming negligible contribution from
the summing junction)....

nice artwork BTW

That, (which can be corrected by moving Q2's collector) plus Q1/Q2 operate
at vanishingly low IC so that the input impedance has nothing to do with
that of a CFB, even at moderate audio frequencies.

Take IC3=1mA, HFE=100 for all the Qs. That makes IC1=25nA and gmQ1=1uA/V and
a total transconductance one third of it or gm = 0.33uA/V.

Not quite, essentially all the incremental base drive voltage is
developed across hie of the first transistor. Your reasoning is wrong
and your estimate is high.

The TLV GBW product is 3MHz, so at 10kHz the input impedance is 3M/300=10K
and inductive (a 0.16H inductance).
With a 3pF input it'll resonate at a low 230kHz and give a 2nd order
roll-off above this frequency.

Yeah- right, i,in= A(jw)*gm*vin or Zin= re,Q1/A(jw), for the TLV274 with
A(jw)=6.28E6/(jw), this becomes Zin= re,Q1 * jw /6.28E6 so that at 1KHz
you have re,Q1E-3. Taking re,Q1=beta^2*0.026/IcQ3,Q this makes Zin about
2.6K at 1KHz. But you have a fairly huge gain of
vout=A(jw)*gm*beta^2*2*Rc*vin, and since gm=1/(re,Q3*beta^2),
Av=A(jw)*2*Rc/re,Q3, so that Yf becomes Av/Rf, and the current divide
ratio is Av/(Av+A(jw)*gm*Rf) or 1/(1+Rf/(2*Rc*beta^2))- this is looking
more like a CMF all the time- a broadband split neglecting a few poles
here and there.

The 3rd stage miller capacitance will also roll-off in that frequency range.

That is very easily taken care of- but it's not my job to educate
you...all you need to know is that the CE rolloff is not an issue.

Hardly a usable CFB amp

Nah- "hardly usable" is what your boss says about you. SPICE does not
agree with your conclusions- what does your IQ non-enhancer,
Mathematica, tell you now? The one drawback to the triple Darlington is
the same peaking effect due to error amplifier impedance growth with
frequency, it is possible to replace Q1 with a PNP emitter follower
feedback to OA input, interchange OA (-) and (+), but then this creates
another headache by introducing a dominant pole in the loop.

 

[The Phantom]

OK, fair enough.

= --------------------------------------------------------------------

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.

I understand now.

By saying:

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

You're absolutely correct. It is really nice, having spent hours on
a really complicated problem, discovering a stupid mistake that would
entail more hours of computation after having discovered the mistake.
to be able to just key in a small change, hit enter, and !bang!, the
final result, a result derived from a large amount of algebraic
drudgery!

Yes it is.

Except that you didn't use that particular formulation. You used a
non-standard (in my experience) expression, namely;

A(s) = WT/((WT/A0 + s)) and this caused your results to contain both
A0 and WT. There's no need for this since WT incudes A0.

(I'm going to use Wp for the pole frequency in the single pole model
for an op amp in what follows.)

A standard expression for op amp gain, (which is exactly equivalent
to yours, but not involving *both* A0 and WT explicitly) such as found
on page 19 of the classic Schaumann and Van Valkenburg, is:

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

another alternative would be:

A(s) = (A0 Wp)/(s + Wp)

Solving the circuit with either of these expressions gives results
that explicitly display the effect of Wp, and only one variable that
involves the DC gain of the op amp.

(I personally like to use the expression:

A(s) = A0/(1 + sT), where T is the time constant of the pole.)

Well, you can't just make A0 and C1 go to the limit and expect the DC
response.

Yes, you can, if you use the second version of the op amp gain I
gave above. If you use the first, standard, expression for A(s), then
you have to make WT go to infinity (but not A0 since it's no longer in
the result expressions) and then C1.

You also have to either make WT go to the limit,

Only if you have used an expression for A(s) which involves *both*
A0 *and* WT, which is not necessary.

or make s=0.

In your Mathematica program, you have these four lines (change mode
temporarily from using s to using p ):

Limit[sol, A0 -> infinity]
Limit[%, C1 -> infinity]
Limit[%, WT -> infinity]
Limit[%%, p -> infinity]

This should be Limit[%%, p -> 0] of course. I proofread all this
several times, and still this slipped by. :-(

The fourth limit operates on the result of the second limit, which
expression already lacks the variable p,

I just noticed that this is not true; I think I must have been
making small changes to the program (fooling around to explore some
things) and so what I was looking at when I said this was not your
original program.

so of course the result of
this limit is the same as the result of the second limit and therefore
accidentally gives the correct DC gain;

It gives the correct result because products such as p*C1 have been
eliminated by the second limit.

but, in general, just letting
p -> 0 won't give the DC gain if capacitors are involved,

I think this is true if any products of p and C1 are present in the
expression before letting p -> 0. If you let p -> 0 after the first
limit of the four above, then you get the result -(R4 + R5)/R3 (WT
drops out, too), which I believe happens as explained below.

 

[Jim Thompson]

Fred Bloggs wrote:


Terry Given wrote: [snip]

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

shouldnt there be something to limit current though? +5V...Vce...Vbe
with nary a resistor in sight. ultimately the base current could be as
high as the opamp output current (assuming negligible contribution from
the summing junction)....

nice artwork BTW

That, (which can be corrected by moving Q2's collector) plus Q1/Q2 operate
at vanishingly low IC so that the input impedance has nothing to do with
that of a CFB, even at moderate audio frequencies.

Take IC3=1mA, HFE=100 for all the Qs. That makes IC1=25nA and gmQ1=1uA/V and
a total transconductance one third of it or gm = 0.33uA/V.

Not quite, essentially all the incremental base drive voltage is
developed across hie of the first transistor. Your reasoning is wrong
and your estimate is high.

And it STILL will go up in flames during power-up. Sheeesh! Such
DESIGN ;-)

[snip]

...Jim Thompson

 

[Winfield Hill]

Jim Thompson wrote...

... Sheeesh! ...

Hey, that's my word!

 

[Jim Thompson]

Jim Thompson wrote...

Hey, that's my word!

Will you share? It's a GOOD word!

...Jim Thompson

 

[RST Engineering \(jw\)]

Jeez, all I asked as the OP was what time it was, I didn't need to know how
to make the watch or the theory of molecular resonance.

I'll ask a HARD question next time.



{;-)

Jim

 

[Terry Given]

"Terry Given" <[email protected]> a écrit dans le message de


Fred Bloggs wrote:


Terry Given wrote:

[snip]

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

[snip]

shouldnt there be something to limit current though? +5V...Vce...Vbe
with nary a resistor in sight. ultimately the base current could be as
high as the opamp output current (assuming negligible contribution from
the summing junction)....

nice artwork BTW



That, (which can be corrected by moving Q2's collector) plus Q1/Q2 operate
at vanishingly low IC so that the input impedance has nothing to do with
that of a CFB, even at moderate audio frequencies.

Take IC3=1mA, HFE=100 for all the Qs. That makes IC1=25nA and gmQ1=1uA/V and
a total transconductance one third of it or gm = 0.33uA/V.

Not quite, essentially all the incremental base drive voltage is
developed across hie of the first transistor. Your reasoning is wrong
and your estimate is high.

And it STILL will go up in flames during power-up. Sheeesh! Such
DESIGN ;-)

[snip]

...Jim Thompson

Hi Jim,

can you please elaborate on why? I presume because of the 2nd transistor....

Cheers
Terry

 

[John Woodgate]

I read in sci.electronics.design that Jim Thompson

Will you share? It's a GOOD word!

Offer him a half-share in 'Snooooort!'

 

[Winfield Hill]

John Woodgate wrote...

Jim Thompson wrote ...

Offer him a half-share in 'Snooooort!'

Plus a half share of Bwaahahahaha!

 

[Jim Thompson]

Fred Bartoli wrote:
"Terry Given" <[email protected]> a écrit dans le message de
Fred Bloggs wrote:
Terry Given wrote:
What about the typo? the 2nd transistor shorts out the +5V supply....
shouldnt there be something to limit current though? +5V...Vce...Vbe
with nary a resistor in sight. ultimately the base current could be as
high as the opamp output current (assuming negligible contribution from
the summing junction).... [snip]
Not quite, essentially all the incremental base drive voltage is
developed across hie of the first transistor. Your reasoning is wrong
and your estimate is high.

"hie"... what BS! Fred must be of the "Kevin School" of transistor
operation ;-)

And it STILL will go up in flames during power-up. Sheeesh! Such
DESIGN ;-)

[snip]

...Jim Thompson

Hi Jim,

can you please elaborate on why? I presume because of the 2nd transistor....

Cheers
Terry

No limit on IB of 1st transistor except output capability of OpAmp

IC of 1st transistor IS limited somewhat by FB resistors, but
substantial IB is introduced into 2nd transistor

Absolutely nothing to limit IC of 2nd transistor except current
crowding

My best guess is that all three transistors will flame or be seriously
damaged when Murphy chooses ;-)

(And the bandwidth will be nothing like claimed.)

What was the original intent (OP)? I didn't get in on this thread at
the beginning.

...Jim Thompson

 

[Fred Bloggs]

Fred Bartoli wrote:

"Terry Given" <[email protected]> a écrit dans le message de

Fred Bloggs wrote:

Terry Given wrote:

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

shouldnt there be something to limit current though? +5V...Vce...Vbe
with nary a resistor in sight. ultimately the base current could be as
high as the opamp output current (assuming negligible contribution from
the summing junction)....

[snip]

Not quite, essentially all the incremental base drive voltage is
developed across hie of the first transistor. Your reasoning is wrong
and your estimate is high.

"hie"... what BS! Fred must be of the "Kevin School" of transistor
operation ;-)

And it STILL will go up in flames during power-up. Sheeesh! Such
DESIGN ;-)

[snip]

...Jim Thompson

Hi Jim,

can you please elaborate on why? I presume because of the 2nd transistor....

Cheers
Terry

No limit on IB of 1st transistor except output capability of OpAmp

IC of 1st transistor IS limited somewhat by FB resistors, but
substantial IB is introduced into 2nd transistor

Absolutely nothing to limit IC of 2nd transistor except current
crowding

My best guess is that all three transistors will flame or be seriously
damaged when Murphy chooses ;-)

(And the bandwidth will be nothing like claimed.)

What was the original intent (OP)? I didn't get in on this thread at
the beginning.

...Jim Thompson

This is not rocket science- we already said that the second transistor C
should be tied to the CE collector. If you're worried by the OA dumping
excessive current into the base then put a 560 in series with its output.

 

[Jim Thompson]

Jim Thompson wrote:

Fred Bartoli wrote:

"Terry Given" <[email protected]> a écrit dans le message de

Fred Bloggs wrote:

Terry Given wrote:

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

shouldnt there be something to limit current though? +5V...Vce...Vbe
with nary a resistor in sight. ultimately the base current could be as
high as the opamp output current (assuming negligible contribution from
the summing junction)....
[snip]

Not quite, essentially all the incremental base drive voltage is
developed across hie of the first transistor. Your reasoning is wrong
and your estimate is high.

"hie"... what BS! Fred must be of the "Kevin School" of transistor
operation ;-)

And it STILL will go up in flames during power-up. Sheeesh! Such
DESIGN ;-)

[snip]

...Jim Thompson

Hi Jim,

can you please elaborate on why? I presume because of the 2nd transistor....

Cheers
Terry

No limit on IB of 1st transistor except output capability of OpAmp

IC of 1st transistor IS limited somewhat by FB resistors, but
substantial IB is introduced into 2nd transistor

Absolutely nothing to limit IC of 2nd transistor except current
crowding

My best guess is that all three transistors will flame or be seriously
damaged when Murphy chooses ;-)

(And the bandwidth will be nothing like claimed.)

What was the original intent (OP)? I didn't get in on this thread at
the beginning.

...Jim Thompson

This is not rocket science- we already said that the second transistor C
should be tied to the CE collector. If you're worried by the OA dumping
excessive current into the base then put a 560 in series with its output.

That's still insufficient. Good engineering practice would limit base
drive from the OpAmp AND also the collector current of the 2nd
transistor.

...Jim Thompson

 

[Rich Grise]

John Woodgate wrote...

Plus a half share of Bwaahahahaha!

Dibs on a half share of ROFLMAOPIMP!

;-)

 

[Winfield Hill]

Rich Grise wrote...

Dibs on a half share of ROFLMAOPIMP!

PIMP?