Maker Pro
Maker Pro

Somebody explaining this design?

E

Eeyore

Jan 1, 1970
0
Jim said:
Now I would like to see a computation of the gain of this "marvy"
amplifier.

Good Lord !

That's easy. Can't do it yourself.

Graham
 
E

Eeyore

Jan 1, 1970
0
Phil said:
"Eeysore = know nothing pommy charlatan and plagiarist."

Plagiarist ?

It was you who copied 'my' mic amp design !

Graham
 
E

Eeyore

Jan 1, 1970
0
Jim said:
Well? Anyone? Mathematical solutions only... no simulations please
;-)

...Jim Thompson

Are you not even going to bother trying yourself ?

Graham
 
E

Eeyore

Jan 1, 1970
0
Fred said:
The circuit is a primitive paradigm, it would have been a simple matter
to include feedback.

With only a single-ended output ? Do go on.

Graham
 
P

Phil Allison

Jan 1, 1970
0
"Eeysore = know nothing pommy charlatan and plagiarist."

Plagiarist ?


** Claiming other people's design work as being yours.

That is plagiarism.

Being a charlatan is a corollary.



........ Phil
 
E

Eeyore

Jan 1, 1970
0
jure said:
what I see is a differential pair of PNP Sziklai connected
transistors (AKA "complementary Darlington" ),
with a Pi emitter tail to the positive rail. Also, there is the
standard feature of the resistor parallel to the "BE" of the second
transistor of the darlington pair.

The differential pair is out of the feedback loop, introducing
offsets, etc.

It's not a DC amplifier hence that's irrelevant.

what is the linear range for this pair ? => dynamic input range ?

Bloody good.

would it be better to close the loop to the bases and have a fully
differential system.

More modern versions close the loop to the emitters of the input devices
actually.

Graham
 
E

Eeyore

Jan 1, 1970
0
Phil said:
"Eeysore = know nothing pommy charlatan and plagiarist."


** Claiming other people's design work as being yours.

Where did I say I *designed* it ?

Graham
 
P

Phil Allison

Jan 1, 1970
0
"Eeysore = know nothing pommy charlatan and plagiarist."

Plagiarist ?


** Claiming other people's design work as being yours.

That is plagiarism.

Being a charlatan is a corollary.




........ Phil
 
T

Tony Williams

Jan 1, 1970
0
Winfield Hill said:
Anyway, Jim's comment, "Looks like crap to me ;-)", implies
an incorrect analysis, as several people correctly saw. In
a differential amplifier the addition of two transistors and
resistors in this way makes a dramatic improvement. If the
amplifier doesn't have feedback, as in this case, some trick
like this is essential. Graham says he designed this scheme
into Studiomaster's 8-4 rack mixer in 1979, but it's an old
trick, going back to the 60's at least, SFAIK.

ISTR seeing it in about 1972, in a high performance
jfet input instrumentation amplifier that we made,
(not my design). The jfet sources were fed by about
1mA constant current npn's, 500uA up the jfet and
500uA to the feedback pnp's. I think the pnp's had
their emitters up on the +15v rail, with resistors
from the jfet drains to their bases. I don't recall
any series resistors between the drains and the inputs
of the opamp.
 
J

Jim Thompson

Jan 1, 1970
0
On Sat, 10 Mar 2007 15:28:19 -0700, Jim Thompson

[snip]
That was to lower the noise, this is to lower the high-level
signal distortion, without excessive feedback. Of course
it's better. http://sound.westhost.com/project66.htm

As Eeyore said, "Q1/Q2 and Q3/Q4 are 'compound pairs' that
have very high linearity compared to a single transistor."
[snip]

OK. Back from lunch and purchasing a new power supply for my PSpice
machine. Fan on the old one developed a bad case of the "grindies".

Now I would like to see a computation of the gain of this "marvy"
amplifier.

Win?

...Jim Thompson

Well? Anyone? Mathematical solutions only... no simulations please
;-)

...Jim Thompson

Still no response after 24 hours.

Where are all the "experts" when a real question is asked ?:)

...Jim Thompson
 
P

Phil Allison

Jan 1, 1970
0
"Jim Thompson"
Still no response after 24 hours.

Where are all the "experts" when a real question is asked ?:)


** Dear, oh dear -

the posturing, PITA, fascist cretin cannot see the one that has been posted
already.




......... Phil
 
E

Eeyore

Jan 1, 1970
0
Jim said:
Still no response after 24 hours.

Where are all the "experts" when a real question is asked ?:)

Phil Allison replied.

As did I to the effect of "if you can't work that one out you're damn useless".

Graham
 
J

Jim Thompson

Jan 1, 1970
0
Isn't it just the Pullup-R divided by the Tail-R,
times the final Feedback-R divided by the Pullup-R,
probably with a x2 in there somewhere?

So the Pullup-R cancels, leaving the overall Gain
as the ratio of final Feedback-R/Tail-R.

Pullup-R's DON'T cancel.

...Jim Thompson
 
J

John Fields

Jan 1, 1970
0
Phil Allison replied.

As did I to the effect of "if you can't work that one out you're damn useless".

---
Since Jim posed the question it's not for him to answer it until and
if he wants to if no one steps forward with the answer. Which isn't
what you did, so the assumption must be made that you couldn't work
it out which, therefore, makes your statement: "if you can't work
that one out you're damn useless" true.
 
J

Jim Thompson

Jan 1, 1970
0
---
Since Jim posed the question it's not for him to answer it until and
if he wants to if no one steps forward with the answer. Which isn't
what you did, so the assumption must be made that you couldn't work
it out which, therefore, makes your statement: "if you can't work
that one out you're damn useless" true.

I worked in out with the sweep of the one functioning eyeball, and
promptly announced that it was crap ;-)

...Jim Thompson
 
E

Eeyore

Jan 1, 1970
0
John said:
---
Since Jim posed the question it's not for him to answer it until and
if he wants to if no one steps forward with the answer. Which isn't
what you did, so the assumption must be made that you couldn't work
it out which, therefore, makes your statement: "if you can't work
that one out you're damn useless" true.

Look sweetie pie, I know that damn circuit inside out and backwards.

If I was to post a response it would take the fun out of it for those not familiar
with it. Anyway Allison already gave us his answer.

Graham
 
T

Terry Given

Jan 1, 1970
0
Eeyore said:
John Fields wrote:




Look sweetie pie, I know that damn circuit inside out and backwards.

If I was to post a response it would take the fun out of it for those not familiar
with it. Anyway Allison already gave us his answer.

Graham

thats not a very good excuse. others here happily share their expertise
on subjects they know well - why wont you do the same?

Cheers
Terry
 
E

Eeyore

Jan 1, 1970
0
Terry said:
thats not a very good excuse. others here happily share their expertise
on subjects they know well - why wont you do the same?

To be honest it's already been answered. There's nothing I can add other than perhaps an
improvement which I've already mentioned.

For clarification that's to 'link out' C2, C3, R10 and R11 and scale R12 and R13 to 12k.
You need to add an output coupling cap since the TL071 output will sit at around +/-
0.5V typically.

1% resistors in the signal path are a good idea too btw.


Graham
 
T

Tony Williams

Jan 1, 1970
0
Pullup-R's DON'T cancel.

Well, let's do some sums. Admittedly on a simplified and
idealised circuit (to make life easy for me).
R2a and R2b are the Pull-up R's in question.

-------+-------------+---Vs
| |
\ \ R3b
/R2a R2b/ +-/\/\---+-->Vout
\ \ | |
| | | _ |
| +---------+--|- \ |
| | | >-+
Vx---> +-------------|---------+--|+_/
| | |
\|/I1 I2\|/ \
| | /R3a
|/ \| \
Vin ---|npn npn|----. |
|\e e/| | |
0v--- | R1 | ---+--+---0v
+----/\/\-----+
| |
\|/ I I \|/
Constant current

I1 = (I + Vin/R1) and I2 = (I - Vin/R1).

First find out where I1 sets the opamp +ve input (Vx).

I(R3a) = (Vs-Vx)/R2a - I1 = (Vs-Vx)/R2a - (I + Vin/R1).

Vx = I(R3a)*R3a = [ (Vs-Vx)/R2a. - (I + Vin/R1) ]*R3a

The opamp negative feedback also sets the -ve input at Vx.

So I(R3b) = (Vs-Vx)/R2b - I2 = (Vs-Vx)/R2b - (I - Vin/R1).

So V(R3b) = I(R3b)*R3b = [ (Vs-Vx/R2b) - (I - Vin/R1) ]*R3b.

Voltage out, Vout = Vx - V(R3b).

Vout = [ (Vs-Vx)/R2a - (I + Vin/R1) ]*R3a
- [ (Vs-Vx)/R2b - (I - Vin/R1) ]*R3b

Now make R2a=R2b=R2 and R3a=R3b=R3, and cancel things.

Vout = -2.Vin*(R3/R1).
 
F

Fred Bloggs

Jan 1, 1970
0
Tony said:
Pullup-R's DON'T cancel.


Well, let's do some sums. Admittedly on a simplified and
idealised circuit (to make life easy for me).
R2a and R2b are the Pull-up R's in question.

-------+-------------+---Vs
| |
\ \ R3b
/R2a R2b/ +-/\/\---+-->Vout
\ \ | |
| | | _ |
| +---------+--|- \ |
| | | >-+
Vx---> +-------------|---------+--|+_/
| | |
\|/I1 I2\|/ \
| | /R3a
|/ \| \
Vin ---|npn npn|----. |
|\e e/| | |
0v--- | R1 | ---+--+---0v
+----/\/\-----+
| |
\|/ I I \|/
Constant current

I1 = (I + Vin/R1) and I2 = (I - Vin/R1).

First find out where I1 sets the opamp +ve input (Vx).

I(R3a) = (Vs-Vx)/R2a - I1 = (Vs-Vx)/R2a - (I + Vin/R1).

Vx = I(R3a)*R3a = [ (Vs-Vx)/R2a. - (I + Vin/R1) ]*R3a

The opamp negative feedback also sets the -ve input at Vx.

So I(R3b) = (Vs-Vx)/R2b - I2 = (Vs-Vx)/R2b - (I - Vin/R1).

So V(R3b) = I(R3b)*R3b = [ (Vs-Vx/R2b) - (I - Vin/R1) ]*R3b.

Voltage out, Vout = Vx - V(R3b).

Vout = [ (Vs-Vx)/R2a - (I + Vin/R1) ]*R3a
- [ (Vs-Vx)/R2b - (I - Vin/R1) ]*R3b

Now make R2a=R2b=R2 and R3a=R3b=R3, and cancel things.

Vout = -2.Vin*(R3/R1).

It's simpler to look at the collector circuits as Norton current sources
with shunt impedances the R2's. Then each is driven by a differential
current of +/-Vin/R1 which is an equivalent voltage drive of
+/-Vin*R2/R1 in series with R2 into the DA. The DA only responds to the
equivalent differential input, which is then 2*Vin*R2/R1, with a gain of
R3/R2, for a Vout=2Vin*R3/R1.
 
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