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
Jim said:Now I would like to see a computation of the gain of this "marvy"
amplifier.
Phil said:"Eeysore = know nothing pommy charlatan and plagiarist."
Jim said:Well? Anyone? Mathematical solutions only... no simulations please
;-)
...Jim Thompson
Fred said:The circuit is a primitive paradigm, it would have been a simple matter
to include feedback.
Plagiarist ?
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.
what is the linear range for this pair ? => dynamic input range ?
would it be better to close the loop to the bases and have a fully
differential system.
Phil said:"Eeysore = know nothing pommy charlatan and plagiarist."
** Claiming other people's design work as being yours.
Plagiarist ?
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.
On Sat, 10 Mar 2007 15:28:19 -0700, Jim Thompson
[snip][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."
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 said:Still no response after 24 hours.
Where are all the "experts" when a real question is asked ?
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.
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.
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.
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
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?
Pullup-R's DON'T cancel.
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).