Improved mic amp posted on Rapidshare

[Eeyore]

Eeyore a écrit :


LOL!
That's your understanding of transistors noise mechanism that need to be
*refreshed*
You've still not correctly explained where does the re noise come from,
which explains why this noise source is half what you think it is.

Eh, to answer this question you have time to draw a book from your
library and read the chapter

Hey. You're here.

If you're such an expert why not tell us ?

Graham

 

[Eeyore]

Is there something about that obsolete NJM amp I'm not aware of?

The fact it's not obsolete ?

You do nothing but dig ever bigger holes for yourself !

Graham

 

[Eeyore]

I could
say make Ie1=2mA so that Ie2 then has to =2mA and then the error amp
pulls the remaining 1mA from each 6.8K putting the OA outputs at about
-6.8V and the pnp collectors at -17+3=-14V. There's an answer. But
(17/4.7K-Vout,IC1a/6.8k)R19=(17/4.7K-(-)Vout,IC1a/6.8K)*R18=>Vout,IC1a=0.
So no currents shunted thru the 6.8K's and Ie's=17/4.7K or 3mA or
whatever. I just noticed this when I saw he dropped the (+) input of
IC1B to GND- wherever that input is referenced determines the
output...details, details, details...

You're an idiot.

Graham

 

[Eeyore]

Jim Thompson a écrit :
Uh?
OK, since Graham wouldn't say it anyway.

What makes all this work as stated is that the output CM is fixed to
ground by IC1B.

You have any better ideas for it ? That's simply once again 'one way' of looking at
it.

Graham

 

[Fred Bloggs]

Fred Bartoli wrote:




You have any better ideas for it ? That's simply once again 'one way' of looking at
it.

Graham

You're no longer entertaining or convincing, just another evasive and
pretentious poof.

 

[Fred Bartoli]

Eeyore a écrit :

Hey. You're here.

If you're such an expert why not tell us ?

Here you go.

re isn't a resistance but the reciprocal of the BJT transconductance.

re= kT/(q.Ic)

By summing it with the other true resistors you imply that re noise is
Johnson noise, which it isn't. It's the noise induced in the noiseless
re (re isn't a real resistor) by the collector current shot noise, that
you see.

So en(re)^2 = re kT/(q.Ic) . 2q Ic = 2 re kT

If it had been Johnson noise you'd have had 4 kT re.

This is why the re *equivalent* noise resistance is half re and not re
as you incorrectly stated.

QED

 

[Fred Bloggs]

View in a fixed-width font such as Courier.


.
.
.
.
.
. I
. / \
. -- -->----------+------------+----6.8K ---.
. \ / | | |
. | | (-) |
. I | +------|\ |
. / \ | | >---+-R-+---R-----.
. -- -->---+-------------------+------|/ | |
. \ / | | | (+) | |
. R18 R19 | | (-) |
. | | | +-----|\ |
. | | | | >-+ V
. | | | VR--|/ |
. --- --- | (+) |
. '----6.8K------------------'
.
.
.
.
. Vi>--- R19-----+----6.8K ---.
. | |
. Vi=I*R19=I*R18 | (-) |
. +------|\ |
. | >---+ 2VR-Vo .------.
. Vi>----R18 ----+------|/ | | |
. | (+) | | R
. | | |(-) |
. | '----R-----+--|\ |
. | | >-+ Vo
. | VR--|/ |
. | (+) |
. '----6.8K----------------------'
.
.
.
.
. Vi*6.8+(2VR-Vo)*R19=Vi*6.8+Vo*R18
.
. or
.
. Vo=VR
.
.

Technically the Vi=I*R18-17V, makes no difference to the analysis of Vo
but is good to know for the Vicmr for the OA.

 

[Fred Bartoli]

Fred Bartoli a écrit :

Eeyore a écrit :

Here you go.

re isn't a resistance but the reciprocal of the BJT transconductance.

re= kT/(q.Ic)

By summing it with the other true resistors you imply that re noise is
Johnson noise, which it isn't. It's the noise induced in the noiseless
re (re isn't a real resistor) by the collector current shot noise, that
you see.

So en(re)^2 = re kT/(q.Ic) . 2q Ic = 2 re kT

Oops, was a bit too fast for Graham.

Should have detailed it:
en(re)^2 = (re sqrt(2q Ic))^2 = re kT/(q.Ic) . 2q Ic = 2 re kT

 

[Fred Bartoli]

Eeyore a écrit :

You have any better ideas for it ? That's simply once again 'one way' of looking at
it.

So where's your other way?

 

[Eeyore]

You're no longer entertaining or convincing, just another evasive and
pretentious poof.

Whilst you're incapable of providing a working circuit.

Graham

 

[Eeyore]

Well, I am still tripping on narcotics. Let's see... at extreme opamp
swing, with lower feedback resistors, could you zener a b-e junction
and reverse the feedback phase?

You know I've forgotten the precise details now (I need to put something down on paper) but
basically yes, you get latch-up. I only saw it at power-on. You've made me think about it
more now.

Graham

 

[John Larkin]

Not as such.



I use +/- 17 volts to obtain maximum headroom (typically +22dBu single ended) without
pushing the op-amps to their absolute maximum (18V) ratings. A number of other
manufacturers use 16V. I reckon if you're going to use more than 15V why not do it in
style ?



That's why those diodes are there.

Still, 17 volts or so differential is scairy. I know that zenering a
b-e junction permanently wrecks beta... I wonder if it affects nf as
well?

Come on, you'll get it. The big clue is in the change in the value of feedback R. I've
almost given it away.

Well, I am still tripping on narcotics. Let's see... at extreme opamp
swing, with lower feedback resistors, could you zener a b-e junction
and reverse the feedback phase?

John

 

[John Larkin]

Minor tweak:


John

 

[Jim Thompson]

Late at night, by candle light, Jim Thompson

Late at night, by candle light, Jim Thompson
<[email protected]> penned this immortal
opus:

On Sun, 18 Mar 2007 12:03:58 +0100, Fred Bartoli
[snip]

Details please, instead of 'did you heard of feedback'.

No doubt the donkey copied it from somewhere. It's far too clever for
HWB (he who brays ;-)

...Jim Thompson

So why don't you come up with why it won't work? Are you just another
pussy-foot dive expecting others to do it so you get to jump on them
in case they get something wrong?

- YD.

YD, Was that statement directed at me?

I'm not the one who said it didn't work... that was Fred Bloggs.

Oh well, what with everyone doing a song and dance about it who's to
know.

The circuit does work, but if I just spit out the answer what good
would that do YOU?

I understand how it works, YOU will only understand if you put forth
some thought process yourself.

Or are you just another one of those "cut-and-paster" circuit
"designers"... totally out-to-lunch if you can't find it analysed in
AoE or similar circuit compendium ?

Quoting my father, "The only things that are learned well are those
things that are learned the hard way".

...Jim Thompson

I did get around to it. Initially not in a quite formal way but after
some thought and doodling the light lit up. Some posts helped too.
Just got pissed at all the prancing around about it.

- YD.

Doesn't take any fancy footwork at all, just some FUNDAMENTALS...

(1) Check to see that loops are truly negative feedback... INN of IC1A
back around to INN of IC1A... one inversion; INP of IC1A thru IC1B
back around to INP of IC1A... one inversion.

(2) IC1A can only be "happy" if INP (+IN) and INN (-IN) are EQUAL
voltages.

(3) This can only be true if the currents thru R18 and R19 are EQUAL.

(4) Which means that the right-hand-ends of R6 and R9 must be at the
same potential.

(5) The only way (4) can be true is if the potential is ZERO (observe
unity feedback around IC1B).

(6) As a sanity check make sure that if either OpAmp is railed (plus
or minus) the inputs to IC1A remain in the allowed common mode range
(and the PNP's aren't saturated). Confirmed.

Q.E.D.

...Jim Thompson

 

[Fred Bloggs]

Doesn't take any fancy footwork at all, just some FUNDAMENTALS...

(1) Check to see that loops are truly negative feedback... INN of IC1A
back around to INN of IC1A... one inversion; INP of IC1A thru IC1B
back around to INP of IC1A... one inversion.

(2) IC1A can only be "happy" if INP (+IN) and INN (-IN) are EQUAL
voltages.

(3) This can only be true if the currents thru R18 and R19 are EQUAL.

(4) Which means that the right-hand-ends of R6 and R9 must be at the
same potential.

(5) The only way (4) can be true is if the potential is ZERO (observe
unity feedback around IC1B).

(6) As a sanity check make sure that if either OpAmp is railed (plus
or minus) the inputs to IC1A remain in the allowed common mode range
(and the PNP's aren't saturated). Confirmed.

Q.E.D.

...Jim Thompson

All they've done is scarf a popular INA topology. See the literature..

 

[Jim Thompson]

Jim Thompson a écrit : [snip]

The only requirement is that any circuit gyrations must stay within
the common-mode range of IC1A. In this case they do.

Yep.

Though I have seen cases where that restriction is not necessarily
required either.

I'd be curious. Can you cite one?

Built a 5V LDO using a 741 and a 2N2907 back in 1970 (hybrid).

Powered up, 741 inputs were at ground (most negative supply), so
outside the proscribed operating conditions.

However output of 741 was always LOW under these conditions, causing
the LDO to self-start.

To be found, to this day, in the TOW missile ;-)

===

Another, of course, is the case of one input to an LM339 inside the
proscribed common-mode range, and the other input is way above rail.
Output reflects the comparison properly in spite of this.

...Jim Thompson

 

[Jim Thompson]

All they've done is scarf a popular INA topology. See the literature..

Not surprising. But I wouldn't know without looking. In a way I'm
better off being ignorant of the way things are done by other
companies.

My most famous case was when I was hired by Silicon Systems to make a
pin-for-pin replacement for a National Hard-Drive Controller Chip...

Not only was I ignorant of HD controllers, I had never even seen
National's schematics.

So I did a design based just on specifications.

My design out-performed National's and the chip was smaller ;-)

...Jim Thompson

 

[Eeyore]

All they've done is scarf a popular INA topology. See the literature.

TI's INAs are actually a knock-off of this one.

The design as described was around in 1987 for sure.

Graham

 

[Rich Grise]

Jim Thompson a écrit : [snip]

The only requirement is that any circuit gyrations must stay within
the common-mode range of IC1A. In this case they do.
Yep.

Though I have seen cases where that restriction is not necessarily
required either.

I'd be curious. Can you cite one?

Built a 5V LDO using a 741 and a 2N2907 back in 1970 (hybrid).

Powered up, 741 inputs were at ground (most negative supply), so
outside the proscribed operating conditions.

However output of 741 was always LOW under these conditions, causing
the LDO to self-start.

To be found, to this day, in the TOW missile ;-)
===
Another, of course, is the case of one input to an LM339 inside the
proscribed common-mode range, and the other input is way above rail.
Output reflects the comparison properly in spite of this.

...Jim Thompson

Do you mean "prescribed"?
http://www.google.com/search?hl=en&q=define:+proscribe

;-)

Cheers!
Rich

 

[Eeyore]

Do you mean "prescribed"?
http://www.google.com/search?hl=en&q=define:+proscribe

;-)

Well spotted !

Graham