Improved mic amp posted on Rapidshare

[Eeyore]

Opamp common-mode?

Not as such.

The rails aren't identified, but I assume they
aren't extreme. The -17 need not be that big.

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 ?

Hey, with the gain pot set low, can't a 48 volt zot still zener one of
the transistors pretty hard?

That's why those diodes are there.

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.

Graham

 

[Jim Thompson]

On Sun, 18 Mar 2007 12:44:25 -0800, John Larkin

[snip]

Hey, with the gain pot set low, can't a 48 volt zot still zener one of
the transistors pretty hard?

John

I didn't know anteater "zots" were measured in volts ;-)

...Jim Thompson

 

[Eeyore]

This is true, also, as is the previous statement.

Exactly. The two statements are effectively identical. It's just a case of where
you're choosing to look.

Graham

 

[Fred Bloggs]

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




Quite obviously you didn't understand me. You too are part of the
pussy-foot dance diva club.

- YD.

Not at all- I have answered all questions put to me about my gained up
mod's to the original circuit.

 

[Fred Bartoli]

Jim Thompson a écrit :

Eeyore a écrit : [snip]

Can we agree that the bases of the input tansistors are held at close to 0V
(I'll ignore Ib x 3k if that's ok with you) ? Vbe of the input devices is ~
550mV so the emitters are at a DC potential of +0.55V

Ok. You can even neglect Vbe. It's not the interesting point there.

Feedback action holds the outputs of the 2 op-amps at close to 0V too

True.

No. Not the feedback loop *alone*. You have to add one condition to make
that happen. That's the one I think Fred overlooked and Graham could
have so easily told us.

No. The feedback loop is differential and ensures that collector
currents are equal, i.e. IC1A in+ and in- are at the same potential.

This is true, also, as is the previous statement.

You're still missing the point. At least I think it's not clear to you.

I know you'll then just say 'it's bloody obvious'. But then, why didn't
you spotted where Fred missed the point. That would've been very easy to
anyone understanding why.

[snip]

...Jim Thompson

 

[Jim Thompson]

Late at night, by candle light, Fred Bloggs <[email protected]> penned
this immortal opus: [snip]

Nah- you need to work on improving your English comprehension and
learning how to read your newsreader...

Quite obviously you didn't understand me. You too are part of the
pussy-foot dance diva club.

- YD.

Not at all- I have answered all questions put to me about my gained up
mod's to the original circuit.

You haven't retracted, "That much is obvious but your differential
feedback only forces equality and not any particular Ic,Q for the
pnp's. So where is the magic I have missed? Did you ever build that
circuit?"

...Jim Thompson

 

[Fred Bartoli]

Eeyore a écrit :

Wheras your arithmetic needs some polishing.

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

 

[Jim Thompson]

Jim Thompson a écrit :

Eeyore a écrit : [snip]
Can we agree that the bases of the input tansistors are held at close to 0V
(I'll ignore Ib x 3k if that's ok with you) ? Vbe of the input devices is ~
550mV so the emitters are at a DC potential of +0.55V

Ok. You can even neglect Vbe. It's not the interesting point there.

Feedback action holds the outputs of the 2 op-amps at close to 0V too

True.

No. Not the feedback loop *alone*. You have to add one condition to make
that happen. That's the one I think Fred overlooked and Graham could
have so easily told us.

[snip]

Methinks YOU yourself aren't quite understanding the circuit.

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

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

...Jim Thompson

 

[Fred Bloggs]

HWB scarfed it from someone else's design.

He's clueless.

Don't anyone answer. I've said too much already.

Maybe we can get both HWB and Fred Bloggs simultaneously ;-)

...Jim Thompson

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

 

[Fred Bloggs]

Late at night, by candle light, Fred Bloggs <[email protected]> penned
this immortal opus: [snip]

Nah- you need to work on improving your English comprehension and
learning how to read your newsreader...


Quite obviously you didn't understand me. You too are part of the
pussy-foot dance diva club.

- YD.

Not at all- I have answered all questions put to me about my gained up
mod's to the original circuit.

You haven't retracted, "That much is obvious but your differential
feedback only forces equality and not any particular Ic,Q for the
pnp's. So where is the magic I have missed? Did you ever build that
circuit?"

...Jim Thompson

Well- I trusted you knew what you were talking about so went a micron
deeper into what the OA outputs were and voila- there it was. It is not
the differential feedback per se that determines the operating point, it
is the reference level about which that differencing occurs that
determines it.

 

[Jim Thompson]

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

Come on, Fred... Show us your analysis of "Improved mic amp posted on
Rapidshare"... no hand-waving... just an exact analysis ;-)

...Jim Thompson

 

[Fred Bloggs]

Come on, Fred... Show us your analysis of "Improved mic amp posted on
Rapidshare"... no hand-waving... just an exact analysis ;-)

...Jim Thompson

I just did, the equation
(17/4.7K-Vout,IC1a/6.8k)R19=(17/4.7K-(-)Vout,IC1a/6.8K)*R18 is the
condition the feedback imposes on the input of IC1A. The Vbe's are
neglected but other than that, exact enough.

 

[Fred Bartoli]

Jim Thompson a écrit :

Jim Thompson a écrit :

On Sun, 18 Mar 2007 20:37:44 +0100, Fred Bartoli

Eeyore a écrit :
[snip]
Can we agree that the bases of the input tansistors are held at close to 0V
(I'll ignore Ib x 3k if that's ok with you) ? Vbe of the input devices is ~
550mV so the emitters are at a DC potential of +0.55V

Ok. You can even neglect Vbe. It's not the interesting point there.

Feedback action holds the outputs of the 2 op-amps at close to 0V too
True.

No. Not the feedback loop *alone*. You have to add one condition to make
that happen. That's the one I think Fred overlooked and Graham could
have so easily told us.

[snip]

Methinks YOU yourself aren't quite understanding the circuit.

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.
OUT+ + OUT- = 0
The feedback loop ensures that
OUT+ - OUT- = 0 (at rest, obviously)

Only both conditions make OUT+ = OUT- = 0

Change the output CM and the input bias point move with it.

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?

 

[John Larkin]

Until those 6k8s get reduced (say to 3k3) in a drive for lower minimum gain.

Can you see where the problem occurs. It had me puzzled for a bit.

Graham

Opamp common-mode? The rails aren't identified, but I assume they
aren't extreme. The -17 need not be that big.

Hey, with the gain pot set low, can't a 48 volt zot still zener one of
the transistors pretty hard?

John

 

[YD]

Late at night, by candle light, Jim Thompson

Who dat? I don't recognize that name.

...Jim Thompson

Google is your friend.

- YD.

 

[YD]

Late at night, by candle light, Jim Thompson

Late at night, by candle light, Jim Thompson

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.

 

[Fred Bloggs]

Come on, Fred... Show us your analysis of "Improved mic amp posted on
Rapidshare"... no hand-waving... just an exact analysis ;-)

...Jim Thompson

This is the best simplification I can come up with:
View in a fixed-width font such as Courier.

 

[Fred Bloggs]

This is the best simplification I can come up with:
View in a fixed-width font such as Courier.


.
.
.
.
.
. I
. / \
. -- -->----------+------------+----6.8K ---.
. \ / | | |
. | | (-) |
. I | +------|\ |
. / \ | | >---+---------.
. -- -->---+-------------------+------|/ | |
. \ / | | | (+) R R
. R18 R19 | | (-) |
. | | | +-----|\ |
. | | | | >-+ Vo
. | | | 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
.
.

Oops:
View in a fixed-width font such as Courier.

 

[Eeyore]

Not at all- I have answered all questions put to me about my gained up
mod's to the original circuit.

Your circuit however is simply a bad joke.

Graham

 

[Eeyore]

Jim Thompson a écrit :

No. Not the feedback loop *alone*. You have to add one condition to make
that happen. That's the one I think Fred overlooked and Graham could
have so easily told us.

What part of op-amp use and behaviour are you unfamiliar with ?

Graham