Improved mic amp posted on Rapidshare

[Jim Thompson]

[snip]

Just come up with the friggin' gain equation and be done with it. If
you won't maybe someone else can. Maybe you're afraid of someone
correcting you.

You do happen to have time enough for posting here, and a lot of it,
how about doing something useful with it. If you already know what
you're on about the write-up shouldn't take more than five minutes of
your oh-so-precious time.

Bunch of supposedly grownups pussy-footing around making themselves
look silly.

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

 

[YD]

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.

 

[YD]

Late at night, by candle light, Eeyore

You mean the various voltages and currents ?

Graham

No, speeds, weights and rate-of-climb. What did you expect?

- YD.

 

[YD]

Late at night, by candle light, Jim Thompson

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]

Late at night, by candle light, Eeyore

You really are a bunch of silly old women.

Graham

And you're making like Zarah Leander about it.

- YD.

 

[Jim Thompson]

Late at night, by candle light, Eeyore


And you're making like Zarah Leander about it.

- YD.

Who dat? I don't recognize that name.

...Jim Thompson

 

[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.

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

 

[Eeyore]

Eeyore a écrit :

Yes. What's the mechanism that precisely bias the circuit.

Just 3 phrases!

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

Feedback action holds the outputs of the 2 op-amps at close to 0V too (and
there's no inter-emitter DC path) so the emitter current is set by (17-0.55) /
4k7 + (-0.55)/6k8.
= 3.5mA - 80 uA = 3.42 mA.

Collector potential is set by Ic ( with a frac tion of a percent of Ie ) i.e
3.42 mA times the collector load resistance of 1k5 ( 3.42*1.5 = 5.13V ) above
the -17V supply = -11.87V.

Graham

 

[Eeyore]

Just come up with the friggin' gain equation and be done with it.

Is *that* what you wanted to know !

It's ((6k8*2)+ Rv +13 +Z.470uF)) / (Rv +13 + Z.470uF) for the differential output.
For single ended (more normal) it's half that figure.

And yes, 470uF is getting a bit small. DC coupling would be better.

Graham

 

[Eeyore]

HWB scarfed it from someone else's design.

Almost everything in the world falls into that category.

However I have a last ace up my sleeve that AFAIK has never previously been
published or built.

Tomorrow maybe. It really is a cracker.

Graham

 

[Eeyore]

Late at night, by candle light, Eeyore penned this immortal opus:

No, speeds, weights and rate-of-climb. What did you expect?

I'm just amazed anyone needs to ask. To me it's obvious at a glance.

Graham

 

[Eeyore]

Late at night, by candle light, Jim Thompson penned this immortal
opus:

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?

Almost certainly.

The Bloggs and Bartoli characters undoubtedly fall into that category.

Graham

 

[Eeyore]

Is there a problem? I don't see it. R1 and R2 force current into the
transistor emitters, a tiny bit of which is stolen by the 6.8K
feedback resistors. Differential dc gain is nearly 0. It looks
perfectly DC-boring to me.

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

 

[John Larkin]

Eeyore a écrit :

Don't need an in depth analysis. Just answer *precisely* Fred's
question. It should take no more than 2 or 3 phrases.
So, restating it: how is the PNP's operating point well defined?

Is there a problem? I don't see it. R1 and R2 force current into the
transistor emitters, a tiny bit of which is stolen by the 6.8K
feedback resistors. Differential dc gain is nearly 0. It looks
perfectly DC-boring to me.

John

 

[Fred Bartoli]

Eeyore a écrit :

Almost certainly.

The Bloggs and Bartoli characters undoubtedly fall into that category.

I do understand how it works, but you seem to still miss something about
noise calculation

 

[Fred Bartoli]

Eeyore a écrit :

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

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

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.



(and

 

[Jim Thompson]

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. 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

 

[Eeyore]

Eeyore a écrit :


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

A different way of saying the same thing. IC1A in+ and in- are *always* at (nearly)
the same potential until feedback action breaks down !

There's more than one way to skin a cat as the proverb says.

Just because you want to analyse it differently doesn't make your way 'right' and my
way 'wrong'. Feedback is required to hold in+ and in- at the same potential. You're
just fixating about one point in the loop for no reason.

Graham

 

[Eeyore]

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

It's perfectly clear to me. How do you reckon I design stuff ?

Graham

 

[Eeyore]

Eeyore a écrit :

I do understand how it works, but you seem to still miss something about
noise calculation

Wheras your arithmetic needs some polishing.

Graham