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Cute amplifier - bootstrapped

F

Fred Bartoli

Jan 1, 1970
0
Jim Thompson a écrit :
On Tue, 22 Jan 2013 11:49:10 -0800 (PST),


Larkin is just trying to pick a fight... ignore him... he's clueless.

The point is that the "bootstrap" did NOT buy more gain, it did buy
some additional bandwidth.

Huh???
The bootstrap buys gain where it bootstraps. (read the bootstrap low
corner frequency is way too high for you to see it effective at low
frequency)
Just boost the bootstrap value to 100u or 1000u and see the difference.
 
J

Jamie

Jan 1, 1970
0
Jim said:
I don't think I was the one who introduced the word "audio".

I just tried to introduce some engineering into the discussion, but
failed... seems this is a "faith" group rather than engineering.

How do you explain the following simulations...


http://www.analog-innovations.com/SED/James_Arthur_Bootstrap_Cute_Amplifier-Current_SED.pdf

I did properly guess that the input device was actually a capacitive
transducer (C2, when I saw the 40KHz peak), and James acknowledged
that.

The "bootstrap" doesn't buy any net "gain", but it does buy some
bandwidth.

The "discussion" does buy the entry of every punk in the world who
can't even do Ohm's Law, but needs to hear his head roar ;-)

...Jim Thompson

That must make you feel right at home :)

Jamie
 
R

rickman

Jan 1, 1970
0
Any AC coupled amp is a differentiator.

That's not true. A high pass filter is not the same as a
differentiator. Even I know that. They may be approximately the same
in a limit condition, but they won't behave the same in a real circuit.

Rick
 
J

Jon Kirwan

Jan 1, 1970
0
I don't think I was the one who introduced the word "audio".

I just tried to introduce some engineering into the discussion, but
failed... seems this is a "faith" group rather than engineering.

How do you explain the following simulations...


http://www.analog-innovations.com/SED/James_Arthur_Bootstrap_Cute_Amplifier-Current_SED.pdf

I did properly guess that the input device was actually a capacitive
transducer (C2, when I saw the 40KHz peak), and James acknowledged
that.

The "bootstrap" doesn't buy any net "gain", but it does buy some
bandwidth.

The "discussion" does buy the entry of every punk in the world who
can't even do Ohm's Law, but needs to hear his head roar ;-)

...Jim Thompson


Just found this on page 85 of "Small Signal Audio Design",
2010, by Douglas Self:

V+
|
|
|
\
/ R4
\ 47k
/
| C2
| || 47u
+---||---,
| || |
| | V+
\ | |
/ R2 | |
\ 47k | |
/ | |
| | |
| | |/c Q1
+------------|
| | |>e
| | |
| | | C1
| | | ||
| | +----||----+--OUT
| '-----+ || |
C4 | | |
|| R6 |/c Q2 | |
IN---||----/\/\--+------| | |
|| 68k | |>e | \
| | | / RLoad
| | | \ ""
| | R5 | /
+----------------/\/\---+ |
| | 220k | |
| | | |
| +------, | |
| | | | gnd
\ | | |
/ R7 \ | \
\ 82k / R3 --- C3 / R1
/ \ 33k --- 47u \ 2.7k
| / | /
| | | |
| | | |
| | | |
gnd gnd gnd gnd

Described as a 2-transistor shunt feedback stage with
bootstrapping added to the first stage to improve linearity.

Jon
 
J

Jon Kirwan

Jan 1, 1970
0
Nonsense.

You love writing equations but if you don't understand them then it's
just obfuscation and, in a bootstrap, the C is large so that it is,
for all practical purposes, an AC short at the frequencies of
interest.



If you had bothered to read the description you might have gleaned
that the 'purpose' was to "improve linearity" and it does that by
increasing gain so there's lots of negative feedback.


That's a bold statement coming from someone who still thinks shorting
the input signal to ground is a good idea.

Here are links to the surrounding pages of Self's book. The
gain is supposed to be about 3 in all cases, I gather. He
starts with a simple amp and makes incremental mods,
displaying some graphs to illustrate the discussion:

http://www.infinitefactors.org/docs/Small-Signal-Self-p81.png
http://www.infinitefactors.org/docs/Small-Signal-Self-p82.png
http://www.infinitefactors.org/docs/Small-Signal-Self-p83.png
http://www.infinitefactors.org/docs/Small-Signal-Self-p84.png
http://www.infinitefactors.org/docs/Small-Signal-Self-p85.png
http://www.infinitefactors.org/docs/Small-Signal-Self-p86.png
http://www.infinitefactors.org/docs/Small-Signal-Self-p87.png
http://www.infinitefactors.org/docs/Small-Signal-Self-p88.png

I think this is fair use of the copyrighted material. It
should provide much better context.

Jon
 
J

Jasen Betts

Jan 1, 1970
0
How it works, and why, has already been explained to you. It's also
been explained that shorting the input signal to ground is not a
'good' simulation and since you can't grasp any of it there's not much
reason to bother with you any more.


If the capacitor is a transducer the input is the base of Q1

why fo you think the Norton equivalent circuit for the transducer
is unacceptable?

âš‚

--- news://freenews.netfront.net/ - complaints: [email protected] ---
 
T

Tim Williams

Jan 1, 1970
0
flipper said:
So even if one knows nothing about 'capacitive transducers' you can
still ask yourself: would anyone design a 'capacitive transducer' in
which there's essentially 'no signal' at the intended frequency of
operation?

Just to stir the pot, ;-)

When I hear "capacitive transducer", I think "capacitance as a function of
x", usually x being (direct or acoustic) linear or angular displacement,
humidity, ionic, etc.

A variable capacitor, biased from a polarizing voltage, will indeed
generate a signal, but the signal from, say, a 100pF air variable being
turned a few degrees at 100Hz, is microscopic. Trying to measure changes
in humidity by the change in voltage, at constant charge, is absurd.

Acoustic transducers with external polarization (rather than electrets or
peizos) do work reasonably at high frequencies (10s-100s kHz), which would
be more what we're talking here, but I'd hardly call such a source "no
signal", even though it might be mV, even uV ripple out of a V bias. It's
simply what it is. The voltage might even be much weaker than the air
variable's case, but because the frequency is higher, the impedance is
lower and easier to deal with.

Most of these transducers, in any case, are measured parametrically, with
an RC or LC oscillator, or a C bridge, where the signal is stronger or
easier to detect (frequency measurements being the most accurate method
available).

Tim
 
T

TralfamadoranJetPilot

Jan 1, 1970
0
Your _main_task_ is to swat anybody who doesn't treat you with the
respect you feel you deserve? I feel so let down--here I thought you
were a dispassionate seeker after Engineering Truth. ;)

Cheers

Phil Hobbs

I thought ET was like big time wrestling... |-P

(Oh... wrong "ET") tee hee hee
 
J

Jasen Betts

Jan 1, 1970
0
I would have no problem with an 'equivalent' model but, as implemented
by JT, they are obviously not 'equivalent' since they produce wildly
different results, which is pretty much the definition of 'not
equivalent'.

What an acceptable model 'is', regardless of whether it's Thevenin or
Norton, is precisely the question at hand.

Now, consider his 'current source' model. It's an AC current source
into a capacitor to ground, the junction of which is the 'signal' into
the base of Q1. However, at 40 kHz 98% of the current source's AC
signal goes through the cap to ground and, just so there's no
confusion, that is entirely independent of amplifier topology.

noy having examined the circuit in detail 98% seems quite high,
what's the resitance of the base-emitter junction?

The norton equivalent models uses the available (short-circuit) current
as the input while the thevenin model uses the open circuit voltage.
So even if one knows nothing about 'capacitive transducers' you can
still ask yourself: would anyone design a 'capacitive transducer' in
which there's essentially 'no signal' at the intended frequency of
operation?

The current that is "lost" doesn't cost anything, it's just not realised.
Furthermore, does it makes sense to you that a 'capacitive
transducer' has maximum output at low frequency but virtually none at
the frequency of operation? So why would anyone excite it at 40 kHz
when there's 16 dB more signal at 100 Hz?

I don't see how that follows. obviously the current needs to be scaled
with the frequency if you want to match a fixed-voltage thevenin
equivalent.
A simple sanity says the 'current source' model doesn't work

if the model really shows 98% of the current flowing through
the capacitor a lower resistance Q1 would seem advantageous.
 
M

MrTallyman

Jan 1, 1970
0
if the model really shows 98% of the current flowing through
the capacitor a lower resistance Q1 would seem advantageous.


Shouldn't that say "the current coupled by the capacitor" as opposed to
flowing through?

Read "coupled" as 'shunted to ground'.

Maybe that will help you see it.
 
R

rickman

Jan 1, 1970
0
In real life, there is no such thing as a true differentiator. The
circuit that James posted is AC coupled, and JT (and not I) called it
a differentiator... just above.

My point was that all AC coupled amps behave sort of like
differentiators. James' circuit is no different.

Ok, I'll bite, why is there no such thing as a "true" differentiator?
The op-amp circuit above is a pretty good example of a "true"
differentiator in my opinion.

I was referring to *your* comment that "Any AC coupled amp is a
differentiator". That is not correct, period.

Rick
 
R

rickman

Jan 1, 1970
0
That's stupid.

You're being Larkinesque evasive (and obfuscative... have you been
taking lessons ?:)

Write the equations. Already!

Tick-tock.

You're going down. It always happens to sycophants, you take the
fall, Larkin walks away>:-}

...Jim Thompson

My dear GOD! You guys are all what, 12 years old???

I bet you and Larkin get together at night and watch SuperTroopers!

"Bite it, rook!"

Rick
 
R

rickman

Jan 1, 1970
0
A true diff would need infinite bandwidth.


It is by Jim's standards for a differentiator.

So all of your statements are wrong. A differentiator is a
differentiator even if it is not "perfect" with ideal components.

AC coupled amps are not "sort of like differentiators". They are like
low pass filters. The only way to consider them differentiators is in
the limit as your signal amplitude diminishes to zero. In that case
your circuit is also not doing anything useful "in the limit".

I see why you have so much trouble with so many people here.

Rick
 
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