On Mon, 12 Mar 2007 23:56:08 GMT, "colin"
wrote:

"John Larkin" wrote in message
.. .
On Mon, 12 Mar 2007 17:34:09 GMT, "colin"
wrote:


"Jim Thompson" wrote in
message news On Mon, 12 Mar 2007 17:14:30 GMT, "colin"
wrote:

"Jim Thompson" wrote in
message ...
On Mon, 12 Mar 2007 17:00:54 GMT, "colin"
wrote:

"Jim Thompson" wrote
in
message ...
On Mon, 12 Mar 2007 16:32:05 GMT, "colin"
wrote:

"Jim Thompson"
wrote
in
message ...
Is S.E.D actually sci.electronics.DUMMIES ??

When I saw this original post....

From: "powerampfreak"
Newsgroups: sci.electronics.design
Subject: Somebody explaining this design?
Date: 9 Mar 2007 12:08:35 -0800
Organization: http://groups.google.com
Lines: 11
Message-ID:
.com

I commented, "Looks like crap to me ;-)"

The response from the hot air crowd, you know, the ones posing as
guru's, was an implication that I was incorrect.

So I posed a simple question... do a hand analysis of the gain of
the
circuit.

It's a simple analysis (if you're not a faker :-)

It's been about 40 hours since I posted that request/taunt.

Nary a peep.

So I think this is TRULY...

sci.electronics.DUMMIES ;-)

I'd not noticed that post till now when I had to go search for it,
I dont look here for a few days sometimes,
I could end up spending most of the day reading and answering posts
here,
wich I feel like ive ended up doing on a few occasions,
It amases me how the likes of Win and several others make so many
posts
wich
go into such detail,
maybe they read/think/type a lot faster than me ?

but this looks like a piece of pie to me,
the input stage is just a differential pair made with Sziklai
darlington
the rest is just taking the ratio of the right resistors .... there
easy
as
cake.

Colin =^.^=


So write an equation. Then the defect/poor-"design" will be
obvious.

Wel ok if you insist, im kinda busy with my lightspeed converter so il
just
give it a quick going over,
in the first stage ignoring emitter resitance as its a darlington,
current is due to the input voltage apearing accros the resistance
between
emitters,
wich is (r2+r6) in parallell with (R9+vr1)
this current flow through the collector load
consisting of (r4+r8) in parallel with (R10,R11)
and produces a voltage - good old ohms law again,
this differential voltage is amplified by the op amp stage in the
ratio
of
its feedback resistors,
ie r13/r10. to produce a single ended op voltage.

see - its just a question of resistance ratios.
now when do I get my pie or cake ?

Colin =^.^=

So write it as a single equation. You're waffling ;-)

jeez, you want me to expand out simple ohms law and stuff ?
...
ree = 1/(1/(r2+r6)+1/(r9+vr1))
rcc = 1/(1/(r4+r8)+1/(r10+r11))
gain = rcc/ree * r12/r11

I should of just said the gain is 42 that would of been easier,
ot would probably be right for some value of vr1 too !

Colin =^.^=


Write down A SINGLE EQUATION showing that the gain is "42" ;-)

(1/(r2+r6)+1/(r9+vr1)) / (1/(r4+r8)+1/(r10+r11)) * r12/r11

g=42 when vr1=675r

Ok, enough Im realy bored with this now,
why cant we talk about something more difficult like trying to measure the
speed of light in a Single direction ?

Colin =^.^=


Get an rf signal generator, 100 MHz maybe, near an oscilloscope. Run a
long coax to a laser (pointer type is fine) from the generator. Run
another long coax from a pin photodetector (with maybe a NON AGC
amplifier) back to the scope. Start with the laser close to the pin
and measure phase shift. Move them apart, ditto.

This would work with a pulse generator, measuring arrival time, too.

but the signal traveling down the coax is governed by the speed of light,
by the time the received signal is brought next to the transmited signal it
has undergone a roundtrip,
moving the optical devices apart just alters the total trip, not just the
trip in one direction only.

Colin =^.^=


The coaxes add a constant delay, which you just subtract out.

C = deltaDistance/deltaTime

Past that, you're getting philosophical.

John