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.
John