Stupid RF newbie has a question...

[Dave]

Tim and I already answered your specific question about '||'. That
means 'in parallel', and I gave some examples.

As for '&', now that I see the document, it plainly means simply
'and'! The formula was presumably badly typeset, and should have had a
blank space on both sides of the '&'.

But isn't that obvious from the context? I don't think you're being
careful enough in your reading. In your reply to Paul Burridge you
said: "There is no Rb in the circuit illustration, and R1 is *not*
parrallel with R2."

Directly above the formula in question, the author says:
"The Thevnin equivalent of VCC, R1 and R2 in circuit # 3 is Vin and Rb
in circuit # 2." If you scroll back up to Circuit 2, which you
presumably read earlier, you'll see Rb.

Note that some care is needed when using symbols like 'RB'; in his
Circuit 2 the 'B' is actually a little smaller than the 'R', but it's
not well-positioned. I'd have written 'Rb'. That's particularly
sensible when you have no subscript symbols available, as is the case
in these newsgroup threads.

As for R1 || R2, Thevenin is about *equivalences*. Presumably you've
read up on his Theorem? (One of hundreds of relevant links is:
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/thevenin.html )

So, the 'R1||R2' part of the formula is meant to be read as 'the value
of R1 and R2 in parallel.

The author has also consistently mis-spelt 'Thevenin'.

Yeah, okay, I see it now. Damn I've gotten dense. Sorry to be such a
bother and so hard to convince. I haven't even touched any of this in over
20 years, and that has apparently affected my ability to reason and actually
think. Can't tell you all how much I appreciate your patience. And your
persistance. I've got a lot of catching up to do. Thank you.

Dave
[email protected]

 

[Dave]

Um, please don't slap me, but I think the ampersand means, "and", and the
"in parallel" sign means "in parallel". It's a sentence, telling you that
the voltage at Vin is voltage-divided from Vcc by R1 and R2, and the
effective base input resistance for the signal is the value of the two
resistors in parallel, because from the signal's POV, Vcc and ground are
the same point. This would be the "input impedance" of the circuit, which
is the load seen by the driving source.

Well, that's a seat-of-the-pants kinda guy's description.

Hope This Helps!
Rich

Yeah, and that's what everybody's been telling me. I just couldn't get the
cobwebs out of my head and see it. Now, however, day is beginning to dawn.
I have a sleep disorder and have essentially been asleep for ten years. Now
a new drug I am taking is affecting that. Can't believe how dense I've
become. Have a long way to go to get back to where I once was.

Dave
[email protected]

 

[Dave]

R1 and R2 ARE in parallel for ac signals - each start off the same
point (the base) and go to a low-impedendace point (Vcc for R1,
GND for R2). That makes them in parallel.

I'm afraid the ampersand in this context just means "and". The author
was showing how to migrate from Circuit #2 to Circuit #3, and this
resulted in the two equations linking the components in Circuit #3
with the components in Circuit #2.

In Circuit #2, the base of the transistor "sees" Rb whereas in Circuit
#3 it sees R1 || R2. For circuits #2 and #3 to be equivalent, then,
we require Rb = R1 || R2 - and that's one of the equations he's listed.

To equate the DC voltages at the base in Circuits #2 and #3, he's
ignored the drop across Rb and said that in Circuit #2, the voltage
is Vin, whereas in Circuit #3, the voltage is Vcc divided down by
R1 and R2, ie Vcc * R2/(R1+R2) - and that gives you his second
equation, Vin = Vcc * R2/(R1+R2).
If you then solve the two equations he's shown (which happen to
be separated by an ampersand!), then you get:
R1 = Vcc * Rb / Vin
R2 = vcc * Rb / (Vcc - Vin)

Yeah, I can see that now. Thanks for the help. I've go a lot of relearning
to do, after not even touching any of this in over 20 years.

Dave
[email protected]

 

[Dave]

[please snip better!]

Okay, that I can see. Crap. Confusing.

Thanks for your help. Now if I can just figure the rest out...

There might be something in here that might help:
http://www.google.com/search?q=thevenin+equivalent&btnG=Search+the+Web

Good Luck!
Rich

Thanks for the link. I'll try not to bother any of you again, at least not
right away. Much appreciate the help.

Dave
[email protected]

 

[Terry Pinnell]

Yeah, okay, I see it now. Damn I've gotten dense. Sorry to be such a
bother and so hard to convince. I haven't even touched any of this in over
20 years, and that has apparently affected my ability to reason and actually
think. Can't tell you all how much I appreciate your patience. And your
persistance. I've got a lot of catching up to do. Thank you.

Don't worry, most of us have BTampersandDT <g>.

 

[Paul Burridge]

Um, please don't slap me, but I think the ampersand means, "and", and the
"in parallel" sign means "in parallel". It's a sentence, telling you that
the voltage at Vin is voltage-divided from Vcc by R1 and R2, and the
effective base input resistance for the signal is the value of the two
resistors in parallel, because from the signal's POV, Vcc and ground are
the same point. This would be the "input impedance" of the circuit, which
is the load seen by the driving source.

Rich, I may be full of sh*t (it has been known) but AIUI (or not) the
overall input impedance of the stage is R1||R2||ZQ where the last term
is the input impedance of the tranny (approximately Beta x Re+RE)
where Re is the dynamic resistance of the B/E junction and RE is the
emitter resistor (where used). Of course, I stand to be corrected (as
usual).

 

[Paul Burridge]

Yeah, okay, I see it now. Damn I've gotten dense. Sorry to be such a
bother and so hard to convince. I haven't even touched any of this in over
20 years, and that has apparently affected my ability to reason and actually
think. Can't tell you all how much I appreciate your patience. And your
persistance. I've got a lot of catching up to do. Thank you.

I can certainly sympathise with that! I was in your position about 3
years ago and I've *still* got a lot of catching up to do. :-(

 

[Tim Wescott]

Dave wrote:

-- snip --

Well, that sounds plausible until you look at the circuit issustration over
the text. There is no Rb, and R1 and R2 are not in parrallel. Please see
my reply to Paul Burridge.

-- snip --

Dave
[email protected]

It _has_ been a while for you. The network composed of Vcc, R1 and R2
looks to the base of the transistor like a lower voltage source in
series with a single resistor (Thevinen's Equivalent). The effective
voltage is just the divided Vcc, and the effective resistance is R1 in
parallel with R2.

 

[Dave]

Dave wrote:

-- snip --
-- snip --


It _has_ been a while for you. The network composed of Vcc, R1 and R2
looks to the base of the transistor like a lower voltage source in
series with a single resistor (Thevinen's Equivalent). The effective
voltage is just the divided Vcc, and the effective resistance is R1 in
parallel with R2.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Yes, I see that now. (shaking head) Somehow I just didn't realize how much
I have forgotten. I guess that still being able to figure out and fix
things around the house gave me a certain comfort level that was deceptive.
But, as I mentioned elsewhere, I've been asleep for ten years, and before
that only swapped boards for twelve years. I was remembering the kid who
taught himself the fundamentals in high school (where they had no official
electronics course) and then graduated from trade school (didn't have the
time or money for college) having only missed one question on one test one
time (and that out of carelessness.) I do apologize for being such a
bother, and I do appreciate everyone's patience. I really will try not to
bother you again, at least not right away. In the meantime, I have a
lot of re-reading to do.

Dave
[email protected]

 

[Rich Grise]

Rich, I may be full of sh*t (it has been known) but AIUI (or not) the
overall input impedance of the stage is R1||R2||ZQ where the last term
is the input impedance of the tranny (approximately Beta x Re+RE)
where Re is the dynamic resistance of the B/E junction and RE is the
emitter resistor (where used). Of course, I stand to be corrected (as
usual).

Yes, you're absolutely right. I'm real shaky in that aspect of knowing
what a circuit is doing; to me it's kind of a "black box." But yes,
the input impedance is as you've said; maybe Rb then only means the
effective base bias resistance, or something like that. Like in almost
everything else, I've learned just enough to sling some bullshit and
get in trouble. ;-)

Cheers!
Rich