QUOTE TO READ
Rx=a+b*c+1
In the above equation for output impedance, if you want to assume that 'c' is zero,
then the output impedance becomes a+1.
END QUOTE TO READ
Ratch,
If you read the above QUOTE TO READ tell me what the output impedance is.
Is it "Rx" or something else ??
MrAl,
Yes, I did define R[T]=R[c]||R[L], ignoring h[oe]. And I am saying that Rout, the output
impedance, is -R[T]||R[c], where -R[T] comes from the denominator of the transfer function when
the denominator is set to zero. I am NOT saying that the output impedance is R[T]||R[c].
Well if you want to say NOW that the output resistance Rout=RT||Rc that's fine, but
dont say that i misinterpreted your information when you presented it the way you did.
You solved for RT which was not the output resistance yet you stated clearly that is
was, so there was no other way to 'interpret' that. The attachment is a snap shot
of your own written attachment. You are clearly referring to the 'above' equation
when you talk about output impedance, and you even go as far as to calculate the
"output impedance" from the equation for -RT with hie equal to zero, and then state
that is the output impedance. See how misleading that can be?
I am not putting R[c] in parallel with R[T], I am putting R[c] in parallel with -R[T], where
-R[T] is obtained from the denominator of the transfer function when the denominator is set to
zero.
Yes, NOW you are putting Rc in parallel with -RT, but that's not what you did before as
the attachment clearly shows, and that's the reason for the misunderstanding. That's
what i am trying to get through to you.
R[L] is the load resistance. It is included within R[T]. It is not used for the output
impedance, but it will be used for the input impedance. The OP wanted to calculate that, too.
We never got around to doing that yet, although the formula is in the attachment on post #40.
Yeah, what is that "-Rg" supposed to be anyway, and why are you solving for it?
I dont get that at all, sorry.
Sure, but R[L] is not the unknown, it is a given. It is not used for the output impedance, but
for the imput impedance.
Huh??? It is a given? And where is it 'given' ?
No, you intrepreted what I wrote wrong, that is why you are confused.
Sorry have to chuckle a little here, because there was no other way to interpret it.
If you read my above QUOTE TO READ and the attachement, you can not 'intepret' it
any other way. Furthermore, if you calculate the output impedance with the 'new'
method you've described of putting Rc in parallel with -RT then you dont get the
result you quoted in the attachement when hie is assumed to be zero. So there is
no way you could have meant that RT was not the output impedance.
No, not the R[T] from my definition, -R[T] calculated from the denominator of the transfer
function when the denominator is set to zero. Until you understand the previous sentence, you
will be confused.
It is starting to look like you learned this a long time ago and then forgot how to
do it exactly, and that's ok really, nothing wrong with that. I am refreshing in
several areas myself.
Perhaps it be better to start with a different fresh problem. Something fairly simple, but not
trivial. I will let you choose one.
Ratch
Ok, i choose we do an exercise on writing what you really mean rather than writing something
else. That would make things go a lot smoother
Now getting back to the 'other' equation, the one above -RT which is -Rg.
Are you going to come back here now and say that that equation TOO isnt REALLY
for -Rg, but is REALLY for something else, or what?
Why are you solving for -Rg anyway?