Ok great...
Then perhaps you would be good enough to go through a numerical example for oh say, the 2N3904 transistor. This is a common transistor for which data is available on the web. You could also state what value you intend to use for hie to calculate "RT".
I take it "RT" is what you are calling the original circuits output resistance then Rout? If not then explain how you get the value of Rout, and what numerically it comes out to in your numerical example.
Hi Jony,
How did you determine that Rout was 280 ohms?
I used this equation to find HieAlso, how did you measure an hie value of 2.2k ?
Here you have the three technique that came to my mind to measure hie in this circuitAlso, what technique do you use to measure hie in a given circuit like we are looking at with the original transistor circuit?
I build this circuit in LTspice And I use AC sweep to find Rout vs frequency
View attachment 61078
And Rout vs frequency
I change the BJT to 2N3904 and for F = 1KHz Rout = 210.718Ω
I used this equation to find Hie
Hie ≈ re * (β +1) ≈ 26mV/Ic*(β +1)
Of course to find Hie I can also use the LTspice (Hie = 3.378KΩ).
Or we can use the this datasheet
https://www.electro-tech-online.com/custompdfs/2012/02/2N3904.pdf (page 5)
And when I use the simulation data for Hie and Hfe and use Ratch equation to find Rout
Rg = 10K; RF = 47K; Hie = 3.38K Rc = 4.7K; Hfe = 300;
Rout = Rc || (Rg*Rf + Rf*Hie + Rg*Hie)/(Hie + Rg + Rg*Hfe) = 4.7KΩ||219.9Ω = 210Ω
It seems that Ratch equation is correct.
Yes, that's what i thought...doesnt 1020 ohms seem a little too high, like more than two times the true value? That's similar to what i was getting using your formula, so something is either missing or something else must be wrong.
And when I use the simulation data for Hie and Hfe and use Ratch equation to find Rout
Rg = 10K; RF = 47K; Hie = 3.38K Rc = 4.7K; Hfe = 300;
Rout = Rc || (Rg*Rf + Rf*Hie + Rg*Hie)/(Hie + Rg + Rg*Hfe) = 4.7KΩ||219.9Ω = 210Ω
It seems that Ratch equation is correct.
What we need to do is set this up in spice like Jony did and MEASURE hie.
Just as a reminder, the output impedance is NOT the impedance with the load resistor RL connected (and RL is not Rc, Rc is the collector resistor which goes to Vcc).
And I'm also in the team that thinks that Rc is a part of a amplifier not a load.
That is why I includes Rc in my Rout calculations.
MrAl,
I believe a little more clarification is in order. Why the big push on measuring the parameters of the 2N3904, when we all do not have the same transistor? The h[ie] of that transistor can vary from 1k to 10k and the beta from 100 to 400. So why can't we just agree on some parameter values and calculate the output impedance from there? I designated R[T] as the parallel combination of R[c] and R[L]=1000. In accordance with the General Immittance Theorem, when the denominator the transfer function is set to zero and and -R[t] is found, that value is the output impedance. -R[t] used in this manner for the output impedance is only a symbolic artifice, and not a real value. I perhaps should have included the value of R[c] in my node equations. Then R[t] would have only have been R[L]. Anyway, my figures say 385 ohms for the output of the transistor without R[c] in parallel. When R[c] is put in parallel, the value drops to 356 ohms. If I had included R[c] in my node equations, I would have gotten 356 ohms directly without calculating its parallel effect afterwards.
Ratch
"Why the big push on measuring the parameters of the 2N3904, when we all do not have the same transistor?"
Actually THAT is the reason why we DO have to measure parameters so this question doesnt make that much sense. If we all had the same transistor we would all have the same hie. If you want we can all use a pure linear model though, in which case a simple agreement would suffice as you say. I was using 4000 ohms for hie, and 180 for hfe because that is what i measured with the Spice model i have.
Sorry but you seem to be a bit hard to talk to when discussing these technical issues because you keep either changing your mind or you are misspeaking or something. This makes it take a lot longer to figure out what is going on.
For example, first you said that Rf did not induce negative feedback, then you changed it later to say that it actually does.
So you see how this can get confusing? I realize that it's not as easy to talk about these kinds of topics on the internet but the more care that goes into making a statement about fact the less time it takes to figure out what is going on, like when one equation is different than the other.
Before i sum up with the equation that has shown itself to be within 700 micro ohms of the exact output resistance, i'll back up a little to your equation which you nicely presented back in this thread in a pdf file, and show you one of the main reasons for the confusion.
If you view the attached gif file, you'll see i posted a quick snapshot of your equation and circuit. One thing you've shown here is that RT is the parallel combination of Rc and RL.
Now if we want to equate a parallel resistance Rp to two other resistors R1 and R2, we of course have:
Rp=R1*R2/(R1+R2)
In the case of RT, your pdf showed this:
RT=1/hoe||Rc||RL
and we agreed that we would ignore 1/hoe, so we are left with:
RT=Rc||RL
which of course means:
RT=Rc*RL/(Rc+RL)
Now if we solve for RL, we get:
RL=(Rc*RT)/(Rc-RT)
and here we can see that we dont get RL by putting Rc in parallel with RT.
See that now?
But if you do want to put Rc in parallel with RT anyway, then you should at least state the reason why it is valid to do such a thing and we can go from there, no problem.
Now the equation i wanted to present is this one:
Rout=(Rc*Rf*Rg+hie*Rc*Rg+hie*Rc*Rf)/(Rf*Rg+hfe*Rc*Rg+Rc*Rg+hie*Rg+hie*Rf+hie*Rc)
and that includes hie which of course has to be measured because it varies quite a bit with bias point selection.
Now this last equation computes an Rout that is within 700 micro ohms of the measured value with a perfectly linear model (like the hybrid model) and the only reason for the discrepancy is probably because of the accuracy involved in measuring the values of the voltages and currents. This uses the previously mentioned voltage source in series with a resistance as the input model rather than just the voltage source alone (which isnt too bad really i guess).
My challenge to you now is to see if you can match that equation with your own derivation
Yes, very good. But if you made it more clear in your attachment (and you still havent actually, you just refer to the theory without explanation, still, after i've told you several times) we would have reached the same equation long long ago.
You wrote (without 1/hoe): RT=Rc||RL
and now you are saying Rout=RT||Rc when you NEVER showed that in your attachment, nor anything like that.
So explain where you got the idea to put Rc in parallel with RT, and also explain what happened to RL, and what exactly is RL then?
Also, why mention that RT=Rc||RL if you never intend to use that?
RL appears to be the load, which would mean if we knew what RT was we would calculate RL from:
RL=(Rc*RT)/(Rc-RT)
In short, you wrote one thing and meant another thing several times now, so you cant expect anything but confusion.
So you are still left to explain by what mechanism you are allowed to put Rc in parallel with RT. If you want to refer to something on the web that would be fine with me.
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].
Yes, NOW you are putting Rc in parallel with -RT, but that's not what you did before asI 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.
Yeah, what is that "-Rg" supposed to be anyway, and why are you solving for it?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.
Huh??? It is a given? And where is it 'given' ?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.
Sorry have to chuckle a little here, because there was no other way to interpret it.No, you intrepreted what I wrote wrong, that is why you are confused.
It is starting to look like you learned this a long time ago and then forgot how toNo, 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.
Perhaps it be better to start with a different fresh problem. Something fairly simple, but not
trivial. I will let you choose one.
Ratch
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 ??
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?
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.
Yeah, what is that "-Rg" supposed to be anyway, and why are you solving for it?
I dont get that at all, sorry.
Huh??? It is a given? And where is it 'given' ?
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.
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.
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?
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