MrAl - thank you for the long answer.
I am very sorry, but I must state that - again - there is a big misunderstanding - or better: misinterpretation ?

Therefore, at first it seems that it is necessary to clarify the difference between static and dynamic/differential resistances:

(a) A static resistance (symbol R) is - according to Ohms law - a ratio of the two DC values R=V/I
(b) A dynamic/differential resistance (lowercase letters r) is the ratio of two infinitesimal small values r=dV/dI (or r=v/i).
That means: The quantity r is the slope of the corresponding non-linear function V=f(I).
In most cases, we have (example: Diode) I=f(V) and the slope is called "conductance g".
Of course, for a fixed DC bias point on the non-linear I-V characteristic the dynamic/differential quantity (r or g) has also a fixed value!!

That is the background for two comments to your last post:
1) (Quote:" 're' may or may not be dynamic, depending on what we are doing.")
No - re is always (per definition) a dynamic (differential) quantity because it is the inverse of the slope Ic=f(Vbe) at a fixed DC current.

2) (Quote: "Also, note that if you argue that 're' does change then that means that you dont really accept the value of Vbe")
There is no sentence in the whole discussion in which I have argued "that re does change...".
Instead, all my calculations were based on a fixed value 1/gm (=re) derived from the collector current Ic I have computed before (using the temperature voltage of Vt=26mV). As you know, the expression gm=Ic/Vt is the well-known (sufficiently good) approximation for the slope of the exponential Ic=f(Vbe) characteristic.

I hope I could clarify some misunderstandings (perhaps based on different definitions?).
Regards
LvW

Hello again and thanks for the reply,

We disagree on something here but i think it is simply because we are looking at the same thing differently.

The value for 're'. Let's concentrate on that for a minute.
First, when we apply a DC voltage for +Vcc, the circuit goes into a steady state mode after a few seconds or less, and the theoretical rationale is that at 't' (time) approaching infinity the voltage become constant. This puts a constant voltage across each and every element, including the capacitors. We can look at this in two different ways. One, where the cap voltages become constant, or Two, where we simply open circuit them. In both cases, we get the DC bias point analysis and that leads to a certain emitter current i call "iE" and i use lower case "i" because the upper case looks like a dang lower case "L" or a numerical "1" and i dont want that to happen. If you want to call it "Ie" that's up to you, but you see it is harder to read in some fonts.
OK, so we have the DC bias point established, and a value for 're' rendered, which if you want to call "Re" now that is just fine with me. The point here is that we MIGHT consider 're' or "Re" to be constant now. We dont HAVE to, but if we do, we find a simplification, and that simplification is that we no longer have to take averages of two readings to get the gain.

But let me back up a minute because i already said that in a previous post.
Let me now say that with that constant and/or dynamic 're' or 'Re', if we apply a VERY VERY small signal, on the order of 1e-6 volts to the input, we see a change in 're' that is SO SMALL that it becomes insignificant to even consider it in any way that will affect the calculation of the emitter current EVEN THOUGH the output voltage changes. We can say this the other way too, that the emitter current changes SO LITTLE that the calculation of 're' will be very insignificant, and yet the output voltage will go up (or down) to new level. If we were lucky enough to have zero volts out BEFORE this small input change then if we see 2e-6 on the output, then the gain is then:
G=Vout/Vin=2e-6/1e-6=2
and of course if it went down then it would have been -2 instead.
If we go back and check on 're', we see that it hardly changed at all and thus we can consider it constant for this test.

This is the way a lot of these problems work though, not just this one. The dynamic resistance of a simple diode is a simpler example. When we bias the diode and see a resistance of say 10 ohms, if we change the current through the diode by a very small amount we dont see the resistance change by much, and if the resistance does not change much then we may not see much difference in the observed variable when we either allow it to change or keep it constant. Granted, it depends on the application, but if the resistance is swamped by some other resistance like 100 ohms, then the resistance of the series combo goes from 110 to maybe 110.1 and often that's not significant.

Is it possible you are using a different model than i am, as Jony suggests?
What model are you using this time, and where is your 're' placed? I would like to verify that i use the same model or if different, than i would be more than happy to switch models to match yours.

 

MrAl - thanks again.
Regarding your first sentence I am happy to say: No, we do not disagree at all.
I support everything you wrote.
I never have suggested to assume a (slightly) varying value for "re".
As I have mentioned in my former post, the quantity 1/gm=re is nothing else than the inverse slope of the Ic=f(Vbe) curve at the fixed bias point (Ic) and as such it also has a fixed and finite value. No disagreement.
By its nature, the slope is always a differential quantity which we here in Germany sometimes also call "dynamic" quantity.
I think it's the word "dynamic" that made you probably think that I wouldn't consider the slope to be constant.
Most probably, this was the background of the misunderstanding between us.

As far as your last question is concerned: Of course, both models give the same result.
But I must admit that (as I have mentioned before) I do not like the term "re" and the corresponding re-model. I know that we can use this model - however, it contains the quantity 1/gm=re as a resistor symbol - but "re" is in fact NOT a two-pole resistor (not an internal emitter resstor) but a "transfer-resistor" because it connects the voltage between two nodes (VBE) with a current through another node (IC ).
And - as we have seen in our discussion - using the symbol "re" can cause severe misunderstandings, in particular because there are other external parts (RE1, RE2, RE).
Therefore , I prefer the Pi model and the corresponding form of the gain function: A=gmRc/(1+gmRE)=Rc/[(1/gm) + RE] .
(By the way, for calculating the bias point for the circuit under discussion I didn`t use any model - it is a standard procedure using superposition) .

Regards
LvW

 

Hello again LvW,

Thanks for clarifying some of this stuff.

You said you did not use a model for the bias point, but how is that possible? Surely there must be a circuit.
Also, what are you using for 'gm'?

 

Hello again LvW,

Thanks for clarifying some of this stuff.

You said you did not use a model for the bias point, but how is that possible? Surely there must be a circuit.
Also, what are you using for 'gm'?

I have used, of course, the original circuit diagram. May be that during calculation I had the equivalent diagram (model) in my mind, but i did not use it explicitely. It was not the first time I have analyzed such a circuit (I was teaching electronics for 25 years).
In post#4 I have already mentioned the value of gm:
Transconductance gm=0.0642 A/V (corresponding to Ic=1.67mA).

 

However, what did you get for 're' in that model?

Are you asking for this?
re = 25mV/1.89453mA = 13.195Ω

however in the simulation i got (using that value of 're'):
0.001853598

Strange, because I use this model in the simulation

And got this result Ic = 1.88823mA

 

I have used, of course, the original circuit diagram. May be that during calculation I had the equivalent diagram (model) in my mind, but i did not use it explicitely. It was not the first time I have analyzed such a circuit (I was teaching electronics for 25 years).
In post#4 I have already mentioned the value of gm:
Transconductance gm=0.0642 A/V (corresponding to Ic=1.67mA).

Hello,

And where did you get the 1.67ma from?

What i am doing is trying to track down why three or four of us here all get difference answers

 

Are you asking for this?
re = 25mV/1.89453mA = 13.195Ω





Strange, because I use this model in the simulation

And got this result Ic = 1.88823mA

Hi,

Ok thanks, and where do you place 're' then?

What i am doing is trying to track down why three or four of us here all get difference answers

 

Hello,

And where did you get the 1.67ma from?
What i am doing is trying to track down why three or four of us here all get difference answers

Of course, I have computed the DC bias current - using standard procedures based on the given data.
In addition, I have simulated the circuit (BC548) with results as given in my post#4:

IE=1.675 mA (Sim: 1.785mA)

By the way - one reason for slightly different answers may be the temperature voltage: 25 resp. 26mV

 

Of course, I have computed the DC bias current - using standard procedures based on the given data.
In addition, I have simulated the circuit (BC548) with results as given in my post#4:

IE=1.675 mA (Sim: 1.785mA)

By the way - one reason for slightly different answers may be the temperature voltage: 25 resp. 26mV

Hi,

Yes, 25 vs 26, but i saw that already and changed to what you guys had been using which was 25.

Now you gave two results up above for IE.
So which one are you actually using?
And when you say 'sim', which circuit are you using? If you look at Jony's reply you have the choice of 3 different ones i think.

I think i am getting close to an answer for all this.

 

Hi,
Yes, 25 vs 26, but i saw that already and changed to what you guys had been using which was 25.
Now you gave two results up above for IE.
So which one are you actually using?
And when you say 'sim', which circuit are you using? If you look at Jony's reply you have the choice of 3 different ones i think.
I think i am getting close to an answer for all this.

When I say "sim" - I am referring to PSpice simulation using the BC548 model and, of course, the given resistors.
This applies to the DC collector current aswell as the gain value (in post 4).

 

When I say "sim" - I am referring to PSpice simulation using the BC548 model and, of course, the given resistors.
This applies to the DC collector current aswell as the gain value (in post 4).

Hi again,

Thanks. So i will take it you are using values in post 4.

Now which of the three circuits are you using to get the DC bias point, one of the three in Jony's post with the three circuits?
Either first, second, or third.

 

Now which of the three circuits are you using to get the DC bias point, one of the three in Jony's post with the three circuits?
Either first, second, or third.

I am not quite sure if I understood your question - so let me answer this way:
All three "equivalent circuit diagrams" are well-known and can be found in most textbooks.
But one should realize that these diagram only reflect the well-known set of equations which are always used to analyze such a classical gan stage as shown in the task description. It is just a visualization of these equations - nothing else and not more.

So - it is not necessary to "use" these equivalent circuits if I one is familiar with these equations.
Hence, to see the original circuit diagram and to understand the working principle is sufficient.
That was the way I have treated the gain stage under discussion.
And - as I have mentioned in a former post, for finding the DC bias point I have used the superposition principle because it is a linear circuit (we always assume that its behaviour is linear) with two DC sources (Vcc and Vbe).

 

Ok thanks, and where do you place 're' then?

At the emiter, look here:

And the Transistor voltage gain is :

Av1 = - (Rc||RL)/(Re1 + re) * Hfe/(Hfe + 1) = - 15.4371V/V

and the whole amplifier gain is :

Av = Uout/Ug = Rin/(Rg + Rin) * Av1 = - 9.9363V/V

Where:

Rin = Rth||(hfe +1)*(Re1 + re) = 9.03161kΩ

Rth = R1||R2

 

At the emiter, look here:

And the Transistor voltage gain is :

Av1 = - (Rc||RL)/(Re1 + re) * Hfe/(Hfe + 1) = - 15.4371V/V

and the whole amplifier gain is :

Av = Uout/Ug = Rin/(Rg + Rin) * Av1 = - 9.9363V/V

Where:

Rin = Rth||(hfe +1)*(Re1 + re) = 9.03161kΩ

Rth = R1||R2

Hi,

Ok thanks, and what circuit did you use to determine the value of 're' then?

 

I am not quite sure if I understood your question - so let me answer this way:
All three "equivalent circuit diagrams" are well-known and can be found in most textbooks.
But one should realize that these diagram only reflect the well-known set of equations which are always used to analyze such a classical gan stage as shown in the task description. It is just a visualization of these equations - nothing else and not more.

So - it is not necessary to "use" these equivalent circuits if I one is familiar with these equations.
Hence, to see the original circuit diagram and to understand the working principle is sufficient.
That was the way I have treated the gain stage under discussion.
And - as I have mentioned in a former post, for finding the DC bias point I have used the superposition principle because it is a linear circuit (we always assume that its behaviour is linear) with two DC sources (Vcc and Vbe).

Hi,

I appreciate your reply but when we analyze a circuit and asked about it we dont say, "Well i analyzed the circuit", because that does not help the other guy understand the procedure you used.
What i need to know is what circuit you used to determine 're', or at least how you determined what 'IE" was.

The crux to all this i think is how we determine what 'IE' is (or a related parameter).
If you want to know why we got different results you have to give me all the information you used to determine your result or else i can not follow your line of reasoning. It could be the most traditional way to do this, but i will never know that if you dont get more specific and detailed. You see how Jony points out exactly what he is using each time. This is not about generalities now, it is about getting more detailed.

 

Hi,
I appreciate your reply but when we analyze a circuit and asked about it we dont say, "Well i analyzed the circuit", because that does not help the other guy understand the procedure you used.
What i need to know is what circuit you used to determine 're', or at least how you determined what 'IE" was.
The crux to all this i think is how we determine what 'IE' is (or a related parameter).
If you want to know why we got different results you have to give me all the information you used to determine your result or else i can not follow your line of reasoning. It could be the most traditional way to do this, but i will never know that if you dont get more specific and detailed. You see how Jony points out exactly what he is using each time. This is not about generalities now, it is about getting more detailed.

First answer:
For determining "re" I do not need any circuit . I know that the transconductance gm (in your terms 1/gm=re) is related to the collector current by the well known equation IC/Vt with Vt=25....26mV.

Second answer:
Surely, I will here not repeat the whole calculation - therefore ust some hints:
For finding the DC bias point it is the first and main task to find the DC voltage VB at the base node.
Then, because of VBE=0.65V, we have the emitter DC voltage across RE=RE1+RE2 and the corresponding current IE (which gives IC and the transconductance gm ...and then the signal gain).
For finding the voltage VB there are two basic alternatives:
* Applying the basic laws (KCL,KVL) we arrive - as far as I remember - at 4 equations with 4 unknowns. This set of equations can be solved,
* A similar method (applying KCL, KVL) is the superposition method which splits the whole calculation into 2 separate steps:
VB1=f(VCC) with VBE=0 and VB2=f(VBE) with VCC=0. Then: VB=VB1+VB2.
This is the classical way of solving a linear circuit using superposition

* The only "critical point" for both methods is the knowledge that the emitter resistor RE enters the calculation with (1+beta)*RE (because the current in RE is not IB but IE).

I only can repeat that the calculation contains no hidden "secrets" . Analyzing such a circuit needs not more than the knowledge how to apply basic formulas for circuit analyses.
Any further question?

For your convenience, I repeat the results I have given already in post #4:
(In brackets: PSpice simulation based on BC548)
IE=1.675 mA (Sim: 1.785mA)
IC=1.67mA (Sim: 1.776 mA)
Transconductance gm=0.0642 A/V
Vbasis=1.605 V (Sim: 1.696 V)
Input resistance at the base node (incl R1||R2): 9.1 kOhm
Total gain: A=-9.78 (Sim: -9.355).

 

First answer:
For determining "re" I do not need any circuit . I know that the transconductance gm (in your terms 1/gm=re) is related to the collector current by the well known equation IC/Vt with Vt=25....26mV.

Second answer:
Surely, I will here not repeat the whole calculation - therefore ust some hints:
For finding the DC bias point it is the first and main task to find the DC voltage VB at the base node.
Then, because of VBE=0.65V, we have the emitter DC voltage across RE=RE1+RE2 and the corresponding current IE (which gives IC and the transconductance gm ...and then the signal gain).
For finding the voltage VB there are two basic alternatives:
* Applying the basic laws (KCL,KVL) we arrive - as far as I remember - at 4 equations with 4 unknowns. This set of equations can be solved,
* A similar method (applying KCL, KVL) is the superposition method which splits the whole calculation into 2 separate steps:
VB1=f(VCC) with VBE=0 and VB2=f(VBE) with VCC=0. Then: VB=VB1+VB2.
This is the classical way of solving a linear circuit using superposition

* The only "critical point" for both methods is the knowledge that the emitter resistor RE enters the calculation with (1+beta)*RE (because the current in RE is not IB but IE).

I only can repeat that the calculation contains no hidden "secrets" . Analyzing such a circuit needs not more than the knowledge how to apply basic formulas for circuit analyses.
Any further question?

For your convenience, I repeat the results I have given already in post #4:
(In brackets: PSpice simulation based on BC548)
IE=1.675 mA (Sim: 1.785mA)
IC=1.67mA (Sim: 1.776 mA)
Transconductance gm=0.0642 A/V
Vbasis=1.605 V (Sim: 1.696 V)
Input resistance at the base node (incl R1||R2): 9.1 kOhm
Total gain: A=-9.78 (Sim: -9.355).

Hi,

That's very nice of you.
I just need one more thing...
After you calculate the value of 're', then i guess you enter it into the circuit and proceed with the AC analysis?
Note this is a very simple question i dont need any values here.

I have some interesting news after this i think.

 

Yes - that`s correct.
The AC analysis contains three basic steps:
* Finding the signal input resistance r,in at the base node (incl. R1 and R2); this necessary because of the source resistor RG.
* Computing the gain (referenced to the base node) using the formula A1=gm(RC||RL)/[1+gm(RE1)]
* Computing the total gain A=A1[r,in/(r,in+RG)]

 

Ok thanks, and what circuit did you use to determine the value of 're' then?

I use the circuit from post #45 and thevenin's theorem
Ib = (Vth - Vbe) )/(Rth + (Hfe + 1)*(Re1 + Re2))
Ic = β*Ib
Ie = (β + 1)*Ib

Where:
Vth = Vcc * R2/(R1 +R2)
and
Rth = R1||R2

 

I use the circuit from post #45 and thevenin's theorem
Ib = (Vth - Vbe) )/(Rth + (Hfe + 1)*(Re1 + Re2))
Ic = β*Ib
Ie = (β + 1)*Ib

Where:
Vth = Vcc * R2/(R1 +R2)
and
Rth = R1||R2

Hi,

Ok thanks to you too.
And so you calculate 're' from 0.025/Ie right?
Then, insert 're' into the AC circuit and do the AC analysis?
Are those two right?