Common Transistor Amplifier:- doubts on Input impedance

Rin(base),ac =~Hfe(rtr) rtr= transresistance
rtr =0.026 V/IC = 0.026 V/1 mA = 26ohms
rtr=26 Ohms
so the Rin = 1/R1+1/R2+1/100(26)

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Its correct. Input (base) resistance = r(pi) =Vt/Ib= beta*re =beta*(Vt/Ic)
Where Vt =thermal voltage = 26mV at room temperature

---------- Post added at 18:07 ---------- Previous post was at 17:58 ----------

but i think HFE is AC current gain ( differential from beta )

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You are right but some thing left in the hfe/beta definition that I already wrote in #17 comment. H-parameter model are defined for low frequency only and valid for dc also. But not for high frequency (then you need to add capacitor model).
If you are not concerned with frequency then using term Beta is perfectly fine. And if you see Hfe curve (i posted grapg #17) hfe = ice/Ibe is more or less constant over the amplifying area. of course the hfe value may change with temperature.

Qube said:

Thanks for verifying LVW..
i guess,i have figured out the input impedance a little but not perfect..I simulated that circuit in LTSpice, with a 0.1V Ac signal,i got clean amplification at the collector...Now i want to learn more about output impedance... plz give me a clue,i will research it and try to learn

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Hi Qube,

I am very sorry but I have to correct myself.
In your first posting the emitter resistor was ac-bypassed by a capacitor.
Thus, I have (blindly) assumed the same mehod for your last circuit.
Therefore, when there is no capacitor in parallel to Re the value for Rin is larger than calculated by you:

As before: h11=h21/g=h21*Vt/Ic=100*0.026/0.001=2.6 kohms (with your terms: rtr=1/g ; g=transconductance)

and now: rin,base=h11+h21*Re=2.6 k + 100*1k =102.6 kohms.

Finally: Rin=R1||R2||(rin, base).

Please, take into account that in case of an emitter capacitor the value of Ce should be larger as C1 by a factor of at least 100.
Only in this case the input capacitor C1 (together with Rin) determines the highpass corner frequency.

LvW said:

Hi Qube,

I am very sorry but I have to correct myself.
In your first posting the emitter resistor was ac-bypassed by a capacitor.
Thus, I have (blindly) assumed the same mehod for your last circuit.
Therefore, when there is no capacitor in parallel to Re the value for Rin is larger than calculated by you:

As before: h11=h21/g=h21*Vt/Ic=100*0.026/0.001=2.6 kohms (with your terms: rtr=1/g ; g=transconductance)

and now: rin,base=h11+h21*Re=2.6 k + 100*1k =102.6 kohms.

Finally: Rin=R1||R2||(rin, base).

Please, take into account that in case of an emitter capacitor the value of Ce should be larger as C1 by a factor of at least 100.
Only in this case the input capacitor C1 (together with Rin) determines the highpass corner frequency.

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yes,i had read that bypass capacitor not only increases the gain of the transistor,but also decreases the input impedance..

And im not getting the terms h11 and h21 that you have used here...

before Rin(base),ac=hfe(rtr)

when bypass cap is not present , Rin(base),ac= hfe(rtr)+(R1*RE)???? i assume you got 1K from Re,but is the 100 Hfe??
Rin(base),ac=100(26)+(100*1K)

am i right??

Qube said:

yes,i had read that bypass capacitor not only increases the gain of the transistor,but also decreases the input impedance..

And im not getting the terms h11 and h21 that you have used here...

before Rin(base),ac=hfe(rtr)

when bypass cap is not present , Rin(base),ac= hfe(rtr)+(R1*RE)???? i assume you got 1K from Re,but is the 100 Hfe??
Rin(base),ac=100(26)+(100*1K)

am i right??

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Yes, I think you got it .

I like to give some explanations to the terms I've used:

Normally, the small-signal low frequency behaviour of a BJT is described with the so-called "hybrid parameters".
These parameters can be found in each transistor data sheet and are given for the CE configuration only..
h11= input resistance
h21= signal current gain (sometimes also called hfe)
h12= backward gain (very small, can be neglected)
h22= output conductance.

It can be shown that:
forward transconductance g=h21/h11=Ic/Vt (Ic= collector DC current and Vt=thermal voltage constant=26 mV) .
Thus, you can express the input resistor at the base node as h11=h21/g=h21*Vt/Ic .

So Rin(base)=(hfe*Rtr)+(hfe*Re) //when Bypass capacitor is not present

So Rin(base),ac = 102K

There fore Rin=R1||R2||Rin base

after solving the equation

Rin= 12K ohms

Cin=1/(2*3.14*100hz*12000);= 0.13uF
and the Cin will be 0.13uF
So the input impedance of that circuit would be 12Kohms...

Qube said:

Rin= 12K ohms
Cin=1/(2*3.14*100hz*12000);= 0.13uF
and the Cin will be 0.13uF
So the input impedance of that circuit would be 12Kohms...

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Yes, correct.
The calculated value for Cin is the MINIMUM value Cmin to allow a highpass corner that is equal to or below 100 Hz.

I think your attachment is wrong . because you wrote hfe and the under that , DC current gain !!! but i think HFE is AC current gain ( differential from beta )

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I didn't wrote anything, because the diagram is copied from a NXP datasheet. I agree, that hFE is the usual term for small signal current gain. The difference to large signal gain is however small for a curve like above. For the present discussion, it doesn't really matter.

LvW said:

Yes, correct.
The calculated value for Cin is the MINIMUM value Cmin to allow a highpass corner that is equal to or below 100 Hz.

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ty LvW, now we have the Rin,we can calculate the Cin using Cin=1/(2*3.14*Hz*Rin)

But to calculate the Cout,we need the Output impedance of this circuit... so how do i calculate the Rout??

I have attached the edited image with correct value so far i have got...

Output resistance of BJT device is generally defined by Early voltage,VA, and that gives r0 = VA/Ic. Va is in the range of >100Volts (-ve). that gives r0 in the range of 100's of killo-ohm. and most of the time neglected.
If device is operating in the active region, Ib can be neglected and Ic will be equal to Ie and Load resistance can be calculated by Vcc/IC. In saturation, Ib becomes significant, and you need to get the value of Ic= Ie-Ib. And in the same way , can be calculated for Rload.

1. There's no Cout in your circuit at all. But I guess, you are talking about an output coupling capacitor to get a pure AC aoutput voltage.
2. In many cases, the load impedance will be higher than the output impedance and Cout can be calculated based on the load impedance only.
3. If this is not the case, you can assume that the output impedance is almost equal to Rc. With a simplified transistor model, the output resistance of the common emitter stage r0 is assumed as infinite. The error introduced by this assumption is most likely smaller than the tolerance of available capacitors.

LvW,

question: Did you create the (funny) document Qube_A.pdf ? Do you expect somebody will read it?

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Sorry I did not answer sooner. By funny, I guess you mean the inclusion of a lot of terms that were not really relevant. For that I apologize. You are right in avering that for this circuit, whose input and output are isolated by a current generator, it is easier to not worry about how the output affects the input and vice versa. However, that may not be the case for a feedback circuit that does indeed connect all parts of the circuit. I guess I wanted to show how the equations could be set up, and how they "blow up" if simplifications are not applied and undefined terms are added in.

1.) Do you really believe that the "best way" to find input/output impedances of a BJT amplifier is to use the corresponding transfer function? What is the advantage if compared with the direct and classical methods?

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In short, less work. If you have to find the transfer function. then you already have computed the input and output impedance. Of course if the circuit is well known, and the impedances can be seen by inspection, then the General Immittance Theorem (GIT) is slower.

From the engineering point of you I disagree with "don`t forget that the source impedance affects the output impedance.....". I think this influence which is caused by the parameter h12 (backward gain) can and should be neglected because it influences the result certainly by less than 1% only. The same applies to the influence the load impedance has on the input impedance.

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The load impedance does affect the input impedance and vice versa. In the problem of this thread, its influence was very small. It still was in effect without my using h12. The difference in value was out to the 6th decimal place for a output impedance between 0 and infinity. The 1200 ohm output resistor stabilized everything. In other circuits, who knows?

You shouldn't forget that in the area of electronics no formula is correct by 100% - and it is good engineering practice to introduce simplifications and to neglect some effects that are well below the limits of other unavoidable uncertainties (tolerances of passive parts and active BJT parameters) .

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Agreed. Next time I will do that.

Ratch

FvM said:

1. There's no Cout in your circuit at all. But I guess, you are talking about an output coupling capacitor to get a pure AC aoutput voltage.
2. In many cases, the load impedance will be higher than the output impedance and Cout can be calculated based on the load impedance only.
3. If this is not the case, you can assume that the output impedance is almost equal to Rc. With a simplified transistor model, the output resistance of the common emitter stage r0 is assumed as infinite. The error introduced by this assumption is most likely smaller than the tolerance of available capacitors.

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Of course, I agree to everything - except point 2. (typing error?): I think, Cout can be calculated based on the output resistance of the stage (not the larger load resistance).

---------- Post added at 09:43 ---------- Previous post was at 09:35 ----------

Ratch said:

........... I guess I wanted to show how the equations could be set up, and how they "blow up" if simplifications are not applied and undefined terms are added in.

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Hi Ratch,
thanks for your detailed reply and the explanation/motivation to show us such a large gain expression (containing nearly all terms that theoretically may have an influence).
This clearly demonstrates that it is absolutely necessary to neglect all terms that have an influence smaller than uncertainties and errors due to practical restrictions (temperature, tolerances,...).

So now im clear that Zi = R1||R2||Rin(base),ac
Rin(base),ac=(hfe*Rtr)+(hfe*Re)

It is best to have Zout << ZIn by the factor of 10 in audio frequencies

Now how do i calculate the output impedance??

As FvM said, "In many cases, the load impedance will be higher than the output impedance and Cout can be calculated based on the load impedance only."
so if the load is about 3K,then i can assume Zout is 3K can calculate Cout based on Zout which is 3K here

Which resistors affect the output impedance???

Hi Qube
The out put impedance will given by RC in parallel with ro of your transistor . ro will given by Va/IC ( 1/hoe ) . and Va is early voltage .
but if you want consider the capacitors , it will be a bit different .
Best Wishes
Goldsmith

Qube said:

Which resistors affect the output impedance???

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The outpüut cut-off frequency is determined by the time constant Tout=Rp*Cout
with: Cout=coupling capacitor and Rp=Rout||Rload.
Rload= external load resistor
Rout=Rc||ro with Rc=Collector resistor and ro=1/h22=output resistance of the BJT (inverse of the Ic=f(Vce) slope) >>> can be neglected because very large (as mentioned already).

---------- Post added at 16:26 ---------- Previous post was at 14:31 ----------

Correction:

I am sorry, I have to announce an error in my last posting.
The effective resistance that defines the gain value of the stage is Rp=Rout||Rload.
However, Rp is NOT the resistance that defines the cut-off frequency of the output circuitry (high pass behaviour).
Because the capacitor Cout is charged with Rout and Rload in series the correct time constant Tout=1/w,out
is
Tout=Cout*(Rout+Rload).

(This fact was mentioned implicitely already by FvM's post#30 ,point 2)

@LvW
also just to confirm
is it Rin(base),ac=(hfe*Rtr)+(hfe*Re)//When bypass capacitor is absent
or
Rin(base),ac=((hfe+1)*Rtr)+((hfe+1)*Re)//When bypass capacitor is absent

Anyway both will give result which varies a very little..

The first expression in your last posting is the correct one.
LvW

so the out put impedance = Rload= Rc in this below circuit

so output impedance of this circuit is 6K and Cout can be calculated based on Rc??? im i right or i m getting it wrong??

out put impedance = Rload= Rc in this below circuit

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For the shown circuit, it is true.
But Rload may be different or is chosen according to your application.
There is two things, If it is capacitive loading, or resistive loading.
For Capacitive loading , Rload = Rc ( most of the time)( for dc application)
For resistive loading, Rload = Rc||RL

To avoid misunderstandings:
Each transistor stage will work upon a certain load impedance Rload.
For calculating the resulting gain value the parallel combination Rc||Rload is to be considered.
However, for calculating the ouput coupling capacitance (if any) based on a certain edge frequency, the series combination Rc+Rload must be used.