Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

emitter follower ac analysis

Status
Not open for further replies.

yassin.kraouch

Advanced Member level 2
Advanced Member level 2
Joined
Jul 5, 2009
Messages
631
Helped
40
Reputation
80
Reaction score
40
Trophy points
1,308
Activity points
5,094
Hi,
i did not understand why in this schematic they consider that in ac analysis R1 is parallel always to R2 ? emitter follower.png
 

Hi!

Because in AC analysis, also called "small signal analysis", the VDD is connected to the ground as it is a continous signal.
So just imagine to folded the schematic from the top to the bottom. R1//R2...
 

Yassin, imagine a simple resistive divider connected between Vcc and ground.
Seen from Vcc both are in series, but seen from the middle node both are in parallel.
 
Hi!

Because in AC analysis, also called "small signal analysis", the VDD is connected to the ground as it is a continous signal.
So just imagine to folded the schematic from the top to the bottom. R1//R2...
really i am confused, why you said it is connected to ground ?
 

really i am confused, why you said it is connected to ground ?
Hi Yassin
I think "LvW" means the R2 ! it is tied through the ground ! in AC analysis , DC sources are short circuit ! so what will happen ? R1 and R2 both have a common point ! base node ! and then another common point which will produced by shorting the supply . so they are in parallel together .
ok ?

Best Wishes
Goldsmith
 

Let's consider the simple divider R1, R2 with two voltage generator connected to it:

https://obrazki.elektroda.pl/9814407200_1369376798.jpg

We want to find Z=V2/I2 that is the impedance seen vy generator V2.
Appling the basic circuit rules we have:

1) I3=I1+I2
2) V2+R1*I1=V1
3) R2*I3=V2

from 2) I1=(V1-V2)/R1
from 3) I3=V2/R2

substituing in 1) we have:

R1*V2=R2*V1-R2*V2+R1*R2*I2

that is:

(R1+R2)*V2-R2*V1=R1*R2*I2

thus:

(R1+R2)/(R1*R2)*V2/I2-V1/(R1*I2)=1

that is:

(R1+R2)/(R1*R2)*Z-V1/(R1*I2)=1

now we have to apply the contribution of each generator considering the same frequency. But when V2 is generating the frequency "f", V1 at that frequency is generating 0 volt (is a DC generator) so V1(f)=0.

(R1+R2)/(R1*R2)*Z=1

then:

Z=R1*R2/(R1+R2)=R1//R2

Then all generator at a different frequency with respect to that of analysis have to be considered short circuits.
In a similar way you can verify that in case of current generators (insteda of voltage generator) you have to consider that generating different frequency that that of analysis as open circuits (i=0)
 

Hi Yassin
I think "LvW" means the R2 ! it is tied through the ground ! in AC analysis , DC sources are short circuit ! so what will happen ? R1 and R2 both have a common point ! base node ! and then another common point which will produced by shorting the supply . so they are in parallel together .
ok ?

Best Wishes
Goldsmith
Yes i understand what you mean, why we have to consider that DC sources are short circuit ?? this is the main point i didn't understand
 

Yes i understand what you mean, why we have to consider that DC sources are short circuit ?? this is the main point i didn't understand

Because a voltage source is assumed to have an internal source resistance of 0 ohms (otherwise it cannot be considered as a voltage SOURCE).
 

Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top