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.

Help me understand this equation made using KCL

Status
Not open for further replies.

mujee

Member level 1
Joined
Feb 11, 2007
Messages
37
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,286
Activity points
1,585
hey guys
below is the circuit given for node analysis. now using kcl we come up with following equations

GAE1 + G3(E1 - E2) = I1 (1)
GBE2 - G3(E1 - E2) = I2 (2)

i used plus in equation 2 before G3(E1 - E2) but my friend told me that it should be negative.i am too much confused about this minus sign before G3(E1 - E2) . i dont know why it isnt positive the way it is in equation 1

please elaborate as i couldnt understand the basic logic behind this
64_1179770609.png
 

the_edge

Full Member level 3
Joined
Feb 18, 2007
Messages
181
Helped
18
Reputation
36
Reaction score
5
Trophy points
1,298
Location
Belgrade, Serbia
Activity points
2,264
Re: a simple question




From I kirchofs law you can see that:
Ib=I2+I3 => I2 = Ib - I3

Ib = E2Gb
I3=G3*(E1-E2)

=> I2 = E2Gb - G3*(E1-E2)


First equation is not good too...
I1+Ia+I3=0 => Ia+I3=-I1

GaE1 + G3(E1-E2) = -I1
 

v_c

Advanced Member level 2
Joined
Oct 11, 2005
Messages
514
Helped
113
Reputation
226
Reaction score
37
Trophy points
1,308
Location
Chicago, IL, USA
Activity points
8,226
Re: a simple question

Mujee,

it should also be noted that you have to assume E0=0 (datum, ground or reference node, whatever you wish to call it). If this not chosen, you can still write the KCL equations, but you must use the voltage differentials (like E1-E0, E2-E0) to denote the voltage across GA and GB.

For KCL you first have to decide which way you have to take the currents. You can do (a) sum of currents leaving a node = 0, (b) sum of currents entering a node = 0, or (c) sum of currents leaving a node = sum of currents entering a node = 0. You start doing this by assigning currents in the directions that you choose, no matter if the actual current is in the assigned direction or not. If the actual current was going the opposite way that you assigned, you will see that when you do the algebra as the numerical value will be negative.

So assuming that E0=0. Let us write the equation for currents leaving the node marked by E1.
I1 + GA*E1 + G3*(E1-E2) = 0

Now, let us write the KCL equations for currents leaving the node marked by E2.
G3*(E2-E1) + GB*E2 - I2 = 0
The reason you see -I2 instead of I2 is because we chose to write currents leaving the node and current I2 is shown *entering* the node. If you always write the currents leaving the node =0, the voltages at which you are writing the equation will always come first in the differentials. By the way, with the way your circuit is drawn, without a "ground" or "datum" node, your KCL equations would be
I1 + GA*(E1-E0) + G3*(E1-E2) = 0
G3*(E2-E1) + GB*(E2-E0) - I2 = 0

Best regards,
v_c
 

Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Top