A
Active8
 Jan 1, 1970
 0
Followup *was* set for SEB, but I don't care. This is relevant to
both groups, really. This may also be of benefit to beginners
confused with KCL and KVL node and mesh methods  like the poster
with the LED/resistor problem.
I just stumbled across another set of lecture notes and it ocurred
to me that there was something I failed to mention in the original
thread  the order of subscripts with more than one unique
designator.
If you can deal with handwritten notes and have a clue, this looks
pretty good.
http://ececlassweb.ucsd.edu:16080/winter05/ece102/
Reference for this topic is:
http://ececlassweb.ucsd.edu:16080/winter05/ece102/ECE102_S_BJT_largesignal.pdf
I'll refer to page 15 since it's about NPN BJTs. IMO, it's easier to
do the arithmetic mentally  not having to turn the BJT upside down.
The only screw up I see so far is using lower case v for v_XX. It
should be V_XX, thus my "have a clue" qualifier. He's talking about
large signal conditions and leading up to biasing, which sets the
operating point and determines the conditions for small signal
analysis where v_X and v_x are correct. No offence to current
intended by my omission
The point is that the first designator in the subscript is a value
from which you subtract the value of the second subscript.
Note how he uses V_BC. Look at the "inequality equation" (an
oxymoron!) Check his math. It's correct by convention. No confusion
possible. Try it for page 7 on PNPs where he uses both V_BC and
V_CB. Hint: maybe read it standing on your head.
This subscript convention is important  it keeps the signs correct.
You may need to write simultaneous equations or even a netlister.
R_1
___
+___+ node 3  unnecessary, but relevant
  V_3 to the discussion  for this
  lame example  unless you
 .. need to know this voltage
   R_3
  
 ''
 R_2 
 ___ 
node 1 +___+ node 2
V_1   V_2
 
 ..
^ / \   R_4
I_1  ( )  
 \ / ''
  
 
+++

o node 0  reference node
So in a node analysis, for node 1 connected to R_1 and R_2, the
current leaving that node and going into R_2 (charge flow flow?!)
would be
V_1  V_2

R_2
V_1  V_2 is the voltage drop across R_2, V_R2 or V_1,2. V_1 could
be written as:
V_1,0 IOW, V_1  V_0 = V_1  0 = V_1  assuming V_0 is O V which is
what you try to shoot for in these simple problems.
If you do the same for R_1 at node 1,
V_1  V_3

R_1
and simplify the node equation, you'll see that the easiest way to
do it is to take all the resistors connected to node 1 as
contributions to the current out of node_1 caused by V_1 and sum
them
1 1
 +  ... they're on the bottom
R_1 R_2
to write the term for that node. Then take the nodes connected to
those resistors and subtract their contributions. That gives the
term for the other relevant nodes. Make it easier and convert to
conductances (1/R for the uniniated).
sum of current sources in, I_1 = (G_1 + G_2)V_1  G_2*V_2  G_1*V_3
Now you have the first node equation with the node voltages in
order. Then you'd write the other nodes under that so the voltages
appear directly under the same voltage and plug it into your matrice
or do your gaussian elimination without getting all fouled up.
The same conventions apply in all math and science. length_2,1 ...
the distance to point 2, starting at point 1 or x_2  x_1.
You can do the same thing with mesh (loop) analysis.
sum of voltage sources in a loop = sum of voltage drops around that
loop minus any voltage drops associated with that loop that are
caused by another loop current ... (R_1 + R_2 + ... + R_n)I_1 minus
say, R_2*I_2 and R_3*I_3  where the loop currents are, by
convention assumed to be in a clockwise direction. The signs will
work themselves out.
HTH and special thanks to google for archiving all this blathering
in case I loose it and need to retrieve it when I need it for a web
page I'm such a user I should use their free storage space at
www.gmail.com
both groups, really. This may also be of benefit to beginners
confused with KCL and KVL node and mesh methods  like the poster
with the LED/resistor problem.
I just stumbled across another set of lecture notes and it ocurred
to me that there was something I failed to mention in the original
thread  the order of subscripts with more than one unique
designator.
If you can deal with handwritten notes and have a clue, this looks
pretty good.
http://ececlassweb.ucsd.edu:16080/winter05/ece102/
Reference for this topic is:
http://ececlassweb.ucsd.edu:16080/winter05/ece102/ECE102_S_BJT_largesignal.pdf
I'll refer to page 15 since it's about NPN BJTs. IMO, it's easier to
do the arithmetic mentally  not having to turn the BJT upside down.
The only screw up I see so far is using lower case v for v_XX. It
should be V_XX, thus my "have a clue" qualifier. He's talking about
large signal conditions and leading up to biasing, which sets the
operating point and determines the conditions for small signal
analysis where v_X and v_x are correct. No offence to current
intended by my omission
The point is that the first designator in the subscript is a value
from which you subtract the value of the second subscript.
Note how he uses V_BC. Look at the "inequality equation" (an
oxymoron!) Check his math. It's correct by convention. No confusion
possible. Try it for page 7 on PNPs where he uses both V_BC and
V_CB. Hint: maybe read it standing on your head.
This subscript convention is important  it keeps the signs correct.
You may need to write simultaneous equations or even a netlister.
R_1
___
+___+ node 3  unnecessary, but relevant
  V_3 to the discussion  for this
  lame example  unless you
 .. need to know this voltage
   R_3
  
 ''
 R_2 
 ___ 
node 1 +___+ node 2
V_1   V_2
 
 ..
^ / \   R_4
I_1  ( )  
 \ / ''
  
 
+++

o node 0  reference node
So in a node analysis, for node 1 connected to R_1 and R_2, the
current leaving that node and going into R_2 (charge flow flow?!)
would be
V_1  V_2

R_2
V_1  V_2 is the voltage drop across R_2, V_R2 or V_1,2. V_1 could
be written as:
V_1,0 IOW, V_1  V_0 = V_1  0 = V_1  assuming V_0 is O V which is
what you try to shoot for in these simple problems.
If you do the same for R_1 at node 1,
V_1  V_3

R_1
and simplify the node equation, you'll see that the easiest way to
do it is to take all the resistors connected to node 1 as
contributions to the current out of node_1 caused by V_1 and sum
them
1 1
 +  ... they're on the bottom
R_1 R_2
to write the term for that node. Then take the nodes connected to
those resistors and subtract their contributions. That gives the
term for the other relevant nodes. Make it easier and convert to
conductances (1/R for the uniniated).
sum of current sources in, I_1 = (G_1 + G_2)V_1  G_2*V_2  G_1*V_3
Now you have the first node equation with the node voltages in
order. Then you'd write the other nodes under that so the voltages
appear directly under the same voltage and plug it into your matrice
or do your gaussian elimination without getting all fouled up.
The same conventions apply in all math and science. length_2,1 ...
the distance to point 2, starting at point 1 or x_2  x_1.
You can do the same thing with mesh (loop) analysis.
sum of voltage sources in a loop = sum of voltage drops around that
loop minus any voltage drops associated with that loop that are
caused by another loop current ... (R_1 + R_2 + ... + R_n)I_1 minus
say, R_2*I_2 and R_3*I_3  where the loop currents are, by
convention assumed to be in a clockwise direction. The signs will
work themselves out.
HTH and special thanks to google for archiving all this blathering
in case I loose it and need to retrieve it when I need it for a web
page I'm such a user I should use their free storage space at
www.gmail.com