Modified drawing
+V
!
/
\ R1 +V
/ !
\ \
D1 ! /
Vin -!<--+ \ R4
V /
D2--- R3 !
+--/\/\----+---- to one input of the op amp
! the mirror image version goes
/ to the other
\ R2
/
\
!
-V **************** Mirror image along here **************
[...]
what if i would to have a window comparator for 3-6 volts how can i
compute for the values of the resistors
I think the reason you are having trouble is that several of the values
are just "pick one" so you don't see an obvious "this resistor has to be
10K".
Assume R1 is 10K.
Assume R3+R4 is 100K.
Find R2 such that D1 is carrying the same current as D2 with the input
voltage at the positive trip point.
Do the same for the mirror image version.
Calculate the voltage at the junction of R2 and R3 (and R2' and R3') at
the two trip points.
Now you can write equations for what the inputs of the op-amps will be in
each case.
The two inputs to the op-amp just become equal when the voltage passes the
trip point.
This means that you end up with two equations that have R3, R4, R3' and
R4' in them.
You also have R3 = 100K-R4 and R3' = 100K-R4'
This means you have 4 equations and 4 variables. If the gods are smiling
on you, there is a solution that does not require negitive resistors.
