Wow thank a bunchhhhhhh Man,
you are so kind and helpful, thanks again, Now I will go and make several experiments by all your sayings. I have to reread this thread again to full may brain with your valuable info.
So I can use "t=-RC*ln((Vc-Vs)/Vs)" to calculate the elipased time for any volatge of the Cap.
Ok, concerning to the simple flasher circuit which I have put its schematic here,
My relay turns on when the voltage across it reaches to 9V and turns off when the voltage drops to 3V. I want to solve it and find the time constant for charging and discharging proses with this asumation that my 400R coil is a pure resistor with turn on voltage of 9V and turn off voltage of 3V, so I can use this formula "t=-RC*ln((Vc-Vs)/Vs)
" for charging proses and time constant at 9V and the other formulas and your explanations for discharge proses, right.
What's your idea of how much of deviance I will get in reality? it's no matter though because I wanted to learn the BASIC principles behind the circuit, and I could by your help.
Thanks again
Hi again,
Actually the correct formula is:
t=-RC*ln((Vs-Vc)/Vs)
(note the swapping of Vs and Vc), sorry about that oversight on my part.
I *always* test my formulas at least once before posting but this time i
was in a big hurry and didnt check the result. That's what can happen
when we dont check the results of our calculations
Yes, that will tell you the time it takes the capacitor to charge up
to the voltage Vc given RC and Vs (and of course Vc itself) with
the circuit when R=400 and C=1000uf (top switch open).
For example, if Vc is 7.6v and Vs is 12v (same R and C as above) then
t ends up being approximately 0.4 seconds. That means it takes
approximately 0.4 seconds to charge to 7.6 volts. The node voltage
across R2 will be 12 minus that voltage. After 5 time constants
(which would be 5 times 0.4 or 2.0 seconds) the capacitor will be
very nearly charged to the full supply voltage of 12v.
You also bring up a good point about 'calling' the coil a pure resistance,
because if the inductance is small compared to the capacitance (1000uf)
then the circuit will behave more like the coil was a pure resistance rather
than partly inductive. That's a very good point actually and definitely
worth a try for sure.
As far as calculation vs real life circuits, when we do a calculation for a circuit
like this we can get pretty close unless the elements themselves are not
very well within tolerance. For example, if the cap is 10 percent low then
we will get 10 percent faster action, and if 10 percent high then we will
get 10 percent slower action. You also have to be a little careful about
what you connect across the cap to measure the voltage with. If the
impedance is too low it will affect the time constant and the final voltage
quite a bit and could make it work very unlike the calculation. With 400
ohms as the bottom resistor though this isnt likely unless you are using
a really old old meter. Another thing though is the tolerance of the
measuring device...if it is off by 5 percent then the voltage measurement
will be off either plus or minus 5 percent.
If you would like to measure the resistance of your coil i could join you with
the calculations and see if we get somewhat the same results.
I dont mind helping you learn circuit analysis and actually it is nice to see
someone with the interest you have in this stuff. I've actually taught
circuit analysis in the past, but many people dont like the math so they
drop out of it. If you like the math at least a little then you will like
doing circuit analysis, and if you dont you'll be stuck using other people's
circuit analysis software and dealing with that instead, which many
people find a lot simpler
There are a few tricks you can learn that can get you going pretty quick,
again if you dont mind a little math too.