RC

[Electronman]

Thanks a lot sir, you are a genius,

Just another question is why we want to do so? is that due to this fact that the capacitor has a charge voltage and then a discharge voltage or based upon having 2 power sources in the circuit..??

Please tell me if we have to use superposition law for all firs order circuits base up on capacitors ( due to existance 2 sources: a, power supply and b, the capacitor itself)?

Thanks a bunch

 

[Electronman]

Superposition is for when we have 2 or more power sources, why could you use it here in this circuit too?

 

[MrAl]

Hi again,


We are not forced to use superposition, but for the question of how R1 gets in parallel
with R2 to form the time constant R*C1 can be visualized a little easier when looking
at what happens when we do use superposition. See the attached drawing for a
graphical view on this and note the last circuit how R1 and R2 ended up in parallel.

You can use superposition for any circuit like this really. It's a more general principle
that pertains to linear circuits.

Here's the drawing:

 

[Electronman]

Oh, now I understand what you are talking about, you have used the source transfer or something so to make the single sourece to be converted in to 2 and then use superposition law. good job.

sorry but 2 other questions.
Suppose the SW2 (the above switch) is open for hours, and I turn the left SW on (while the capacitor has no charge) and want to calculate the time constant for the capacitor for while the 9V is across it ( 6V is across the R2), Then I am not able to use the the formula (t=RC) so I have to use the first order formula I,e
Vc=Vb(1-e^(-T/RC)) so: 9=15(1-e^(-T/(400x1000uF)),
If I find "T" then It is my time constant i.e when the capacitor reaches to "9V" right?
1: How to convert the above formula so that I could calculate "T"? I know that I have to use "Ln" but can you let me the right formula to do it so that I do not make mistake (sorry my poor math)?
2: do I need to multiple the calculated "T" by 5 after I calcualte it by the formula I asked you?

 

[MrAl]

Hi again,

If you start with a formula like:
Vc=Vs*(1-e^(-t/RC))

and you want to solve for t, you have to manipulate the formula a bit using
algebra and logarithms.

Starting with:
Vc=Vs*(1-e^(-t/RC))

we can first distribute the Vs to get:
Vc=(Vs*1-Vs*e^(-t/RC))

then simplify and drop the outer parens and we get:
Vc=Vs-Vs*e^(-t/RC)

and then subtract Vs from both sides and we get:
Vc-Vs=-Vs*e^(-t/RC)

and then divide both sides by Vs and we get:
(Vc-Vs)/Vs=-e^(-t/RC)

and then multiply both sides by -1 and we get:
(Vs-Vc)/Vs=e^(-t/RC)


and now that we have the exponential part isolated we can take the natural log
of both sides and we get:
ln((Vs-Vc)/Vs)=(-t/RC)

drop the right side outer parens and we get:
ln((Vs-Vc)/Vs)=-t/RC

multiply both sides by -1 and we get:
-ln((Vs-Vc)/Vs)=t/RC

and finally multiply both sides by RC and we get:
-RC*ln((Vs-Vc)/Vs)=t

and swap sides to get the final formula:
t=-RC*ln((Vs-Vc)/Vs)

So now knowing R,C,Vc and Vs, we can calculate t instead of calculating Vc as before.

When you calculate t in this way you do not need to multiply that t by 5 because
that is the actual t that would cause the voltage Vc.

 

[Electronman]

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

 

[MrAl]

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.

 

[Electronman]

Thanks a bunch sir,
Yea why, not, I would like to know how to measure the resistance of the coil.
I am trying your explanations practically and even by a simulator Software (Proteus 7.2).
And I want to improve my design power till next year, So you will see several other circuits here posted by me asking for help to complete their analyze and would be glad if you share your valuable knowladge with me by helping me to solve their problems.

Thanks

 

[Electronman]

Sorry after rereading your post I faced another problem, at post 38 You told that:
""but with the second
part (switch 2 turned on) you would have to calculate the difference between the
12v supply and the R1,R2 node (voltage across C1) and so Vd would come out to
the difference between 12 and the voltage divider 3.72v (final value of node R1,R2)
and so the cap voltage then would be:

Vc=12-(12-3.72)*(1-e^(-t/RC))
""

Can you explain it to me please? Why you did not use the Voltage across t he cap for Vs there? Why "12-3.72" which gives the voltage across the R2??
Is that Vs at that state?

 

[MrAl]

Hi again,


Sometimes the exponential part of the wave is relative to zero volts, and
sometimes it is relative to some other node like the junction of R1 and R2
and so you have to take that into consideration.
If you need more details i can provide more and maybe a waveform or two to
illustrate.

The formula to use then would be:
Vc=Vc0-(Vc0-3.72)*(1-e^(-t/RC))

where Vc0 is the voltage across the cap at the start when switch 2 is closed.
Vc is the cap voltage anytime after that.

 

[Electronman]

No it does not make any sense,
in this formula:
Vc=Vc0-(Vc0-3.72)*(1-e^(-t/RC))

I am not able to understand this factor: ""Vc0-(Vc0-3.72)""
because the cap has an initial voltage then I can almost understand the parameter Vc0, but what about (Vc0-3.72)? If Vc0-3.72 is the Vs for the cap when the R1 is in the circuit then It seems to me that Vs must be the voltage across the R1 i.e 3.72 which feeds the cap and can be considered as Vs?

 

[MrAl]

Hi again,

Let me rewrite the part you are talking about:
"Vc0-(Vc0-3.72)"

to this:
"Vc0-(Vc0-Vn)"

where Vn is calculated as the difference between the 12v supply and the
R1,R2 node (which yes is across the cap):
Vn=Vs*R1/(R1+R2)

Here are all the formulas written out together:
R1=180
R2=400
C=1000e-6
R=R1*R2/(R1+R2)
RC=R*C
Vs=12
Vn=Vs*R1/(R1+R2)
Vc(t)=Vc0-(Vc0-Vn)*(1-e^(-t/RC))

Now lets say we allowed the cap to charge up to 6v with SW1 closed and
then at that point turned on SW2. That means Vc0=6 and the response of
the capacitor voltage after 0.124 seconds would be:

Vc=Vc(t)=Vc(0.124)=4.562 volts approximately.

Note that is the voltage directly across the capacitor again.

BTW Vn comes out to about 3.72v above.

Sometimes when we are given formulas like this they are not that intuitive.
The best way to handle unknown circuits is to analyze them using the more
general techniques of circuit analysis, which are more intuitive as well.


I hope this is a bit clearer now and i think you are getting a better grasp, but
you should realize that as soon as the circuit gets a little more complex these
formulas again will not apply and new formulas would have to be developed.

Any other problems with this circuit?

 

[Electronman]

Hi and thanks,
Vc0-Vn must be equal to Vs in this formula: Vc=Vs(1-e^(-t/Rc) while it seems to me that it does not, right? If so then maybe "Vc(t)=Vc0-(Vc0-Vn)*(1-e^(-t/RC))" Is another independent formula for discharging cap??

 

[MrAl]

Hi again,


When there is no initial voltage on the capacitor we dont have to subtract
anything, but when there is a voltage to start with we do. In your circuit
when SW1 is closed it is assumed that at first there is no voltage across
the cap so we can use the simplest formula. Later when we decide to close
SW2 we have to think about what the cap voltage was just before that.
This is true in another case too if we opened SW1 before the cap was charged
and then later closed it again. The second closing would required taking into
consideration the initial cap voltage unlike before when the cap voltage was zero.

I am uploading a new schematic with the components moved around slightly so
that one end of the cap is grounded. This will make it easier to understand as
well as talk about. Take a look at this new schematic before you read the
rest of this as all of this pertains to that new schematic.

Starting with the initial cap voltage Vc0=0 when switch SW1 is closed the
cap charges like this:

Vc=Vs*(1-e^(-t/RC))

(where Vs is the full source voltage 12v here)
and that is rather simple because there is no cap voltage to start with.

When there is a cap voltage to start with the exponential still looks the same
but it's amplitude is different. In order to take into account the initial cap voltage
we first have to subtract that from the source voltage Vs first, but then
later add it back again. This modifies the formula a little:

Vc=(Vs-Vc0)*(1-e^(-t/RC))+Vc0

Note above that we simply first subtract Vc0 from Vs, multiply that result times
(1-e^(-t/RC)), and then finally add Vc0 back to that result to get the final value
of Vc at time t.

Take a second to note the 'magic' formula "1-e^(-t/RC)" as that is the basic
charging pattern for caps and resistors. The cap charging wave always follows
that basic pattern.

Let me know what you think about all this and we will eventually get to the
discharging part of the problem again. Right now lets concentrate on the
charging with or without an initial cap voltage Vc0 ok?

The new schematic:

 

[Electronman]

Thanks a bunch genius man!

I want to reread the whole of this thread to see if I can understand my problem. Your words are so helpful thanks.

A question which arised to me is: when you killed the voltage supplies ( my battery or my regulated power supply, right?) why could you do so while they have an internal resistance and seem to NOT be constant or ideal suppliers? or maybe I am getting confused again?

 

[MrAl]

Hi again,

It doesnt take a genius to understand this kind of circuit i assure you, just
basic circuit analysis concepts anyone can learn.

When we kill the supply we are assuming that the internal resistance is low
compared to the other resistances of the circuit such as R2 (400 ohms).
This keeps the error down to a very low value.

 

[Electronman]

I Guessed so about killing it. but not sure if we con extent out calculations to a real circuit?

 

[MrAl]

Hello again,


Sure, why not? That's why we do the calculations in the first place.

Try it with a simulator first and see if you can match your calculations up
with your simulations...it's a great way to learn this stuff. I'd be happy to
help if you get stuck somewhere.

 

[Electronman]

Thanks, I think I found the story behind supplies.

Thanks for your kind.

Really sorry but I have another question regarding to our circuit which you calculated it's parameters:
Suppose I am starting to design the circuit, At firs I look at what parameters I currently have,

I have relay's voltages in 2 process (1- voltage which makes my 12V relay on and causes the contacts to stick together Which I can found it by tuning my regulated power supply (it was 6V for my relay). 2- The voltage of the relay when I want it's sticked contacts to be freed which was almost 3V for mine). So I have my Cap voltage in the formula Vc=Vs*(1-e^(-t/RC)) and so can calculate the TIME parameter (i.e "t") for my circuit and in opposite side if I want to decide of the time by myself then I can put the decided "t" to the formula and calculate the cap voltage, right.
Now consider when that I am trying to design the circuit by myself, Then I have to undefined parameters, R and C?! So I have 2 undefined parameters then How can I calculate and find one of parameters "Vc" and or "t" while R and C are undefined????! If you know what I mean?

 

[Electronman]

Sorry for a lot of questions, But I am learning and need to know, besides, Actually this is a simple circuit and so I am able to understand your meanings more easily than more complicated circuits and this helps me to be able to analyze them easily in future.

The other question is how to determine the MAX power draw of 180 ohms resistor to choice a suitable one?

Thanks before hands.

P.s I know that you have +30 years of experience in electronics and so it is very good to benifit of your valuable info.