RC

[MrAl]

Hello there,


If you are looking for something that is highly predictable then try using a
comparator and driver transistor to drive the relay, or skip the relay and
just use a transistor. That would be more reliable too.

At the very least use a transistor with the relay or just use two transistors.
That kind of circuit will be much more predictable too and will not depend on
the type of bulb used as long as it matches the supply voltage.

The reason why your original circuits are not that easy to analyze is because
they contain elements that are not that well defined...you may switch bulbs,
different kind of relays, etc. Different bulbs will cause oscillation at different
frequencies, and so will different relays. That's two elements that are very
hard to define without having the actual components right in front of you.
When you switch to transistors and possibly comparators, you can then work
with components that have characteristics that are very well known and so
they are very well defined and a lot of calculations with these kinds of elements
will be very close to what happens in real life circuits using these elements.
A big advantage too is that you can design the circuit in a way so that the
main element that is harder to define (the bulb) does not change the way
the circuit works any more...the circuit operation will not depend on that
any more and so you can use a wider range of bulb types without changing
anything else and still get the very same operation.

 

[Electronman]

So a question which arises to me is that how designers design circuits? The above is a simple one but seems We have to go with trial and error!.

Another question is do you know any source or book to learn us how circuits are working? I mean a book with circuits and explanation to how they are working (I.e what a practical Capacitor, transistor, resistor... does in the said circuits??

 

[indulis]

While this is a DC circuit, it's not operating in a "steady state"... your interested in what happens in it's transition from one state (off) to another (on). Look at a RLC's response to a "step function"... then you will understand why there isn't a "simple solution" to your question.

 

[EN0]

So a question which arises to me is that how designers design circuits? The above is a simple one but seems We have to go with trial and error!.

Another question is do you know any source or book to learn us how circuits are working? I mean a book with circuits and explanation to how they are working (I.e what a practical Capacitor, transistor, resistor... does in the said circuits??

Designers design circuits primarily by using mathematics and learning the theory. If they know the electronic theory, they know where to apply parts. Then they use math to find the right values for those parts.

 

[colin55]

Designers design circuits by trial and error. How do you think the Chinese designed circuits 50 years ago, when computers and books were not available?
If you think you can design a circuit using a "simulation program," you are fooling yourself.
Getting a circuit to work is much more complex than any simulation software package can provide and even simulation packages for microcontroller programs do not take into account any of the complexities of the input and output devices.
The only way to design a circuit is to build it and test it and then give it to someone else to use. That way you will find the problems and limitations, as they try to destroy it.
When it passes the "idiot test," you can market it.

 

[indulis]

Designers design circuits by trial and error. How do you think the Chinese designed circuits 50 years ago, when computers and books were not available?

I'm sure your not serious when you say... trial and error, and I'd bet there were books in China 50 years ago!!!


Electronic design is all about math.

 

[colin55]

I attend an inventors association.
Why does it take an electronics inventor years to "perfect" his design?
When you ask for a quote for a particular design, why does it cost thousands of dollars and 3 months?
If all it takes is "maths" to design a project, why does the printed circuit board company get prototype after prototype before the final design?
Why is it that the final design is identified as rev 1.6?
Tell this to the 70 inventors at our meetings . . . . all it takes is “maths” to design a project!!!!!!!
You are fooling yourself . . . you obviously haven't designed a project in your life.

 

[MrAl]

So a question which arises to me is that how designers design circuits? The above is a simple one but seems We have to go with trial and error!.

Hi again,


That's not entirely true. The reason why this circuit is a little more difficult
to nail down is because you have given us so many variables and i dont think
you realized the consequences of doing that.
When a designer goes to design a circuit like this, they often start with
parts that they are already familiar with from past projects, and also have
more well defined objectives and also have data sheets on actual part numbers
that they can work with.
Here, you say you want it to work with different bulbs at least, so that
presents a different problem than if you knew exactly what bulb you
wanted to use from the start. However, given that same objective
a designer would have to handle that the same way: by designing a circuit
that doesnt depend on the bulbs characteristics, and this may lead to
a slightly more complicated circuit than if the bulb characteristics are
already known and will always be the same.
It's sort of like anything else...if you say you want to carry some water to
some other location we cant tell you how big of a bucket you need...we
would also need to know how much water you intend to carry. It's that
simple really.
A general rule might be that the more well defined the problem is the less
complex the circuit can be, but the less well defined (part values can change)
the more complex it usually becomes so that it can automatically adjust to
the changes that come later.
Maybe a good case in point is a solar array collector, where the user wants it
to track the sun position and also track the maximum power point of the
array. If the sun didnt move relative to the earth it would be cake, but it
does move and so the circuit and entire system becomes more complex.
The solar array may also become a little dirty and change the amount of
power available.

These kinds of problems have been dealt with in the past many times for
many kinds of circuits, and there is even some nomenclature that has come
up for these kinds of situations. Generally the change in parameter is viewed
as a 'disturbance' to the system, and the system being designed is designed
ahead of time to be able to automatically adjust itself to make up for that
disturbance, usually using some kind of feedback system to measure the
output and compare to some reference and induce changes that will keep
the output following the reference as well as needed.
The whole of these techniques are addressed in what has become known
as "Control Theory", and this theory can make itself useful even in simple
applications like the one you are suggesting here.

So it is not that the circuit here can not be done more simply, it's just that
when you want it to do more than just work with specific parts it's going
to take a little more circuitry to do the job correctly.
This happens with many many circuits so dont get too alarmed over it.

 

[Electronman]

Hi again,


That's not entirely true. The reason why this circuit is a little more difficult
to nail down is because you have given us so many variables and i dont think
you realized the consequences of doing that.
When a designer goes to design a circuit like this, they often start with
parts that they are already familiar with from past projects, and also have
more well defined objectives and also have data sheets on actual part numbers
that they can work with.
Here, you say you want it to work with different bulbs at least, so that
presents a different problem than if you knew exactly what bulb you
wanted to use from the start. However, given that same objective
a designer would have to handle that the same way: by designing a circuit
that doesnt depend on the bulbs characteristics, and this may lead to
a slightly more complicated circuit than if the bulb characteristics are
already known and will always be the same.
It's sort of like anything else...if you say you want to carry some water to
some other location we cant tell you how big of a bucket you need...we
would also need to know how much water you intend to carry. It's that
simple really.
A general rule might be that the more well defined the problem is the less
complex the circuit can be, but the less well defined (part values can change)
the more complex it usually becomes so that it can automatically adjust to
the changes that come later.
Maybe a good case in point is a solar array collector, where the user wants it
to track the sun position and also track the maximum power point of the
array. If the sun didnt move relative to the earth it would be cake, but it
does move and so the circuit and entire system becomes more complex.
The solar array may also become a little dirty and change the amount of
power available.

These kinds of problems have been dealt with in the past many times for
many kinds of circuits, and there is even some nomenclature that has come
up for these kinds of situations. Generally the change in parameter is viewed
as a 'disturbance' to the system, and the system being designed is designed
ahead of time to be able to automatically adjust itself to make up for that
disturbance, usually using some kind of feedback system to measure the
output and compare to some reference and induce changes that will keep
the output following the reference as well as needed.
The whole of these techniques are addressed in what has become known
as "Control Theory", and this theory can make itself useful even in simple
applications like the one you are suggesting here.

So it is not that the circuit here can not be done more simply, it's just that
when you want it to do more than just work with specific parts it's going
to take a little more circuitry to do the job correctly.
This happens with many many circuits so dont get too alarmed over it.

Hello and thanks to you and to all,
You explained it completely to me and cleared me so now I noticed several advices by that.

I was amazed when I noticed that my RC circuit is not an RC circuit but is an RLC circuit, so I tried to remove several components specially the bulbs and the discharging resistor, then I posted it here hopping that somebody will tell that I am dealing with a first order circuit and can use the known formula for RC time constant circuits and can use Vc=Vs(1-e^(-t/RC) to find the time constant for the circuit, then see how can solve the next problem I.e when bulbs and discharge resistors come in, but I told that it is not possible to simply find the Time constant for the circuit because it is an RLC circuit, while I do not know any formula to find the time constant for an RLC circuit!?
No one asked me to give some clearness about the components I used even no one asked me that for the charging circuit containing just the Cap and Relay's wound.

I wish there was a book with practical circuits explaining how they are designed? When no one respond me for a such book I thought engineers use softwares to do so, so they are not able to good define the ciruits

 

[MrAl]

Hello again,


Actually there is a sort of time constant for the RLC circuit too, which is:
R/(2*L)
but instead of that indicating how the current decays it instead indicates how
the envelope of the current decays. It's almost the same unless you end up
with an oscillator, which your circuit is not in most cases.
Notice also that R/(2*L) doesnt help that much here anyway, because we dont
know what your coil inductance is, and also that inductance is very nonlinear
which varies as the armature pulls in. Also, for circuits like this it would be
a little rare to consider the inductance anyway and rather go with the known
resistance and nominal coil voltage and take it from there.

If you want to learn how to design circuits then you need to learn first how to
analyze circuits using various circuit analysis techniques. You then start to
analyze all the circuits you can find that are the type you are most interested
in. Doing this gives you a great sense of what works and what doesnt, and
how to proceed with a design of something. Sometimes it's like inventing but
other times it simply means following someone else's design procedure where
almost everything is mapped out ahead of time. It's always good to be able to
analyze circuits though so that you can test the design on paper before you
start building it up.

You might start by telling us what you have learned in the past such as
algebra, trig, geometry, calculus, differential equations, etc., etc.
I think people here will be able to help you from there.

 

[12345678901234567890]

MRAl

VERY, VERY , VERY, GOOD explanations.

 

[MrAl]

MRAl

VERY, VERY , VERY, GOOD explanations.

Hello again,


Well thank you 12345...etc. I hope that helped some.

 

[indulis]

I attend an inventors association.
Why does it take an electronics inventor years to "perfect" his design?
When you ask for a quote for a particular design, why does it cost thousands of dollars and 3 months?
If all it takes is "maths" to design a project, why does the printed circuit board company get prototype after prototype before the final design?
Why is it that the final design is identified as rev 1.6?
Tell this to the 70 inventors at our meetings . . . . all it takes is “maths” to design a project!!!!!!!
You are fooling yourself . . . you obviously haven't designed a project in your life.

What do you mean by “electronic inventor”?? Are you defining that as someone who is developing leading edge technology or someone who is trying to come up with just another “me too mouse trap”?? Those are two very different environments. Development and general design work just aren’t the same and never will be.

Now, as for,

You are fooling yourself . . . you obviously haven't designed a project in your life.

What's obvious is that you don’t know me, what I do for a living or for how long I’ve been doing it, so that is a rather arrogant comment on your part.

 

[Electronman]

Ok, what about this simple one?

Can we find the final voltage of the capacitor after the Switch of 180 ohms Resistor is closed.

 

[Electronman]

Can somebody suggest me how to solve the above circuit?
It is not a simple RC circuit at some times (when we close the SW of 180Ohms resistor)

 

[MrAl]

Hi,

Are you saying that the first lower switch is closed already and then you close the
upper switch and you want to know the voltage across the capacitor after a
very long time has passed?
If so, it is just the voltage across the resistor as long as both switches stay closed.


Just to note, it helps if you label the switches too so we can refer to them
as SW1, SW2, etc.

 

[Electronman]

Hi,

Are you saying that the first lower switch is closed already and then you close the
upper switch and you want to know the voltage across the capacitor after a
very long time has passed?
If so, it is just the voltage across the resistor as long as both switches stay closed.


Just to note, it helps if you label the switches too so we can refer to them
as SW1, SW2, etc.

I think I will end up with just a voltage divider via resistors, and the cap will have the volatge across the 180R resistor and has nothing to do even if it has alery been charged via the 400R resistor?

Can we find any time constant for the cap and see whenit reachs to voltage of 180R?

besides what about when the cap is charging but not full charged then how can I calculate the time constant for this circuit?

Thanks

 

[MrAl]

Hello again,


The time constant of this circuit isnt too hard to calculate, but you have to realize
that the more complicated the circuit gets the harder it starts to become to calculate
the time constant and you eventually have to resort to more sophisticated circuit
analysis techniques that you would have to learn. Some of the techniques are not
that difficult however but they do take a little while to learn.

As you know, for a single resistor and capacitor R and C the time constant is
simply R*C. For this simple circuit the time constant can still be calculated by
multiplying R times C, but before that we have to calculate what R is.
For this circuit, R is simply the parallel combination of R1 and R2:
R=R1*R2/(R1+R2)
and so again the time constant is:
R*C.

As you also know the voltage across the capacitor in a single RC circuit is:
Vc=Vs*(1-e^(-t/RC))

but there is another thing to think about with the two resistor circuit, and that is the
source voltage level Vs.

With a single R and C we usually take the voltage to be the full battery voltage,
but for this circuit because there is a divider we also have to calculate a new
voltage value. Since we want the voltage across the capacitor, we first calculate
the divider voltage Vd:
Vd=Vbatt*R1/(R1+R2)
so
Vd=12*180/(180+400)

and now we have the right voltage to use for the equation:

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

with R calculated above as the parallel combination of R1 and R2.

Note that Vc is now the voltage across the capacitor, with the reference
node being the junction of R1 and R2. Thus if we used a meter to try to see
this change we would put the black lead on the junction of R1 and R2 and the
red lead on the +12v power supply positive terminal.

If you want to calculate the voltage at the node junction of R1 and R2 with
reference to ground, then you would change Vd to be the voltage across
the 400 ohm resistor due to the voltage divider instead of across the 180
ohm resistor.

Note the above analysis is for when both switches are turned on at the same time.
If you want to calculate what happens when switch 1 is turned on first and switch
2 is turned on later, you have to first calculate the single RC circuit and then add
the second resistor later when switch 2 is turned on. When switch 1 is turned on
the cap starts to charge up with full battery voltage until it reaches the full 12v.
Later when switch 2 is turned on the cap starts to discharge until it reaches the
final value determined by the voltage divider.
The first part is easy because it's just a single RC circuit, 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))

where again R is the parallel combination of R1 and R2 and the time constant again
is R*C.

The above assumes that the capacitor C1 has been allowed to charge to the full voltage
of 12v before the second switch was closed. If the capacitor was not allowed to charge
for at least 5 time constants or so, then the initial cap voltage is not 12v when switch
2 is closed so we have to modify the formula to account for this by changing the 12v to
whatever the cap voltage was just before switch 2 is closed:

Vc=Vc0-(Vc0-3.72)*(1-e^(-t/RC))

so if the initial cap voltage is 6v then:

Vc=6-(6-3.72)*(1-e^(-t/RC))

Some simplification of this last formula might be possible also.

 

[Electronman]

Hello again,


The time constant of this circuit isnt too hard to calculate, but you have to realize
that the more complicated the circuit gets the harder it starts to become to calculate
the time constant and you eventually have to resort to more sophisticated circuit
analysis techniques that you would have to learn. Some of the techniques are not
that difficult however but they do take a little while to learn.

As you know, for a single resistor and capacitor R and C the time constant is
simply R*C. For this simple circuit the time constant can still be calculated by
multiplying R times C, but before that we have to calculate what R is.
For this circuit, R is simply the parallel combination of R1 and R2:
R=R1*R2/(R1+R2)
and so again the time constant is:
R*C.

As you also know the voltage across the capacitor in a single RC circuit is:
Vc=Vs*(1-e^(-t/RC))

but there is another thing to think about with the two resistor circuit, and that is the
source voltage level Vs.

With a single R and C we usually take the voltage to be the full battery voltage,
but for this circuit because there is a divider we also have to calculate a new
voltage value. Since we want the voltage across the capacitor, we first calculate
the divider voltage Vd:
Vd=Vbatt*R1/(R1+R2)
so
Vd=12*180/(180+400)

and now we have the right voltage to use for the equation:

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

with R calculated above as the parallel combination of R1 and R2.

Note that Vc is now the voltage across the capacitor, with the reference
node being the junction of R1 and R2. Thus if we used a meter to try to see
this change we would put the black lead on the junction of R1 and R2 and the
red lead on the +12v power supply positive terminal.

If you want to calculate the voltage at the node junction of R1 and R2 with
reference to ground, then you would change Vd to be the voltage across
the 400 ohm resistor due to the voltage divider instead of across the 180
ohm resistor.

Note the above analysis is for when both switches are turned on at the same time.
If you want to calculate what happens when switch 1 is turned on first and switch
2 is turned on later, you have to first calculate the single RC circuit and then add
the second resistor later when switch 2 is turned on. When switch 1 is turned on
the cap starts to charge up with full battery voltage until it reaches the full 12v.
Later when switch 2 is turned on the cap starts to discharge until it reaches the
final value determined by the voltage divider.
The first part is easy because it's just a single RC circuit, 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))

where again R is the parallel combination of R1 and R2 and the time constant again
is R*C.

The above assumes that the capacitor C1 has been allowed to charge to the full voltage
of 12v before the second switch was closed. If the capacitor was not allowed to charge
for at least 5 time constants or so, then the initial cap voltage is not 12v when switch
2 is closed so we have to modify the formula to account for this by changing the 12v to
whatever the cap voltage was just before switch 2 is closed:

Vc=Vc0-(Vc0-3.72)*(1-e^(-t/RC))

so if the initial cap voltage is 6v then:

Vc=6-(6-3.72)*(1-e^(-t/RC))

Some simplification of this last formula might be possible also.

Hello And thanks a bunch for your very very helpful posts.

Very complete explanation thanks for it too, but one thing that I am not able to understand is Why there is a PARALLEL combination of R1 and R2 in this circuit? I am not able to see any though!?

 

[MrAl]

Hello again,


That's a good question and one that is a little hard to understand at first, but
a simple application of superposition helps to explain a little bit.

One way to fully analyze this circuit is to use superposition where the two
sources to be 'killed' are actually one and the same source, the 12v supply.
That's what we will attempt now...

The first step would be to realize that the 12v source feeds BOTH the resistor
R1 and the capacitor C1. Simply put, they both connect to that 12v supply.
To make this circuit simpler to analyze, we can use superposition by performing
a few simple manipulations on the circuit to allow the individual analysis of two
parts of the circuit (which are simpler than the whole) and then add the results
together to get the final analysis. This means there will be two steps where
we manipulate the circuit and one final extra step where we simply add the two
results together:

[1]
Disconnect the left side of C1 and short it to ground (which BTW puts it in parallel
with R2) and take the response at the R1,R2 node. This gives us a voltage we will
call simply Va. So far though the analysis is incomplete, but we have simplifed the
circuit a little.
[2]
Reconnect the capacitor as it was before to the 12v supply, then disconnect the
left side of R1 and short that lead of R1 to ground instead of to the 12v supply.
Note that doing this puts R1 in parallel with R2, which is the key to this question of
how R1 gets in parallel to R2. Then, find the response at the R2,C1 node (same
node as before but now the left side of R1 is shorted to ground). We'll call this
new voltage response Vb.
[3] (final step)
The only thing left to do now is to add Va plus Vb, and that gives us the total
response, but note that in order to get Vb we had to put R1 in parallel with R2.

Another way to find out how R1 gets in parallel with R2 is to do a full analysis
of the circuit and look at the time constant in the resulting formula. It's the
denominator D of the exponent of e^(-t/D) which results. From that we can
get the value of the equivalent resistor R in the exponential part of the formula
which written out that way comes out to e^(-t/RC). If we did this numerically
we could simply divide the denominator by the capacitance C, but if we did this
algebraically we would actually end up with R*C in the denominator, and R would
come out equal to the parallel combination of R1 and R2.

I know this sounds a little strange at first, but after you do a few problems like
this on paper it starts to make a lot of sense.




If instead of using superposition we could do a full analysis like this...


First, transform all the elements into their impedance equivalents:
R1=>R1
R2=>R2
zC1=1/(s*C1)

Next, calculate the parallel combination of R1 and zC1 and call this
new impedance Zu:
Zu=R1/(s*C1*R1+1)

Next, calculate the voltage divider effect of combined impedance Zu and R2:
V=E*R2/(Zu+R2);

which comes out to:
V=E*(s*C1*R1*R2+R2)/(s*C1*R1*R2+R2+R1)

Now we want the step response so multiply that by 1/s:
V=(1/s)*E*(s*C1*R1*R2+R2)/(s*C1*R1*R2+R2+R1)

which comes out to:
V=E*(s*C1*R1*R2+R2)/(s^2*C1*R1*R2+s*R2+s*R1)

Now transform to the time domain using the inverse Laplace transform, and we get:
V=E*(R1*e^(-t*(R2+R1)/(C1*R1*R2)))/(R2+R1)+E*R2/(R2+R1)

and now we concentrate on the exponential part:
e^(-t*(R2+R1)/(C1*R1*R2))

and note that this is the same as:
e^(-t/(R*C1))

where R*C1 is the time constant if we substitute:
1/((R1+R2)/(R2*R1))

with R, and that:
1/((R1+R2)/(R2*R1))=(R1*R2)/(R1+R2)

and finally we note that that last result is equal to R:
R=(R1*R2)/(R1+R2)

which is again the parallel combination of R1 and R2. Thus, R*C1 is the time constant
of the circuit and R equals the parallel combination of R1 and R2.