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 by trial and error. How do you think the Chinese designed circuits 50 years ago, when computers and books were not available?
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.
MRAl
VERY, VERY , VERY, GOOD explanations.
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.
You are fooling yourself . . . you obviously haven't designed a project in your life.
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.
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.
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