Wrapping my head around "Time Constant"

Thread Starter

cheddy

Joined Oct 19, 2007
87
I'm learning about capacitors now and I understand all the equations about charging current and the exponential curve of voltage over time. The chapter is starting to have me calculate the time constant for various different experiments.

While I can blindly follow the equations and come up with solutions that are right I still can't wrap my head around WHY if you multiply the resistance and the capacitance you arrive at a number (tau) that just happens to equal the time it takes to reach 63% of the maximum voltage.

It just doesn't seem intuitive and it's bothering me. Is it simply too complicated for a beginner to understand so it's not explained in depth or am I missing something?

http://www.kpsec.freeuk.com/capacit.htm#charging
 

Thread Starter

cheddy

Joined Oct 19, 2007
87
After reading through the relevant sections in the All About Circuits E-book I am still having the same difficulty understanding.

For example:
http://www.allaboutcircuits.com/vol_1/chpt_16/4.html



Tau = R x C is stated as fact without any derivation. I accept it is true because experimentation shows it to be true but I am having a problem grasping why it's true.

More generally, it doesn't seem intuitive to me if you multiply Ohms and Farads you arrive at a time in seconds. A time which represents how long it takes to reach ~63% of the maximum voltage.
 

Salgat

Joined Dec 23, 2006
215
Capacitance,
C = Q/V

Q = coulombs
V = voltage

RC = R*Q/V
R/V = I^(-1)

I = Q/T
Inverse of I = T/Q

T/Q * Q = T

If you substitute and simplify you eventually reach T. The time is essentially derived from the relationship of current and time, its reasonable not to see this at first.
 

Dave

Joined Nov 17, 2003
6,960
it doesn't seem intuitive to me if you multiply Ohms and Farads you arrive at a time in seconds.
Ignore this bit. GS3 has given you the proper answer in post #7.

Tau = R x C is stated as fact without any derivation. I accept it is true because experimentation shows it to be true but I am having a problem grasping why it's true.
I think this is essentially the case. The response of an RC circuit is given in terms of the solution to the LTI differential equation dV(t) = -aV(t), where a is the exponential decay and is the reciprocal of the time constant (this is intuitive since increasing the time constant results in a reduction in the rate of decay).

The general solution of this equation gives you the equation you are familiar with for an exponential decaying function: V(t) = Vo(t)exp[-at]

Empirically, a is equal to 1/RC, i.e. increasing the value of R or C will result in a reduction in the rate of exponential decay - this is intuitive since increase the either of these two values results in an increase in the "resistance" (different from the resistance R) of the circuit to changes in V. Therefore, by reasoning from the above argument the time constant is equal to RC.

Some scientist will probably go back to first principles and derive this from Maxwell's Equations, however I don't have time right now! - Edit: This scientist is Salgat! (see above).

Dave
 

Dragon

Joined Sep 25, 2007
42
cheddy,

Its good for you that you are analysing things on mathematical lines rather than accepting them as such. However, dont fret too much if you just cannot find a logical explanation/derivation for a particular concept. At times you just have to wait until you are exposed to the complexities of engineering mathematics.

A 'text' book on ciruit analysis when I first studied the subject cleared such concepts for me. I remember that it explains in detail exactly what you are looking for. Try loaning it from a library, or better still, buy one. I referenced to it for most part of my engineering degree.

Electric Circuits Fundamentals by Sergio Franco.

Also, you might want to consult Microelectronics by Sedra/Smith. (The part that deals with basic power supply designs. Thay have derivations on the concept of smoothing capacitors which rely extensively on TAU conept);)
 

GS3

Joined Sep 21, 2007
408
The time constant is not strictly time, but a value that is used with time to describe a particular behaviour. I know some literature describes time constant in units of second, but I find this terminology both contradictory and confusing.
The time constant is time, meaning it has the dimension of time. Check out Wikipedia and you can see the dimensions of the Farad: (S^4 * A^2) / (M^2 * Kg) and of the Ohm: (M^2 * Kg) / (S^3 * A^2). Multiply one by the other and you get pure, unadulterated, seconds of time.
 

Dave

Joined Nov 17, 2003
6,960
The time constant is time, meaning it has the dimension of time. Check out Wikipedia and you can see the dimensions of the Farad: (S^4 * A^2) / (M^2 * Kg) and of the Ohm: (M^2 * Kg) / (S^3 * A^2). Multiply one by the other and you get pure, unadulterated, seconds of time.
I stand whole heartedly corrected. Indeed there is some basis for RC being referred to as time and as such is not just a convention due to its relationship with the time variable, t, in RC applications.

Indeed the situation is the same for the L/R time constant. Henrys are V.s/A and Ohms are V/A, therefore L/R = (V.s/A)/(V/A) = s.

Thanks for the correction.

Dave
 

Nomad

Joined Oct 21, 2007
43
well i think so Brain, but me and Pippi Longstocking! What would the children look like?! NARF!
 

Dragon

Joined Sep 25, 2007
42
Wow! The perfect analogies very clearly explain the concepts which mathematics cannot address alone. Even most high profile books miss this point. Well done AAC!
I wish I knew about AAC during my undergrad years! :rolleyes:
 

Dave

Joined Nov 17, 2003
6,960
Wow! The perfect analogies very clearly explain the concepts which mathematics cannot address alone. Even most high profile books miss this point. Well done AAC!
I wish I knew about AAC during my undergrad years! :rolleyes:
Thanks for the comments, the authors of this book and the many people who have given time to proofread and amend the text have done a great job.

Feel free to let others know about AAC so that they may benefit from the resources it has to offer.

Dave
 
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