chandra999777
New Member
why a capacitor time constant is considered as 63.2% only
There is no "peak" to the exponential rise. The exponential rise is a smooth curve that starts at zero and asymptotically approaches the final value, with the rate of rise continually slowing. The 63.2% level is just a point on that curve which occurs at the one time-constant (RC) time.It's considered 63.2% because that's the 'peak' of the exponential rise, after that point the voltage rises much slower, and before the rise is too fast. It's not ALWAYS considered the constant, just a common one. You can set a voltage comparator to any percentage of line voltage you want using different resistors. 63.2% is also the best point to avoid noise.
The RC rate of rise is continually slowing in a smooth fashion. There is no point of abrupt change in an exponential rise.Crutschow, it's the mid point at which the rate of change goes from fast to slow rising?
e is derived from the point on an exponential curve where the slope is equal to 1 so it's not just some arbitrary number someone thought up. That's what I was trying to say =)In an RC circuit, the value of the time constant (in seconds) is equal to the product of the circuit resistance (in ohms) and the circuit capacitance (in farads), i.e. τ = R × C. It is the time required to charge the capacitor, through the resistor, to 63.2 (≈ 63) percent of full charge; or to discharge it to 36.8 (≈ 37) percent of its initial voltage. These values are derived from the mathematical constant e, specifically 1 − e − 1 and e − 1 respectively.
I think I understand what you are trying to say but it's still a little confusing.Ahh thank wikipedia, this explains what I was trying to say (rather poorly) a whole lot better.
e is derived from the point on an exponential curve where the slope is equal to 1 so it's not just some arbitrary number someone thought up. That's what I was trying to say =)