# capacitor charging

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#### chandra999777

##### New Member
why a capacitor time constant is considered as 63.2% only

#### crutschow

##### Well-Known Member
You can't have a capacitor time-constant by itself, it also requires a resistor. A time-constant is simply a convenient engineering definition which, for a resistor and capacitor, is the resistor value multiplied by the capacitor value (R X C). If you apply a step voltage to a resistor connected to a capacitor (other cap lead to ground), then the capacitor voltage will rise to approximately 63.2% (actually 1 - 1/e) of the step voltage in one time-constant of time.

##### Banned
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.

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#### crutschow

##### Well-Known Member
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.
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.

#### k7elp60

##### Active Member
If I remember correctly the 62.5% is only when the capacitor is being charged thru a resistor. I you charge the capacitor with a constant current then the charge voltage increase linearly.

#### Speakerguy

##### Active Member
Yes, if you charge the capacitor with a current source the voltage ramps up linearly. If you charge it with a voltage source through a resistor the voltage goes up as 1 - e^-t/RC.

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##### Banned
Crutschow, it's the mid point at which the rate of change goes from fast to slow rising?

#### crutschow

##### Well-Known Member
Crutschow, it's the mid point at which the rate of change goes from fast to slow rising?
The RC rate of rise is continually slowing in a smooth fashion. There is no point of abrupt change in an exponential rise.

##### Banned
Ahh thank wikipedia, this explains what I was trying to say (rather poorly) a whole lot better.
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.
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 =)

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#### crutschow

##### Well-Known Member
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 =)
I think I understand what you are trying to say but it's still a little confusing.

e is not derived from the exponential curve, it is used to describe it. As speakerguy79 noted, the curve is 1 - e^-t/RC where e is the base for the natural logarithm. At the arbitrary time of one time constant (t = RC) that point on the curve equals 1-(1/e) which is 0.632 or 63.2%.