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Why do we use .63 as the times the toatal voltage change...

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helpmeplz

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I was reading a how-to-electronics book for begginers, and when they were measuring capacitors and graphing them (power source 9V battery).
So why oh why do they say .63? The book says (and I quote):

"We can calculate any rising exponential curve's time constant, where a value of 0.63 times the total voltage change cuts the time axis at a time equalling the curve's time constant."

WTF??!! Does this mean? Where the hell did they get this? Then it says:


"...when the capacitor voltage is about 5V7 (that is .63 times the total voltage change of 9V)..."


Please help me understand this concept of measuring capacitors and graphing them. Thank you for your time.
 

Phasor

Member
We're talking about an RC series circuit here... that is, a resistor in series with a capacitor, and a voltage source across them both.

First - the time constant: If you multiply the capacitance of the capacitor (in Farads), by the resistance of the resistor (in ohms), you end up with a number, whose units are seconds.

Initially, the capacitor is not charged, ie. it has zero volts across it. The capacitor charges up, and the voltage rises.

Now, it happens that, in one time constant, the voltage across the capacitor is 0.63 of the total supply voltage.

Here's a worked example:

A resistor of 1.5k is in series with a capacitor of 100uF. This is placed across a supply voltage of 12V DC.

Time constant = 1500 ohms x 0.0001 F = 0.15 seconds

Voltage across the capacitor after 0.15s = 0.63 x 12V = 7.56V

Now, it gets harder: What is the voltage across the capacitor after 2 time constants?

2 time constants = 2 x 0.15s = 0.30s

V (2T) = 0.63Vs + 0.63(Vs - 0.63Vs)

...which simplifies to...

V (2T) = 0.8631 x Vs = 0.8631 x 12V
V (2T) = 10.3572V

In that second equation, V (2T), the first term (0.63Vs) is the voltage already on the capacitor after the first time constant. The second term, 0.63(Vs - 0.63Vs), is the additional voltage from the second time constant.

Theoretically, the capacitor never charges up to the full supply voltage. However, in practice, we consider the capacitor to be fully charged after 5 time constants.

Does this make things a bit clearer? Let us know if you need more help.
 

helpmeplz

New Member
why 5t? Is that so you fully know that the capacitor has reached capactitance? And is .63 standard? I know 5t is standard. But is .63? Can I use it with all types of capactiors? Thank you for your help and time spent on writing a reply to help me understand basic electronics more. Thank you very much.
 

Gene

New Member
5t is the measurable limit on charging. As Phasor said, the capacitor never quite reaches full charge but it can get close - 5t is very, very close. If you accept that the charge rate is not constant over time and that it is 63% charged at 1t, then also accept that it is about 87% charged at 2t, about 93% charged at 3t, and about 98% charged at 4t. Now, by the time you get to 5t you are about as close to fully charged as you are going to get. (I think it has to do with electron saturation of the plates inside the capacitor)

As to why 63% is the universally accepted standard, I have no idea.
 

Phasor

Member
The 0.63 comes from the equation:

Vc = Vs x (1 - e^-(t/RC))

otherwise expressed as

Vc = Vs x (1 - e^-n)
where n = the number of time constants.
e = the natural base, 2.71828....
Vc = the voltage across the capacitor
Vs = the supply voltage

If you make n = 1, and Vs = 1, you find that the answer is 0.63 (approximately).
 
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