Why is the time constant 63.2% and not 50% or 70%?

Question

I am studying about RC and RL circuits. Why is the time constant equal to 63.2% of the output voltage? Why is it defined as 63% and not any other value?

Does a circuit start working at 63% of output voltage? Why not at 50%?

Answer 1

Other answers haven't yet hit upon what makes e special: defining the time constant as the time required for something to drop by a factor of e means that at any moment of time, the rate of change will be such that--if that rate were continued--the time required to decay to nothing would be one time constant.

For example, if one has a 1uF cap and a 1M resistor, the time constant will be one second. If the capacitor is charged to 10 volts, the voltage will fall at a rate of 10 volts/second. If it's charged to 5 volts, the voltage will fall at a rate of 5 volts/second. The fact that the rate of change decreases as the voltage does means that the voltage won't actually decay to nothing in one second, but the rate of decrease at any moment in time will be the current voltage divided by the time constant.

If the time constant were defined as any other unit (e.g. half-life), then the rate of decay would no longer correspond so nicely with the time constant.

Answer 2

It's built into the mathematics of exponential decay associated with first-order systems. If the response starts at unity at t=0, then after one "unit of time", the response is \$e^{-1} = 0.36788\$. When you're looking at a risetime, you subtract this from unity, giving 0.63212 or 63.2%.

The "unit of time" is referred to as the "time constant" of the system, and is usually denoted τ (tau). The full expression for the system response over time (t) is

$$V(t) = V_0 e^{-\frac{t}{\tau}}$$

So the time constant is a useful quantity to know. If want to measure the time constant directly, you measure the time it takes to get to 63.2% of its final value.

In electronics, it works out that the time constant (in seconds) is equal to R×C in an R-C circuit or L/R in an R-L circuit, when you use ohms, farads and henries as units for the component values. This means that if you know the time constant, you can derive one of the component values if you know the other.

Answer 3

The decay of an RC parallel circuit with capacitor charged to Vo

v(t) = \$Vo(1-e^{-t/\tau})\$ , where \$\tau\$ is the time constant R\$\cdot\$C.

So v(\$\tau\$)/Vo is approximately 0.63212055882855767840447622983854

In other words, the time constant is defined by the RC product (or L/R ratio), and the seemingly arbitrary voltage is a result of that definition and the way exponential decay or charging occurs.

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Exponential decay is common to various physical processes such as radioactive decay, some kinds of cooling etc. and can be described by a first-order Ordinary Differential Equation (ODE).

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Suppose you want to know the time when the voltage is 0.5 of the initial voltage (or final voltage if charging from 0). It is (from the above)

t = -\$\ln(0.5)\tau\$ or about 0.693RC

Either way you do it, some irrational numbers pop up and dealing with RC=\$\tau\$ is the "natural" way.

Answer 4

Just as a complement to the other excellent answers by Dave Tweed, supercat and Spehro Phefany, I'll add my 2 cents.

First a bit of nitpicking, as I wrote in a comment, the time constant is not defined as 63%. Formally it is defined as the inverse of the coefficient of the exponent of an exponential function. That is, if Q is the relevant quantity (voltage, current, power, whatever), and Q decays with time as:

\[ Q(t) = Q_0 \; e^{- k t} \qquad (k>0) \]

Then the time constant of the decaying process is defined as \$ \tau = 1 / k \$.

As others have pointed out, this means that for \$ t = \tau\$ the quantity has decreased by about 63% (i.e. the quantity is about 37% of the starting value):

\[ \frac{Q(\tau)}{Q_0} = e^{-1} \approx 0.367 = 36.7 \% \]

What other answers have only marginally touched is why that choice has been made. The answer is simplicity: the time constant gives an easy way to compare the speed of evolution of similar processes. In electronics often the time constant can be interpreted as "reaction speed" of a circuit. If you know the time constants of two circuits it's easy to compare their "relative speed" by comparing those constants.

Moreover, the time constant is a quantity easily understandable in an intuitive way. For example, if I say that a circuit settles with a time constant \$ \tau = 1 \mu s\$, then I can easily understand that after a time \$3\tau=3\mu s\$ (or maybe \$5\tau=5\mu s\$, depending on the accuracy of what you are doing) I can consider the transient ended (\$3\tau\$ and \$5\tau\$ are the most common choices as rules of thumb for the conventional transient duration).

In other words the time constant is an easy and understandable way to convey the time scale on which a phenomenon occurs.

Answer 5

This comes from the \$e\$ constant value \$1-e^{-1} \approx 0.63\$.