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%?

It coincides with 1/bandwidth and it's the time value in the first order lag \$\frac{1}{1+j\omega\tau}\$ or \$\frac{1}{1+\tau s}\$ . In radioactive decay they use 50% ('half-life'). — Chu, Sep 18 '18 at 15:12
@AndrewMorton: I'm not entirely sure what it says about me that I guessed that would be the answer just from the title. — Ilmari Karonen, Sep 19 '18 at 12:28
how interesting that that constant is so near to golden ratio (61.8%) — code_monk, Sep 19 '18 at 17:58
@code_monk: As interesting as \$e^\pi - \pi \approx 19.999\$? — Nominal Animal, Sep 20 '18 at 12:33
Just a nitpick: the time constant is not *defined* to be 63%. It is defined to be the inverse of the coefficient in the exponent of an exponential function (see the excellent answers in this thread). It just turns out *as a consequence* that the value of the quantity after a time span equal to the time constant is *approximately* (with 2-digit accuracy) 63% of the initial value. — LorenzoDonati4Ukraine-OnStrike, Sep 20 '18 at 18:46

score 78 · Accepted Answer · edited Sep 19 '18 at 12:53

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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.

edited Sep 19 '18 at 12:53

Justin

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answered Sep 18 '18 at 18:28

supercat

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This may well be the best answer, as it answers the question of "*Why?*" in a tangible way, instead of showing "*how*" to calculate it. – Bort Sep 18 '18 at 18:52
Awesome, I can't believe I've never learned this! (BTW, a graph would make this answer even more awesome). – Justin Sep 18 '18 at 19:00
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That's an excellent intuitive insight. +1 – Spehro Pefhany Sep 18 '18 at 19:32
@Justin: I don't have any graphs handy, but would not be offended in the least if someone were to include one whose creator grants permission for such usage. A graph showing X intercepts from multiple points on the curve would be best (to show that the time between any point on the curve and the X intercept is always one time constant). – supercat Sep 18 '18 at 20:17
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"the rate of decrease at any moment in time will be the current voltage" I suppose that while "current" in this context is ambiguous, both meanings work. – Acccumulation Sep 18 '18 at 20:48
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@supercat - I've added a graph of your example. Feel free to suggest any changes to it. – Justin Sep 19 '18 at 12:55

Dave Tweed · Answer 2 · 2018-09-18T15:28:25.823

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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.

edited Sep 18 '18 at 15:28

answered Sep 18 '18 at 14:52

Dave Tweed

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For an exponential decay or rise we should use step response to reduce complexity. So that e−1 is taken into account.Am i right? – Bala Subramanian Sep 18 '18 at 15:49
@BalaSubramanian: yes, right. – Dave Tweed Sep 18 '18 at 16:05
But i have one doubt, for example in designing RC circuit for timer or counter.It discharges and charges at particular time period. Is the time period is same as time constant. Does the required IC or device stops working at 63% of voltage? – Bala Subramanian Sep 18 '18 at 16:13
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@BalaSubramanian: No, not necessarily. Each timer has its own method of picking a threshold. For example, the (in)famous 555 uses 1/3 and 2/3 Vcc as its thresholds, which means that its time intervals are 0.693⋅R⋅C or 1.1⋅R⋅C, depending on the mode of operation. \$-\ln(1/3) = 1.0986\$ and \$\ln(2/3) - \ln(1/3) = 0.6931\$. – Dave Tweed Sep 18 '18 at 16:26

Spehro Pefhany · Answer 3 · 2018-09-18T15:41:24.110

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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.

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).

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.

edited Sep 18 '18 at 15:41

answered Sep 18 '18 at 14:52

Spehro Pefhany

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That is a very rough approximation. – Arsenal Sep 18 '18 at 14:55
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@Arsenal I could use MATLAB and get it to a few thousand decimal places if you'd like. – Spehro Pefhany Sep 18 '18 at 14:56
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@Arsenal, I suppose 22/7 isn't good enough for you either? :D – Sep 18 '18 at 16:20
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22/7 is a terrible approximation to e. 19/7 is much better. – alephzero Sep 18 '18 at 21:41
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@alephzero [Here](http://mathworld.wolfram.com/eApproximations.html) is a better one. – Spehro Pefhany Sep 18 '18 at 21:45
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@SpehroPefhany (wrt to that approximation you linked to) I'm always amazed at how math people like to spend their time (well, I guess crosswords puzzles are too easy for them! :-) – LorenzoDonati4Ukraine-OnStrike Sep 20 '18 at 18:41

LorenzoDonati4Ukraine-OnStrike · Answer 4 · 2018-09-23T10:41:36.300

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.

score 1 · Answer 5 · edited Sep 19 '18 at 20:35

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

edited Sep 19 '18 at 20:35

Solomon Ucko

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answered Sep 18 '18 at 14:52

Matthijs Huisman

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

5 Answers5

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