Por que o tempo é constante 63,2% e não 50% ou 70%?

30

Estou estudando sobre circuitos RC e RL. Por que a constante de tempo é igual a 63,2% da tensão de saída? Por que é definido como 63% e não qualquer outro valor?

Um circuito começa a funcionar com 63% da tensão de saída? Por que não 50%?

— Bala Subramanian
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41

1-e ^ -1 = 0,6321 ...

— Andrew Morton

3

Ele coincide com 1 / largura de banda e é o valor do tempo na primeira ordem de atraso

ou

\frac{1}{1 + j ω τ}

$\frac{1}{1+j\omega\tau}$

. No decaimento radioativo, eles usam 50% (meia-vida).

\frac{1}{1 + τ s}

$\frac{1}{1+\tau s}$

— 18718 Chu

11

@AndrewMorton: Não sei bem o que diz sobre mim que imaginei que seria a resposta apenas do título.

— Ilmari Karonen

4

@code_monk: Tão interessante quanto

?

e^{π} - π \approx 19.999

$e^\pi - \pi \approx 19.999$

— Animal Nominal

3

Apenas um resumo: a constante de tempo não está definida para ser de 63%. É definido como o inverso do coeficiente no expoente de uma função exponencial (veja as excelentes respostas neste tópico). É apenas gira para fora , em consequência de que o valor da quantidade após um intervalo de tempo igual à constante de tempo é aproximadamente (com precisão dois dígitos) 63% do valor inicial.

— Lorenzo Donati apoia Monica

64

Outras respostas ainda não atingiram o que torna e especial: definir o tempo constante como o tempo necessário para que algo caia por um fator de e significa que, a qualquer momento, a taxa de mudança será tal que - se isso a taxa continuava - o tempo necessário para decair para nada seria uma vez constante.

Por exemplo, se alguém tiver uma tampa de 1uF e um resistor de 1M, a constante de tempo será de um segundo. Se o capacitor for carregado em 10 volts, a tensão cairá a uma taxa de 10 volts / segundo. Se for carregada a 5 volts, a tensão cairá a uma taxa de 5 volts / segundo. O fato de a taxa de variação diminuir à medida que a tensão diminui significa que a tensão não decairá em nada em um segundo, mas a taxa de diminuição a qualquer momento no tempo será a tensão atual dividida pela constante de tempo.

Se a constante de tempo fosse definida como qualquer outra unidade (por exemplo, meia-vida), a taxa de decaimento deixaria de corresponder tão bem à constante de tempo.

— supercat
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3

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

Awesome, I can't believe I've never learned this! (BTW, a graph would make this answer even more awesome).

— Reinstate Monica

1

That's an excellent intuitive insight. +1

— Spehro Pefhany

1

"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

11

@supercat - I've added a graph of your example. Feel free to suggest any changes to it.

— Reinstate Monica

49

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}{τ}}

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

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.

— Dave Tweed
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1

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

@BalaSubramanian: yes, right.

— Dave Tweed

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

2

@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

$-\ln(1/3) = 1.0986$ and

\ln (2 / 3) - \ln (1 / 3) = 0.6931

$\ln(2/3) - \ln(1/3) = 0.6931$ .

— Dave Tweed

11

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.

— Spehro Pefhany
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10

That is a very rough approximation.

— Arsenal

1

@Arsenal I could use MATLAB and get it to a few thousand decimal places if you'd like.

— Spehro Pefhany

2

@Arsenal, I suppose 22/7 isn't good enough for you either? :D

— Wossname

3

22/7 is a terrible approximation to e. 19/7 is much better.

— alephzero

2

@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! :-)

— Lorenzo Donati supports Monica

3

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} (k > 0)

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

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 (τ)}{Q_{0}} = e^{- 1} \approx 0.367 = 36.7 %

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

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.

— Lorenzo Donati supports Monica
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-1

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

— Matthijs Huisman
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