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Why is the time constant 63.2% and not 50% or 70%?
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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%?
circuit-analysis circuit-design math time-constant
edited 16 mins ago
bariod
32
asked 2 days ago
Bala Subramanian
129128
add a comment |Â
up vote
23
down vote
favorite
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%?
circuit-analysis circuit-design math time-constant
edited 16 mins ago
bariod
32
asked 2 days ago
Bala Subramanian
129128
38
1-e^-1 = 0.6321...
– Andrew Morton
2 days ago
2
It coincides with 1/bandwidth and it's the time value in the first order lag $frac11+jomegatau$ or $frac11+tau s$ . In radioactive decay they use 50% ('half-life').
– Chu
yesterday
1
@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
yesterday
how interesting that that constant is so near to golden ratio (61.8%)
– code_monk
21 hours ago
2
@code_monk: As interesting as $e^pi - pi approx 19.999$?
– Nominal Animal
2 hours ago
add a comment |Â
up vote
23
down vote
favorite
up vote
23
down vote
favorite
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%?
circuit-analysis circuit-design math time-constant
edited 16 mins ago
bariod
32
asked 2 days ago
Bala Subramanian
129128
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%?
circuit-analysis circuit-design math time-constant
circuit-analysis circuit-design math time-constant
edited 16 mins ago
bariod
32
asked 2 days ago
Bala Subramanian
129128
edited 16 mins ago
bariod
32
asked 2 days ago
Bala Subramanian
129128
edited 16 mins ago
bariod
32
edited 16 mins ago
bariod
32
edited 16 mins ago
bariod
32
32
asked 2 days ago
Bala Subramanian
129128
asked 2 days ago
Bala Subramanian
129128
asked 2 days ago
Bala Subramanian
129128
129128
38
1-e^-1 = 0.6321...
– Andrew Morton
2 days ago
2
It coincides with 1/bandwidth and it's the time value in the first order lag $frac11+jomegatau$ or $frac11+tau s$ . In radioactive decay they use 50% ('half-life').
– Chu
yesterday
1
@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
yesterday
how interesting that that constant is so near to golden ratio (61.8%)
– code_monk
21 hours ago
2
@code_monk: As interesting as $e^pi - pi approx 19.999$?
– Nominal Animal
2 hours ago
add a comment |Â
38
1-e^-1 = 0.6321...
– Andrew Morton
2 days ago
2
It coincides with 1/bandwidth and it's the time value in the first order lag $frac11+jomegatau$ or $frac11+tau s$ . In radioactive decay they use 50% ('half-life').
– Chu
yesterday
1
@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
yesterday
how interesting that that constant is so near to golden ratio (61.8%)
– code_monk
21 hours ago
2
@code_monk: As interesting as $e^pi - pi approx 19.999$?
– Nominal Animal
2 hours ago
38
38
1-e^-1 = 0.6321...
– Andrew Morton
2 days ago
1-e^-1 = 0.6321...
– Andrew Morton
2 days ago
2
2
It coincides with 1/bandwidth and it's the time value in the first order lag $frac11+jomegatau$ or $frac11+tau s$ . In radioactive decay they use 50% ('half-life').
– Chu
yesterday
It coincides with 1/bandwidth and it's the time value in the first order lag $frac11+jomegatau$ or $frac11+tau s$ . In radioactive decay they use 50% ('half-life').
– Chu
yesterday
1
1
@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
yesterday
@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
yesterday
how interesting that that constant is so near to golden ratio (61.8%)
– code_monk
21 hours ago
how interesting that that constant is so near to golden ratio (61.8%)
– code_monk
21 hours ago
2
2
@code_monk: As interesting as $e^pi - pi approx 19.999$?
– Nominal Animal
2 hours ago
@code_monk: As interesting as $e^pi - pi approx 19.999$?
– Nominal Animal
2 hours ago
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
44
down vote
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 Ï„ (tau). The full expression for the system response over time (t) is
$$V(t) = V_0 e^-fracttau$$
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.
answered 2 days ago
Dave Tweed♦
109k9131235
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
yesterday
@BalaSubramanian: yes, right.
– Dave Tweed♦
yesterday
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
yesterday
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. $-ln(1/3) = 1.0986$ and $ln(2/3) - ln(1/3) = 0.6931$.
– Dave Tweed♦
yesterday
add a comment |Â
up vote
39
down vote
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 yesterday
Justin
3,1921022
answered yesterday
supercat
37.3k158107
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
yesterday
Awesome, I can't believe I've never learned this! (BTW, a graph would make this answer even more awesome).
– Justin
yesterday
That's an excellent intuitive insight. +1
– Spehro Pefhany
yesterday
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
yesterday
6
@supercat - I've added a graph of your example. Feel free to suggest any changes to it.
– Justin
yesterday
 |Â
show 2 more comments
up vote
10
down vote
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.
answered 2 days ago
Spehro Pefhany
195k4139388
8
That is a very rough approximation.
– Arsenal
2 days ago
1
@Arsenal I could use MATLAB and get it to a few thousand decimal places if you'd like.
– Spehro Pefhany
2 days ago
2
@Arsenal, I suppose 22/7 isn't good enough for you either? :D
– Wossname
yesterday
3
22/7 is a terrible approximation to e. 19/7 is much better.
– alephzero
yesterday
@alephzero Here is a better one.
– Spehro Pefhany
yesterday
add a comment |Â
up vote
0
down vote
This comes from the $e$ constant value $1-e^-1 approx 0.63$.
edited 18 hours ago
Solomon Ucko
1033
answered 2 days ago
Matthijs Huisman
323
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
44
down vote
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 Ï„ (tau). The full expression for the system response over time (t) is
$$V(t) = V_0 e^-fracttau$$
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.
answered 2 days ago
Dave Tweed♦
109k9131235
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
yesterday
@BalaSubramanian: yes, right.
– Dave Tweed♦
yesterday
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
yesterday
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. $-ln(1/3) = 1.0986$ and $ln(2/3) - ln(1/3) = 0.6931$.
– Dave Tweed♦
yesterday
add a comment |Â
up vote
44
down vote
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 Ï„ (tau). The full expression for the system response over time (t) is
$$V(t) = V_0 e^-fracttau$$
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.
answered 2 days ago
Dave Tweed♦
109k9131235
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
yesterday
@BalaSubramanian: yes, right.
– Dave Tweed♦
yesterday
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
yesterday
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. $-ln(1/3) = 1.0986$ and $ln(2/3) - ln(1/3) = 0.6931$.
– Dave Tweed♦
yesterday
add a comment |Â
up vote
44
down vote
up vote
44
down vote
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 Ï„ (tau). The full expression for the system response over time (t) is
$$V(t) = V_0 e^-fracttau$$
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.
answered 2 days ago
Dave Tweed♦
109k9131235
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 Ï„ (tau). The full expression for the system response over time (t) is
$$V(t) = V_0 e^-fracttau$$
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.
answered 2 days ago
Dave Tweed♦
109k9131235
edited yesterday
answered 2 days ago
Dave Tweed♦
109k9131235
answered 2 days ago
Dave Tweed♦
109k9131235
answered 2 days ago
Dave Tweed♦
109k9131235
109k9131235
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
yesterday
@BalaSubramanian: yes, right.
– Dave Tweed♦
yesterday
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
yesterday
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. $-ln(1/3) = 1.0986$ and $ln(2/3) - ln(1/3) = 0.6931$.
– Dave Tweed♦
yesterday
add a comment |Â
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
yesterday
@BalaSubramanian: yes, right.
– Dave Tweed♦
yesterday
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
yesterday
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. $-ln(1/3) = 1.0986$ and $ln(2/3) - ln(1/3) = 0.6931$.
– Dave Tweed♦
yesterday
1
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
yesterday
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
yesterday
@BalaSubramanian: yes, right.
– Dave Tweed♦
yesterday
@BalaSubramanian: yes, right.
– Dave Tweed♦
yesterday
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
yesterday
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
yesterday
2
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. $-ln(1/3) = 1.0986$ and $ln(2/3) - ln(1/3) = 0.6931$.
– Dave Tweed♦
yesterday
@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♦
yesterday
add a comment |Â
up vote
39
down vote
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 yesterday
Justin
3,1921022
answered yesterday
supercat
37.3k158107
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
yesterday
Awesome, I can't believe I've never learned this! (BTW, a graph would make this answer even more awesome).
– Justin
yesterday
That's an excellent intuitive insight. +1
– Spehro Pefhany
yesterday
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
yesterday
6
@supercat - I've added a graph of your example. Feel free to suggest any changes to it.
– Justin
yesterday
 |Â
show 2 more comments
up vote
39
down vote
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 yesterday
Justin
3,1921022
answered yesterday
supercat
37.3k158107
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
yesterday
Awesome, I can't believe I've never learned this! (BTW, a graph would make this answer even more awesome).
– Justin
yesterday
That's an excellent intuitive insight. +1
– Spehro Pefhany
yesterday
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
yesterday
6
@supercat - I've added a graph of your example. Feel free to suggest any changes to it.
– Justin
yesterday
 |Â
show 2 more comments
up vote
39
down vote
up vote
39
down vote
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 yesterday
Justin
3,1921022
answered yesterday
supercat
37.3k158107
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 yesterday
Justin
3,1921022
answered yesterday
supercat
37.3k158107
edited yesterday
Justin
3,1921022
edited yesterday
Justin
3,1921022
edited yesterday
Justin
3,1921022
3,1921022
answered yesterday
supercat
37.3k158107
answered yesterday
supercat
37.3k158107
answered yesterday
supercat
37.3k158107
37.3k158107
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
yesterday
Awesome, I can't believe I've never learned this! (BTW, a graph would make this answer even more awesome).
– Justin
yesterday
That's an excellent intuitive insight. +1
– Spehro Pefhany
yesterday
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
yesterday
6
@supercat - I've added a graph of your example. Feel free to suggest any changes to it.
– Justin
yesterday
 |Â
show 2 more comments
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
yesterday
Awesome, I can't believe I've never learned this! (BTW, a graph would make this answer even more awesome).
– Justin
yesterday
That's an excellent intuitive insight. +1
– Spehro Pefhany
yesterday
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
yesterday
6
@supercat - I've added a graph of your example. Feel free to suggest any changes to it.
– Justin
yesterday
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
yesterday
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
yesterday
Awesome, I can't believe I've never learned this! (BTW, a graph would make this answer even more awesome).
– Justin
yesterday
Awesome, I can't believe I've never learned this! (BTW, a graph would make this answer even more awesome).
– Justin
yesterday
That's an excellent intuitive insight. +1
– Spehro Pefhany
yesterday
That's an excellent intuitive insight. +1
– Spehro Pefhany
yesterday
1
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
yesterday
"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
yesterday
6
6
@supercat - I've added a graph of your example. Feel free to suggest any changes to it.
– Justin
yesterday
@supercat - I've added a graph of your example. Feel free to suggest any changes to it.
– Justin
yesterday
 |Â
show 2 more comments
up vote
10
down vote
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.
answered 2 days ago
Spehro Pefhany
195k4139388
8
That is a very rough approximation.
– Arsenal
2 days ago
1
@Arsenal I could use MATLAB and get it to a few thousand decimal places if you'd like.
– Spehro Pefhany
2 days ago
2
@Arsenal, I suppose 22/7 isn't good enough for you either? :D
– Wossname
yesterday
3
22/7 is a terrible approximation to e. 19/7 is much better.
– alephzero
yesterday
@alephzero Here is a better one.
– Spehro Pefhany
yesterday
add a comment |Â
up vote
10
down vote
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.
answered 2 days ago
Spehro Pefhany
195k4139388
8
That is a very rough approximation.
– Arsenal
2 days ago
1
@Arsenal I could use MATLAB and get it to a few thousand decimal places if you'd like.
– Spehro Pefhany
2 days ago
2
@Arsenal, I suppose 22/7 isn't good enough for you either? :D
– Wossname
yesterday
3
22/7 is a terrible approximation to e. 19/7 is much better.
– alephzero
yesterday
@alephzero Here is a better one.
– Spehro Pefhany
yesterday
add a comment |Â
up vote
10
down vote
up vote
10
down vote
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.
answered 2 days ago
Spehro Pefhany
195k4139388
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.
answered 2 days ago
Spehro Pefhany
195k4139388
edited yesterday
answered 2 days ago
Spehro Pefhany
195k4139388
answered 2 days ago
Spehro Pefhany
195k4139388
answered 2 days ago
Spehro Pefhany
195k4139388
195k4139388
8
That is a very rough approximation.
– Arsenal
2 days ago
1
@Arsenal I could use MATLAB and get it to a few thousand decimal places if you'd like.
– Spehro Pefhany
2 days ago
2
@Arsenal, I suppose 22/7 isn't good enough for you either? :D
– Wossname
yesterday
3
22/7 is a terrible approximation to e. 19/7 is much better.
– alephzero
yesterday
@alephzero Here is a better one.
– Spehro Pefhany
yesterday
add a comment |Â
8
That is a very rough approximation.
– Arsenal
2 days ago
1
@Arsenal I could use MATLAB and get it to a few thousand decimal places if you'd like.
– Spehro Pefhany
2 days ago
2
@Arsenal, I suppose 22/7 isn't good enough for you either? :D
– Wossname
yesterday
3
22/7 is a terrible approximation to e. 19/7 is much better.
– alephzero
yesterday
@alephzero Here is a better one.
– Spehro Pefhany
yesterday
8
8
That is a very rough approximation.
– Arsenal
2 days ago
That is a very rough approximation.
– Arsenal
2 days ago
1
1
@Arsenal I could use MATLAB and get it to a few thousand decimal places if you'd like.
– Spehro Pefhany
2 days ago
@Arsenal I could use MATLAB and get it to a few thousand decimal places if you'd like.
– Spehro Pefhany
2 days ago
2
2
@Arsenal, I suppose 22/7 isn't good enough for you either? :D
– Wossname
yesterday
@Arsenal, I suppose 22/7 isn't good enough for you either? :D
– Wossname
yesterday
3
3
22/7 is a terrible approximation to e. 19/7 is much better.
– alephzero
yesterday
22/7 is a terrible approximation to e. 19/7 is much better.
– alephzero
yesterday
@alephzero Here is a better one.
– Spehro Pefhany
yesterday
@alephzero Here is a better one.
– Spehro Pefhany
yesterday
add a comment |Â
up vote
0
down vote
This comes from the $e$ constant value $1-e^-1 approx 0.63$.
edited 18 hours ago
Solomon Ucko
1033
answered 2 days ago
Matthijs Huisman
323
add a comment |Â
up vote
0
down vote
This comes from the $e$ constant value $1-e^-1 approx 0.63$.
edited 18 hours ago
Solomon Ucko
1033
answered 2 days ago
Matthijs Huisman
323
add a comment |Â
up vote
0
down vote
up vote
0
down vote
This comes from the $e$ constant value $1-e^-1 approx 0.63$.
edited 18 hours ago
Solomon Ucko
1033
answered 2 days ago
Matthijs Huisman
323
This comes from the $e$ constant value $1-e^-1 approx 0.63$.
edited 18 hours ago
Solomon Ucko
1033
answered 2 days ago
Matthijs Huisman
323
edited 18 hours ago
Solomon Ucko
1033
edited 18 hours ago
Solomon Ucko
1033
edited 18 hours ago
Solomon Ucko
1033
1033
answered 2 days ago
Matthijs Huisman
323
answered 2 days ago
Matthijs Huisman
323
answered 2 days ago
Matthijs Huisman
323
323
add a comment |Â
add a comment |Â
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38
1-e^-1 = 0.6321...
– Andrew Morton
2 days ago
2
It coincides with 1/bandwidth and it's the time value in the first order lag $frac11+jomegatau$ or $frac11+tau s$ . In radioactive decay they use 50% ('half-life').
– Chu
yesterday
1
@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
yesterday
how interesting that that constant is so near to golden ratio (61.8%)
– code_monk
21 hours ago
2
@code_monk: As interesting as $e^pi - pi approx 19.999$?
– Nominal Animal
2 hours ago