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0.693??? RC Time constant

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shaneshane1

New Member
Hi

I know that to find R in a RC Time constant is to do the following

eg: 2(seconds) / 0.000470(farads) / 0.693 :confused: = 6140ohms

what does 0.693 represent, i thought that it might have been 0.7V?
 

dknguyen

Well-Known Member
Most Helpful Member
The way the resistor-capacitor charging equation works out, if you set the time duration to be equal to R*C (the RC time constant which also appears in the equation) then the capacitor will have charged to 63% of the full voltage, or discharge down to 100%-63% = 37% of the full voltage.

The 0.693 number used for 555 timer calculations comes from how the voltage charges back and forth between 1/3V and 2/3V, while the RC time constant is for charging and discharging between 0V and +V.

...I think. I've never actually sat down and did the calculations. THat's just my reasoning. I'd like to know too actually.
 
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Diver300

Well-Known Member
Most Helpful Member
If you want to find the time constant, you don't need the 0.693

The time constant is just RC

The time constant is the total change divided by the initial rate of change.
For example, 470 :mu:F and 10 k:eek:hm: resistor, time constant is 4.7 s.

If the capacitor is charged to 5 V, the current is 5 / 10000 = 0.5 mA
0.5 mA causes the capacitor to change at 0.0005 / 0.00047 = 1.06 V / second.
Time constant is 5 / 1.06 = 4.7 s, and it doesn't matter what starting voltage you chose.

(Time constants apply to all sorts of things, not just electricity)

Because the voltage falls, the current falls, so the rate of change of voltage also falls. It's called exponential decay.

The formula is V = Vstart * e^(-t/RC)
where V is the voltage on the capacitor
Vstart it what it starts at
e is the mathematical constant = 2.717......
t is the time in seconds
R is the resitance
C is the capacitance

The way that time constants are often measure or used is with "half-lives", or the time to get to half the start value.

If the time to get to half the start value is Th, then:-

0.5 = e^(-Th/RC)

This gives Th/RC = 0.693 and that is where 0.693 comes from.
 

audioguru

Well-Known Member
Most Helpful Member
0.693 is used to calculate how long the capacitor charges to 67% of the supply voltage then discharges to 33% of the supply voltage in a 555 astable oscillator circuit.
 

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GuiRitter

New Member
The Secret

hi y'all! i came here from google and i will reveal the secret! i also fought a lot to find out where it came from, i found it and i forgot it... but today i found it again! prepare yourself:

it's the natural logarithm of two!

don't believe me?
Multivibrator - Wikipedia, the free encyclopedia

from windows' calculator: ≈ 0,69314718055994530941723212145818

i hope i made y'all happy!
 
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Abizzle

New Member
So some of these replies helped me, but not really. So I had to keep looking. This thread explained the most. Specifically, user Ernie M.
 

jpanhalt

Well-Known Member
Most Helpful Member
Welcome to ETO

That was 6 years ago. Why are you still concerned about it in light of post #6?
 

Abizzle

New Member
Sorry, I was recently looking into 555 timers and it's astable operation, and I saw the 0.693 number and I looked it up. This post came up first in the results, so I just wanted to share my results in terms of what I found was helpful. And also the fact that I didn't check Wikipedia :banghead:
 

jpanhalt

Well-Known Member
Most Helpful Member
There are always first, second, and n^th order effects. Still, why are you concerned? Do you really think the published equations for the 555 apply to the 10^-4 decimal?
 

Ratchit

Well-Known Member
Hi

I know that to find R in a RC Time constant is to do the following

eg: 2(seconds) / 0.000470(farads) / 0.693 :confused: = 6140ohms

what does 0.693 represent, i thought that it might have been 0.7V?

The rise and fall of voltage in a RC circuit is determined by a differential equation (DE). That is because the current is dependent on the rate of voltage change across the capacitor. Attached is the derivation and solution of the DE. As you can see form the result, the constant 0.6321 comes from the voltage rise in one time constant and the constant 0.3679 comes from the voltage fall in one time constant. Ask if you have any questions.
shaneshane1.JPG

Ratch
 

kinarfi

Well-Known Member
When I was much younger, learning electronics, ".693" was just one of those constant you learned, like the √2 and .707, the reciprocal if √2. Later I saw that .693 number being used in the calculation of the flow of water through a hole in a container, the illustration had a vertical pipe full of water with a hole in the side, at the bottom, and the statement that the pipe would be empty after 5 time constants, calculated by the size of the hole, the diameter, and the height of the water in the tube, just like the charge or discharge of a capacitor. I was impressed that an electronics number was used elsewhere. Back then, we didn't have computers, internet or Google to research such interesting discoveries, so I just took it to be one of those constants that applies to nature. Now , I have the computer, internet, E. T. O., and google, I just googled ".693" and learned a lot. Try it.
Jeff
 

Ratchit

Well-Known Member
When I was much younger, learning electronics, ".693" was just one of those constant you learned, like the √2 and .707, the reciprocal if √2. Later I saw that .693 number being used in the calculation of the flow of water through a hole in a container, the illustration had a vertical pipe full of water with a hole in the side, at the bottom, and the statement that the pipe would be empty after 5 time constants, calculated by the size of the hole, the diameter, and the height of the water in the tube, just like the charge or discharge of a capacitor. I was impressed that an electronics number was used elsewhere. Back then, we didn't have computers, internet or Google to research such interesting discoveries, so I just took it to be one of those constants that applies to nature. Now , I have the computer, internet, E. T. O., and google, I just googled ".693" and learned a lot. Try it.
Jeff

Yes, 0.693 = ln(2), but that constant does not show up in the decay or rise of a RC circuit, which the OP was asking about. e^-1 = 0.37 or (1-0.37) - 0.63 does, however.

Ratch
 

audioguru

Well-Known Member
Most Helpful Member
I think time constants (capacitor charging and discharging times) are easy to see on a graph:
 

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kinarfi

Well-Known Member
Yes, 0.693 = ln(2), but that constant does not show up in the decay or rise of a RC circuit, which the OP was asking about. e^-1 = 0.37 or (1-0.37) - 0.63 does, however.

Ratch
The OP did ask about .693 and Audioguru covered it with graphs, what I learned is that after 5 TC , the charge or discharge was considered done.
If you start a charge at 0 volt with 100 volts applied, after 1 TC1 the voltage is 69.3 volts,
during TC5, the capacitor charges 69.3% of remaining 0.888287 (0.615583) so now the voltage is 99.7273
during TC6, the capacitor charges 69.3% of remaining 0.272704 (0.188984) so now the voltage is 99.91628
The table carries the process out to TC11
693 TC.JPG

Jeff
 

Ratchit

Well-Known Member
The OP did ask about .693 and Audioguru covered it with graphs, what I learned is that after 5 TC , the charge or discharge was considered done.
If you start a charge at 0 volt with 100 volts applied, after 1 TC1 the voltage is 69.3 volts,
during TC5, the capacitor charges 69.3% of remaining 0.888287 (0.615583) so now the voltage is 99.7273
during TC6, the capacitor charges 69.3% of remaining 0.272704 (0.188984) so now the voltage is 99.91628
The table carries the process out to TC11
View attachment 109446
Jeff

Your premise is false and your table is incorrect. Ln(2) =0.693147 is not involved in the energizing or de-energizing of a capacitor. The correct value for your example is 100(1-e^-1) = 63.2121 for the first time constant (TC) and 100(1-e^-5) = 99.3263 for 5 TC's.

Ratch
 
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kinarfi

Well-Known Member
I just don't know, maybe electronics has change since I graduated, we only had slide rules back then and
It was close enough
693 TC#2.JPG
 

Ratchit

Well-Known Member
I just don't know, maybe electronics has change since I graduated, we only had slide rules back then and
It was close enough
View attachment 109454

I did 1/e on my old slide rule and got 0.368, accurate to 3 significant figures. Anyhow, '"good enough" depends on the application. The important point is that you do not think that the rise or decay of a RC circuit is logarithmic instead of exponential.

Ratch
 

AnalogKid

Well-Known Member
Most Helpful Member
I just don't know, maybe electronics has change since I graduated, we only had slide rules back then and
It was close enough
Really? What course were you in where an 8th grade level math problem was allowed to have an error of almost 10%?

ak
 
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