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Lost on RC time constant

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zachtheterrible

Active Member
It is very simple to find how long it will take for the capacitor to charge to one time constant or 63% of the supply: t=RC.

But say that I want to figure out how long it will take for my capacitor to charge up to 20% of my 3 volt supply (.6v) given R and C.

Please help, thanx :lol:
 

lord loh.

Member
the formula is

Vk=Vk0e^-(t-to)/RC


Where

Vk is the voltage across the capacitor.

Vk0 is the initial voltage across the capacitor.

This is the formula for decay.....

But I guess the time to decay 20% would be the time to charge 20%if the R is the same for charging and discharging.....

I tried to form a vague version of subscripts....<sub>subscript</sub> is not working :(
 

audioguru

Well-Known Member
Most Helpful Member
My graph shows 0.22 time constants will have the voltage charge to 20% of the supply voltage. But only with a load resistance that is much higher than the resistor's value.
 
But say that I want to figure out how long it will take for my capacitor to charge up to 20% of my 3 volt supply (.6v) given R and C.

Here you go:
 

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ljcox

Well-Known Member
To calculate the time to charge to a particular voltage assuming the capacitor is initially uncharged, you have to transpose Miles Prower's formula.

t = RC ln {(Vdc/(Vdc - V)}

So for Vdc = 3 V and V = 0.6 V

t = RC ln (1.25) = 0.223 RC

Len
 

zachtheterrible

Active Member
thanx for the help everyone :lol:

But . . . howcome you are all giving me different formulas? I am terrible with learning new equations, so im going to say some values and if someone could plug everything in and do the equation for me as an example, that would be great.

V=3v
Charge to 20% of my power supply
R=1k
C=100uf

Thank you :lol:
 

audioguru

Well-Known Member
Most Helpful Member
Hi Zach,
Your RC time constant is 100ms. It will charge to 20% (0.6V) from a 3V supply in 0.22 of a time constant.
0.22 times 100ms = 22ms. :lol:
 

zachtheterrible

Active Member
thanx audio, it doesnt get any simpler than that :lol: i tried it and it works wonderfully.

thank you for all the help everyone, i guess ill understand that rather complicated math in time . . . :lol:
 

ljcox

Well-Known Member
howcome you are all giving me different formulas?

They are not different formulae, simply different forms of the same.

However, the one posted by lord loh is for the discharge of a capacitor.

But he made an error. He meant t/to, not t-to.

I transposed the one given by Miles Prower to make "t" the subject since that is what you wanted.

Len
 

mstechca

New Member
zachtheterrible said:
But . . . howcome you are all giving me different formulas? I am terrible with learning new equations
you and I are in the same boat.

You know what I do? I go straight to work.

Zach, make sure you know Ohms law because that law can save your semiconductors lives. hehe.

so im going to say some values and if someone could plug everything in and do the equation for me as an example, that would be great.

V=3v
Charge to 20% of my power supply
R=1k
C=100uf

this is how I do it.

according to the "Technical books", The equation for time is:

t = 0.69 * R * C

so for you, it is 0.69 * 1000 * 0.0001 = 0.069

but when time moves on, 20% is 20% of 100 right?

so we do this: 0.069 * 0.20 = 0.0138

lets say 13.8ms.
 

ljcox

Well-Known Member
mstechca said:
zachtheterrible said:
But . . . howcome you are all giving me different formulas? I am terrible with learning new equations
you and I are in the same boat.

You know what I do? I go straight to work.

Zach, make sure you know Ohms law because that law can save your semiconductors lives. hehe.

so im going to say some values and if someone could plug everything in and do the equation for me as an example, that would be great.

V=3v
Charge to 20% of my power supply
R=1k
C=100uf

this is how I do it.

according to the "Technical books", The equation for time is:

t = 0.69 * R * C

so for you, it is 0.69 * 1000 * 0.0001 = 0.069

but when time moves on, 20% is 20% of 100 right?

so we do this: 0.069 * 0.20 = 0.0138

lets say 13.8ms.

This is wrong. The charge/discharge of capacitors is expotential so you can't use linear approximations. 0.69 = ln 2.

The correct answer is t = RC ln (1.25) = 0.223 RC = 22.3 millisec if R = 1k & C = 100 uF.

Len
 

audioguru

Well-Known Member
Most Helpful Member
Instead of the complicated formula of the voltage of a capacitor charging exponentially, I simply look at the curve in a book. It is scaled in time-constants and in percentage of full-charge. It looks exactly like the voltage of a charging capacitor as seen on a 'scope.

Using the graph, my estimate of a 22ms charge time was pretty darn close to the actual 22.3ms. You will rarely find resistors and capacitors with a tolerance so close. :lol:
 

bonxer

New Member
You will understand all of that much better when you take a calculus course. The 'e' is Euler's number, and is approximately 2.71828. It is useful when the slope of a function changes over time proportionally to the value at any given time. Basically, the more stuff there is, the faster the stuff is going to change. :) Large voltage across the cap makes the voltage vs time curve slope sharper. As it discharges, the voltage decreases. So does its rate of discharge, accordingly. Similar deal with charging. Below is a picture of an RC circuit repeatedly charging and discharging. Ignore the scale, I just grabbed it from a google image search.
 

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ljcox

Well-Known Member
Bonoxer's chart shows how the slope of the waveform is much steeper at the start than later in the curve. This illustrates why mstechca's method is not accurate. mstechca assumed that the curve is a straight line.

Len
 

audioguru

Well-Known Member
Most Helpful Member
Replace the charging resistor with a constant current source so the capacitor charges linearly and makes a straight line. Or you could have the cap as the negative feedback in an opamp integrator to make both lines linear and straight. :)
 

mstechca

New Member
How about this:

take the capacitor you got right now, and ditch it.

now take a new capacitor which is 1/5 the value of the old one, and try to get the tolerance as close to 0% as possible.

Use the same resistor.

Doesn't that make sense?

Let's look at it for a sec...

Say that C1 = 5F, C2 = 1F, and R1 = 1 ohm.

Let's ignore the 0.69 in the equation.

So we get C1 * R1 = 5 seconds
C2 * R1 = 1 second

Thats probably the easiest way to go.

Nevermind the complex formulas unless you are going to write an exam.
 

zachtheterrible

Active Member
This is the way that audioguru showed me to do it, and it works very well.

I have:
V=3
R=1k
C=100uf

I want .6v, which is 20% of the power supply (.6/3=.2)

Now, RC=T
1000*.0001=.1 (100ms)
Now I want %20 of my power supply so .1*.2=.02 (20ms)

Someday ill learn calculus :lol:

Isnt that a little bit of a pain mstecha? itll suck too when you have to get a capacitor that is 5 times bigger than the one that you could use, which will take up more space. But it will work :lol:
 

audioguru

Well-Known Member
Most Helpful Member
Mstechca and Zach,
You cannot just use a cap 1/5th the value and expect it to charge to 20% of the voltage as a full value cap, it is just a coincidence.
The cap charges in a curve, it starts charging quickly because the resistor feeding it has a high current due to the resistor having a high voltage across it. As the cap charges, the resistor has less current because it has less voltage across it. With less current charging the cap, it charges slower and makes the curve of voltage over time.

Cap voltage.....Time
........10%........0.1RC
........20%........0.223RC
........40%........0.52RC
........50%........0.71RC
........63%........1.0RC
........80%........1.6RC
........99%........5.0RC
.......100%.......infinity
 

audioguru

Well-Known Member
Most Helpful Member
If you don't want to use a big cap, use a higher resistor value instead, if you don't load it too much by its load. :lol:
 
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