Put t on the left side? Vc = V * (1- e^(-t / R*C))

[blueroomelectronics]

It's high school math and I've simply forgotten how. Yes getting old is fun.

I want t (time) on the left side of the equation and Vc on the right.

Vc = V * (1- e^(-t / R*C))

 

[Roff]

It's high school math and I've simply forgotten how. Yes getting old is fun.

I want t (time) on the left side of the equation and Vc on the right.

Vc = V * (1- e^(-t / R*C))

Vc/V = 1- e^(-t /( R*C))
e^(-t /( R*C))=1-(Vc/V)
t/(R*C)=-ln(1-(Vc/V))

t=-R*C*(ln(1-(Vc/V)))

 

[blueroomelectronics]

Well Roff you've earned a Cricket PCB when they're ready.Thanks.

 

[bloody-orc]

I get something different:

And please. write your formulas mathematically correct!!! there are braces missing all over the place!

 

[Roff]

I get something different:

And please. write your formulas mathematically correct!!! there are braces missing all over the place!

Good catch. The "C"s in the right side of the last two equations should be lower case (ideally, all the lower case "c"s should be subscripted). Those were typos. I have edited my original post to reflect this. My results and yours now match.
Some braces are missing in my equations because I copied and pasted William's equation.
I have edited my original post to reflect these corrections.

 

[ljcox]

Ron,
It would be easier for novices if you eliminated the negative sign.

ie. t = R*C *ln{Vc/(Vc-V)}

 

[Leftyretro]

Ron,
It would be easier for novices if you eliminated the negative sign.

ie. t = R*C *ln{Vc/(Vc-V)}

For us math challenged members what is and where did the "ln" term come from?

Lefty

 

[ericgibbs]

For us math challenged members what is and where did the "ln" term come from?

Lefty

hi,
Look here for explanation:
Its a standard exponential to natural log transform

A couple of simple examples to demonstrate.
Exp(1) = 2.718282
ln(2.718282) = 1.0

Exp(2) = 7.389056
ln(7.389056) = 2.0

**broken link removed**

 

[ljcox]

Consider common logarithms, ie. to the base 10.

If y = 10^x then x = log y

 

[Hero999]

Logs are used to get the exponent, here's how it's done:

[latex]z = x^y[/latex]
[latex]y = log_x(z)[/latex]

To get log to the base x:
[latex]log_xz = \frac{log(z)}{log(x)}[/latex]
It doesn't matter whether you use log to the base e or to the base 10 for this.

As far as e is concerned it should be obvious:

[latex]x = exp{y}[/latex]
[latex]y = log_e(x) = ln(x)[/latex]

 

[Roff]

Ron,
It would be easier for novices if you eliminated the negative sign.

ie. t = R*C *ln{Vc/(Vc-V)}

Len, this won't work. The log of a negative number doesn't exist.

 

[eng1]

Well, 1-(V/Vc) and Vc/(Vc-V) are reciprocal and have the same sign, don't they?

 

[bloody-orc]

eng1. you are wrong sadly (according to my calculations)

 

[Roff]

Well, 1-(V/Vc) and Vc/(Vc-V) are reciprocal and have the same sign, don't they?

You just made me realize that there is another typo in my original post (sheesh!).
t/(R*C)=-ln(1-(V/Vc))

t=-R*C*(ln(1-(V/Vc)))

Should read:
t/(R*C)=-ln(1-(Vc/V))

t=-R*C*(ln(1-(Vc/V)))

I have edited the original post to correct these errors.

 

[eng1]

eng1. you are wrong sadly (according to my calculations)

You've tested for equality, but my point was if 1-(V/Vc) is positive (negative) at any instant, also Vc/(Vc-V) is positive (negative) at the same instant because it's the reciprocal.
Len's equation seems correct to me. I got it from Roff's equation applying the formula ln(1/x) = -1*ln(x)

 

[Roff]

You've tested for equality, but my point was if 1-(V/Vc) is positive (negative) at any instant, also Vc/(Vc-V) is positive (negative) at the same instant because it's the reciprocal.
Len's equation seems correct to me. I got it from Roff's equation applying the formula ln(1/x) = -1*ln(x)

As I pointed out above (we were posting at the same time), my original equation was wrong. Sorry, Len.

 

[eng1]

t=-R*C*(ln(1-(Vc/V)))

Yes, now the argument of the log function is positive

After your correction, an equivalent equation is:
t=RC ln[V/(V-Vc)]

 

[ljcox]

I'm glad we sorted that out in the end!

 

[Roff]

I'm glad we sorted that out in the end!

With apologies to Sir Walter Scott...

Oh, what a tangled web we weave,
When first we practice to... help someone.

 

[blueroomelectronics]

Well I put the formula into Just BASIC

VDD = 5
R = 649
C = .00018
Vc = 3.125
print R*C*(exp(1)*(1-(Vc/VDD)))

Turns out I've got to use the TLC555 version as it will be eaiser to implement.

Shall we start discussing ideal values for R1 & R2