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

blueroomelectronics · 31st August 2007, 03:58 AM

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 · 31st August 2007, 04:56 AM

Quote:

Originally Posted by 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))

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 · 31st August 2007, 05:22 AM

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

bloody-orc · 31st August 2007, 10:40 AM

I get something different:

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

Roff · 31st August 2007, 02:19 PM

Quote:

Originally Posted by bloody-orc

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 · 1st September 2007, 05:03 AM

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

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

Leftyretro · 1st September 2007, 05:46 AM

Quote:

Originally Posted by ljcox

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 · 1st September 2007, 06:52 AM

Quote:

Originally Posted by Leftyretro

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

http://whyslopes.com/freeAccess/natu...exponenti.html

ljcox · 1st September 2007, 10:10 AM

Consider common logarithms, ie. to the base 10.

If y = 10^x then x = log y

Hero999 · 1st September 2007, 11:14 AM

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

$z = x^y$
$y = log_x(z)$

To get log to the base x:
$log_xz = \frac{log(z)}{log(x)}$
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:

$x = exp{y}$
$y = log_e(x) = ln(x)$

Roff · 1st September 2007, 02:02 PM

Quote:

Originally Posted by ljcox

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 · 1st September 2007, 04:05 PM

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

bloody-orc · 1st September 2007, 04:18 PM

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

Roff · 1st September 2007, 04:23 PM

Quote:

Originally Posted by eng1

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 · 1st September 2007, 04:27 PM

Quote:

Originally Posted by bloody-orc

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)

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31st August 2007, 03:58 AM	(permalink)
blueroomelectronics	*Put t on the left side? Vc = V (1- e^(-t / RC))* 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)) __________________ Bill Smart Kits build Smart People http://www.blueroomelectronics.com

31st August 2007, 05:22 AM	(permalink)
blueroomelectronics	Well Roff you've earned a Cricket PCB when they're ready.Thanks. __________________ Bill Smart Kits build Smart People http://www.blueroomelectronics.com

1st September 2007, 05:03 AM	(permalink)
ljcox	Ron, It would be easier for novices if you eliminated the negative sign. ie. t = RC ln{Vc/(Vc-V)} __________________ Len

1st September 2007, 10:10 AM	(permalink)
ljcox	Consider common logarithms, ie. to the base 10. If y = 10^x then x = log y __________________ Len

1st September 2007, 11:14 AM	(permalink)
Hero999	Logs are used to get the exponent, here's how it's done: $z = x^y$ $y = log_x(z)$ To get log to the base x: $log_xz = \frac{log(z)}{log(x)}$ 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: $x = exp{y}$ $y = log_e(x) = ln(x)$ __________________ I also post at the following sites: http://www.stop-microsoft.org http://www.heated-debates.com Screen name: Aloone_Jonez

1st September 2007, 04:05 PM	(permalink)
eng1	Well, 1-(V/Vc) and Vc/(Vc-V) are reciprocal and have the same sign, don't they?