# Question with RC Time Constant

Discussion in 'Electronic Basics' started by Panther, Dec 22, 2005.

1. ### PantherGuest

Basically I've done a practical with an RC network and I've got 2 wires that
touch to "complete the circuit" and provide current to the capacitor. I need
to find out the amount of time they've touched by lookign tat the voltage in
the capactiros. I have the capacitor charge/discharge curves.

my question is can I work out the time using simply time constants? 5TCs =
time to charge a capacitor (almost 100%, irrelevant). So because I have the
capacitor charge curve, say it only had half the voltage o fthe input
voltage, and i look up half on the y axis and find wher eit meets, and look
up the x-axis' time and work out how long that is.

or do i do this: http://www.play-hookey.com/dc_theory/rc_circuits.html

basically use that formula. if i do use it i will get more ducks.

are both these methods correct?
thanks

2. ### Bill BowdenGuest

are both these methods correct?

Seems too complicated. Why not charge the capacitor with a constant
current so the voltage change is linear? That way you would know the
time to charge to 1 volt is 1/10 the time to charge to 10 volts.

You can make a fairly constant current source with one transistor and 3
resistors. But then you have the problem of discharging the cap. Maybe
one more resistor could do that?

-Bill

3. ### John PopelishGuest

You can only estimate a minimum charge time (must be greater than 5*
R*C seconds) once the capacitor reaches essentially full charge
voltage. In order to measure the charge time accurately from the
voltage, the voltage has to still be measurably changing at the moment
you open the switch.

The capacitor charges to 1-(1/e) or about 63.2% of full charge (with e
being the constant, 2.71828) when the charge time has been one R*C.

The formula for voltage versus time, assuming the capacitor starts at
zero, is, 1-e^(-time/R*C) times full charge voltage.

To solve that for time versus voltage, time capacitor has been
charging is -ln(1-V)*R*C seconds, with V being the fraction of full
charge voltage, and R in ohms, C in farads.

For other

4. ### Bob MonsenGuest

Yes, you can do this, but you need to scale the output appropriately. A
generic graph like that can have its time scale marked in time constants,
and its voltage scale go from 0 to 1. In that case, using the y=0 axis as
the initial voltage, you find the point where y = (end -
initial)/(charging - initial). This allows you to determine how many time
constants you've charged for. If a time constant in your circuit is equal
to TC, then the answer is graphical time constants / TC.

--
Regards,
Bob Monsen

Mike is like the first man to discover fire. Fire was there all along
after he showed them how, anybody could use it’¡Ä anybody with sense
enough not to get burned with it.

5. ### Pooh BearGuest

You need this equation that's on that page.

Vc = E(1 - e^(-t/RC))

As suggested though, a constant current source makes it easier since theer will
be a linear rise in the cap voltage voltage with time ( like a scope timebase ).

Make sure the cap is fully discharged before commencing timing though.

Graham

6. ### PantherGuest

I can't change it though I've already done it. Thanks.

7. ### PantherGuest

I can't, I've already done the practical. Thanks anyway.

8. ### PantherGuest

I think it's best if I used the formula (more advanced).

Thanksd

9. ### PantherGuest

Thanks for the help!  