time constant ?

[bodysaffa]

Hi Folks,

Just looking at Time Constant of Capacitors.
T = RC
Easy enough, change R or C and change T but...

What is the effect of voltage?

I've looked at several sources and I don't see any mention of it.

Thanks,
Jay

 

[Nigel Goodwin]

Hi Folks,

Just looking at Time Constant of Capacitors.
T = RC
Easy enough, change R or C and change T but...

What is the effect of voltage?

I've looked at several sources and I don't see any mention of it.

Because it has no effect!.

The charging time is based on the time it takes to reach a certain percentage of the supply voltage - from memory I seem to recall it was something like 60%?. To fully charge the capacitor takes about 5T.

So with a 10V supply it will take T seconds to reach 6V, with a 100V supply it will take T seconds to reach 60V.

 

[zachtheterrible]

how does one do this calculation?

T, is that in seconds?
R, in ohms?
C, in picofarads?

Obviously this is not right cuz 100ohms X 1000pf=100,000 seconds. this i know is not true :lol:

 

[Nigel Goodwin]

how does one do this calculation?

T, is that in seconds?
R, in ohms?
C, in picofarads?

Obviously this is not right cuz 100ohms X 1000pf=100,000 seconds. this i know is not true :lol:

You didn't choose picoseconds, or picoohms, so why choose picofarads?.

It's in farads - the basic formulas usually use the basic values, ohms, volts, amps, farads, henries.

 

[Styx]

R*C give you the time to 63% of full charge or one T period

and as Nigel said 5T is 99% of voltage since a capacitor can (theoretically) never reach the applied voltage


and also your maths is wrong

100R * 1000pF = 100ns for one time period which seems very reasonable for that size cap or 500e-9s to be fully charged

 

[Nigel Goodwin]

R*C give you the time to 63% of full charge or one T period

Thank you :lol: it's been a LONG time since I used it!.

 

[zachtheterrible]

You didn't choose picoseconds, or picoohms, so why choose picofarads?.

as always nigel, very reasonable!!

so if u take ur result (T) and multiply it by 5, thats how long it will take the cap to reach full capacity? (or 99%)

So why the 63%? of what use is that?

thanx, ive been speculating @ this 4 a wile

 

[TheOne]

It has to do with the natural exponential function e^x which governs some functions in nature (decay of radioactive material, loss of heat in a body etc.)and also the charge and discharge rate of capacitors.

Just to show where the 63% comes from:

So e^1 = 2.72, take the reciprocal you get 0.37, which gives a remainder of 0.63 from 1. multiply by 100% = 63%

The typical formula will be in the form- V(capacitor) = E(1 - e^-t/RC)

Interestingly e^x is also the ONLY mathematical function which when differentiated or integrated will give the exact function again!

 

[Styx]

dont forget zero

 

[TheOne]

As far as I know, zero is not a function

 

[Styx]

that may be the case, but the number zero is a special case anyway (as is infinity)

also

int(e^x,y) gives you y*e^x + C

 

[TheOne]

int(e^x,y) gives you y*e^x + C

I thought it was very clear that I was talking about the special property of the function e^x but maybe not..

Put differently "The exponential function e^x is a special function in that it's the only function (up to a scalar multiple) that is the derivative of itself"

or as described on this link

**broken link removed**

Quote "While doing so, you will sort of 'stumble' upon one of the most profound items in all of mathematics; that is the fact that the derivative of e^x is equal to itself. This is the only function with this property (except for the trivial function f(x) = 0),"

Also note that it states the case of, the result of the function f(x) = zero

**broken link removed**

gives a nice interactive illustration for this

 

[Styx]

trivial function or not it is another example of a fn that int,diff is equal to itself.

However where f(x) = 0 has one extra proporty over e^x (and the point that I was making)

the integral of zero w.r.t. ANY variable is zero
the derivative of zero w.r.t. ANY variable is zero

the integral of e^x is only equal to e^x if it is done w.r.t. "x" if it is done against any other variable then it is just e^x multiplied by that variable and not e^x

the derivetive of e^x is only equal to e^x if it is done w.r.t "x" if it is done against any other variable then it is always zero.


My point was just that declaring an integration/differation without no dimention to do it against is meaningless.

anyway a trivial solution is still a solution, sayin there is only one solution is not the solution. As an engineer you can appreciate finding the simplest solution to a problem

I mean you did blast me down with what you call the "trivial" solution, let's just call this revenge

call it quits, both of us know what we are talking about, both of us were right, I was just more right

 

[TheOne]

Styx

No hard feelings I think it was healthy and in good spirit, also to the benefit of other members.

You being more right, oh well lets leave it there....

Engineer??...I hi-jack my dad's pc when he's not at his desk :lol:, I have to go, hear him coming back!

 

[bodysaffa]

time constant

I realized answer right after I sent the question.

But on the other hand the circuit was a relaxation oscillator, with a neon lamp.

So increasing the voltage has the effect of decreasing the resistance and increasing the rate of flashing. The cap fills up faster.

I assume in filter circuits the voltage is held constant, so the variables are just C and R.

jay

 

[Nigel Goodwin]

Re: time constant

I realized answer right after I sent the question.

But on the other hand the circuit was a relaxation oscillator, with a neon lamp.

So increasing the voltage has the effect of decreasing the resistance and increasing the rate of flashing. The cap fills up faster.

I assume in filter circuits the voltage is held constant, so the variables are just C and R.

No, as already explained the voltage makes no difference! - in your neon relaxation oscillator it's the NEON which makes the difference. A neon has a fairly constant trigger point, rather like a zener diode, so it will trigger at the same voltage regardless of the supply voltage. As you increase the supply voltage it will trigger at a lower percentage of the supply, thus increasing the frequency.

 

[Styx]

Exacltly, As Nigel stated

So with a 10V supply it will take T seconds to reach 6V, with a 100V supply it will take T seconds to reach 60V.

Thus lets say the NEON has a threshold voltage of 6V, then for a supply of 10V it will take one RC time constant. If the voltage was 100V then it will cross 6V alot quicker, 10x quicker BUT the RC time constant is still the same

This is why as much info to a problem is needed