Hi all,
I'm wondering if someone can help me tidy up my algebra? My problem concerns the time taken for a capacitor to charge from one voltage to another (I'm sure this must be a classic electrical engineering exam question).
So...
The capacitor C is charged from 0v through a resistor R towards a supply voltage Vmax. At time t the voltage is:
V(t) = Vmax( 1 - e^(-t/RC) ) [1]
...and so the capacitor reaches an arbirary voltage V1 at time t1, where:
t1 = -RC ln( 1 - V1/Vmax ) [2]
...and likewise sometime later it reaches a higher voltage V2 at time t2, where:
t2 = -RC ln( 1 - V2/Vmax ) [3]
the time taken to charge from V1 to V2 is (t2 - t1), or:
RC ln( 1 - V1/Vmax ) - RC ln( 1 - V2/Vmax ) [4]
and removing the common factor of the time constant RC gives:
t1,2 = RC ( ln( 1 - V1/Vmax ) - ln( 1 - V2/Vmax ) ) [5]
Now, I'm pretty confident that this is correct, but is there a more elegent way to express it? Is there a cleaverer way to aproach this problem? I know that at some point, when Vmax >> (V1 and V2) the charge current becomes nearly constant and the time tends towards:
t1,2 = (V2-V1)RC / Vmax [6]
is there a way to see this limit from [5] above?
This isn't really important, but - as often happens - although I can come up with a solution I believe to be correct, my mathematics is not really good enough to give me a neat and tidy answer.
Thanks!
Having looked at this again, the simplification of the term within the ln() that I was looking for is:
(1 - V1/Vmax) / (1-V2/Vmax) = (Vmax-V1) / (Vmax-V2)
so the entire expression [5] from my original post becomes:
t1,2 = RC ( ln( 1 - V1/Vmax ) - ln( 1 - V2/Vmax ) ) [5]
t1,2 = RC( ln( (Vmax-V1) / (Vmax-V2) ) ) [6]
When x is close to 1, ln(x) can be aproximated as (x-1) - this gives us the linear aproximation for when V1 and V2 are very close.
Sometimes I almost enjoy maths... I just wish it wasn't so hard.
Having looked at this again, the simplification of the term within the ln() that I was looking for is:
(1 - V1/Vmax) / (1-V2/Vmax) = (Vmax-V1) / (Vmax-V2)
so the entire expression [5] from my original post becomes:
t1,2 = RC ( ln( 1 - V1/Vmax ) - ln( 1 - V2/Vmax ) ) [5]
t1,2 = RC( ln( (Vmax-V1) / (Vmax-V2) ) ) [6]
When x is close to 1, ln(x) can be aproximated as (x-1) - this gives us the linear aproximation for when V1 and V2 are very close.
Sometimes I almost enjoy maths... I just wish it wasn't so hard.
MrAl and Mr Ratch are maths gurus methinks.
spec
dt=(Vc2-Vc1)*RC/(Vm-Vc1)
...the problem it that, quite often, so am I.I am unclear as to what you are trying to do.
Thanks for the contunied interest folks.
Looking at MrAl's work above:
I've statisfied myself that I can reach this same result from equation [6] in my post #7 above:
t1,2 = RC( ln( (Vmax-V1) / (Vmax-V2) ) ) [6]
Using the aproximation ln(x) ≈ x-1 when x is close to 1
t1,2 = RC( (Vmax-V1) / (Vmax-V2) - 1 ) [7]
which can be re-written as
t1,2 = RC( (V2-V1) / (Vmax-V1) ) [8]
(hope you don't mind me continuing to use my nomenclature - it just makes it more consistent with my original question).
To my mind, this is again an improvement on [6] in terms of clarity and readability. It's immediately obvious the that time interval varies in direct proportion to the voltage interval and in inverse proportion to the difference between the threshold voltages and the supply.
It's also very satisfying when the same result can be reached in two different ways; MrAl's derivation is the way we'd all probably aproach working out a "back of an envelope" estimate, and it's reassuring to see that it talles up with the more formal aproach.
Ratchit has raised an important point here though:
...the problem it that, quite often, so am I.
The bigger picture its that I'm trying to improve my design skills on the theoretical level and move away from the "cut and try" aproach. I've always imagined that a propper and rigerous design aproach would start by modeling the circuit algebraically at a high level (assuming the perfect components that spec mentions in #4), before moving on to simulations and more practical considderations. I'm not a trained engineer, so I don't know to what extent professionals work like this in real life.
My problem with maths is that I often don't know what sort of expression I'm expecting at the end of a derevation, I don't recognise the common patterns and I'm not able to reach for the common techniques and identities used to deal with them.
As MrAl says, I suppose it's all down to practice.
Not that I thought there was anything wrong with the way you aproached it (like I said, it seemed like the common sense aproach to me) - it's more that I wanted to prove that the aproximation I'd been talking about earlier actually gave the expected results.Oh so you just wanted to go at this in a different way then? That's cool i guess, whatever you like.
Not that I thought there was anything wrong with the way you aproached it (like I said, it seemed like the common sense aproach to me) - it's more that I wanted to prove that the aproximation I'd been talking about earlier actually gave the expected results.
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