hi all,
i'm confused...
what would be the output voltage of an ideal capacitor across a constant current source?
tnx

 

It would be potentially infinite. The voltage across a capacitor is proportional to the integral of the current I, times time. Since the current is constant it may be taken outside the integral. If the lower limit of integration is considered time t = 0. then:

V = (1/C) * I * t
V -> Inf as t -> Inf

 

hi all,
i'm confused...
what would be the output voltage of an ideal capacitor across a constant current source?
tnx

Papabravo is correct.

In the real world however, it is tough to find an infinite voltage source so the maximum voltage that will be developed across your capacitor will be governed by the voltage being used to produce your constant current source.

hgmjr

 

An ideal capacitor will not get fully charge, that means as time increases as current enters to it, the voltage also increases. In this way, as time approaches infinity, the voltage across it will also be to infinity. In graph, a constant current flowing into a capacitor will result also to a constantly increasing potential at its plates. Since a capacitor can be likened to a tank, then a constant current flow into it will just accumulate charges, and when you accumulate that constant input, it will result to a linear graph.

ex. 1 1 1 1 1 1 1 1 1 ---> this serves as the constant current per time (t).
1 2 3 4 5 6 7 8 9 --> this serves as the voltage across the capacitor.

thus, since our current input is constant through time, then the increase on the capacitor voltage is also constant, which relates in mathematical way as a constant slope or in a simple way a line. by the way, integration is just like summing all the values.

on the other hand, a real-world capacitor cannot charge infinitely, they have a rated maximum voltage across it, so a constant current source flowing into it would either blow-up the capacitor-this happens if the current source used has
a clamping voltage greater than the maximum input voltage of the used capacitor or if that clamping voltage is lesser than the maximum input voltage of the capacitor, then it will just settle to the clamping voltage value. This means that the system is on equilibrium, such that it cannot push more current to the capacitor.

 

Let's assume the cap is not charged, and at some time zero you connect the cap to the current source.

The equation of a cap is:

I = C x (delta V) / (delta t)

where: I = current, C = capacitance, V = voltage, t = time

Actually it is a differential equation (a calculus thing). In english this simply means the current thru a cap is equal to the rate of change of the voltage across it (times the capacitance).

If we re-arrange to solve for delta V:

delta V = ( I / C ) x (delta t)

It’s now safe to cancel out (or ignore) the deltas:

V = ( I / C ) x t

C is a constants and you just told us you made I also a constant. So this equation this just says the voltage increases linearly with time.

It just ramps up from zero to some great number as time increases.

The voltage is only infinite after an infinite amount of time has passed.

 

I don't think the OP is still around (since 2006 post) to really care.

 

It was in fact the OP's final post on this board, but it was among my first of many. Whoda thunk it!

 

However ErnieM has corrected a misunderstanding by markcyriljubay.

But welcome mark and keep them coming. AAC is a great and friendly place.



Many circuits rely on capacitors supplied from constant current generators.
It is greatly used to linearise otherwise exponential ramps.

 

However ErnieM has corrected a misunderstanding by markcyriljubay.

I did?

I used the fundamental definition of capacitance formula, while mark used the expression for charge to voltage for a cap.

I forget which is the chicken and which is the egg, but one formula is derived from the other, yes?

I thought we just said the same thing two different ways.

(Just checked: mark's form is fundamental, mine is derived from his.)

 

They say the same thing in two different ways: integration of a derivative.