Charging time of capacitor circuit?

I need to develop a circuit to measure charging time of capacitor
I am kind of less in practical knowledge (this is my first project)
So far I have thought of giving a reference input and capacitor voltage to comparator
then there is counter that will count the number of time it will go high
then the time divided by the no. of times it will go high will give me the charging time that will be displayed on LCD
Is this method Okay? Will it work?

Can any one please draw me a detailed circuit daigram
for the same PLEASE!!!

Charge the capacitor via a constant-current source (CCS) and measure the time it takes for the voltage across the capacitor to charge from 0V to the full supply voltage. For example, if you have a CCS operated from a 5V supply, then you need to measure the time taken for the voltage to reach 5V (or somewhat less depending on the voltage lost across the CCS).

The capacitance is then, given by, I(t) = C * (dV/dt) where dV is the change in voltage (5V) and dt is the time taken.

Rearranging, C = I(t) * dt / dV

So if the CCS is for 1mA and the final capacitor voltage is 5Vm and time taken is 1s then,

C = 1mA * 1s / 5V = 2 x 10^-4 F or 200 uF

Note that you need CCS to work at much lower currents in order to measure smaller capacitances.

I know that T = CR (time constant = capacitor rating times resistance) but I do not know how to calculate (theoretically) the time taken for the final ~36.8% of the capacitor to charge.

The charge is an exponential function of time:
Q=Qmax[1−e(−t/RC)]

The time constant tells you the shape (how sharply curved it is) of the exponential charging curve--it gives you a handy measure of how fast the capacitor will charge. Examine that charging equation again. Let's say that the time constant RC = 10 seconds. That means that 10 seconds after you apply the voltage, the fraction of charge on the capacitor will be:
1−e−(10/10)=1−e−1=1−0.368=0.638

That means it will have 63.8% of its final (total) charge.

What about after 20 seconds (2 time constants)? The fraction of charge will be:
1−e−(20/10)=1−e−2=1−0.368∗0.368=0.865

thus 86.5% charged.

As time goes on, each additional second adds less and less total charge.
For briefly explanation visit http://actpcb.com/capacitor-calculator operation.

Well, for the purpose of determining charging time, a constant-current source is much better than a simple resistive source - though it is a bit more complex physically, it makes the reading of capacitor charge time much easier, as it is now simply a linear function of time. However, yes, given a particular RC circuit, we have the 63.8% charge and then extrapolate that to get to the final time taken to charge.

When a battery is connected to a series resistor and capacitor, the initial current is high as the battery transports charge from one plate of the capacitor to the other. The charging current asymptotically approaches zero as the capacitor becomes charged up to the battery voltage. Charging the capacitor stores energy in the electric field between the capacitor plates. The rate of charging is typically described in terms of a time constant RC.
Thanks!!!!