Question: I have a state diagram that is consisted of 3 states and in reset it comes to state s0 then if an event happen on start signal goes to state 2 and statye there for 15 clock sycle and then after that goes to state 3 and then back to state s0.
My question is that when VHDL coding how could I do the 15 clock delay of state 2.
I know that a counter needs to be there but it is not clear for me how to do the VHDL coding.
Could any one please give me some help.
Thank you

Answer: 
Lets look at this another way. What is the voltage at the terminals of this circuit:

![schematic](https://i.stack.imgur.com/ZC860.png)

[simulate this circuit](/plugins/schematics?image=http%3a%2f%2fi.stack.imgur.com%2fZC860.png) – Schematic created using [CircuitLab](https://www.circuitlab.com/)

Will the voltage drop? Simple answer: no. Long answer: Well there is no load on it, so no current flowing, so by Ohm's law, no voltage drop across the resistor.

The same is true for your capacitor, there is no current flow, so no voltage drop across it's reactance.

Now what a capacitor will do in an AC circuit if you have a load on it, is to limit the current flowing, which it can do without dissipating large amounts of power like a normal resistor would - it stores and releases charge. That doesn't mean to say it won't heat up, there are losses in it which limit how much power it can safely transfer. It will also change the power factor a long way from unity which can be bad for whatever is supplying your mains voltage (increased losses in transformers and transmission lines, etc.).

Using a capacitor in this way means you would be building a potential divider circuit with whatever you connect, so it won't be a very good regulator. With little load the voltage will go up (no load = open circuit = full mains voltage), and if you put a large load the voltage will drop as more and more voltage is dropped over the capacitor (so high enough rated X series capacitor!).
A capacitor "drops voltage" only if it is used as a voltage divider in conjuction with another impedance. You would have gotten the same result if you had used a resistor in this configuration. In this case, the second impedance is the input impedance of your voltmeter, which is probably 10MΩ or more.