I have a project in my electronic class that I need a little help in.
The following is the exact description for the assignment.

COMPUTER POWER- ON RESET SWITCH. For the following two assignments, start by connecting a SPDT switch so that its common terminal puts out either +5 V OR 0 V, as shown (see the attached picture on the left).

Now connect this output to your diode circuit, designed and constructed so that its output voltage will rise slowly (approximately 63% in 10 seconds) when the switch is turned to the up position, but will drop instantly to nearly 0 v when the switch is then turned to the down position.

Now do same as above, but the output voltage rises instantly when the switch is turned to the up position, but drops slowly when the switch is then turned to the down position.

Again, the diagram to the left below is what the instructor gave us to work with. The diagram to the right is what I think I need to complete the assignment. 63% is equivalent to two LED diodes that range from (1.7V -2.0V). As far as the slow rise in voltage, would that mean I would also need to add a capacitor? Any info would be great. I am going to lab tomorrow night to complete this assignment and would like to be prepared to construct the circuit before coming to class.

 

Hint:
A resistor can be used to control the flow of current into or out of a capacitor and hence can determine the time it takes for the cap to charge or discharge.
A diode allows current to flow in one direction only (the direction the 'arrow' points).

 

So I need all three: resistors, capacitors and LED diodes (Diodes is the main topic right now)?. I understand the concept of the diode but with the rises slowly and drops instantly i can not seem to understand. I understand the drop instantly part.

Thanks for helping out. Below is the latest I put together it shows the 63% which would be around 3.2v (which it is). To the right is a capacitor that is not connected.

 

Do you understand that a resistor in series with a capacitor will produce a rate-of-rise in the voltage with a time-constant of R*C?
Do you know what a time-constant is?

 

it shows the 63% which would be around 3.2v (which it is)

I don't know how you managed to find anywhere in that circuit with a voltage of 3.2V! How do you propose to create the slow rise?

 

c c cccc

 

Do you understand that a resistor in series with a capacitor will produce a rate-of-rise in the voltage with a time-constant of R*C?
Do you know what a time-constant is?

Yes the rise or fall of an output signal dependent on
circuitry from start to finish correct?

 

The functional relationship between the input voltage and the capacitor voltage is described by a first order differential equation, which has an exponential function as it's solution.

 

The functional relationship between the input voltage and the capacitor voltage is described by a first order differential equation, which has an exponential function as it's solution.

Thanks