Alarm Circuit Capacitor Discharge

Discussion in 'Homework Help' started by p75213, Oct 26, 2011.

  1. p75213

    Thread Starter Member

    May 24, 2011
    Can somebody explain how this circuit works? I can't see how the capacitor is doing anything. It simply charges up and that's it. I can't see how it discharges. The capacitor must fire because without it there is not enough current through the alarm to set it of.
  2. Jon Wilder

    New Member

    Oct 25, 2011
    Someone didn't notice the variable R. ;)

    In order for current within a circuit to change, one of two things must happen -

    1) Voltage must change

    2) Circuit resistance must change

    Well, we know that our voltage is constant since we have a 9V battery powering it. This means that option 1 doesn't apply.

    However, we have a variable R in the path that exists across the capacitor (i.e. the "discharge" path). Since its value is undefined, it can be anything at any given time.

    So...that being said...let's take a look at the circuit.

    You have both a charge AND a discharge circuit. The "charge" circuit is comprised of -

    The battery

    The 10K Resistor

    The Switch

    The discharge circuit is comprised of -

    The 4K resistor (alarm)

    The undefined variable R

    When the switch closes, it charges up the 80uF capacitor. Let's assume that our variable R is at infinite ohms. While at infinite ohms, the resistance is too high for the 80uF capacitor to discharge. Therefore, no discharge current exists.

    Our alarm sounds when the discharge current exceeds 120uA.

    The charge circuit has charged our 80uF capacitor up to the supply voltage of 9V.

    Now...let's assume that our variable R has now dropped to a value of 0 ohms. At 0 ohms, the 4K ohm alarm circuit will see the 9V that exists across the charged capacitor. Let's use Ohm's Law to calculate the alarm circuit current -

    9V / 4K Ohm = 2.25mA

    At a circuit current of 2.25mA, does the alarm sound?
  3. p75213

    Thread Starter Member

    May 24, 2011
    Are you saying that the alarm is sounded by the variable resistor dropping in ohms?
  4. Jon Wilder

    New Member

    Oct 25, 2011
    That would be correct. ;)
  5. Adjuster

    Well-Known Member

    Dec 26, 2010
    I'm sorry, but the variable resistor will not be adjusted during the actual discharge, as that would make the delay impossible to calculate, and the example gives numerical answers.

    What actually happens is that after the switch closes, the capacitor voltage builds up until there is enough voltage to drive a current i = 120μA through the 4kΩ "alarm".

    The capacitor voltage won't get to 9V, but may get high enough to reach this threshold. Let's see, how much current will flow in the end?

    1. With R=0, we have a total circuit resistance of 10kΩ+4kΩ=14kΩ, final current I=V/R, so the current will eventually reach ?
    2. With R=6kΩ what is the final current?
    For each of the two endpoint currents, find the fraction of the endpoint representing 120μA. So, if the endpoint current were 600μA, (it's not, this is just an example), the fraction would be 120μ/600μA = 0.2

    Next, for R=0 and R=6kΩ, work out the effective time constant, C times the parallel combination of 10kΩ plus (R+4kΩ)

    Now find out how many time constants it takes in each case. The current ratio of actual to final currents i/I will be the same as the capacitor voltage ratios v/V.

    v/V = 1-exp(-t/RC), t = ?