How Long Does the Buzzer Sound in a Timing Circuit?

[Masafi]

In a timing circuit, an electronic processor operates so that the buzzer sounds when Vc is greater than Vs.

The switch S is normally open. Explain in detail what happens in the circuit after the switch S is closed for a moment then opened again. Your answer should include an appropriate calculation and a sketch graph.

Answer is:

S closed --> C charges
up to Vs
Instantly/very quickly
S open: discharge starts
Exponential discharge
Vc = Vs e^(-t/RC)
¾ Vs = Vs e^(-t/RC)
ln ¾ = -t/RC (1)
t = 29.7 s

Buzzer sounds for 29.7 s

However, shouldn't the answer be 2 x 29.7, as the buzzer also sounds during the time the capacitor is charged and Vc > 0.75Vs?

 

[jbsteele]

Yes, the answer should be 2 x 29.7 seconds. When the switch is closed, the capacitor begins charging to Vs and Vc starts to increase. When the switch is opened, the capacitor starts to discharge exponentially and Vc decreases. The buzzer will sound when Vc is greater than 0.75Vs. Since the buzzer sounds during the time the capacitor is charging and discharging, the total time for the buzzer to sound is 2 x 29.7 seconds.

 

[kerimek]

Yes, you are correct. The buzzer will sound for a total of 2 x 29.7 s, as it will sound during both the charging and discharging phases of the capacitor. This is because in a timing circuit, the buzzer is triggered when Vc is greater than Vs, regardless of whether Vc is increasing or decreasing.

To further explain, when the switch S is closed, the capacitor C starts to charge up to Vs very quickly. This is because the capacitor acts as a storage device for electric charge, and when connected to a voltage source (in this case, Vs), it will charge up until it reaches the same voltage as the source.

Once the switch S is opened, the capacitor starts to discharge through the resistor R in an exponential manner, following the equation Vc = Vs e^(-t/RC). This means that the voltage across the capacitor decreases over time, and eventually reaches 0.75Vs at time t = 29.7 s, as shown in equation (1).

At this point, the buzzer will sound for the first time, indicating that Vc is greater than Vs. However, the capacitor is still discharging, and the voltage continues to decrease. Eventually, it will reach 0 V, at which point the buzzer will stop sounding.

But, as you correctly pointed out, the switch S was only closed for a moment, and it will be opened again after the capacitor has fully discharged. This means that the capacitor will start to charge again, and the whole process will repeat itself. This time, however, the buzzer will sound for the second time after the capacitor has charged up to 0.75Vs, which will take another 29.7 s.

Therefore, the total time for which the buzzer will sound is 2 x 29.7 s, as it will sound during both the charging and discharging phases of the capacitor. This can also be seen on the sketch graph, where the buzzer sounds twice, once during the charging phase and once during the discharging phase.