Alarms play an important role in our daily life.It assists us from waking up daily morning to carry out chores inside the schedule.Here is a similar minutes alarm circuit that built around famous versatile chip 555 with time limitations up to only 10 minutes.
WORKING OF TEN MINUTES ALARM CIRCUIT:
The circuit was powered by simple 9v battery and you can of course use any alternative source.The 555 chip was wired in the monostable mode in the above circuit.The minutes alarm circuit was designed in such a way the buzzer will sound when the output from 555 is low.As you can see the buzzer was connected to pin 3 of 555 to sink the incoming current into the pin.
The trigger pin was connected in R1 and C1 which keeps it in the high state.Whereas RV1, R2 and C3 forms the timing element for this alarm.When the Circuit is turned on the electrolytic cap C3 starts charging through RV1 and R2.When the capacitor voltage reaches the threshold value (2/3 Vcc) the internal flip-flop in the 555 IC got reset, thus in turn the output changes its state to 0 or low state.The buzzer gets activated and starts producing the alarm sound to the user.
As stated above RV1, R2 and C3 form the timing elements of this circuit.The timing factor can be determined by the formula
T = 1.1 * R * C
= 1.1 * (RV1+R2) * C3 (According to our circuit)
= 1.1 * 1.1M * 220uF = 266.2 sec
The time range is around 4 and half minutes approx.However while implementing it in real-time the electrolytic capacitor starts leaking charge, therefore we will get around ten minutes approximately while RV1 is in max position.This time range can be altered by adjusting the RV1 value.
Hi Frank!Just a doubt about the timing factor: the discharge rate time constant for an RC is R*C, while you scale it by 1.1.Is that 1.1 scaling factor due to the threshold value which is in your case 2/3 Vcc instead of the usual 63.3% Vcc?
what I should change to make it sound only 3 seconds?thanks in advance