Electrical Engineering - Increasing the Time an Alarm will Buzz

In summary: According to your schematic, you are using a supply voltage of 18V. Correct? So where did the "6V" figure come from? What is special about the "5V" figure? Where did the "9V" come from?
  • #1
Mosaness
92
0

Homework Statement



The problem is to create a small alarm that will buzz approximately 6 seconds before it stops buzzing. I managed to create a working alarm, only problem is mines will buzz for around 1.5 seconds not six.

Homework Equations



τ = RC
V = V0e-t/τ


The Attempt at a Solution



Req. = 400Ω + (80Ωll100Ω) ≈ 444.44 Ω
VTH = 6V * (400Ω/480Ω) = 5V

Substituting this into equation #2 above gives 5V = 9V*e-6/τ
Solving for T yields:

Actually, I just lost myself in my work. Can someone help me.
 

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  • #2
Mosaness said:

Homework Statement



The problem is to create a small alarm that will buzz approximately 6 seconds before it stops buzzing. I managed to create a working alarm, only problem is mines will buzz for around 1.5 seconds not six.

Homework Equations



τ = RC
V = V0e-t/τ


The Attempt at a Solution



Req. = 400Ω + (80Ωll100Ω) ≈ 444.44 Ω
VTH = 6V * (400Ω/480Ω) = 5V

Substituting this into equation #2 above gives 5V = 9V*e-6/τ
Solving for T yields:

Actually, I just lost myself in my work. Can someone help me.

Where is the buzzer? How is it turned on and off?
 
  • #3
berkeman said:
Where is the buzzer? How is it turned on and off?

The buzzer has a resistance of 400 ohms so that's what was used. It is turned on when someone opens the window.
 
  • #4
Mosaness said:
The buzzer has a resistance of 400 ohms so that's what was used. It is turned on when someone opens the window.

And how do you intend to connect the buzzer and window switch to your circuit that you've drawn? How does the circuit reset when the window is later closed?

BTW, it would be better to use an NPN transistor as part of the circuit that drives the buzzer. Do you know why?
 
  • #5
berkeman said:
And how do you intend to connect the buzzer and window switch to your circuit that you've drawn? How does the circuit reset when the window is later closed?

BTW, it would be better to use an NPN transistor as part of the circuit that drives the buzzer. Do you know why?

Oh...the class I am taking is a basic class. I've never even heard of a NPN transistor. I made a PSPICE design. The switch was connected right next to the 100Ω.
 
  • #6
I did a calculation earlier but I slightly messed up. I obtained Vthevenin by the equation 18V * 80/180 --> which gave me 8V.

I then set that so:

8V = 18V*e^-6/tau and solved for tau, which gave me tau = 7.3989

Using that tau I solved for C, which gave me C = .0166 F
 
  • #7
Mosaness said:
I did a calculation earlier but I slightly messed up. I obtained Vthevenin by the equation 18V * 80/180 --> which gave me 8V.

I then set that so:

8V = 18V*e^-6/tau and solved for tau, which gave me tau = 7.3989

Using that tau I solved for C, which gave me C = .0166 F

Why not just use a "normally open" (NO) switch in series with the buzzer, a resistor and a cap? Or even just the switch, buzzer and cap...?
 
  • #8
berkeman said:
Why not just use a "normally open" (NO) switch in series with the buzzer, a resistor and a cap? Or even just the switch, buzzer and cap...?

Well I just went on the way I know a few people did. And that seems to be giving me an appropriate time...
 
  • #9
Mosaness said:
Req. = 400Ω + (80Ωll100Ω) ≈ 444.44 Ω
VTH = 6V * (400Ω/480Ω) = 5V

Substituting this into equation #2 above gives 5V = 9V*e-6/τ
Req I understand. VTH I don't.

According to your schematic, you are using a supply voltage of 18V. Correct? So where did the "6V" figure come from? What is special about the "5V" figure? Where did the "9V" come from? What operating parameter of the buzzer are you working with, in trying to determine how long it will actually "buzz"?
 

1. How can I increase the time an alarm will buzz?

In order to increase the time an alarm will buzz, you will need to adjust the timing circuit of the alarm. This can be done by changing the resistor or capacitor values in the circuit to alter the frequency of the alarm's oscillations. Alternatively, you can add a timer circuit to the alarm that will extend the time before the alarm sounds.

2. What are the factors that affect the buzzing time of an alarm?

The buzzing time of an alarm is primarily affected by the timing circuit components, such as resistors and capacitors, as well as the power supply voltage. Other factors that can impact the buzzing time include temperature, humidity, and any external interference.

3. Can I increase the buzzing time without altering the alarm's circuit?

Yes, it is possible to increase the buzzing time without altering the alarm's circuit. One way to do this is by using an external timer circuit that is connected to the alarm. Another option is to use a relay or transistor to control the power supply to the alarm, allowing you to extend the buzzing time.

4. Is it safe to modify the alarm circuit to increase the buzzing time?

Modifying the alarm circuit can be dangerous if you do not have the proper knowledge and experience with electrical engineering. It is important to follow safety precautions and consult a professional if you are unsure about making modifications to the circuit. Alternatively, you can use external components to increase the buzzing time without altering the alarm's circuit.

5. Can I control the buzzing time of an alarm remotely?

Yes, it is possible to control the buzzing time of an alarm remotely by using wireless communication and a microcontroller. This allows you to adjust the timing circuit of the alarm without physically altering it. However, this requires a more advanced understanding of electrical engineering and programming.

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