KB buzzer circuit project

Discussion in 'The Projects Forum' started by jlaw, Jun 6, 2010.

  1. jlaw

    Thread Starter New Member

    Feb 27, 2010
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    Hey I've pretty much completed a design project building a type of quiz buzzer for a school project. It is comprised of a lockout circuit so only one person can be buzzed in at a time and a time which counts down from 9 to 0. The main problem I have now though is how to supply power to it. It is designed to operate on 3.5 amps and at least 9 volts. I do have a battery holder that can provide this but I'm worried that this thing may at some point go up in flames. Could someone take a glance at my circuit and tell me if I have anything to worry about? (and if there is any way to simplify the circuit it would be great to know as well!) :)
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    The buzzer should go between the 9 volt supply and the FET drain. Placing it in the source to ground leg will cause problems. The FET conducts fully when the gate is 10 volts above the source. With the buzzer dropping voltage, the G-S voltage will diminish, and tend to shut it off.

    How does the circuit indicate who pressed his switch first?

    What circuit element is going to draw 3.5 amps of current?
     
  3. jlaw

    Thread Starter New Member

    Feb 27, 2010
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    0
    Oh sorry I should have said around 20-35 mA :cool: I forgot to draw it but the same lines that lead to the diodes at the bottom also lead to leds. whichever switch is pushed on will have its respective led turned on and if the second switch is pressed afterward the led will remain low because its circuit is turned off. Thanks! I'll try the FET thing now!
     
  4. jlaw

    Thread Starter New Member

    Feb 27, 2010
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    0
    Another question. I have a seven segment led display and its packaging says its maximum ratings atre 30mA and 2V however it wont light up under these conditions (frustrated, I hooked it to about 5V and 140mA and it glowed well with no problems). Will it hurt the display to put it at these kind of higher values? :)
     
  5. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    Those figures apply to the LED, in that the device will need those 2 volts across it to go into conduction, and need current to be limited to 30 ma, or it will burn up.

    If you are using 9 volts, you determine the resistance by subtracting the LED voltage from the 9 volt figure. That leaves you with 7 volts. 30 ma is right at the upper limit, so use 20 ma as a desired current Ohm's law (R = E/I) gives us 350 ohms. That is not a standard value, so you can choose 330 for a bit more current, or 360 for a bit less.

    The power is only 140 mw, so a 1/4 watt resistor will work.

    If you want to learn a bit more, here is a link into our Ebook - http://www.allaboutcircuits.com/vol_3/chpt_3/12.html
     
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