Very simple questions about transistors, voltage drop, and why Im killing transistors

Discussion in 'The Projects Forum' started by poopscoop, Dec 12, 2012.

  1. poopscoop

    Thread Starter Member

    Dec 12, 2012
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    I am an extreme beginner, and started experimenting and reading 2 days ago. Any input is appreciated, and please don't assume I know anything because I don't. Attached is my circuit diagram.

    The idea is that while the switch is open, the final transistor will be closed(Non conducting). When the switch is open, the second transistor conducts, pulling the base of the third transistor to ground and making it non-conductive. When the switch closes and current flows, it pulls the second transistor's base to ground, shutting it completely, forcing the third transistor's base to a positive voltage, opening the gate and allowing the full current (Minus the voltage drop of the transistor) into the buzzer.

    For a reason I don't understand, when I didn't have a resistor on the base of the first transistor I kept burning it out. I have no idea why 9v unimpeded would destroy that transistor. Any ideas?

    If I add another transistor between the second and the third, effectively making a darlington pair out of the final 2, would the final transistor attached to the buzzer conduct more completely and increase the current available?

    Is there a way to replace the final transistor with a FET? Would there be any advantage to doing that?

    Thanks a ton,
    Poopscoop
     
  2. antonv

    Member

    Nov 27, 2012
    149
    27
    The transistor's base-emitter junction is a diode, for your purposes it does not limit current and it is fully on at 0.6 V or so. With 9 V applied it will allow as much current to flow as the battery will supply, and so be destroyed.

    Without knowing how much current the buzzer needs and what kinds of transistors you are using it's hard to say if another would help.

    A MOSFET can certainly be used but again, it may be better or not depending on the size of the buzzer.
     
  3. poopscoop

    Thread Starter Member

    Dec 12, 2012
    139
    16
    Piezo Buzzer, 6-12VDC, Don't know the current off the top of my head but it produces 75DB at 12V, so not an absurdly high current by any means.

    Transistors are
    First 2: 2N4401
    Final: 2N3904
     
  4. MrChips

    Moderator

    Oct 2, 2009
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    You don't need three transistors. One transistor will do.
    Make sure you put a resistor in series with the base otherwise you will blow the transistor.
     
  5. poopscoop

    Thread Starter Member

    Dec 12, 2012
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    When I do that the "leakage" is too high, and the buzzer hums. I experimented with one transistor and even with the base completely disconnected from everything I still had around 2 volts across the transistor.
     
  6. MrChips

    Moderator

    Oct 2, 2009
    12,421
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    Did you connect the base to ground through a resistor?
    Also try with the buzzer on the collector side of the transistor.
     
  7. ScottWang

    Moderator

    Aug 23, 2012
    4,853
    767
    If you have no idea about the transistor, then you can use the circuits as below to do the testing and measuring.

    Adjust VR1 to get the appropriate value of resistors, and measure them, and change to the fixed value.

    [​IMG]
     
    absf likes this.
  8. wayneh

    Expert

    Sep 9, 2010
    12,094
    3,033
    + 1
    You can use a rough rule of thumb that each transistor "switch" can give a current magnification of about 10X. So for instance 10mA to the base of one transistor can fully turn on a 100mA current path. That's probably in the range that's plenty for your buzzer. If you needed, say 10 amps, then you might need another two levels.

    A MOSFET is a great choice for a switch since it doesn't require any current on its gate to turn fully on. But it does need at least about 10V for a normal (not "logic level") FET and that would not be a good choice for a 9V battery, which will be running at maybe 7V when it's aging. A logic level MOSFET solves that because it requires a lower voltage on its gate to turn fully on.

    There is little advantage to a FET in this circuit, since it doesn't save any components, maybe a resistor or two at most. If you needed 10A output, it would be a big advantage since you could do that with a single MOSFET.
     
  9. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    Your output transistor is an emitter-follower, not a switch. Its base resistor value is much too high so it causes a huge voltage loss.

    An NPN transistor used as a switch has the beeper connected from its collector to the positive supply. If its base current is 1.5mA then the beeper gets about 8.9V.

    Assume the piezo beeper draws 15mA at 9V. Assume the 2N3904 transistor has its minimum hFE of 100.

    Then look what happens in your circuit:
     
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  10. poopscoop

    Thread Starter Member

    Dec 12, 2012
    139
    16
    I appreciate the input.

    I cannot seem to make this work with only one transistor. With the base completely unplugged I'm still getting current flowing from collector to emitter. The only way I can get the transistor to completely stop conducting is to put it on the collector leg of another transistor, as shown in my diagram.

    How do I get the transistors to shut off other than they way I'm doing it now?
     
  11. poopscoop

    Thread Starter Member

    Dec 12, 2012
    139
    16
    I think I figured it out. I connect the base to ground through a resistor.
     
  12. Audioguru

    New Member

    Dec 20, 2007
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    896
    Of course.
    A transistor without a resistor or another transistor pulling its base to ground amplifies its own leakage current. There will be a very high leakage current if the collector and emitter pins of the transistor are swapped. Look at the photo of the transistor's pins in its datasheet.
     
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