Battery getting hot

Discussion in 'General Electronics Chat' started by Katherine1, May 19, 2013.

  1. Katherine1

    Thread Starter New Member

    Mar 1, 2013
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    I did a circuit using 9V battery. However, the battery becomes hot very quickly. May I know what;s the problem?

    Attached is the schematic circuit. Is there a possibility that I have drawn wrongly which leads to short circuit?
     
  2. #12

    Expert

    Nov 30, 2010
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    You have the battery attached backwards.

    It probably killed the IC, too.
     
  3. PackratKing

    Well-Known Member

    Jul 13, 2008
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    I'll concur on both points...
     
  4. Katherine1

    Thread Starter New Member

    Mar 1, 2013
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    Does that mean I just need to change the terminal of the battery and IC?
     
  5. DerStrom8

    Well-Known Member

    Feb 20, 2011
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    There are actually a few errors in that schematic, but #12 pointed out the main one. The battery is backwards. Also, pin 4 should be connected directly to +. Right now you're "resetting" the chip at the same time that you "set" it. You can connect it directly to positive in order for the chip to time out on its own, or you can connect it via a separate pull-up resistor to a second switch, that when pressed would ground it and reset the chip before the timing period ends.
     
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  6. DerStrom8

    Well-Known Member

    Feb 20, 2011
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    No, you need to connect the entire battery the other way around (reverse the + and - wires on the battery).
     
  7. #12

    Expert

    Nov 30, 2010
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    Thanks DS8. I gave it a quick scan and still missed the reset pin. Many eyes catch more mistakes.
     
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  8. Katherine1

    Thread Starter New Member

    Mar 1, 2013
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    Oh really, thx for pointing out for me. This is my first time doing a circuit. I'll try to reverse the battery and connects pin 4 to the positive. One more problem, the switch i use is a 4 leg push button (https://www.sparkfun.com/products/9190) . when i press it, it pops back. Does that mean it cant be used in my case?
     
  9. DerStrom8

    Well-Known Member

    Feb 20, 2011
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    I see you're using that pushbutton as the trigger. That is exactly the type of button you want. The 555 will be set with only a single input pulse. Pressing and releasing the button will be perfect for the trigger.
     
  10. LDC3

    Active Member

    Apr 27, 2013
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    Another problem is that you are putting about 8 V on the speaker (which is usually 8 ohms). This would result in a 1 amp current. Much more than the battery can supply. I think a typical 9 V battery can supply 0.1 amps for a very short time.
     
  11. Wendy

    Moderator

    Mar 24, 2008
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    The transistor has no current limiting for the transistor. That is a dead short to the output of the 555.

    A transistor BE junction looks like a diode, a forward biased diode. Put a 1KΩ resistor between pin 3 and the Base of the transistor.
     
  12. DerStrom8

    Well-Known Member

    Feb 20, 2011
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    Oops, forgot to put that in my post listing the problems. Thanks Bill :p
     
  13. DerStrom8

    Well-Known Member

    Feb 20, 2011
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    It should still work though, it just won't be as powerful.
     
  14. timescope

    Member

    Dec 14, 2011
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    BPR1 could be one of those piezo buzzers with a built in oscillator that only requires a DC supply to produce a sound (a cylindrical black thing with a hole on top and two pins on the base)

    The circuit is configured as a re-triggerable monostable by connecting the reset and trigger together: when SW2 is pressed, the capacitor is discharged and a new timing cycle begins when it is released.

    One possible problem would be the leakage current of the 1000uF capacitor : if it is too large, pin 6 will never reach the threshold of 2/3 of the supply voltage.

    Timescope
     
  15. LDC3

    Active Member

    Apr 27, 2013
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    Well if this is the case, most piezo buzzers have a current limit under 50 mA. With no other resistance, the current would be way to much for it.
     
  16. timescope

    Member

    Dec 14, 2011
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    mouser piezo buzzer datasheet

    These devices have an inbuilt transistor oscillator circuit that regulates the current. The above device consumes 8mA (max) at 9v and can be used from 3v to 30v.

    Timescope
     
  17. Katherine1

    Thread Starter New Member

    Mar 1, 2013
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  18. timescope

    Member

    Dec 14, 2011
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    Use the "Estimated Service" information in the datasheet.

    First you must implement the corrections suggested and get the circuit working properly then measure the current consumption.

    1. Correct the battery polarity
    2. Add resistor in series with the transistor base
    3. Replace the 555 timer as it is probably damaged

    Timescope
     
  19. LDC3

    Active Member

    Apr 27, 2013
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    From the datasheet, if the load is 620 ohms with an average voltage of 7 volts, the average current would be about 11.3 mA. A load of 180 ohms has an average current of about 38.9 mA.
     
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