Connect loudspeaker to 555 output

Discussion in 'The Projects Forum' started by TsAmE, Sep 14, 2010.

  1. TsAmE

    Thread Starter Member

    Apr 19, 2010
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    I got a small loudspeaker and tried connecting it to the output of a 555 (pin 3), but I am not sure which pin of the speaker to use (1, 2, 3 or 4).
     
  2. Wendy

    Moderator

    Mar 24, 2008
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    1 and 2 are the same pins, as are 3 and 4.
     
  3. SgtWookie

    Expert

    Jul 17, 2007
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    You should also use a capacitor in series with the speaker to block the DC current path. Otherwise, you may burn up the speaker coil.

    Use a 10uF to 470uF aluminum electrolytic cap. Positive lead goes to the 555 output.
     
  4. Wendy

    Moderator

    Mar 24, 2008
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    I had already suggested that here.

    Volume control 555-speaker circuit?

    I don't know if the OP decided to ignore the advice or is using an old drawing.

    As Wookie says, you need to isolate the DC component, it is nothing but bad for the speaker.
     
  5. TsAmE

    Thread Starter Member

    Apr 19, 2010
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    I was just using this old drawing to show that the speaker was connected to a 555, but I am taking your advice, thanks.
     
  6. Wendy

    Moderator

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    One additional comment, the reason there are 4 points is 2 of them are where the speaker wires go in, and the other two are for connecting wires. If you look close you can tell which is which.
     
  7. TsAmE

    Thread Starter Member

    Apr 19, 2010
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    Do I have to connect all 4 pins? Or can I connect either 1 and 2 or 3 and 4 like you said? If so which part is - and +?
     
  8. Wendy

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    There really isn't a + or -, though they can be marked as such. If you connect them backwards the sound waves will be inverted, it is hard for humans to tell the difference either way.

    Like I said, you are looking at terminals, solder points for the device. The wires from the speaker connect to one pair, you connect to the other. If you unsolder a wire from the speaker and it comes free it gets a bit difficult, since you have to put it back.
     
  9. windoze killa

    AAC Fanatic!

    Feb 23, 2006
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    The 2 inner pins (2 and 3) are already connected to the coil of the speaker. Connect GND to one of the outer pins (pin 1) and the 555 through a capacitor to the other outer pin (pin 4).
     
  10. Audioguru

    New Member

    Dec 20, 2007
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    Usually a 555 is overloaded if it drives an 8 ohm speaker.
    With a new 9V battery the output of the 555 will be 2V peak or more. 2V/8 ohms = 250mA but the max allowed current from a 555 is only 200mA.
     
  11. TsAmE

    Thread Starter Member

    Apr 19, 2010
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    I was using Yenka to test the circuit and tried connecting the speaker to the 555 with the capacitor like you said, but the 555 blew up and the following message was displayed: The power dissipated was 1.29W. The maximum rating is 600 mW. Why is this?
     
  12. Wendy

    Moderator

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    9V/2 is 4.5 P-P. RMS in a square wave is the same as P-P, and the real voltage with the output max being 7.8V the P-P will be 3.9V. With a 8Ω such as the speaker you will have 7.8V/8Ω, or almost an amp out of the 555 (max current limit is 200ma), and a power rating of 3.9V²/8Ω, or 1.9W.

    You could add a 33Ω resistor to bring the current down the 555 specs, or you can add an amplifier similar to this.

    [​IMG] [​IMG]
     
  13. Audioguru

    New Member

    Dec 20, 2007
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    The 555 is not an audio power amplifier. Its max allowed output current is only 200mA.

    With the 9V supply the output swing as shown on the datasheet is typically 5.8V p-p when the output current is 100mA. Your LDR causes the duty-cycle to vary so sometimes the output is trying to feed pulses that are nearly 5.8V/8 ohms= 725mA which is much too high causing the output swing to be less which causes most of the signal to be across the output transistors of the 555 at the very high current which causes a lot of heat.

    Simply add a resistor in series with the speaker to reduce the max current. Then the resistor will get hot and share the heat.

    With a high value resistor instead of the LDR which makes a square-wave output then the positive swing and negative swing are about 2V each. Then the output current is 2V/8 ohms= 250mA and the power dissipated in the 555 is (4.5V - 2V) x 250mA= 625mW which makes it too hot. The excess current also might destroy it.

    Look at the datasheet of the LM555. The 8-pins DIP package is rated to dissipate a max of 1180mW but the tiny surface-mount package can dissipate 613mW.
     
  14. TsAmE

    Thread Starter Member

    Apr 19, 2010
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    Thanks for the info. I am curious, howcome there is AC current flowing in this circuit, if it is connected to a DC supply (battery) and not an AC supply?
     
  15. Wendy

    Moderator

    Mar 24, 2008
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    The capacitor was to separate the two. AC and DC are combined all the time, since most circuits are DC. Look up "bias" as it applies to electronics.
     
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