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Connecting a 9v AC Circuit to a 9v Battery

  1. Feb 15, 2012 #1
    I have taken apart a speaker set which uses a transformer to transform the 230-240v mains it receives into 9v AC 1A. I want to connect this circuit to a 9v battery. Nothing complicated-the standard Duracell one. Does anyone have any advice/comments on whether this will work or not?

    Regards and Many Thanks in advance,
  2. jcsd
  3. Feb 15, 2012 #2
    Seeing as how no one's answered this, I'll take a shot. Are you wanting to step up the voltage on a 9 volt battery with the audio transformer? If so, I don't think what you are trying to do will work for two reasons: First, in order to get anywhere near 230 volts, you'd need to put in 9 volts at one amp, which I don't think a standard 9 volt battery will put out. Secondly, it needs to be in an AC form for the transformer to continuously work (otherwise you'd get a brief voltage spike from the output, then nothing), as transformers need a constantly expanding or collapsing magnetic field to operate. What exactly is it you intend to do here?
  4. Feb 15, 2012 #3


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    It sounds as if you want to convert the output of this transformer to 9 volts DC so that you can use it instead of the 9 volt battery?

    If so, yes, it may just be possible depending on how much current is being drawn.

    If you rectify the output with a bridge rectifier and then filter it with a large electrolytic capacitor you would get about 11 volts DC out.
    This may be enough to run a 9 volt regulator although a little bit more voltage would be better..

    First, it would be necessary to measure the actual voltage coming from the transformer. Typically, these give more voltage than it says on the label.
    Last edited: Feb 15, 2012
  5. Feb 15, 2012 #4


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    Note: If you connect the output directly to a 9V battery it will alternately charge and discharge the battery 60 times per second. The battery will act like a big heating element, probably bursting into flame in a few minutes. I wouldn't advise this.

    If you want to charge the battery you'll need to be more specific than "standard Duracell one" is it a rechargeable alkaline? a lithium ion? a non-rechargeable alkaline?

    Are you trying to charge it? Just seeing what happens? trying to replace it or supplement its output?
  6. Feb 15, 2012 #5

    jim hardy

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    ""I want to connect this circuit to a 9v battery""

    which circuit? The transformer or the speaker set?
  7. Feb 15, 2012 #6
    There is 240v going into the transformer and 9v 1 A coming out. i want to replace the transformer with a 9v battery so i can run this device from batteries rather than from the mains. Thanks guys!!
  8. Feb 15, 2012 #7
    I don't know what your "speaker set" actually is or how it's designed to work, but AC and DC are two totally different things.

    Also, 1A of current is probably more than a 9V battery can supply... I am not sure what the internal resistance is but I would be surprised if a 9V can support a discharge rate of 1A. The capacity of the battery is only ~500mAH so even if it could supply that current, it would be totally dead within 30 minutes. And more likely, it would power the device for less time before performance degrades and it no longer works.

    Maybe 15-20 minutes of runtime.
  9. Feb 16, 2012 #8
    Just replace the transformer with the battery.
    If the voltage is the same, take care on the maximum current that must be delivered by the batteries (i.e. Duracell 9V MN1604B1 maybe you need to use two in parallel).
    Last edited by a moderator: Feb 16, 2012
  10. Feb 16, 2012 #9
    I'm not sure about the circuitry you're dealing with, but it may be possible to get it to run (not at it's best, but run) with less than 1 amp. If so, you're still looking at making a device that will convert the DC to AC at the same frequency produced by the transformer. If that is the case, you're looking at creating an inverter. It is possible to make a simple one with just two small (.22 uF capacitors), a potentiometer, and a 74AC240 octal buffer chip that will produce a square wave AC output. Unfortunately, such buffer chips at most can only take 7 volts (6 volts is recommended). Thus, if you were to use this method, you'd have to route the output to transistors that will in turn be able to switch on and off the heavier power load.

    As a side note, a circuit utilizing the more common 555 timer can be made to produce an AC output, however, careful calculations must be made regarding the size of the resistors for it to produce a near 50% duty cycle (when the alterations between + and - are equal).
    Last edited: Feb 16, 2012
  11. Feb 16, 2012 #10


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    Ahhh, I misunderstood. As mentioned already, it may or may not work depending on two things. The first is how much current the circuit draws (the 1 A is the max output of the transformer, the circuit may draw much less).

    The second issue is how the circuit uses that input voltage. 9V AC is oscillating at an average voltage of 9V +/- and can have a peak voltage as high as 1.414 times 9V i.e. up to 12V. The circuit will want a DC voltage so it will rectify and regulate the input. If it uses full wave rectification then replacing AC with DC input should work either polarity but if it is half-wave you will need to try attaching DC both ways to see which one matches up.

    Attaching a 9V battery will not damage the circuit so go ahead and give it a try. If it doesn't work try reversing the battery connection. If you get nothing but the battery warms quickly you're supplying insufficient current. If you get nothing either way and the battery isn't discharging then you may need a bit more voltage, try up to 12V DC input but no higher.

    If you tell us what type of speaker circuit we may be able to estimate better the power it draws. Note voltage * current = power. Typically a 9V alkaline has 500-600mAh which comes to about 5 watt hours. If your circuit draws 1 watt the battery should handle it for a few hours. If it draws the full 1A rating of the transformer then you're talking 9watts which will deplete the battery in about 30 minutes. Probably less given that's a bit beyond the battery's rating and it will be burning nearly half its power through resistive heating. More likely the voltage drop at that heavy a current will be insufficient to run the circuit.

    You may want to build a battery pack with 6 AA, C, or D cells connected in series. You'll get respectively 5x 40x or 60x the lifetime and better supplied current (lower battery resistance).

    Give it a try.
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