I want to power a self drive piezo buzzer with a rated current given as >3mA. http://www.microbuzzer.com/buzzer/spec/TE-HPS12C-H2.5.pdf

I am looking for a small battery I can use to power the buzzer (and drive circuit), but it is hard to get a battery with a Continuous current of 3mA that is not also too large in size for my liking. The buzzer has dimensions of 12*12 H2.5mm and I would like a battery somewhere on the same scale if possible.

The buzzer sound output is given as >75 dB, but I do not mind if it falls below this (say to around 60 dB). For this reason, I am thinking that a smaller battery with a Continuous current lower than 3mA could be used - resulting in an (acceptable) lower sound output. Am I correct in thinking this? I am wondering if anyone can offer any advice/opinions on how low I can go with the Continuous current of the battery. At what level will the battery be unable to power the buzzer? Could I use a battery with a Continuous current lower than 1mA? Is there anything else I should be aware of when selecting the best battery for my purpose? (electrical engineering novice here - my science education stopped after "this is a liquid, this is a solid, this is a gas"... I am serious)

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    \$\begingroup\$ How long are you going to power the buzzer, with what intervals? Just curious.. \$\endgroup\$ – abdullah kahraman Feb 7 '13 at 10:41
  • \$\begingroup\$ My impression is you also need an oscillator/driver for this (passive) piezo buffer. It doesn't make a sound other than a single click when you connect the battery. \$\endgroup\$ – jippie Feb 7 '13 at 11:19
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    \$\begingroup\$ jippie is right you'll need your own driver. There won't be a particular lower current limit, other than just like turning down the volume on a speaker at some point you won't be able to hear it. My gut feel is that at 1mA it would still be above 60dB, but the manufacturer may be able to provide more detailed information. \$\endgroup\$ – PeterJ Feb 7 '13 at 11:41
  • \$\begingroup\$ @OlinLathrop I am sorry, I made a mistake while typing the first sentence. I meant to write "external drive" instead of "self drive". I always knew the buzzer would not oscillate on its own. My apologies. I did read the data sheet (many times) but as I am new to this, I was not completely sure of my understanding of the data. When you say the buzzer will likely draw considerably less than 3mA at the lower end of the operating voltage range, how much current would that probably be? Something in the area of 0.5mA? Thanks again, sorry about the self drive/ external drive mix up! \$\endgroup\$ – Eddie Feb 7 '13 at 14:38
  • \$\begingroup\$ @abdullahkahraman Thanks for the comment. I want to power the buzzer for three minutes. The intervals may be hours, or >5 min. \$\endgroup\$ – Eddie Feb 7 '13 at 14:40

You link to the datasheet, but apparently didn't read it. All the information you ask for is in there, and it's not hard to find since it is a single page. Here is the electrical characteristics section:

First, note that the operating voltage is 1-25 Vpp. Obviously this thing can be driven over a wide range.

The current is shown as ≤ 3 mA. That would have to apply over the full voltage range, so very very likely it will draw considerably less than that at the lower end of the operating voltage range.

I can't begin to guess how you got the idea this thing would oscillate on its own. The datasheet never mentions that, and in fact makes it quite clear it won't. This is just a basic piezo element. First note that the voltage is specified as peak to peak, so obviously is intended to be AC. Second, there is a frequency response graph:

What could that possibly mean if it generated its own frequency? There would be no frequency to "respond" to.

As usual, you have to actually read the datasheet of any part you intend to use.

  • \$\begingroup\$ @Eddie: There is no way to know from the datasheet how much current it will draw at the low end of the operating voltage range. The easiest way to get some idea is to test it. There will be part to part variation, so also test the full voltage current and scale the results accordingly. That will factor out some of the part variation. Then derate by at least a factor of 2, maybe more if you need to be really certain. Also, try asking the manufacturer. \$\endgroup\$ – Olin Lathrop Feb 7 '13 at 14:13

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