using AC power for piezo buzzer

I want to power a piezo buzzer using 12.8V AC. My circuit knowledge is not much. Can I just put in a diode and a capacitor to convert to DC? I want to use something like one of these buzzers from Radio Shack:

95dB, 7mA@12V DC == 84mW

108dB, 7-14V DC, 150mA max

The smaller buzzer uses only 84 mW, so can I just use one 1N914 diode and a capacitor?

Yes. I figure you'll get around 10 volts DC out of it if my assumptions are good.

I think the 914 is too light though it might work for a while.

Geoff

Use something like a 1N4004 and it won't die charging your capacitor.

I've used the 200mA 1N4148 for power supply circuits, but only when the source impedance was both known and relatively high.

Best regards, Spehro Pefhany

--
"it\'s the network..."                          "The Journey is the reward"
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Actually, the 1N914 and a small cap, maybe 10 uF, would probably work, although a fatter diode might be prudent. The piezo can probably stand

17 volts. Add a resistor in series somewhere if it's too loud.

You might try not using the cap, just the diode and the piezo. That might make an interesting sound.

John

--- For that one, you should wire it up like this, where RL is the buzzer:

E1 E2 E3 / / / AC>---[1N400X>]--+--[R1]----+ |+ | [C1] [RL] | | AC>--------------+----------+

for an invariant 12.8VRMS input, and a .07V drop across the diode,

E2 = E1(sqrt(2)) - 0.7V = (12.8V * 1.4514) - 0.7V ~ 17.4VDC

To drop 17.4V to te 12V the buzzer wants to see, with 7mA of current in the load (the buzzer, RL):

E2 - E3 17.4V - 12V R1 = --------- = ------------ ~ 771 ohms IL 0.007

750 ohms is a standard 5% value and will be fine. It will dissipate:

P = (E2 - E3) * (IL) = 5.4V * 0.007A = 0.038 watts,

so a 1/4 watter will be fine.

For 1V of ripple across C1,

IL dt 0.007A * 0.017s C1 = ------- = ----------------- = 116.6 µF dV 1V

The closest 20% electrolytic is 120µF, so to be sure of getting that you'll need something 20% higher than that to start with. That's

144µF, and the closest 20% to that is 150µF, so 150µF at 25V would be fine. Actually, anything over 144µF would be fine and the higher the capacitance the lower the ripple.

---

--- That's basically the same as the other one, except that you can go up to 14V and it looks like at that voltage you'll need 150mA.

So, looking at this one:

E1 E2 E3 / / / AC>---[1N400X>]--+--[R1]----+ |+ | [C1] [RL] | | AC>--------------+----------+

E2 - E3 17.4V - 14V R1 = --------- = ------------- = 22.6 ohms IL 0.15A

22 ohms is a standard 5% value and it'll need to dissipate:

P = (E2 - E2) * IL = 3.4V * 0.15A = 0.510 watts,

So you should use a 22 ohm 1 watter. (or two 110 ohm half-watters in series)

For 1 volt of ripple out of the cap:

IL dt 0.15A * 0.0167s C = ------- = ----------------- = 0.0025 F = 2500µF dV 1V

In order to get to half that size you could double the ripple frequency to 120Hz by using a full-wave bridge rectifier:

E2 +------+ / AC>---|~ +|---+--[R1]-->>--+ | | |+ | | | [C1] [RL] | | | | AC>---|~ -|---+-------->>--+ +------+

But now E2 changes to 16.7V because of the two-diode drop in the bridge, so you'd have to go back through and work out R1 with that in mind.

A simpler way to go might be to run the buzzer on 12V and use a linear regulator after the cap instead of a resistor, like this:

E2 12V +------+ / / AC>---|~ +|---+--[7812]----+ | | |+ | | | | [C1] | [RL] | | | | | AC>---|~ -|---+-----+------+ +------+

What that'll allow you to do is to use a smaller capacitor, which will generate more ripple, but the 7812 will keep the output steady at 12V as long as you don't let the voltage on the cap get _too_ low.

A 7812 wants to see no less than 14V on its input, so with 17.4V being the peak the cap can charge to, we can stand:

E2 - 14V = 17.4V - 14V = 3.4V

of ripple across the cap.

Assuming 150 mA of current into the load, for 3.4V of ripple we'll need C1 to have a capacitance of:

IL dt 0.15A * 0.0083s C -------- = ----------------- = 0.000366F = 366µF dv 3.4V

for a 20% cap you'd need to get 439µF, so 510µF at 25V would be OK.

IMO, that would be the way to go and you could use it for either buzzer. (Overkill for the little one, though)

---

--- Yes.

-- John Fields Professional Circuit Designer

Would it be okay to use more than one diode wired in parallel?

The 1N914 is supposed to be able to dissipate 250mW continuous, so I thought one would be enough for the smaller buzzer (84mW).

It's a capacitor that makes sound, which makes it a very leaky capacitor.

John

In that arrangement, I'd put maybe a 100K resistor across the piezo; it _is_ a capacitor, you know. :-)

But, yeah - a 2 KHz sonalert warbling at 60Hz. :-)

Cheers! Rich

Thanks for all the replies so far.

The buzzer is to be an add-on to a doorbell. The house's doorbell and thermostat are powered by a transformer whose iron core is 2" x 2 1/4". The doorbell uses a solenoid, and I measured 12.8 VAC across the solenoid while the solenoid is energized. I am in the US, so the transformer is fed by 115 VAC.

NT, does your 17v dc number assume full-wave rectification?

I have an old PC power supply from which I could probably rob a rectifier and capacitors. Would that be better than using a 1N914?

lossy != leaky, but you know that.

Best regards, Spehro Pefhany

--
"it\'s the network..."                          "The Journey is the reward"
speff@interlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com

snipped

This is good stuff, but I think overlooks 2 issues.

The first is transformer regulation. This is a small transformer designed for only occasional duty, so regulation will be poor. Thus we need to allow for 33% regulation. This will take us to well over 20v after rectification.

The 2nd is that the buzzer may consume a chopped current waveform. If this is so, a series dropping resistor will still expose it to the full rectified voltage, and since thats double its rated, it may either die or misbehave. This is why I suggested putting your dropper R between diode and reservoir rather than after it, then keeping the resevoir as small as is acceptable to maintain decent response times. This would make it not so simple to calculate the required resistance, but it would be easy enough to guess then tweak.

Matt, >20v will occur whatever type of rectification is used.

NT

yes, so far.

yes

No. Any moment that your transformer output is higher V than the cap, current will flow into the cap. So the cap charges to the peak of the transformers output waveform, not the average.

Also 12v ac is not 12v peak, its 12v rms. The peak of 12v rms is 12 x

1.414 = 17v.

Also your small transformer will have a fair amount of resistance in its windings, and its designed to give 12v ac on load. So when lightly loaded, it gives much more. Thats what 33% regulation means. The end result is about 23v. The buzzer might survive that but I'm not optimistic.

NT

Despite what some apparently-well-informed posters have written, my simple-minded view is that one diode would chop off everything below

0.7V in the sine wave and that the cap would smooth out the resulting curve, resulting in a (12.8V - 0.7V)/2 DC == 6 VDC average voltage. I'd be happy if somebody would explain why this is wrong.

I presume you are talking about a power supply. You get (peak - (.7 and some loss)),because that flows into your capacitor,with a single phase rectifier you just get a bigger ripple when you start using curent. That might be oke for a small load. Mean value appears only, if you do not use a capacitor, in which case you have a constructed a rather useless supply.

--
You seem to have missed the part where he explained that he measured
the transformer voltage _while loaded_ with the solenoid and got
12.8VRMS.  Such being the case, his peak output voltage could never
exceed (12.8V * 1.414) - 1 diode drop ~ 17.4V

--- In the first place, the 12.8VAC is the AC equivalent of what it would take to heat up a resistor to the same temperature as if

12.8VDC was connected across the resistor. Since AC varies sinusoidally (hopefully) between some positive voltage and some negative voltage, it has to fall below the the DC value some of the time, and when it does less power will be dissipated in the resistor during that time. Consequently, since the AC voltage falls below the DC voltage some of the time, it has to be driven higher than the DC voltage some of the time in order for the temperature of the resistor to be the same in both cases.

To make a long story longer, if you sample the instantaneous voltage at an infinite number of points along a single AC cycle, square each of those samples, add them all up and take the square root of the sum, it turns out that the peak voltage required to heat the resistor up to the the same temperature as with DC is the DC value times the square root of two. That AC voltage is called the Root-Mean-Squared voltage and is implied whenever not specifically denied.

What all that means is that your 12.8VAC is 12.8VRMS, which corresponds to positive and negative peak voltages in the sinusoid of:

Peak = RMS * (sqrt(2)) = 12.8V * 1.414 ~ 18V

So you have a circuit which looks like:

E1 / ACIN>---+ +----+ | | | P)||(S | R)||(E [SOL] I)||(C | | | | ACIN>---+ +----+ | GND

Where, with the solenoid across the secondary, E1 is 12.8VRMS or,

18VPK.

Now, if we add a diode in series with the output of the transformer we'll have:

E1 E2 / / ACIN>---+ +----+--[1N4000>]-->

| | | P)||(S | R)||(E [SOL] I)||(C | | | | ACIN>---+ +----+ | GND

Where the 0.7V dropped across the diode will cause E2 to fall 0.7V below E1. Since we're now talking DC,

E2 = E1 - 0.7V = 18VPK - 0.7V = 17.3VDC

If we now add a cap:

18VPK E2 17.3VDC / / ACIN>---+ +----+--[1N4000>]--+-->

| | | | P)||(S | |+ R)||(E [SOL] [CAP] I)||(C | | | | | | ACIN>---+ +----+-------------+--> | GND

What will happen is that with every pulse coming out of the diode the cap will fill up more and more fully with electrons until the voltage across it equals 17.3V. When that happens, there will be no voltage differential between the diode's cathode and the cap's + input, and it will be as charged as it can be.

-- John Fields Professional Circuit Designer

No- you will blow up the buzzer. You will have to do something like this: View in a fixed-width font such as Courier.

. . . . 1N4004 100 . >---|>|-----/\\/\\---------+------+--->

. | |+ + . ---/ --- . 12V // \\ --- buzzer (7mA) . 12.8VAC Zener --- 100U . 1N4744 | | - . | | . 100 | | . >-----------/\\/\\---------+------+--->

. . . . .

The second 150mA buzzer will be beyond your capabilities at this point.

Okay---thanks to you and others---I see. When there is no drain on the cap, its voltage goes to the peak of the output voltage of the transformer.

Would the buzzer obey Ohm's law? At various voltages, would there be a constant ratio between voltage and current?