using AC power for piezo buzzer

[Matt]

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
http://www.radioshack.com/product/index.jsp?productId=2062404&cp=&kw=buzzer&parentPage=search

108dB, 7-14V DC, 150mA max
http://www.radioshack.com/product/index.jsp?productId=2062396&cp=&kw=buzzer&parentPage=search


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

 

[[email protected]]

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
http://www.radioshack.com/product/index.jsp?productId=2062404&cp=&kw=buzzer&parentPage=search

108dB, 7-14V DC, 150mA max
http://www.radioshack.com/product/index.jsp?productId=2062396&cp=&kw=buzzer&parentPage=search


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

12.8v ac would give you 12.8x1.414-1 = 17v dc. But since this is likely
off a small transformer, you may IRL be looking at 33% regulation. With
a small load that'll take you up to 23v. So no.

If you put the right resistor between diode and cap, that should work.
You'll want to keep the cap as small as possible so the buzzer turns on
and off quickly.


NT

 

[grunt]

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

 

[Matt]

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

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).

 

[Spehro Pefhany]

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).

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

 

[Matt]

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

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?

 

[John Fields]

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
http://www.radioshack.com/product/index.jsp?productId=2062404&cp=&kw=buzzer&parentPage=search

---
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.
---

108dB, 7-14V DC, 150mA max
http://www.radioshack.com/product/index.jsp?productId=2062396&cp=&kw=buzzer&parentPage=search

---
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)
---

 

[John Larkin]

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?

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

 

[Rich Grise]

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

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

 

[John Larkin]

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

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

John

 

[Spehro Pefhany]

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

John

lossy != leaky, but you know that.


Best regards,
Spehro Pefhany

 

[[email protected]]

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
http://www.radioshack.com/product/index.jsp?productId=2062404&cp=&kw=buzzer&parentPage=search

---
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.

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

 

[Matt]

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

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.

 

[Sjouke Burry]

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.

 

[[email protected]]

[email protected] wrote:

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

yes, so far.

and that the cap would smooth out the resulting
curve,
yes

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.

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

 

[John Fields]

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
http://www.radioshack.com/product/index.jsp?productId=2062404&cp=&kw=buzzer&parentPage=search

---
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.

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.

---
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
---

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.

---
17.4V isn't twice 12V, and _If_ the load on the capacitor is
chopped, then the only time the voltage on the buzzer will rise to
the voltage on the capacitor is when the load isn't drawing current.

If that's going to be a problem, then I'd fix it with either a 12V
zener across RL, or by substituting an 78L12 for R1, like this:

E1 E2 E3
/ / /
AC>---[1N400X>]--+--[78L12]--+
|+ | |
[C1] | [RL]
| | |
AC>--------------+-----+-----+

 

[John Fields]

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.

---
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.

 

[Fred Bloggs]

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
http://www.radioshack.com/product/index.jsp?productId=2062404&cp=&kw=buzzer&parentPage=search


108dB, 7-14V DC, 150mA max
http://www.radioshack.com/product/index.jsp?productId=2062396&cp=&kw=buzzer&parentPage=search


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

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.

 

[Matt]

yes, so far.




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.

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.

 

[Matt]

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

John

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