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'[EE] Inductor for piezo'
2006\11\30@030633 by Jinx

face picon face
I need an extra-loud sound from a small piezo disc and thought
I'd give the inductor method a go. Usually a 1k resistor across the
piezo as a non-capacitive load for the transistor produces a loud
enough beep. This mail from a while back says how

======

Wed, 16 Jun 2004 09:12:57, from Matt Pobursky

1. Find the capacitance of the piezo device, almost always listed in
the spec sheet.
2. Calculate the device's capacitive reactance at the frequency you will
be driving the device.
3. Calculate an inductor value for the equivalent inductive reactance.
4. Place an inductor of said value in parallel with the piezo device.
5. Drive the whole parallel piezo/inductor circuit with a FET/transistor
rated ~10x higher voltage than your voltage source.

======

My piezo, 27mm diaphragm, is around 15nF and resonates at 4.6kHz

Capacitive reactance Xc = 1/(2*pi f C), f = 4600, C = 15nF

Xc = 1/(6.28 * 4600 * 15*10^-9)
= 1/0.00043332
= 2308 ohms

Inductive reactance XL = 2*pi f L, XL = 2308, f = 4600, 2*pi = 6.28

L = XL/(2*pi f)
= 2308/(6.28 * 4600)
= 0.080H

80mH seems rather large. I was expecting perhaps a couple of
hundred uH (no evidence to presume that, apart from physical
size). Are my units and the method correct ?

TIA

2006\11\30@042254 by Mike Harrison

flavicon
face
On Thu, 30 Nov 2006 21:05:58 +1300, you wrote:

{Quote hidden}

This is definitely in the ballpark. Piezos are very complex loads - far from a simple capacitance,
so it is usually easiest to do it by trial and error & measurement rather than calculating.
In practice the value is not especially critical unless you're trying to squeeze the last drop of
efficiency, and the value will often be simply 'the biggest in the range' or whatever form factor
inductor you use. Even 1mH gives a decent voltage boost.
Epcos do a nice range of axial leaded parts which are ideal for this :

100uH :
http://uk.farnell.com/jsp/endecaSearch/partDetail.jsp?SKU=518300

If you need small, look at inductors aimed at EL backlight inverters  - these will be in the right
inductance range - 10s of mH.

Note that if the piezo gets disconnected, the voltage can easily hit the hundreds, which can kill
teh transistor. I usually use an MPSA42 which can take 300V - needs fairly strong base drive (2K2
from 5V) as gain is low.  Standard NPNs are usually OK as long as the piezo stays connected.

Another good option if space is limited is a flyback converter + H-bridge - using two SOT-23 dual
transistors for the H-bridge saves space - dual digital-transistors save even more - component count
ends up as 2xDTR, 1 transistor, 1 diode, & inductor. Only snag is you need 2 PIC pins to dribe the
H-bridge and 1 for the flyback PWM. Drive the flyback switching transistor from a PWM peripheral if
available at, say 100KHz ( can easily be bit-bashed if no PWM available) , the inductor will be in
the 100uH area, and you can control volume by adjusting the PWM ratio. A bit more complex but very
cheap & small. This solution works especially well at low supply voltages.




2006\11\30@170200 by Jinx

face picon face
Thanks for the considered reply Mike. I'm sure I saw a
circuit once with a little180uH axial inductor that claimed
the noise was ear-splitting. I've a few inductors around
that so I'll experiment with them and frequency, see if I
can find the "ouch" note


'[EE] Inductor for piezo'
2007\02\02@050632 by Jinx
face picon face
part 1 709 bytes content-type:text/plain; (decoded 7bit)

I'm playing about with the attached to get more volume from
a piezo disk. It's based on a solar-powered white LED driver
posted previously. 'Enable' is a low from the PIC to turn the
circuit on. This is done just before the beep is required to let
the reservoir cap charge, then a tone is sent from the PIC into
the piezo switch transistor

Attached (a) to share and (b) for a question. What is the function
of the inductor that the 5V goes through ? I presume it's a filter
of some sort to stop the 50V pulses going back into the PSU but
I'm not sure how/why. I substituted it with a diode but that merely
put 4.3V on the 220uF cap and the circuit didn't oscillate at all


part 2 2201 bytes content-type:image/gif; (decode)


part 3 35 bytes content-type:text/plain; charset="us-ascii"
(decoded 7bit)

2007\02\02@061431 by Michael Rigby-Jones

picon face


>-----Original Message-----
>From: spam_OUTpiclist-bouncesTakeThisOuTspammit.edu [.....piclist-bouncesKILLspamspam@spam@mit.edu]
>On Behalf Of Jinx
>Sent: 02 February 2007 10:05
>To: Microcontroller discussion list - Public.
>Subject: Re: [EE] Inductor for piezo
>
>
>I'm playing about with the attached to get more volume from
>a piezo disk. It's based on a solar-powered white LED driver
>posted previously. 'Enable' is a low from the PIC to turn the
>circuit on. This is done just before the beep is required to
>let the reservoir cap charge, then a tone is sent from the PIC
>into the piezo switch transistor
>
>Attached (a) to share and (b) for a question. What is the
>function of the inductor that the 5V goes through ?

That inductor is what actualy provides the voltage step up.  The transistor connected to it pulls a current through the inductor to ground, and when the transistor is switched off the voltage on the inductor rises until the 1N4148 diode is biased into conduction.  Ignoring all the other circuitry, just look at the inductor in the 5v line, the diode and the middle transistor and you will see this is a classic "boost" switcher topology.

I suspect other inductor is simply to provide enough phase shift to get the circuit to self-oscillate.

Regards

Mike

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2007\02\02@063239 by Jinx

face picon face
> >function of the inductor that the 5V goes through ?
>
> That inductor is what actualy provides the voltage step up.  The
> transistor connected to it pulls a current through the inductor to
> ground, and when the transistor is switched off the voltage on
> the inductor rises until the 1N4148 diode is biased into conduction

Ah, thanks. I thought the other inductor was the booster, but now
I see it explained as an inductor switched on the 5V line that makes
sense. So, if another source of transistor drive were available (eg
PWM from the PIC that's already there) the L R and C around the
switching transistor could be dispensed with ?

2007\02\02@064152 by Michael Rigby-Jones

picon face


{Quote hidden}

Yes, but:

With a fixed pulse width (i.e. no feedback) the voltage would vary with load.  With no load the voltage could get very high.

You would need to make sure the "on" time of the switching transistor is not long enough to saturate the inductor which would cause high currents (only limited by inductor resistance) to be drawn from the 5v rail.

Regards

Mike

=======================================================================
This e-mail is intended for the person it is addressed to only. The
information contained in it may be confidential and/or protected by
law. If you are not the intended recipient of this message, you must
not make any use of this information, or copy or show it to any
person. Please contact us immediately to tell us that you have
received this e-mail, and return the original to us. Any use,
forwarding, printing or copying of this message is strictly prohibited.
No part of this message can be considered a request for goods or
services.
=======================================================================

2007\02\02@065749 by Matt Pobursky

flavicon
face
On Fri, 02 Feb 2007 23:05:22 +1300, Jinx wrote:
> I'm playing about with the attached to get more volume from a piezo
> disk. It's based on a solar-powered white LED driver posted
> previously. 'Enable' is a low from the PIC to turn the circuit on.
> This is done just before the beep is required to let the reservoir
> cap charge, then a tone is sent from the PIC into the piezo switch
> transistor
>
> Attached (a) to share and (b) for a question. What is the function
> of the inductor that the 5V goes through ? I presume it's a filter
> of some sort to stop the 50V pulses going back into the PSU but I'm
> not sure how/why. I substituted it with a diode but that merely put
> 4.3V on the 220uF cap and the circuit didn't oscillate at all

Michael Rigby-Jones reply was spot on -- it's a basic boost converter
with on/off control.

I would also note that the method used to enable the circuit has one
potential drawback. The enable transistor method implemented could be
a fairly serious power hog, depending on your power requirements and
the source of your 5V (i.e. battery powered circuit). It will draw
about 500uA continuously when the circuit is "disabled", i.e. the
BC547 enable transistor is turned "on". The discharge path is through
the two 470uH inductors in series, through the 10K resistor, through
the BC547 transistor to ground. Assuming the DC resistance of the two
inductors is relatively small (a few 100 Ohms or less) then you
basically have a 10K resistor hung off your 5V supply to ground
whenever the circuit is disabled.

If power consumption when the piezo ckt. is not enabled is a concern,
I would either switch the whole subcircuit ground or better yet
implement a zero power high side 5V switch to the circuit with a PNP
transistor or P-Fet.

Gahhh... all I've been doing the past several years is really low
power battery operated circuits, it seems.

Matt Pobursky
Maximum Performance Systems



2007\02\02@091629 by Harold Hallikainen

face
flavicon
face
As others have pointed out, it's a boost converter driving the piezo.

I recently had to drive a piezo from a PIC running on a 3V lithium
battery. I got an adequate level by connecting the piezo in bridge mode
between two PIC output pins, making one go high when the other goes low
and vice versa. It's certainly not 50V though!

Harold


--
FCC Rules Updated Daily at http://www.hallikainen.com - Advertising
opportunities available!

2007\02\02@100557 by Gordon Williams

picon face

----- Original Message -----
From: "Matt Pobursky" <EraseMEpiclistspam_OUTspamTakeThisOuTmps-design.com>
To: "Microcontroller discussion list - Public." <piclistspamspam_OUTmit.edu>
Sent: Friday, February 02, 2007 6:57 AM
Subject: Re: [EE] Inductor for piezo


> Michael Rigby-Jones reply was spot on -- it's a basic boost converter
> with on/off control.
>
> I would also note that the method used to enable the circuit has one
> potential drawback. The enable transistor method implemented could be
> a fairly serious power hog, depending on your power requirements and

Wouldn't there be two other places were you would be loosing much more than
500 uA?

It is disabled with "Enable" high (5V?) so wouldn't you be loosing about 4mA
on the enable line?

Also if the PIC continues to pulse the transistor on the right hand side or
leaves the line high, you are going to loose another 2 to 4mA there?

Interesting circuit and something that I might be able to use.

Regards,

Gordon Williams

2007\02\02@114046 by Bob Blick

face picon face
> I'm playing about with the attached to get more volume from
> a piezo disk. It's based on a solar-powered white LED driver
> posted previously. 'Enable' is a low from the PIC to turn the
> circuit on. This is done just before the beep is required to let
> the reservoir cap charge, then a tone is sent from the PIC into
> the piezo switch transistor
>
> Attached (a) to share and (b) for a question. What is the function
> of the inductor that the 5V goes through ? I presume it's a filter
> of some sort to stop the 50V pulses going back into the PSU but
> I'm not sure how/why. I substituted it with a diode but that merely
> put 4.3V on the 220uF cap and the circuit didn't oscillate at all
> --

Hi Jinx,

The inductor you speak of is required.

You could replace the vertically oriented inductor with a resistor.

Cheers,

Bob


2007\02\02@124426 by Matt Pobursky

flavicon
face
On Fri, 2 Feb 2007 06:11:59 -0800 (PST), Harold Hallikainen wrote:
> As others have pointed out, it's a boost converter driving the
> piezo.
>
> I recently had to drive a piezo from a PIC running on a 3V lithium
> battery. I got an adequate level by connecting the piezo in bridge
> mode between two PIC output pins, making one go high when the other
> goes low and vice versa. It's certainly not 50V though!

I have this requirement come up all the time.

Usually the client doesn't think the buzzer is loud enough with a
bridged connection as you describe and I end up with either a
"booster coil" across the piezo (cheap but the coil must be sized to
the particular piezo you use) or a boost regulator supplying the
voltage to the piezo.

Matt Pobursky
Maximum Performance Systems


2007\02\02@124852 by Matt Pobursky

flavicon
face
On Fri, 2 Feb 2007 10:01:37 -0500, Gordon Williams wrote:
{Quote hidden}

Yep, that's an even bigger power waster. I tend to use lots of
jellybean N-Fets instead of bipolar transistors as switches to ground
in battery powered designs for this reason. I use a lot of P-Fets to
switch high side power because of low power losses and near zero
current switching.

> Also if the PIC continues to pulse the transistor on the right hand
> side or leaves the line high, you are going to loose another 2 to
> 4mA there?

Well that's where good firmware design comes in -- don't leave the
line high when the buzzer is supposed to be off! I would assume that
you would notice the buzzer beeping if you let the drive signal pulse
continuously... And you are right, there are two places you will have
losses when the drive transistor is "on": the collector current and
the base drive current.

> Interesting circuit and something that I might be able to use.

It is an interesting circuit although I would probably make a few
changes to it for my typical piezo buzzer design requirements.
Digital volume control is a spec I run into all the time and once
you've got to add that it changes the picture quite a bit. Most
everything I do these days runs at 3V as well... so that might impact
some of the circuit design. I also worry about component tolerances
when you have a design that relies on discrete analog component
values for it's basic operation. Many analog passive components
aren't very precise to begin with (inductors, transistor gains,
etc.). Many times you are better off to use a slightly more expensive
integrated part like a switcher chip that is guaranteed to work "by
design" even thought the BOM cost is slightly higher. Circuits with
less components are always more reliable mechanically (less solder
joints to fail) and cost less to assemble.

I really enjoy looking at other people's designs. Once I figure out
how their design works and maybe get some insight into why they did
things the way they did I usually ask "OK, now how would I have done
it?"

Matt Pobursky
Maximum Performance Systems


2007\02\02@134631 by Harold Hallikainen

face
flavicon
face

{Quote hidden}

Coil across the piezo? I wonder what the impedance (R+jX) of the piezo is
at the driving frequency. PERHAPS, if it's reactive, a series reactor
(capacitor or inductor) would increase the current through the device,
increasing the power. Or, if the resistance is high, a T or pi network
could transform it down to something that matches the output drive
capability of the PIC. Of course, the PIC can only ouput so much voltage
at so much current, so there may just not be enough power available on a
pin to drive it. Adding an external driver would then be needed.

In the application I did, they wanted a hiss sound (like white noise). I
used the PIC to generate the burst of noise, put that on one pin and the
inverse on another pin with the piezo between the two. With reasonable
acoustic design, this was loud enough when run on a 3V lithium battery.

Harold

--
FCC Rules Updated Daily at http://www.hallikainen.com - Advertising
opportunities available!

2007\02\02@142345 by David VanHorn

picon face
>
>
> Coil across the piezo? I wonder what the impedance (R+jX) of the piezo is
> at the driving frequency. PERHAPS, if it's reactive,


It's plenty reactive, it's a capacitor. R is relatively small in relation to
jX
The parallel inductor forms a parallel resonant tank (we hope) whose
impedance at resonance is high. If the resistance of the inductor is
relatively low, then the circulating currents can build up fairly high,
until the input energy is balanced by resistive losses and losses due to
moving air.

a series reactor
> (capacitor or inductor) would increase the current through the device,
> increasing the power. Or, if the resistance is high, a T or pi network
> could transform it down to something that matches the output drive
> capability of the PIC.


Transformers work too.

Of course, the PIC can only ouput so much voltage
> at so much current, so there may just not be enough power available on a
> pin to drive it. Adding an external driver would then be needed.


100mW or more, depending on the pin. 20mA at 5V gets you to 100mW

2007\02\02@160658 by Jinx

face picon face
> Coil across the piezo?

Hi Harold, if you look up this document, page 6

http://www.murata.com/catalog/p15e6.pdf

they suggest using 20-50mH to raise the voltage and SPL, rather
than the 1k resistor in page 3 diagrams

p19e, p37e and p57e are also about characteristics of and driving
piezos

2007\02\03@025854 by Vasile Surducan

face picon face
Joe, I'll go further with the question. What the hack need this
circuit an enable input as long the driving signal could be tied to
ground by command ?
Only a series inductance with the piezo, computed on the resonant
frequency of the piezo and one diode to protect the CE of the signal
driver transistor against inverse voltage is required here. The rest
is unuseless.

A piezo may sound loudly if it's mounted into a resonant chamber.
Such chamber could increase the noise from 50dB to more than 90dB
measured at 1 m in front of the piezo keeping the same driving
voltage.
All piezo sirens used in alarm systems are based on this tehnique.
But most of them run at 12V because the resonance is producing a
higher voltage on piezo. BTW, standard piezo are almost dying at 50V
ac across them. You may hear the effect, it's a very high frequency
wistle over the driving frequency range.

Vasile

On 2/2/07, Jinx <@spam@joecolquittKILLspamspamclear.net.nz> wrote:
{Quote hidden}

> -

2007\02\03@193022 by Jinx

face picon face
part 1 2762 bytes content-type:text/plain; (decoded 7bit)

Hi Vasile

> Joe, I'll go further with the question. What the hack need this
> circuit an enable input as long the driving signal could be tied to
> ground by command ?

What I'm doing is experimenting with the circuit to find out what's
going to be useful in different applications

> Only a series inductance with the piezo, computed on the resonant
> frequency of the piezo and one diode to protect the CE of the signal
> driver transistor against inverse voltage is required here. The rest
> is unuseless.

Now, you say series, and I've seen that mentioned a couple of times
elsewhere, but parallel seems more common. For example Murata
suggest a few 10s of mH in parallel with their elements. You'll see in
the attached schematic how the voltage has been boosted by the
inductors, in this case small transformers/chokes. This sound bomb
is very loud (105dB), and took two towels to muffle so I could take
some readings

Also attached is a little logic problem. It's the footprint of one of the
transformer-like inductors, showing the 0V and piezo connections
in-circuit. For the life of me I can't figure out how it might be wound
and internally configured. I thought it might be a small impedance-
matching audio transformer, but there's no isolation. There seem to
be basic windings with 45 ohm DC resistance, perhaps some centre-
tapping, but beyond that I don't know. Anyone have a clue ?

Using this method

http://www.thekeeser.com/Electronics%20info/measure_an_unknown_inductor.htm

I get 37mH at 4580Hz from A or B to 0V pin, with C=33nF, the
approximate capacitance of the piezo, so I guess no matter what
that 'transformer' actually is, it's the effective inductance that's
important, and 37mH is definitely in the suggested range

> A piezo may sound loudly if it's mounted into a resonant chamber.
> Such chamber could increase the noise from 50dB to more than
> 90dB measured at 1 m in front of the piezo keeping the same driving
> voltage

That's quite true. SPL is very dependent on mounting and enclosure.
I'd say, from my tinkering, that there's more to gain from driving with
the correct frequency and mounting them properly than from simply
increasing the voltage

> All piezo sirens used in alarm systems are based on this tehnique.
> But most of them run at 12V because the resonance is producing a
> higher voltage on piezo. BTW, standard piezo are almost dying at 50V
> ac across them. You may hear the effect, it's a very high frequency
> wistle over the driving frequency range.

The small ones I have do sound being over-driven. However, note
that the sound bomb does have a fairly square-ish 50V as the output
and the tone is penetratingly clear


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2007\02\04@034408 by Vasile Surducan

face picon face
On 2/3/07, Jinx <KILLspamjoecolquittKILLspamspamclear.net.nz> wrote:
{Quote hidden}

It could be a serial resonance oscillating circuit either a parallel
resonant circuit, or a transformer. You agree the loud is proportional
with the energy absorbed into the piezoelectric dish, and with the
amount of absorbed energy which is transformed in mechanical vibration
of the piezo (with a proportionality factor k, which unfortunately is
much lower than 1)

The serial resonance is known as current resonance (the equivalent
resistance of series LC at resonance is zero) while the parallel
resonance is known as voltage resonance (the equivalent parallel
resistance of LC at resonance is infinite). The parallel resonance
could look better with higher supply voltage but is not a general
rule.

The loudest piezos I've seen where driven with a transformer and where
sinking quite large current (from 300mA to 1A at 12V)


>
> Also attached is a little logic problem. It's the footprint of one of the
> transformer-like inductors, showing the 0V and piezo connections
> in-circuit. For the life of me I can't figure out how it might be wound
> and internally configured. I thought it might be a small impedance-
> matching audio transformer, but there's no isolation. There seem to
> be basic windings with 45 ohm DC resistance, perhaps some centre-
> tapping, but beyond that I don't know. Anyone have a clue ?

Imagine an autotransformer (probably called self transformer in
english). There is one coil between A and B and another coil between B
and C.  AB and BC are series connected. Piezo is connected between A
and C. C is connected to ground and B is connected to the colector of
the driving transistor.
This is doubling  the voltage on the piezo even there is no resonance.
Changing the frequency will change also the amplitude of the signal as
long one frequency is equal with the resonant frequency of the
transformer-piezo assembly.
You may get interesting chirps based only on the frequency change
arround the resonant value.


{Quote hidden}

> -

2007\02\15@030605 by Jinx

face picon face
part 1 542 bytes content-type:text/plain; (decoded 7bit)

Just to wrap it up - attached is what I've settled with

Soft - enough to be noticed more than a few yards away
Loud - ear-poppingly nasty

The initial 30mm disc was salvaged from an old Christmas
card BTW. It shall Jingle Bells no more. The 22mH is from
RS, 191-1210. Price not too hideous for a one-off but I'll
get bulk somewhere else

With a 40mm disc (ex-phone) at ~ 2.5kHz the Soft sound
carries further. Up close the Loud is not pleasant and I did
feel giddy for a moment as my ears were assaulted



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