Electronics Forum / Basics / June 2007

 I have an electronic thermometer that uses a piezo speaker for an

It will perobably be AC.
(remove the piezo and see if it works with DC (some do)
use the thermometer battery.

if it's AC use capacitors between the thermometer circuit and the
rectifier, this way the relay and the thermometer can both be run from
the same power source with short-circuiting one of the piezo wires.

Is there a better way? A simpler

                                             |
 something like this perhaps            +----+
                                        |    |
                                       /     |
    ||                               |/      |
----||--+-->|---------+---+-[100]----|       |
    ||  |             |   |          |\|    / 2N2222A
        +--|<----.    |   |           ~\  |/
    ||           |    |  ===            --|  
----||--+-->|----|----'   | 100nF         |\|
    ||  |        |        |                ~\
        +--|<----+--------+------------------+
   22nF    1N914

Firstly, the signal depends on the piezo device itself.

Two types -

Transducer -
http://www.maplin.co.uk/Module.aspx?TabID=1&DOY=24m6&ModuleNo=3202&criteria=

Buzzer -
http://www.maplin.co.uk/Module.aspx?ModuleNo=3211&criteria=piezo&doy=24m6

If it is a piezo transducer then you need to feed a signal of the frequency
you want to sound  (typically a square wave).

However, it is a buzzer, then you just feed a DC voltage becuase the unit
has the electronics to turn it into a sqaure wave.

Either way, these signals will be very low current and not enough to power a
relay coil.

You will need to add a transistor stage to drive the relay coil.

If it is a transducer, its not true AC - just DC pulsed on an off at the
desired frequency.  So a simple RC circuit will probably be sufficient prior
to the transistor.

Cheers,

Nigel

Hi, Eric.  I'm going to assume your thermometer has a 3V battery -- if
it's different, it will change the way you look at this problem.

Most small piezo beepers are driven directly by two logic outputs from
the microcontroller -- one on each side of the piezo element.  To turn
on the beeper, the red is made logic high (3V) while the black is
logic low (0V).  It then reverses polarity 6000 times a second or so,
depending on the resonant frequency of the element.  To the element,
it looks like a 6V peak-to-peak square wave, which should be more than
enough to drive many small elements.

You can use a small bridge rectifier to give you a 4.6VDC "ON" signal,
which should be plenty to drive a darlington transistor.  That should
be able to turn on a reasonably-sized relay (opf course, you're
assuming the thermometer battery isn't connected to anything else
here, and you're using another power supply to drive the relay).

Here's one way to solve the problem (view in fixed font or M$
Notepad):

(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

You can use a TIP110 TO-220 NPN darlington to drive up to 2A of relay
coil current.

A simpler way (assuming a manual reset button is OK with you) is to
use a sensitive gate SCR in place of the darlington.  That will mean
you can lose the 1uF filter cap:

(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

This is pretty simple.  If minimum component count is important, you
can replace the resistors and silicon with a solid state relay that
operates on a 3VDC input signal (you'd still need the 1uF filter
cap).  However, that will load down the thermometer battery more than
these setups (the darlington only uses 1/3mA or so of battery current
-- the SCR about twice that).  A SSR input will use several mA at
least.

If you have a 1.5V supply, it becomes a little more complicated,
because you only have a 3Vp.p. square wave.  You'll then have to use
schottky diodes and possibly a voltage doubler, especially if you want
to drive a logic level MOSFET instead of a darlington transistor or
SCR.

Good luck
Chris