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Transistor circuit what works when it shouldn't? Why?

I'm looking to build a circuit that will alert me to the location of a lost model airplane based on modifying a commercially available lost plane finder. I de-soldered the wimpy 5 VDC piezo alarm off the board of the commercial unit. I probed the two pads where the piezo alarm was powered and confirmed I get a steady on-off pulse of 5VDC with a one-second interval. (I did a test with a small reed relay and got enough current to throw the relay and close a circuit on a much more powerful 12VDC piezo siren (this unit has a piezo driver and just needs

12VDC and it starts hollering!)

I would have just stuck with this however my goal was to get all the goodies back into the black case the loud piezo siren. Space is tight so I looked to use a switching transistor instead. I hooked up the commercial lost plane finder so that it had power and as I was completing the wiring it just started working! My intention was to connect the emitter to the negative side of the Piezo siren/lost plane finder (labeled simply as "on/off 5 VDC) in the photo)

Now I'm confused why this is working and am hesitant to start hooking up more wires when its working already.

How can this circuit work with the emitter not connected to anything?

Here's a link to the photo I took of a schematic I have drawn of the current working config.

formatting link

Any thoughts on this?

Thanks, Lawrence

Reply to
Tristar500
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That is one cute circuit. The transistor is conducting through its base-collector junction, which is basically a diode. Your batteries are in series so that you actually have 17 volts (nominal).

Reply to
kell

+12---------------Piezo------+ c| 330 / +5------[R1]-------+------b| NPN | \\

-(5)-----+ 10K [R2] e| | | |

-(12)----+---------+---------+

Reply to
ehsjr

Sorry, I don't understand. I checked the voltage with an analog meter at the 12 Volt piezo and it's 12 volts so I dont see that the piezo is "seeing" anything more then just the 12 volts powering it.

Any other ideas anybody?

Lawrence

Reply to
Tristar500

A previous poster is correct, in that you have simply forward-biased a PN junction (the B-C path), and thus have no switching control. Now, could you clarify that schematic? What is the 5V on-off thing you have drawn there? You say it's the piezo alarm, but you have a second box labeled "pieza siren." What is really going on there?

--
Al Brennan

"If you only knew the magnificence of the 3, 6 and 9,
 then you would have a key to the universe." Nicola Tesla
Reply to
Kitchen Man

The 5 volt on -off thing is a commercially available lost plane finder. It works well and also alerts you to low battery voltage etc. It just isn't all that loud. As I mentioned, I removed the piezo off the board of the unit and are now using the 5 volt signal that used to power the little piezo to switch on a much bigger noise maker.

My intention was to use the switching transistor as a relay.

Does this make sense?

Reply to
Tristar500 

--
Yes. Your circuit should be hooked up like this: (View in a
non-proportional font like Courier
 
                          
                          +-------+
                          |       |
                       [SIREN]    |
                          |       |+
                          C     [12V]
           +----[1KR]---B       [BAT]
          +|              E       |-
        [PLANE ]          |       |
        [FINDER]          |       |
          -|              |       |
           +--------------+-------+
Reply to
John Fields

HI John, I tried again and imported your schematic into word and this time I could decipher. I got it wired up and it works! Brillant!

Thanks John, and thanks everybody for helping me through this.

Next project will be to build my own lost plane finder from scratch. I plan to have it start making noise once the transmitter is shut off an it looses signal.

Reply to
Tristar500

A small current at the base of a transistor controls a larger current through the emitter-collector. In your diagram, a large current exists at the base, and there is no emitter current. Your transistor is acting as a diode. That is an undesireable situation for what you want.

The base current of a transistor should be kept small - too much current through the base will burn the transistor out. There is nothing in your existing setup to protect the base from too much current.

The diagram I drew shows the transistor properly configured as a switch, with a small base current (that current is limited by resistor R1) that controls a large emitter-collector current.

I added some insurance in the circuit with R2. It is very likely not needed, but insures that the base of the transistor is connected to something (ground in this case) in the event that your switched 5V circuit leaves it "floating".

Ed

I checked the voltage with an analog meter

Reply to
ehsjr

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