Buzzer voltage drop

The voltage drop across it is the voltage you supply to it. If you have 12 volts available and want 6 volts to appear across the buzzer while it draws 25 ma, assume the buzzer looks like 240 ohms and put another 240 in series with it. They (the buzzer and the resistor should split the 12 volts into two approximately 6 volt drops.

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John Popelish

Of course, I left a gotcha in there to see if you would ask the next question. The buzzer draws some pulsed waveform that averages 25 ma when you apply 6 volts across it. There may be large instantaneous current variations. You may need to add a capacitor in parallel with the buzzer to soak up all those variations and hold something close to the average drop across the buzzer to make ir work correctly.

I would guess that 10 microfarads or so might be enough.

Capacitors are sort of like fly wheels (if voltage is like rotational speed and current is like torque) in this application. Internal engines put out an average torque at some speed but without a flywheel, the speed varies all over the place each turn. :)

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John Popelish

That is he general idea.

Leds have very large current swings for very small voltage changes, so they are not modeled by ohm's law, very well.

Normally you need some resistance in series with LEDs to make sure the current is controlled. But the buzzer may perform that resistor function, approximately. But it would worry me.

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John Popelish

I have an electronic buzzer (ICC pn: BS2316L-06) I am going to put it into a circuit with some LED's and other components. It draws 25ma and is rated at 6V, how do I know what the voltage drop across the buzzer is? Is it 6V? I need to know this so I can calculate what size resistor to put in the circuit.

Thanks Pat

The 25mA listedfor the electromechanical buzzer is an average -- sometimes a lot more, sometimes less. It will work with a series resistor, but not too well.

Let's assume you're running off a 13.8V supply. In order to get it to run well, you might want to use a zener like this (view in fixed font or M$ Notepad):

13.8V + | 1N4737A | Vz=7.5V/-/ ^ | | | | .---| '---| | .---| | | '---| | | ___ |/ o-|___|-o-|2N3904 Control 10K | |>

Signal .-. | 10K| | | | | | '-' | | | === === GND GND created by Andy´s ASCII-Circuit v1.24.140803 Beta

Good luck Chris

Thanks, I understand that better. I am an ME and I struggled in our 1 required EE class!

OK, thanks. Let's forget about the cap for now so I don't have an overload. Let's say my simple circuit consists of a 12V source, an LED that requires

25ma and has a Vf of 3.2v, and the buzzer. To figure out the correct resistor the calculation would be 12-(3.2+6) /.025 = 112 ohms. Is this correct?

Also, theoretically if the my source voltage was 9v would I not need a resistor at all?

Don't leave me any gotchas! I need it explained to me like I was 5.

Patrick