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Buzzer current limiting resistor value

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BarakG:
Hello

I would like to know how did the circuit designer of the circuit in the image calculated resistor R100 value.



I have tried to figure it out by myself but I'm not sure about it. I would like to know if my calculations are right.
My guess is that the 100ohm resistor is to limit the current to ~20mA.

buzzer's datasheet http://www.puiaudio.com/pdf/SMT-0540-S-R.pdf

The buzzer's datasheet says that Vp-p = 3V, Imax = 100mA.

To use Ohm's law to calculate R100 we have to 'translate' Vp-p to Vrms (kind of avg voltage) so,
Vrms = 0.35355 * Vp-p = 1.06V.

Then by Ohm's law:

R100 = (3.3 - 1.06) / 0.020 = ~100Ohm.

My calculations are right?

ArthurDent:
This is the way I see it.

The driving signal is specified as a square wave on the data sheet and it is used to turn the MOSFET on or off like a switch. The spec sheet says the buzzer has a resistance of 12 ohms with a maximum current rating of 100 MA. With the series resistor of 100 ohms and the buzzer resistance of 12 ohms, when the MOSFET is on, you have 112 ohms across a 3.3 VDC supply which gives you about 29 MA. That gives about a .35 VDC drop across the buzzer and about a 2.9 VDC drop across the resistor.

To drive the buzzer with the maximum current you would need a 21 ohm resistor, in series with the 12 ohm buzzer, or 33 ohms across the 3.3 VDV supply to give you 100 MA. Apparently this would produce far to loud a tone for this application so the circuit design drives the buzzer at less than a third of the maximum current.

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