The following circuit was taken from the following source:
I would like to understand whether it is better to apply a 4000Hz square wave or a 4000Hz sine wave to this buzzer.
In addition, why will it be better or worse?
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\$\begingroup\$ Joey, is this post answered now or is there still something you need clarifying. \$\endgroup\$Andy aka– Andy aka2025-02-17 16:32:17 +00:00Commented Feb 17 at 16:32
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I would like to understand whether it is better to apply a 4000Hz square wave or a 4000Hz sine wave to this buzzer.
PWM is a square wave with a variable duty cycle (that imparts amplitude information) so, it would be problematic to somehow make a weird shaped sinewave that has the same ability to pass PWM information to what is a digital drive circuit (the PNP transistor). Of course the sinewave would be pure at a PWM duty of 50% but, it would be weird shaped for different PWM duty cycles.
Additionally, because the piezo is tuned to be resonant by the series inductor and, that tuning makes it really easy to use an on-off output from that PNP transistor, it makes no electrical sense to try and directly drive the piezo buzzer with a sinewave of variable amplitude (what PWM brings to the party if you do the analysis).
And finally, the use of the inductor (making the circuit resonant) means that you get voltage amplification and the buzzer can be made to sound louder for a smaller power-supply rail.
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\$\begingroup\$ Yes I do. Reading my post again, I see I may not have worded it correctly. When I said apply a square or sine wave, I meant, if you remove the drive circuitry why cant you just apply a PWM wave to the piezo at its resonant frequency. Apologies for the confusion. \$\endgroup\$JoeyB– JoeyB2025-02-17 17:50:53 +00:00Commented Feb 17 at 17:50
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1\$\begingroup\$ @JoeyB you shouldn't apply PWM square wave directly to a piezo because the piezo is highly capacitive and that will wreck your signal. That's why we use an inductor so that it forms a tuned circuit with the piezo capacitance. Does that make more sense? \$\endgroup\$Andy aka– Andy aka2025-02-17 18:04:20 +00:00Commented Feb 17 at 18:04
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\$\begingroup\$ Why would this occur "the piezo is highly capacitive and that will wreck your signal". How does a capacitor 'wreck' the PWM signal. I know at low frequencies the cap is seen as high impedance and up to the caps resonant frequency the impedance will decrease as the frequency increases. So you saying depending on the frequency the piezo will either act as high impedance path (high Vdrop across the piezo) or the peizo will be a short circuit and short the PWM to ground...? \$\endgroup\$JoeyB– JoeyB2025-02-17 19:14:46 +00:00Commented Feb 17 at 19:14
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1\$\begingroup\$ @JoeyB if you apply a square wave or a rectangular PWM wave the rapid edges of the signal cause a massive current surge. Lets say you have 100 volts per microsecond edges and a 20 nF capacitor, C dv/dt will be 2 i.e. 2 amps will be needed to drive the capacitor. Capacitor formula: \$i = C\dfrac{dv}{dt}\$ <-- it's all about the signal edges and not the frequency of the square wave. \$\endgroup\$Andy aka– Andy aka2025-02-17 20:15:43 +00:00Commented Feb 17 at 20:15