I have this circuit below where I want to drive a buzzer that is 4 V and has a max current of 30 mA here is the link Click Here. I'm working with a VCC supply voltage of about 24 V. I know there are voltage regulators or that I could implement a voltage divider with a Zener diode. However, I was wondering if I could just use a single Zener diode with a reverse voltage of 16-20 V and 1-1.5 W and thus drop the voltage to around 4-8 V for the buzzer to operate. In my circuit below the NPN transistor base is controlled by a 555 timer output. I need help to know if this implementation is safe to use or if it's good enough to work with no problems?
Based on buzzer's datasheet max current and voltage you can find the maximum Power needed.
P= V * I= 8(V) * 0.03(A)= 0.24 Watt.
For 24V as Vcc you have 5v on the buzzer. So 24-5= 19V. If you have a resistor with value 1k you will get 19mA current flowing to your buzzer. If you limit the current then your buzzer will be protected. But your resistor's power has to be at least -->19v * 20mA = 0.38Watt. I would suggest to add some more resistors in parallel if you don't have any resistors 0.5W ++ in order for this schematic to work.
I hope this helps.
You could still have thermal issues with derating the Zener in an enclosure above 25'C, unless designed properly such as using 2 in parallel. Always derate power by 50% when you want it to run at 50% of the max temperature rise as MTBF drops 50% for every 10'C rise.
If you operate more than 5V it will draw proportionally more current and fail sooner from greater stress even though it will work at 8V, it does not say for how long.
Rated at 150 mW @ 5V / 30 mA = 167 Ohms with a 16V drop.
But at this resistance 48 mA @ 8V or 400 mW ! with a 16 V drop.
An LM317 adj regulator would be better in a TO-220 , set with 2 resistors to optimal voltage.
However, at one octave up and 7 dB louder you can get a piezo 24V buzzer https://www.digikey.com/en/products/detail/cui-devices/CPI-137-24T/11480419