0
\$\begingroup\$

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?

enter image description here

\$\endgroup\$
3
  • \$\begingroup\$ Your NPN is upside down. \$\endgroup\$
    – winny
    Oct 25 at 6:36
  • \$\begingroup\$ BJT connection aside that plan seems good to me. You'll dissipate about 600mW and I hope you'll not sound that buzzer forever. \$\endgroup\$ Oct 25 at 6:51
  • \$\begingroup\$ A 16V zener doesn't drop the voltage by 16V, it blocks everything above 16V and takes it down to 16V and the excess energy transforms to heat. \$\endgroup\$
    – Lundin
    Oct 25 at 9:14
1
\$\begingroup\$

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.

I suggest you to simply do this schematic: Buzzer 24V->5V schematic

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.

\$\endgroup\$
4
  • \$\begingroup\$ I see, so you think is better to use resistors at least .5W instead of a single Zener diode. \$\endgroup\$
    – Citi
    Oct 25 at 21:10
  • \$\begingroup\$ Yes, because as mentioned bellow, zener diode get's really hot. I've tried using them when I was younger without taking into consideration the heat dissipation and I ended up with a burned pcb :). Also you can not parallel Zener diodes because they don't have the same electrical characteristics. Using parallel Zener ends up in always 1 of them working and heating until the next one takes place (This is not going to work in more strict electrical specifications). \$\endgroup\$ Nov 15 at 6:32
  • \$\begingroup\$ I implemented the above configuration and the buzzer sounds very low, almost not sound at all. I have 24v as the input and a 1k resistor. \$\endgroup\$
    – Citi
    Nov 16 at 8:06
  • \$\begingroup\$ Does your buzzer need PWM signal in order to work ? or it is working only with power ?? If it is the first case you have to adjust the PWM frequency in order to get the one that the buzzer responds to better. And an other thing that you can try is to increase the current flowing through the buzzer buy using a lower value of resistors. BUT this will increase the power on the resistor also so you have to take in mind the heat dissipation and resistor's watt specifications. \$\endgroup\$ Nov 18 at 7:31
1
\$\begingroup\$

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.

enter image description here

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

\$\endgroup\$
1
\$\begingroup\$

At 30 mA you can just power the buzzer from the 555's output (assuming the 555 is running from a suitable voltage)

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.