# How can I power a 5 V - 7 V buzzer with a 24 V power supply?

I'm working with a 24 V power supply. I know I could use a voltage regulator to 5 V to power the buzzer (see image below), but I wanted to experiment to see if could make it work with just a resistor rated for 1 W at 26 mA.

I know the efficiency of using a resistor to drop a voltage is very low, but when I implemented the circuit below, and placed my multimeter red and black leads, I got about 18 V on the buzzer positive side, which is not what I expected to see.

I was wondering why I'm not seeing the values I calculated.

For instance, resistor = Vcc - (Vbuzzer + NPN drop voltage)/ test current => 24V - 5.7V / 26mA = about 703 &ohms.) Therefore, I should be seeing a voltage drop at the buzzer of about 5.7 V not 18.3 V on my multimeter and I should be seeing 18.3V across the resistor, not 5.7V.

I'm wondering why I don't see the right values in the right places. The buzzer is rated for max. 30 mA, 5 V - 7 V input voltage and is an internally driven buzzer. The buzzer will not be ON for long periods.

I have also tested the circuit by just using the resistor and the buzzer without the npn transistor and still see this behavior that I don't seem to understand.

• Can you pop the buzzer datasheet link into your question? There are many types and the coil type, for example, interrupt the supply when the coil is energised. Your current measurement is probably an average. Does it give a DC resistance reading between the terminals? Commented Aug 22, 2022 at 7:43
• What is the voltage from collector to emitter of the transistor? What signal is driving the base? An oscilloscope trace would help greatly to diagnose this mystery. :) Commented Aug 22, 2022 at 8:10
• A resistor drops voltage when current traverses it. If the current is not stable, then voltage drop will be unstable too. I suspect your buzzer consumes current with a frequency of 2.4 kHz, so you should use a scope to verify what's happening and calculate the value of the resistor. Commented Aug 22, 2022 at 8:15
• What happens if you replace D2 with a 100 uF capacitor?
– Jens
Commented Aug 22, 2022 at 13:45
• @Transistor I put the buzzer datasheet on my question. I also measured the DC resistance on the terminals and it was about 7.2Mohms
– Citi
Commented Aug 29, 2022 at 5:52

I suspect that your buzzer may be a mechanical type similar to an electric bell. These cut their own supply when the coil energises and thus set up an oscillation. The sound is characterised by a "buzz" rather than a "beep" tone.

Figure 1. Animaton by ЮК on Wikmedia Commons.

If it is this type of buzzer you may be able to measure the DC resistance between its terminals.

A datasheet link may help clarify.

The effect you are seeing may be because you are measuring average current. The load is inductive and the applied voltage will affect the current and voltage response due to $$\ V = L \frac {di}{dt} \$$.

If you decide that it is a buzzer type as I have described then measure the current when operating from 24 V. If it's the same as the the 5 V value (26 mA ± a bit) then I'd say it will be OK with your present design.

• A 2.4 kHz buzzer is almost certainly not an electromechanical type, which are probably 20-100 Hz or so. Commented Aug 22, 2022 at 8:06
• I have added the buzzer datasheet to my question, but it doesn't mention specific DC resistance
– Citi
Commented Aug 24, 2022 at 6:54
• @Transistor when I have measured the current when operating from 24V just using a 680 ohms resistor and just the buzzer without the npn, I see only a total current of about 10ma, nothing close to 26-30ma that I expected, and the voltage drop of the buzzer is about 17.2V
– Citi
Commented Aug 29, 2022 at 6:05
• @Transitor, I have tried to do some calculations but I can't find why this behavior happens. The buzzer is an active magnetic self-driven.
– Citi
Commented Aug 29, 2022 at 6:06