I have a circuit that uses an LED and a simple buzzer.

I've created a basic circuit (below) and used a speaker to represent the buzzer.

I know two things about the buzzer :

  1. Rated Voltage: 12V; Range of Voltage: 3-24V
  2. Rated Current: 15mA at 12VDC

I'm not sure if that means max current but I'll just assume it does.

I use Ohm's law to calculate 3V / .015 = 200 Ohms and I place a 200 Ohm resistor in the circuit and I'm lucky enough because the LED will illuminate with that current. But I have a couple of questions.

Voltage Drop of LED

But that is probably incorrect, because the voltage drop of the LED is approx 2V, right? I probably need to calculate 1V / 0.15 and use a 60 Ohm resistor (or something close).

Would the buzzer work since it requires min of 3V and the LED drops the voltage by 2V?
I thought that voltage is variable within a circuit, but current is the same throughout?

Can you help me understand the steps I would take to calculate the proper resistor for this circuit and if it is possible to power both the LED and buzzer from the 3V power source?


simulate this circuit – Schematic created using CircuitLab

  • 1
    \$\begingroup\$ you are misunderstanding the buzzer rating ..... if you connect 12 V to the buzzer (without a resistor), it will draw approximately 15 mA \$\endgroup\$
    – jsotola
    Commented May 30, 2019 at 22:44
  • 1
    \$\begingroup\$ If the Buzzer requires a minimum of 3V and LED has a Vf of 2V then your circuit will not work. You need an absolute minimum of 5V and no resistor. \$\endgroup\$ Commented May 31, 2019 at 1:00
  • \$\begingroup\$ @jsotola Thanks for clarification on that. \$\endgroup\$
    – raddevus
    Commented May 31, 2019 at 2:00

1 Answer 1


One thing about your current calculation, a comment also mentioned it but that is to give you a rough estimate of the impedance of the buzzer, if you calculate you get 12/15mA which is about 800 ohms. Now this is not a straight up resistor and it is not exactly the same at all voltages, but as a rough estimate for current and voltage drop it should be close enough.

so assuming your LED is perfect you could get 3V at the buzzer, but since it is not this is not gonna happen you are not complying with the minimum voltage your buzzer wants(3V) so maybe it will not work.

LED forward voltage or voltage drop depends on many things, color is a big factor since you are saying 2V my guess would be that you are working with a red or green one. However, a blue diode drop up to 3.5V so if you have info on your diode look up its datasheet online and see the number for forward voltage.

what you need to do is provide 3V + Led forward voltage as the input to insure 3V are delivered at the buzzer.

because of that somewhere around 5V(add a bit of a safety margin) as the input would be best, you have 800 ohm as the buzzer impedance, it should draw about 3.75mA. I think that is enough for the LED to shine just fine(it will not be super bright), your voltage source needs to be able to provide this amount of power plus the losses on the diode(current will be slightly higher than the number I provided)

finally, there are cases would people add an extra resistor like you did, I will list some:

1) diode short circuits, this would cause the source to apply full voltage to the buzzer: since your buzzer can handle up to 24V this is okay.

2) buzzer short circuits and causes a current draw that could damage my voltage source or LED, here you add a limiting resistor that takes into consideration this possibility and limits current accordingly.

Hope your project goes well.

  • \$\begingroup\$ This was a fantastically comprehensive answer. Thanks very much for all that info. I will wait a bit and then mark this as answer. Thanks again. \$\endgroup\$
    – raddevus
    Commented May 31, 2019 at 1:59

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