0
\$\begingroup\$

Using the GPIO on a BBB I'm attempting to detect a change in voltage coming from a DC source that drives a buzzer when the machines cycle is complete. I've located the DC connections from the machines control board to the buzzer. When testing + to ground I get 13.2V. When testing + to the buzzer I get 6v/6ma. When the buzzer activates it spikes up and down 5 times from 6v/6ma to 10v/20ma. So my question is how do I build a circuit using these two contacts that is reduced to approx 2.5v & < 5ma as to not blow out the GPIO?

I'm very new to this but based on searching the web my initial thought was a voltage divider, since it seems like a simple circuit to create that I can easily calculate the resistor sizes for......BUT I'm concerned about the spikes as the buzzer operates. In my testing of these contacts with my multimeter several times the meter indicated an overload.

I've also read about Triacs and thought that might isolate the spike but the problem is even when the machine is idle/sleeping the voltage between the + and buzzer lead is still 6v so the triac will never reset.

Thank you

Scott

\$\endgroup\$

1 Answer 1

Reset to default
1
\$\begingroup\$

Use a resistor from the buzzer, maybe 1kohm to 10kohm and feed it to a zener diode. The zener will limit the voltage seen by your chip. If your circuit runs from 3.3 volts choose a 2.7 volt zener diode. If it runs from 5volts choose a 4.7volt zener.

Try googling zener voltage regulator.

\$\endgroup\$
4
  • \$\begingroup\$ After researching zener diodes I believe I understand how they will fit my need. I want to confirm that in reading the data sheets for zener diodes that they seem to have a fairly narrow range of operation, for example a BZX79C2V7 2.8v will operate from 2.5v - 2.9v, anything higher and it will fail...is that correct? So using your example if I put a resistor in front of the zener to reduce the voltage wont I also reduce my current under the zener 5mA zener current requirement? \$\endgroup\$
    – ScottEH
    Commented Feb 3, 2014 at 21:52
  • \$\begingroup\$ The resistor protects the zener from taking excessive current and the zener protects the IO pin from excessive voltage. You need to look at the 2.7V zener's data sheet and see what it says about current. \$\endgroup\$
    – Andy aka
    Commented Feb 3, 2014 at 22:26
  • \$\begingroup\$ It's difficult to track down the data but you are looking 2.5V to 2.9V at 5mA and 5mA through a 1k resistor is 5V so already you have made some headroom in protecting your circuit BUT the device is 500mA and can take probably 50mA all day and probably not rise above 3.0V!! How much headroom do you have now at 50mA - there would need to be 50V across the 1k resistor. \$\endgroup\$
    – Andy aka
    Commented Feb 3, 2014 at 22:36
  • \$\begingroup\$ Below probably about 2.3V there is hardly any conduction from the zener - you need to find a device with a more detailed specsheet that shows the exact characteristic. \$\endgroup\$
    – Andy aka
    Commented Feb 3, 2014 at 22:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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