Timeline for Reduce voltage to read buzzer with GPIO
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 4, 2014 at 3:45 | vote | accept | ScottEH | ||
| Feb 3, 2014 at 22:38 | comment | added | Andy aka | 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. | |
| Feb 3, 2014 at 22:36 | comment | added | Andy aka | 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. | |
| Feb 3, 2014 at 22:26 | comment | added | Andy aka | 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. | |
| Feb 3, 2014 at 21:52 | comment | added | ScottEH | 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? | |
| Feb 2, 2014 at 7:42 | history | answered | Andy aka | CC BY-SA 3.0 |