0
\$\begingroup\$

I want to drive a buzzer for 500 ms when reed switch opens, in simple schematics, when reed switch is closed it keeps Q1 pulled high (maybe I need a resistor too) but the problem with the design is that it will keep buzzing till the switch closes again,

I can't use 555 as my system is battery operated and have Vin range of 3-4.2 V, can I make a simple one shot buzzer when switch opens out of simple transistors, capacitors combination here? It is supposed to be a door alarm.

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
1
  • \$\begingroup\$ I think a 555 won't help you much here, see Michal's answer instead, but do note that not all 555 are NE555. There's much less power-hungry CMOS variants like TLC555, which can also work at lower supply voltages. \$\endgroup\$ Commented Nov 29, 2023 at 10:42

3 Answers 3

2
\$\begingroup\$

Use high beta transistor like >250.

enter image description here

Edit: Remove 10k and instead 1k use 10k. It will drain the battery less.

Edit2: You didn’t mention the current draw of buzzer. If the problem is with too low driving current use a PNP to get higher gain. enter image description here

Theoretically 2k2 and 4k7 are no needed. Just connect base of PNP with collector of NPN because maximum of PNP base current should not exceed 100mA. Use 600mA PNP type like 2n2907 or bc639.

\$\endgroup\$
5
  • \$\begingroup\$ Brilliant, I expect beep to be very short, will bigger cap at base help? I guess I can take it from here, thanks \$\endgroup\$
    – asim
    Commented Nov 29, 2023 at 11:17
  • \$\begingroup\$ Bigger cap is good approach. \$\endgroup\$ Commented Nov 29, 2023 at 12:51
  • \$\begingroup\$ Yep, the pulse is way too short to be audible, unable to fix it, tried bigger cap, Darlington pair and lower input resistor \$\endgroup\$
    – asim
    Commented Nov 29, 2023 at 19:43
  • \$\begingroup\$ Lower R means shorter pulse but drives Bjt with higher current. Darlington decreases the effective cap swing since it takes 1.4v so it’s not suitable. \$\endgroup\$ Commented Nov 29, 2023 at 19:52
  • \$\begingroup\$ added new question at electronics.stackexchange.com/questions/691268/… \$\endgroup\$
    – asim
    Commented Nov 30, 2023 at 19:59
0
\$\begingroup\$

Everything has been said on this topic. I would like to add that regardless of the circuit in which you use a bipolar transistor, it is necessary to protect the base with a resistor.

\$\endgroup\$
0
\$\begingroup\$

If you have access to a larger cap, such as 1000 uF or 2200 uF, you can eliminate the transistor. SPST switch, Piezo, capacitor, and a medium-value resistor to discharge the cap when the button is released. The beeper volume will trail off, sorta kinda like a chime.

Note, these parts are from my design library. Your cap would need only a 10 V rating.

enter image description here

\$\endgroup\$

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.