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I need to implement the short beep sound at PSU being switched On/(Off - if easy). Are there any simple circuits for this problem? The piezo buzzers are preferable of their size and PCB mountable option. But is there any way to run them without controllers? Thanks!

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  • \$\begingroup\$ I need a short beep, lets say 1/5 of second (really short). What about piezo buzzer, I have 20pcs of TDK PS1240P02BT. And the circuit must beep during On/Off operation. \$\endgroup\$ – Roman Feb 23 '16 at 17:18
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    \$\begingroup\$ The world needs more beeps sounds \$\endgroup\$ – biggvsdiccvs Feb 23 '16 at 22:21
  • \$\begingroup\$ My curiosity is getting the better of me. Why do you need a beep when it's switched on or off? \$\endgroup\$ – Bernhard Hofmann Feb 24 '16 at 14:05
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    \$\begingroup\$ I understand your question! The russians say: "why the heck needs a goat accordion?". But imagine the blind man who switches the PSU on (he can find the push button), but he can not see the LEDs. How the blind man can understand that the power is on/off? The problem now is to change the frequency of sound on switch off (major notes for On and minor notes for Off). \$\endgroup\$ – Roman Feb 24 '16 at 16:54
  • \$\begingroup\$ That actually makes it easier in some ways - two different buzzers. Might also be audibly different with the same buzzer due to charge/discharge curve shape for the capacitor. \$\endgroup\$ – Ecnerwal Feb 24 '16 at 18:57
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Building on soosai steven's answer, if you use a DPDT relay, you only need one capacitor:

schematic

simulate this circuit – Schematic created using CircuitLab

When the relay pulls in, the capacitor charges through the buzzer. When the relay drops out, the capacitor discharges through the buzzer again. The diode prevents the capacitor from partially discharging through the relay (and any other loads) before the relay drops out.

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  • \$\begingroup\$ I already gave a +1, but I'd do it again if I could for introducing me to the circuitlab editor. Thank you. ☺ \$\endgroup\$ – Bernhard Hofmann Feb 24 '16 at 14:09
  • \$\begingroup\$ I'm having a thought that this should also be do-able in the same general manner but with diodes rather than the relay. But I haven't fleshed it out enough to see if that won't really work, and this will. \$\endgroup\$ – Ecnerwal Feb 24 '16 at 18:54
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What you are asking can be done easily.. What you need are two 1000uF capacitors, a piezo buzzer, a 1 Amp diode and a single pole relay with normally closed contact.

The hand drawn simple schematic explain the buzzer operation as intended

During power up buzzer sounds momentarily until the first capacitor fully charged. During power down the second capacitor discharge through the buzzer again sounds the buzzer momentarily.

The first cap may need a high ohm resistor in parallel to discharge it to prepare it for next power up buzz.

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  • \$\begingroup\$ And where is the buzzer? \$\endgroup\$ – Roman Feb 23 '16 at 19:19
  • \$\begingroup\$ In this circuit, the buzzer would be connected to the power supply through the relay. Meaning that it would sound on power-up, but would need a second power supply on power down...along with the short lifespan of a relay interfering if it will be activated more than a fes times/day. \$\endgroup\$ – Robherc KV5ROB Feb 23 '16 at 20:24
  • \$\begingroup\$ May I ask you to provide the complete circuit to test it in LTSpice? \$\endgroup\$ – Roman Feb 23 '16 at 21:45
  • \$\begingroup\$ @Roman My, aren't we demanding? Please keep in mind that all help you are given on webistes such as this is unpaid assistance you're getting & don't be so demanding with people who are going out of their way to make your life easier. \$\endgroup\$ – Robherc KV5ROB Feb 23 '16 at 22:58
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    \$\begingroup\$ My hand drawn diagram is not great and I intentionally didn't label anything. Pls put some effort to crank your brain to relate the circuit to it's descriptions to understand which us which and how it works. Robherc, it does not need a second power supply, only the second capacitor is needed. Such a simple circuit still need to stimulated in LTSpice? \$\endgroup\$ – soosai steven Feb 23 '16 at 23:44
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If you use a piezo buzzer/beeper with an integral oscillator/driver, this is easy and will beep on power-up.

Beep on power-down is more difficult, so if you don't need it...

enter image description here Here's the LTspice circuit list just in case you want to play with the circuit:

Version 4
SHEET 1 880 680
WIRE -320 64 -528 64
WIRE -272 64 -320 64
WIRE 48 64 -272 64
WIRE 128 64 48 64
WIRE 176 64 128 64
WIRE 224 64 176 64
WIRE 48 80 48 64
WIRE -320 112 -320 64
WIRE 176 112 176 64
WIRE -272 176 -272 64
WIRE -240 176 -272 176
WIRE 48 176 48 144
WIRE 48 176 -16 176
WIRE -320 240 -320 192
WIRE -272 240 -320 240
WIRE -240 240 -272 240
WIRE 176 240 176 192
WIRE 176 240 -16 240
WIRE -320 304 -320 240
WIRE -272 304 -272 240
WIRE -240 304 -272 304
WIRE 224 304 -16 304
WIRE -528 320 -528 64
WIRE 176 336 176 240
WIRE 128 368 128 64
WIRE 128 368 -16 368
WIRE -528 448 -528 400
WIRE -320 448 -320 368
WIRE -320 448 -528 448
WIRE 48 448 48 176
WIRE 48 448 -320 448
WIRE 176 448 176 400
WIRE 176 448 48 448
WIRE -528 512 -528 448
FLAG -528 512 0
FLAG 224 64 MCU_Vcc
FLAG 224 304 PIEZO_Vcc
SYMBOL Misc\\NE555 -128 272 M0
SYMATTR InstName U1
SYMBOL res -304 96 M0
WINDOW 0 44 46 Left 2
SYMATTR InstName Rt
SYMATTR Value 240k
SYMBOL cap -304 304 M0
WINDOW 0 46 33 Left 2
WINDOW 3 28 61 Left 2
SYMATTR InstName Ct
SYMATTR Value 1µ
SYMBOL cap 192 336 M0
WINDOW 0 -37 30 Left 2
WINDOW 3 -34 60 Left 2
SYMATTR InstName C2
SYMATTR Value 10n
SYMBOL res 160 208 M180
WINDOW 0 47 73 Left 2
WINDOW 3 35 45 Left 2
SYMATTR InstName R1
SYMATTR Value 100k
SYMBOL cap 64 80 M0
WINDOW 0 -37 30 Left 2
WINDOW 3 -41 60 Left 2
SYMATTR InstName C1
SYMATTR Value 100n
SYMBOL voltage -528 304 R0
WINDOW 0 16 103 Left 2
WINDOW 3 24 96 Invisible 2
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V2
SYMATTR Value PULSE(0 9 .1 10ms)
TEXT -512 304 Left 2 ;9V
TEXT -504 480 Left 2 ;tout = 1.1RtCt
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  • \$\begingroup\$ What is MCU_Vcc? \$\endgroup\$ – Roman Feb 23 '16 at 19:51
  • \$\begingroup\$ @RobhercKV5ROB: Sorry, but no. It was an [my] error of assumption and it's now fixed. \$\endgroup\$ – EM Fields Feb 23 '16 at 20:55

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