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  • I would like to remove the buzzer on an Intel mainboard, and replace it with an LED with a resistor. Is there a standard voltage that I can assume that comes to the two ends of the buzzer?

  • Why was a buzzer used in the design of computer mainboards instead of an LED?

Update: My buzzer is labeled HYCOM HY-05, datasheet here.

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    \$\begingroup\$ RE the 1st bullet. The part number (p/n) might be written on the buzzer itself. If you look up the p/n on the web, it might tell you the voltage and the type of buzzer (internal oscillator or not). Also, add the buzzer's p/n to your post. Photo of the buzzer on the board might help too. \$\endgroup\$ Jul 22 '12 at 19:24
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I'm not aware of a "standard" voltage as such, but it's pretty likely to be 3.3V or 5V (possibly 12V). The easiest way would just be to measure the voltage across it when it operates, then select your LED + series resistor accordingly.
Bear in mind it could be driven by PWM if it has no internal oscillator, so a scope would be best to test with. PWM will still work for the LED though, but the brightness then depends on the duty cycle.

For example, if we have a standard red LED with a Vf of 2V, and we aim for a ~10mA operating current @ 3.3V. This should be sufficient current for a basic indicator LED.
If the voltage is 3.3V we get (3.3V - 2V) / 0.010A = 130 ohms for the series resistor.
At 5V, we get (5V - 2V) / 0.010 = 300 ohms.
At 12V we get (12V - 2V) / 0.010 = 1k ohms.

Note that depending on the type of buzzer, there may be a series resistor present already so you might be able to fit the LED directly.
The way to test this would be to either check the PCB, find the resistor and read the value, or measure voltage as above, add a resistor of known value in place of the buzzer and measure the voltage drop across it.
Then do (Vsupply - Vr) / (Vr / R) = Runknown.

For example if the supply is 5V and you use a 1000 ohm resistor and measure 4V across it:
(5 - 4) / (4 / 1000) = 250 ohms. Again, bear in mind it may be PWM.

EDIT - now we have the part number, we can see it is a 5V, 50mA electromagnetic buzzer, and it is driven by a 2.4kHz square wave (not DC)
This is fine, just size the resistor for half that of DC to get the same brightness (as the duty cycle is 50%, it means the power is 50% compared to DC)
So if you size for 20mA, you will get the equivalent of 10mA. This should be plenty bright enough.

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  • \$\begingroup\$ Wouldn't there be a huge difference in current that can be sourced from a pin that drives a piezo speaker vs. a pin that drives a buzzer with internal oscillator? I think the use of a piezo speaker is highly likely (dating from the days that we didn't have proper sound cards). \$\endgroup\$
    – jippie
    Jul 22 '12 at 19:35
  • \$\begingroup\$ There's not a huge difference in current between a small electromagnetic buzzer and a piezo (you can get both types that run from e.g. 2mA) Also both types come in versions that have an internal oscillator or require a PWM. I think it is likely to be a piezo too, so it's probably pretty low current, and possibly already has a series resistor present. I'll add this to the answer. \$\endgroup\$
    – Oli Glaser
    Jul 22 '12 at 20:07
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A little bit of history about the second point: Original PC didn't have a sound card and there were no complicated speaker systems we have today. Instead it only had a single weak small speaker to provide sound and was called "PC speaker". The output was created by programming the Intel 8253 Programmable Interrupt Timer. It's counter 2 was connected to the speaker and could be used to play music or in some games made by high sourcerors familiar with the esoteric secrets of Black Magic even produce speech.

Over time sound cards appeared that could produce more complex sounds and that lead to use of external speakers with PC. As those speakers evolved, the PC speaker degenerated. Today, it's usually just a buzzer and isn't as supported in operating systems as it used to be. For example, Windows 98 could use PC speaker on some computers as a normal sound output for all audio events, while new versions of Windows can only beep. The only remaining use of it seems to be to notify user of some low level problem which prevents computer from booting normally.

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  • \$\begingroup\$ +1 for reminding me of those funny little speakers with the terrible sound quality. \$\endgroup\$
    – Oli Glaser
    Jul 22 '12 at 20:29
  • \$\begingroup\$ @Oli Glaser I have stockpile of those speakers I got from some decommissioned computers and put them in every new system I make :). I simply can't call a computer PC if it doesn't have one of those speakers. I sometimes even like to connect them to audio output when I play DOS games. \$\endgroup\$
    – AndrejaKo
    Jul 22 '12 at 20:31
  • \$\begingroup\$ Hehe, you still play the old DOS games? I bet I'd remember a few of those... I have to try and stop myself doing stuff like that cos I'd never get any work done otherwise - this place is bad enough for that :-) Agreed about the use for warning on boot - this makes sense since the sound drivers are not loaded yet. \$\endgroup\$
    – Oli Glaser
    Jul 22 '12 at 23:15
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To answer your second question: a PC motherboard is usually placed in an enclosure, and since most of those are not transparent you would see little of the LED. Also a buzzer will always signal you when you're around, a LED only when you look at it. The buzzer makes more sense.

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    \$\begingroup\$ But other things like power or drive activity are indicated by LED, and I don't see why these are different. And I think LED can be actually placed on the pc case pulled from headers on the mainboard. \$\endgroup\$
    – sawa
    Jul 22 '12 at 19:06
  • \$\begingroup\$ @sawa I (vaguely) remember reading somewhere that buzzers were put there as feedback for blind people. \$\endgroup\$
    – m.Alin
    Jul 22 '12 at 20:25
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    \$\begingroup\$ @m.Alin - Yes, for warning them that they forgot to turn their monitor on ;-) \$\endgroup\$
    – stevenvh
    Jul 22 '12 at 20:26
  • \$\begingroup\$ @stevenvh That's funny! :-) My memory might be failing me; Andreja's explanation sounds more credible. \$\endgroup\$
    – m.Alin
    Jul 22 '12 at 20:30
  • \$\begingroup\$ @m.Alin: don't say "that's funny", say "LOL", or "LMAO". This is the internets! :) \$\endgroup\$ Jul 23 '12 at 10:06
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I Think it's around 3.3 - 5.0 Volts only because at any lower voltage you wouldn't have a response and at any higher the passive buzzer will overheat.

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