-1
\$\begingroup\$

I heard somewhere that computer soundcard can't output high current. That is why computer speaker has amplifier inside them and require some external power to power up the amplifier.

But what about headphone? I have a headphone that is around 8 ohm impedance (tested it with my multimeter). When i plug it in my computer, it work, why? (I mean plug it directly into my soundcard) I mean, these are normal headphone, with no external power and amplification.

Can i break my computer by doing so?

Onboard soundcard from my Asrock Fatal1ty professionnal Z77.

Im trying to understand how speaker/amplificator work in order to make some sound using an microcontroller.

\$\endgroup\$

closed as off-topic by Leon Heller, Samuel, Scott Seidman, Nick Alexeev, Dave Tweed Jan 11 '14 at 22:22

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Questions on the use of electronic devices are off-topic as this site is intended specifically for questions on electronics design." – Leon Heller, Samuel, Scott Seidman, Nick Alexeev, Dave Tweed
If this question can be reworded to fit the rules in the help center, please edit the question.

3
\$\begingroup\$

Typical full level line out voltage from a sound card is about 1V RMS and this can drive an 8-ohm load, delivering a power of: -

\$\dfrac{V^2}{R} = \dfrac{1}{8} = 0.125 watts\$.

Headphones are generally accepted to need between 10 mW and 2 W (wiki source) and the logarithmic average of that is 0.141 W so that seems to stack up.

make some sound using an microcontroller

To make a sound that is a reproduction of music (suitable for headphones) requires the ability to deliver about the same power undistorted. 1V RMS has a peak-to-peak level of 2.828 volts and this is not an unreasonable level to expect from a digital pin that was pulse width modulated but, the current into 8 ohms would peak at: -

\$\dfrac{1 volt}{8 ohms} = 176mA\$

This means you need an amplifier that bridges between your microprocessor and your headphone. A typical MCU pin isn't going to want to supply more than a few mA without the voltage drooping and this will start to clip the music (causing distortion).

I guess, anyone interested in doing this would possibly look at using something like an LM386: -

enter image description here

There are plenty of other options including quite a few that can run directly from a 3.3V supply.

\$\endgroup\$
  • \$\begingroup\$ should i put a capacitor in serie with the input? If yes, how can i know which capacitance do i need? Also 1/8 != 176mA Never mind, found out... Converted that 1 volt to 1.41 volt (peak to peak) \$\endgroup\$ – user1115057 Jan 11 '14 at 18:41
  • \$\begingroup\$ It might be a good idea just in case so to speak. Probably 10uF will do the job and if you lose a little bass then try 100. \$\endgroup\$ – Andy aka Jan 11 '14 at 18:47
  • \$\begingroup\$ What is that amplification circuit? (Not the chip, but the name of the circuit? Buffer, current amplificator, amplificator clas XYZ?) And from which voltage should i drive this? (Output from a 5V mcu or 3.3V) \$\endgroup\$ – user1115057 Jan 11 '14 at 18:53
  • \$\begingroup\$ check the data sheet for best answer but I believe 4v is minimum. It's a power amp. \$\endgroup\$ – Andy aka Jan 11 '14 at 19:14
  • \$\begingroup\$ Just checked, the maximum is 0.4V for the input, so i can't drive it from an mcu \$\endgroup\$ – user1115057 Jan 11 '14 at 19:18
4
\$\begingroup\$

It all depends on what you mean by "high power". Your soundcard can drive 8\$\Omega\$ headphones but the power level is so low that you have to hold them right next to your ears to hear the sound. If you want everyone in an auditorium to hear the sound (and hear it well) you need an external amplifier that can deliver more power.

No, you won't break your computer by plugging in the headphones.

\$\endgroup\$
  • \$\begingroup\$ @WayfaringStranger Point? Read second paragraph of cited article? \$\endgroup\$ – Joe Hass Jan 11 '14 at 15:09

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