3
\$\begingroup\$

I'm using an audio op amp (LME49726) to combine two audio sources together, to be listened to with headphones at its output. Whenever I plug in the headphones, there is a large voltage drop in the supply voltage and it doesn't work. What could be the problem with having a headphone load at the output? What op amp parameters would this involve? I have also tried a general purpose op amp with the same result. This is datasheet of the audio op amp: http://www.ti.com/lit/ds/symlink/lme49726.pdf Here's the simple circuit: enter image description here

\$\endgroup\$
9
  • 2
    \$\begingroup\$ Second line of the datasheet : "Easily drives 2 kilohm loads". What impedance are your headphones? If they are less than 2 kilohms ... no go. \$\endgroup\$
    – user16324
    Commented Nov 9, 2016 at 18:33
  • \$\begingroup\$ I didn't realize that meant you cannot drive a load below 2k...is there an actual parameter in the characteristics that indicates this? \$\endgroup\$
    – cheeto
    Commented Nov 9, 2016 at 18:39
  • \$\begingroup\$ What are you using for a power supply? And a schematic of your circuit would be helpful too. \$\endgroup\$
    – brhans
    Commented Nov 9, 2016 at 18:41
  • 1
    \$\begingroup\$ *is there an actual parameter * Forget the parameters, just read what this opamp is designed for. Does it say "headphone amplifier" anywhere ? I could not find it. Obviously this amplifier is not meant for driving headphones. \$\endgroup\$ Commented Nov 9, 2016 at 18:49
  • 1
    \$\begingroup\$ And next time, include the schematic from the start as then we could have immediately pointed out that you need a DC blocking capacitor at the output ! \$\endgroup\$ Commented Nov 10, 2016 at 6:58

2 Answers 2

8
\$\begingroup\$

That particular op-amp is capable of hundreds of mA output current. I think you should be able to deafen yourself if the headphones are the normal rare-earth ~30\$\Omega\$ variety.

You seem to have identified part of the problem- the supply voltage drops. You need to capacitively couple the headphones. What you have now attempts to put 2.5VDC across the headphones, which is at least 83mA, thereby dragging down your supply.

Try a 100uF electrolytic in series with the output (+ side to the op-amp, - to the headphones).

\$\endgroup\$
6
  • 1
    \$\begingroup\$ Ah that's the problem, I forgot to do that, thanks!! \$\endgroup\$
    – cheeto
    Commented Nov 9, 2016 at 19:05
  • \$\begingroup\$ @cheeto if that kind of cap is too bulky or you don't have one at hand, you can also stabilise the 2.5 V level with another OP and use it as the headphone's ground. Or implement a proper push-pull by inverting the signal once more. But that's arguably overkill for headphones. \$\endgroup\$ Commented Nov 9, 2016 at 20:45
  • \$\begingroup\$ Why does the output coupling capacitor need to be as high as 100 uF and be electrolytic? What about a 10 uF ceramic cap? \$\endgroup\$
    – cheeto
    Commented Nov 10, 2016 at 5:09
  • \$\begingroup\$ It does not need to be electrolytic, but 100uF is pretty small and maintains bass response. You can work out the -3dB frequency. 10uF ceramic would limit the low end response and probably add a lot of audible distortion, but if you can't hear any difference why not. \$\endgroup\$ Commented Nov 10, 2016 at 5:18
  • \$\begingroup\$ Would you calculate the output coupling capacitor you need for your desired frequency response through the 1/(2*pi*RC) formula, with R being the headphone impedance? \$\endgroup\$
    – cheeto
    Commented Dec 21, 2016 at 21:53
1
\$\begingroup\$

If your power supply voltage is dropping then it indicates that the power supply is not capable of supplying the current to the load.
Many headphones are only 8 ohms. Suppose you had an output signal level into the headphones of 1V you would need to supply up to 1V/8 ohms = 125mA. If your signal level is higher you will need correspondingly more current. For example at 4V you would need 0.5A.

The LME49726 is capable of driving an 8 ohm load at signal levels up to 2.8V with a 5V supply and is easily capable of driving typical headphones.

The parameters that would typically be considered when driving headphones from an op-amp would be...

1) Voltage headroom from each supply vs load current (also referred to output swing).

2) Maximum load current.

3) Maximum allowed wattage based on ambient temperature, junction temperature, and thermal resistance (theta-JA). The power dissipation is approximately...
W = Is * VCC + |Iout| * (VCC-Vout) when sourcing current out of the output pin.
W = Is * VCC + |Iout| * Vout when sinking current into the output pin.

4) Also with respect to audio outptut quality you would consider CMRR, input offset voltage, gain bandwidth product, and linearity.

\$\endgroup\$
8
  • 1
    \$\begingroup\$ Most headphones are 30 ohms, speakers mostly 8 ohms. \$\endgroup\$ Commented Nov 9, 2016 at 18:49
  • \$\begingroup\$ If this op amp should be cable of driving headphones, I wonder why I'm having this problem. I attached a drawing of the op amp configuration... \$\endgroup\$
    – cheeto
    Commented Nov 9, 2016 at 18:53
  • \$\begingroup\$ I am very curious where you found that the LME49726 can drive 8 ohms because I don't see that anywhere in the datasheet. The lowest impedance they mention is 600 ohms. Perhaps I am overlooking some information ? \$\endgroup\$ Commented Nov 9, 2016 at 18:53
  • \$\begingroup\$ @FakeMoustache It says on page 4 of the datasheet that the output swing is VDD-0.33V with a 16 ohm load. It also says that it can typically drive 350mA. So as long as you keep the output swing far enough from the rails it can drive the load. \$\endgroup\$
    – user4574
    Commented Nov 9, 2016 at 19:05
  • \$\begingroup\$ OK I see that, but since the distortion specs. are only for a 600 ohm load I have my doubts if this opamp has proper performance (low enough distortion) to actually drive 16 ohms. I'd still choose a headphone amp to drive a headphone. This opamp is still a line driver. \$\endgroup\$ Commented Nov 9, 2016 at 20:11

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.