Requirements: I would like to design a circuit that uses an in-amp, and it needs dual supply (+/- 3V at least). I want it to use two lipo cells so it can be recharged. The output from the circuit is a headphone (earphone). I can have 32 ohm headphones, but that is the max. (There are 600 ohm phones out there but they are too expensive for this project.) This is an educational project for myself, and I would like to build the output stage using discrete components, or maybe simple opamps. My problem is with the output stage. The 32 ohm speaker at 3.7V would have about 100mA peak current, which is too much for a headphone. I could add a 330 ohm resistor in series, or lower the gain of the amp, but then most of the output power becomes heat - not a very good design for a battery operated device. Using a class D amp would probably solve the problem, but I don't think I capable of designing a class d amp. How others do this?

  • \$\begingroup\$ There are single chip D-Class amplifiers available. From what I have seen of them, they are simple to use. I've never built one, though, so I can't tell if they are as simple to use as the datasheets make it appear. The ones I've seen were about the size of an opamp. \$\endgroup\$ – JRE Aug 6 '17 at 11:15
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    \$\begingroup\$ Why not use 2 9V battery with 4556 to learn first? \$\endgroup\$ – Jason Han Aug 6 '17 at 15:12
  • \$\begingroup\$ Actually, NJM4556 can be driven from +-2V, and it is more common and much cheaper than LM4880. 70mA into 150Ohm load is not bad at all! Thanks for the tip! \$\endgroup\$ – nagylzs Aug 6 '17 at 16:25

Let's consider the headphones.

  • Earbuds. No need for a dedicated amp since earbuds audio quality is terrible anyway.
  • In-ear rubber-nipple thingies. I picked a random one off the net: Sennheiser CX300 is 18 ohms and delivers a whooping 118dB for 1V RMS.
  • Real headphones. OK, let's pick some not too expensive ones... HD4.20 again 18 ohms and 118dB for 1V RMS.
  • Some cheaper ones HD2.10 26 ohms and 110dB/1VRMS

Sooo... if you use the latter ones, listen to an average level of 80dB which is already loud, with peaks around 100dB (if your music is not too compressed), then:

  • RMS voltage will be -30dBV, ie 31mV RMS
  • Peak voltage around -10...-3 dBV (like half a volt)
  • Peak current won't be huge either, like 30mA.

Now, let's look for an opamp.

The "maximum output current" specification is useless here. What we want is an opamp that can output the required current with the least amount of performance degradation. An opamp picked at random may have good distortion figures, but when driving relatively high currents (for an opamp) with low headroom, it's an entirely different story.

For example, TL072 isn't so bad with a 10kOhm load, but if asked to drive any kind of actual current, its performance simply ceases to exist.

The 32 ohm speaker at 3.7V would have about 100mA peak current, which is too much for a headphone.

You will go deaf before the voice coil burns.

The resistor which everyone puts at the output of headphone amps actually degrades performance, but it has a rather important purpose: protect the output stage when the output is shorted.

...and every time you insert a 3.5mm jack into the connector, the output will short during insertion. This is a "feature" of this connector, you gotta live with it. If the connector is mis-inserted, the short can be permanent. You must select an opamp that will survive this, so you can get rid of the resistor for better sound.

(Explanation: headphone impedance varies with frequency like every loudspeaker. With a resistor in series you got a voltage divider. Voltage on headphone thus depends on its impedance at the frequency of interest. So you get bumps and dips in the frequency response).

You mention a discrete amp. With such a low supply voltage, discrete design is very complicated. An opamp would offer much better performance.

For example,

OPA1688 is a nice opamp with high output current and rail to rail output; it is specified to drive headphones. Beware the input is not really rail to rail, it should not go above Vcc-2V, but this should be OK.

If you wanna go real fancy, you can use a nested feedback with OPA1652 as frontent, and a beefy fast opamp as the output driver.

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  • \$\begingroup\$ Well I forgot to mention that this is not for listening to music, and sound quality is of less importance. But your comments about the series resistor got me thinking. Is it true that, if I'm willing to burn most of the power on a huge series resistor, then I can make the frequency response close to ideal? E.g. put a 10k resistor in series with the phone, so the frequency response becomes much more flat. (I know this is not closely related to the question.) \$\endgroup\$ – nagylzs Aug 6 '17 at 13:57
  • \$\begingroup\$ Higher resistor will make the amplifier look more like a current source. Since transducers are optimized for a voltage source amplifier, higher output impedance will give you worse frequency response... \$\endgroup\$ – peufeu Aug 6 '17 at 14:18
  • \$\begingroup\$ Very interesting! Thanks, it makes sense this way. Is there a graph where I can see the SPL vs voltage and SPL vs current, and compare them (for an "average" transducer). I guess the graph would not be a line, but I never understood the exact correlation. \$\endgroup\$ – nagylzs Aug 6 '17 at 16:41
  • \$\begingroup\$ If you have frequency response and impedance data, you can take the transducer frequency response, which was taken at constant RMS voltage, and multiply this by its impedance at each frequency... you'll get the response in constant RMS current mode. \$\endgroup\$ – peufeu Aug 6 '17 at 18:47

With a combined rail of 6 volts and (say) 1 volt dropped in the transistor driver stage when producing a high or a low output, the maximum voltage across your load will be about 4 Vp-p. For a sinewave this is an RMS voltage of 1.414 volts and the power into a 32 ohm load will be 62.5 mW.

If you think this is too high you can use a step down transformer to efficiently drop the output voltage. From looking at headphone specs, about 20 mW is around typical max power so I would use a series 33 ohm resistor to approximately quarter the power into the headphones. Power wastage is significant as a percentage but how significant compared to the watt hours deliverable by the batteries?

Also worth factoring in is the type of sound you are listening to. For instance, if music, then although the peak power may appear high, the average power will be ten times lower or more. A peak of 15 mW can easily be an average power of less than 2 mW. Put it into context is my advice.

If I were building this I would use a single battery and an LM4880 amplifier like this: -

enter image description here

It'll work down to 2.7 volts.

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  • \$\begingroup\$ Yes, 62mW is way too much for a headphone. Dual power supply was a requirement, bu I can use just one half of it for the outut stage. \$\endgroup\$ – nagylzs Aug 6 '17 at 14:00

How others do this?

by first trying to figure out what your goals are. For example, are you trying to design a balanced-in/balanced-out headphone driver for a low or high sensitivity phone....

the best (ac/sound) quality would come from CFB opamps. you will find them cheap in used modems - they were widely used as xdsl drivers. no amp, diy or otherwise, will come even close to those little beasts.

as to output attenuation, I would put a 110R resistor on the output to protect for shorts.

if you need additional attenuation, put it on the input. unless you have some low-sensitivity / high-impedance phones (very rare), those amps should just be followers.

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