I'm thinking about the example of two output jacks on the back of a computer, one for line out and one for headphones. I understand that the line out in generally an "unamplified" signal and is intended to be fed to an amplifier with a high input impedance, and the headphone output is amplified by the computer sound card and fed to a lower impedance speaker.

But since what we want is the max voltage transfer to the "load" (the amp or the headphone speaker), then wouldn't it make sense for both to be low impedance outputs? Because the lower the impedance on the output jack the larger the relative voltage will be on whatever is on the other end of the signal. But in reality the line-out impedance is usually somewhat higher than a headphone output. Is there some reason for this design?


2 Answers 2


Both are low impedance outputs, with some differences:

For a line out, output current is negligible so it is not a selection criteria for the opamp driving it. In addition, a low value series resistor (50...100 ohms) may be added to the output as a cheap way to protect the opamp against short circuits.

The output impedance of the source and the input impedance of the receiver form a voltage divider. As long as source impedance is negligible relative to receiver input impedance (for example 100 ohms vs 100k), source impedance pretty much doesn't matter in the context of audio. Maybe you lose 0.1dB of gain, not a problem.

The headphone output needs an output amplifier with a lot more output current capability. For example, with 32 ohms headphones, even a 2 Volts output means 62mA current, which exceeds the capability of most opamps. Loudspeakers and headphones have frequency-dependent impedance, so the voltage divider with the driving source impedance now becomes a problem as it becomes frequency dependent, which will change the frequency response. So a series resistor cannot be added to protect against shorts. This means you need an opamp with beefy output current and full short circuit protection. One of the interesting properties of audio jacks is that the tip and rings of the plug rub against the contact springs of the female jack during insertion and extraction, so it is best to assume anything can short with everything.

When the manufacturer cheaps out on the opamp's current capability, you get an output jack that sounds fine when used as a line out, but sounds terrible as headphone jack.

  • \$\begingroup\$ Wouldn't 2 volts into a 32 ohm headset be a little excessive? That's 0.125 watts. Many headphones can produce 90 to 100 dB of SPL at measly 0.001 watts, and not all can even handle 0.125W without burning. \$\endgroup\$
    – Justme
    Nov 16, 2023 at 0:11
  • \$\begingroup\$ At 1kHz, but if you want to squeeze some bass out of earbuds... \$\endgroup\$
    – bobflux
    Nov 16, 2023 at 0:41

The point is you don't need as low impedance on line out when it is driving a 10kohm input, while the headphone output needs low impedance as it must drive 16 ohm headphones.

Also the difference is in voltages. A modern line out could be as high as 2 VRMS or 5.6Vpp. That is only 0.2mA RMS or 0.56Vpp. So it can work with maybe +/- 5V or less and takes very little current. Headphones could be 16 ohms and can provide ear-deafening SPL at 1 milliwatts, which only requires about 130 mV RMS at 8mA RMS. So voltage is much lower and current is much higher. Transistors intended to drive that headphones are much larger so if you don't need them to drive line outputs, you save silicon area if you only have headphone driver on one output. In reality you don't use such high volume so the voltages and currents are lower.

Then there is the problem of coupling. If we assume that a sound chip only uses single 5V supply voltage, it means that to provide AC out, the IC must internally use 2.5V mid-voltage bias and remove it so it does not destroy the headphones or the amplifier. To use say 20 Hz cut-off, the 10kohm load of line input only needs only a small capacitor in the order of 1 uF. The 16 ohm headphone requires a big hefty 470uF capacitor, but depending on how well designed it is, cheaper designs have smaller capacitor and better designs have larger.

But in practice these values are so large there is no room for such large caps and smaller are used, or some buffer circuitry, external or internal to the audio IC, to make some bipolar audio supplies with capacitive charge pumps, or use positive and negative voltages from the supply directly, so that 1uF can be used to remove DC and then audio is buffered with an amplifier circuit. Unless of course Class D driver is used for headphones.

Since the line putput has higher voltage levels, it is easier to keep noise small enough to be OK. The headphone voltage levels are so low you have to be careful with the driver circuits that you keep the noise level down an don't amplify the noise too much.

So the point is, the line output is cheap and unproblematic, while driving headphones isn't, so you don't really want to use a second headphone driving circuitry just for driving line outputs.

If line output has 1kohm output impedance driving a 10kohm load, there's about 90% voltage being bridged. In practice more, as line output impedances are lower and line inputs higher. Maybe 500 ohms and 20 kohms, so beyond 97% anyway. So it does not need to be 99.99%.

Which is why the headset output must have output impedance somewhere below 1 ohms, as 1 ohm output driving a 16 ohm load is only 94% voltage bridging.


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