Simple Headphone System

My friend wants to connect multiple headphones to a power amp in the easiest and safest way possible. There are many ways to create a headphone system, depending on needs, budget and reliability. This is ONE of many ways.

My friend is using Sennheiser HD-280 cans, which are rated @ 1/2 watt and 64 ohms. I am not sure if 500 milliwatts (mW) is the max power or the nominal (expected safe) power. I did my calculations based on a 100 watt per channel amplifier, which is overkill but what is available (translation: laying around).

The power formula is: WATTS = VOLTS (squared) / OHMS

0.5 = volts (squared) / 64 (we’re solving for volts)

64 * 0.5 = 32, the square root of which = 5.65-volts RMS (Vrms), the amount of voltage required to drive the cans at their rated wattage.

If the power amp is rated for 100 watts into 8-ohms, then 100w = voltage (squared) / 8-ohms = 100 * 8 = volts squared = 800, the square root of which is 28-volts RMS. We now know the amp’s voltage output at rated power and load along with the voltage required for the headphones to dissipate 500mW. The ratio of these two can be converted into the amount of attenuation (in dB) as well as helps us ballpark the series resistor required to safely drive the headphones.

The formula to convert a voltage ratio to the more familiar decibels is…

dB = 20 log (new volts / reference volts)

The resistive ratio will be 5.6 / 28 = .2 (the log of which is -0.69897000433601880478626110527551) times 20 = -13.9 dB (it’s good to know the attenuation in a familiar quantity).

That same 0.2 ratio, in resistive terms = 64 / 64 + ‘X’ (‘X’ is the series resistor we want to find).

We need to isolate and solve for X. To do so we need to get it out of the denominator by swapping 0.2 with ’64 + x’ so that 64 + ‘x’ = 64 / 0.2 = 320 -64 = ‘x’ = 256-ohms.

We then need to calculate the current through the resistor in series with the headphones, which is the same current through the headphone voice coil. 28-volts from the amp will be across 256-ohms + 64-ohms, that’s 28-Vrms / 320 = .0875-amps. Power dissipated by the 256-ohm resistor that’s in series with the headphones = current (squared) * ohms = .0875Amps (squared) * 256-ohms = 1.96 watts. This resistor needs ‘headroom’ so it will provide reliable service and have an adequate safety margin (a.k.a. no fire). (You may not find these exact values but anything close will work, for example 250 ohms @ 5 watts.

Just to confirm that I made no mistakes, let’s calculate the power dissipated by the headphones = .0875amp (squared) * 64 = .49 watts, which is close enough to 1/2 watt.

My resistor values should get us in the ballpark, tho you could figure out your amp’s power and work backwards as I did to know the range (and for the sheer enjoyment of doing the exercise). To test using the amplifier in this example, have your DAW generate 1kHz @ 0dB full scale, send that to your power amp and adjust the gain so that a clean 28-volts appears at the speaker terminals.

My friend was also concerned about whether the amp’s outputs were balanced or ‘bridging,’ reading on a message board that a bridging amp would not like each channel’s ‘low side’ tied to the sleeve of a quarter inch jack. To test, connect a speaker to the amp, play some music at a low-to-moderate level, lift the PLUS wire from its terminal and connect it to a screw on the chassis. If no signal, you’re good to go. When doing the same to the black wire, the signal level should not change – this assumes the MINUS or ‘low’ side of the amp’s output terminals is (indirectly) connected to chassis ground.

Last point is to buy some 256-ohm 3 to 5 watt resistors – whatever is common – then prototype, test and adjust to taste. I would also get some 64-ohm 1/2 watt resistors wired to the 1/4-inch jack normals so that, when no phones are plugged in, the amp sees the same load. This way, plugging and unplugging cans doesn’t change anyone’s level.

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