Phon.Audio
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Headphone

Drive capability

Both voltage and current must clear what your peaks demand.

Simple

Can the headphone amp supply both the "push" (voltage) and the "flow" (current) your headphones need to hit your target volume on peaks? Different headphones run into different walls: some need lots of voltage, some need lots of current.

The theory

Power is voltage × current, but headphones don't draw them in the same proportion. The engine computes the power needed for your peak SPL from the headphone's sensitivity, then derives the voltage (V = √(P·Z)) and current (I = V/Z) that power demands, and checks both against the amp's max voltage and max current. Passing requires clearing both limits — failing either means underdriven peaks.

For experts

Voltage and current limits on the V-I planeAn amplifier's voltage and current limits form a rectangle on the voltage-current plane. A 300 ohm dynamic headphone's steep load line exits through the voltage wall, while a 32 ohm planar's shallow load line exits through the current wall.04080120current (mA)0 V2 V4 V6 VV max 3 Vrms — voltage wallI max 70 mA — current wallthe amp's clean operating region300 Ω dynamic (steep)needs 5.4 V @ 18 mA — starved of voltage32 Ω planar (shallow)needs 2.8 V @ 88 mA— starved of currentBoth points are ~250 mW — the same milliwatts, two different walls

This two-bottleneck model is what single "power" specs miss. High-impedance dynamics (300 Ω Beyerdynamics, 600 Ω vintage) are voltage-hungry: at 300 Ω you might need 5–7 Vrms but trivial current, so they starve on phone outputs and battery dongles that top out around 2 V. Low-impedance planars (Audeze, HiFiMan at 32 Ω and low sensitivity) flip it: modest voltage but serious current, tens to hundreds of mA, which is why they sound limp on otherwise-fine amps that can't source current. The cruel case is a low-impedance, low-sensitivity planar that's both demanding — and it's exactly the combination a one-number "milliwatt" rating hides, because the same mW can be reached by very different V/I pairs. Checking voltage and current separately is the only honest way to predict whether an amp actually drives a given headphone.

V=PZ,  I=PZ

The same milliwatts can be a voltage problem or a current problem — an amp must clear both.

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