0
\$\begingroup\$

The maximum output voltage of my audio source is 3 Vpp and the maximum output current of my amplifier (op-amp AD4075-2) is supposed to be 45 mA.

What would be the ideal impedance and sensitivity of headphones under these conditions? Is it correct to use Ohm's law to calculate the ideal impedance as 66 Ω (3/66=0.045)? Or should this be rather be calculated in Vrms? In that case it would be 23 Ω (1.06/23.5=0.045).

Also, how can I calculate how sensitive the headphones should be (dB/mW), to reach 116 dBSPL at max. current (45 mA) and voltage (3 Vpp)?

\$\endgroup\$
2
  • 1
    \$\begingroup\$ 116 dB(SPL) - painfully loud: permanent hearing damage in a few minutes. why do you want this? \$\endgroup\$
    – Jasen Слава Україні
    Commented Mar 26 at 3:19
  • \$\begingroup\$ There are several online calculators for headphone power vs voltage, current, sensitivity. This one looks pretty good: gadgetmates.com/headphone-power-calculator. \$\endgroup\$
    – td127
    Commented Mar 26 at 4:18

1 Answer 1

Reset to default
3
\$\begingroup\$

Loudspeakers are fairly ohmic, and the current limit is probably a time domain limit so, no don't use RMS figure.

bridge tied load isn't going to work with ordinary headphones so I'm assuming bipolar drive through a capacitor. meaning 3V pk-pk but 1.5V amplitude

45mA at 1.5V is about 33 ohms.

32 ohms is a common headphones impedance, so add 1.5 ohm in series. or just turn the volume down a bit, that is reduce the amplitude sufficient to not exceed 45mA.

Sensitivity, assuming 32 ohm headphones, you have about 65mW of power, so work backwards from there,

65 is 18 decibels, so to get 116dB(SPL) at 65mW input you need an sensitivity of 116-18 = 98 db/mW (which means 98dB(SPL) at 1mW)

These numbers seem reaslistic in the sub-$10 range.

\$\endgroup\$
2
  • \$\begingroup\$ If I enter the data in a headphone calc, 33ohm on 98db/mW at 116dB/mW require 63.10mW and 43.7mArms indeed, as you indicate. But the calc refers to 1.44Vrms (4Vpp) instead of 3Vpp. Why is that? I interpret that with "1.5V amplitude" you meant 1.5Vpeak above. ref.: headphonesty.com/headphone-power-calculator \$\endgroup\$
    – Domingo
    Commented Mar 26 at 7:00
  • \$\begingroup\$ oops, that's for square wave. other waveforms will have less power. \$\endgroup\$
    – Jasen Слава Україні
    Commented Mar 26 at 12:04

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.