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I am using headphones with sensitivity 102 dB SPL / 1 Vrms with impedance of 300 ohms. For a particular volume knob setting on my sound card embedded with a headphone pre-amplifier, I would like to determine the output in dB SPL from the headphones.

I played out a calibration tone of 1 kHz of peak amplitude 1 V (RMS 0.707 V). Using a multimeter, I measured the Open Voltage (Vo) from my sound card for different values on the volume knob. I then measured Voltage with the load (VL) connecting headphones to the soundcard for the same values on the volume knob.

I would like to determine the output impedance of the sound card (pre-amplifier) along with the Voltage input in order to get the power of the amplifier in driving the headphones. I am stuck here, can anyone help? (I can use the power to determine the value in db SPL for a particular volume knob value.)

Open Voltage (Vopen) Voltage across load (VL) Source Resistance (Rs)
0.0795 0.068 50.73
0.1 0.085 52.94
SamGibson
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Tanmayee Pathre
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  • Many multimeters can accurately measure 50Hz or 60Hz but not 1000Hz. A volume control is usually logarithmic to match the logarithmic sensitivity of our hearing. Then the control turned down to half produces about 1/10th the level of maximum. But maximum might be clipping so you cannot assume the level when it is turned down. – Audioguru Aug 03 '21 at 17:53
  • But then how do I go around with the measured values to determine the sound pressure in dB SPL? – Tanmayee Pathre Aug 04 '21 at 12:51

2 Answers2

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Since you already measured how much Vrms is sent to phones when they are connected, you can just convert that Vrms to decibels by using the 1Vrms as reference, and since you know the reference level of 102 dB SPL at 1 Vrms, just add the reference decibel level to your calculated decibels.

E.g if you measure 1 Vrms going to phones, you know that equals 102 dB SPL. 2 Vrms is double so +6 dB to reference. 0.5 Vrms is -6 dB to reference.

You don't need the phone impedace, output impedance or open voltage levels for this.

Justme
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  • Thanks for your answer. So when I measure 1 Vrms going through my headphones that equals 102 dB. My sound card (headphone amp) also mentions that Voltage gain = +6dB (optimised for loudness). Does this mean when I measure voltage or rather when the calibration tone (1 Khz) is played out, it has a gain of +6 dB. – Tanmayee Pathre Aug 03 '21 at 14:36
  • If you measure 1 Vrms out then it's 1 Vrms out what you measure. And 1 Vrms is 102 dB, regardless of any internal gain or volume slider that happens before you measure 1 Vrms. – Justme Aug 03 '21 at 17:12
  • This means at 0.5 Vrms, it should be 96 dB. However, when I hear the audio or tone. It doesn't sound as if its 96 dB SPL. It is at comfortable normal conversation level. Do I need to subtract the value in dB of the attenuator too? – Tanmayee Pathre Aug 03 '21 at 20:09
  • What attenuator do you mean? – Justme Aug 03 '21 at 21:12
  • Attenuator meaning the knob which is used to control the volume. I have measured voltages at different values on the knob. – Tanmayee Pathre Aug 04 '21 at 12:40
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The output impedance of your sound card headphone should be close to zero but you can check.

  • You have the open circuit voltage, \$ V_O \$.
  • You have the voltage into 300 Ω, \$ V_{300} \$.
  • The output impedance and the speakers form a potential divider.

$$ V_{300} = \frac {300}{R_S + 300} V_O $$

Therefore the source resistance will be given by

$$ R_S = \frac {300}{V_{300}} V_O - 300 $$

I'm not sure how much value this will be as the impedance will be frequency dependent. You could try different values such as 100 Hz, 1 kHz and 10 kHz to see. Remember that most multimeters are designed for power frequencies and won't be much use above 100 Hz.

Transistor
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