Measuring the impedance of a speaker

I want to measure the impedance of a speaker (by different frequencies). The idea to test it was the following, I have the follow circuit:

Now, in order to find the impedance I measured the voltage across the speaker and over the potentiometer, if they are different then I turn the potentiometer as much that it will read the same voltage that is standing over the speaker. Then I get (using formula's):

$$\underline{\text{U}_\text{sp}}=\underline{\text{U}_\text{R}}=\underline{\text{I}_\text{sp}}\cdot\underline{\text{Z}_\text{sp}}=\underline{\text{I}_\text{R}}\cdot\underline{\text{Z}_\text{R}}=\underline{\text{I}_\text{in}}\cdot\underline{\text{Z}_\text{sp}}=\underline{\text{I}_\text{in}}\cdot\text{R}\space\Longleftrightarrow\space\underline{\text{Z}_\text{sp}}=\text{R}$$

But I think I'm measuring the absolute value of $\underline{\text{Z}_\text{sp}}$.

Question: Can I use this method? And if I can am I measuring $\left|\underline{\text{Z}_\text{sp}}\right|$?

3 Answers

Assume that you have a conventional permanent-magnet eight-ohm loudspeaker. Your measurements will be roughly close to absolute value of Zspkr. Be aware that your results will change upon mounting (or dis-mounting) your speaker in an enclosure. Here is a simplified equivalent circuit for a loudspeaker at low audio frequencies, where cone motion is uniformly in and out:

simulate this circuit – Schematic created using CircuitLab R1 is the resistance of the wire on the speaker's voice coil that reacts with the magnetic field produced by the permanent magnet. R1's resistance of about six ohms is common for a speaker considered to be " 8 ohms". You can measure this resistance with an ohmmeter.
Since your speaker operates like a motor, it can generate voltage as well as respond to applied current. The components R2, C1, L1 are not real, but they model the back-emf effect of the voice coil's motion in a magnetic field. They are also affected by the mechanical mounting compliance of the voice coil, and the mass of the moving parts. This is a simplified model, because the power transfer from electrical current to the motion of air is not shown - I'm assuming that you want to see the impedance between the speaker's two terminals.
Your loudspeaker goes through a low-frequency resonance (perhaps below 400 Hz) where C1 and L1's impedance cancel, leaving you with a measurement of R1 + R2. You might measure about 20 ohms at resonance using your method.
Above this frequency, your measurements will yield a lower value than at resonance, which dips close to eight ohms, then rises again at still higher frequencies. This model is not adequate to properly represent speaker impedance for these frequencies, deviating considerably at high frequency.
Resonance modelled by R2, C1, L1 is considerably affected by the air path/load on the speaker cone. Model value for R2 also depend on the permanent magnet's strength.
Your measurement method gives results that are very nearly correct at three frequencies where phase angle between voltage and current is near zero:

• at low frequency near 0 Hz
• at resonant frequency
• above resonance, where impedance dips to a minimum

To measure speaker impedance more accurately, you would measure voltage from one speaker terminal to the other, and you would measure speaker current, paying attention to the phase relationship between these two measurements. A two-channel oscilloscope would be a useful measurement tool.

The better way is if you have a sweep tone gen (like Audacity in log scale) then use a large series R >>100 ohms, such that current is constant and does not change much with freq, then scope voltage on speaker, Z(f)=U(f)/I for I = constant average. then record with AUX input and observe on Audacity

(menu) > Generate > Chirp and choose options shown below.

• Oke thanks, But my question was if I can use this method? Or is it wrong (measuring the resistance that the potentiometer is standing on)? Commented Jan 11, 2017 at 20:39
• its perfectly ok to match Z with R for 50%, and you can also measure phase shift V vs I Commented Jan 11, 2017 at 20:41
• Why the 50%???? Commented Jan 11, 2017 at 20:43
• equal to 1/2 of total from amp. Commented Jan 11, 2017 at 20:43
• I dont know what that means? When I set the voltage the same and measure the resistance that the potentiometer is standing on, is that then the same as the absolute value of the impedance of the speaker or not? Commented Jan 11, 2017 at 20:45

That method will work, with some limitations.

1. The resistance of the potentiometer needs to be in the same range as the expected impedance. You won't want to use a 500k potentiometer when you are looking at a speaker at around 8 Ohms. You wouldn't be able to adjust the pot accurately enough to do any good.

2. The potentiometer needs to be able to handle as much power as your amplifier will provide to the speaker. If you are driving an 8 Ohm speaker with 2 Watts, then the pot must also be capable of 2 Watts. Better to use a low power amplifier to do your test. Maybe use a large resistor in series with the amplifier output to reduce the power going to the speaker+pot so as not to burn out the pot.