Measuring speaker driver with high impedance source

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Usually I measure driver impedance just by sweeping with my function gen. It has 50ohm output impedance and to make the measurement easier I add 50ohms in series, so a total of 100ohms. I use a PC scope and with the peak hold on the FFT I get decent sweeps.

Question is: When having the high series resistance, and thus very low damping factor on the speaker driver, doesn't this shift fs up/down, make the Q appear larger than it is, etc?
Is it more accurate using a power amplifier with a very small value series resistor and then measure both the voltage and current?
Does it matter? Will either method be accurate enough?
 

PRR

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> doesn't this shift fs up/down

You *have* the test rig right there. You can apply 50 Ohms, 100 Ohms, or 25 Ohms with trivial re-connection. fs should not shift. Q comes out in the math.

> measure both the voltage and current

It's Ohms Law either way.
 
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If the series resistor is large enough ( 500R to 1k ), then Vdriver x ( Rseries / Vgen )
is essentially equal to the driver impedance.

If Rseries = 1k and Vgen = 1V, then numerically the driver impedance in Ohms = Vdriver x 1000.
 
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The peak of resonance is affected by Q somewhat - at low Q it will shift a bit (as well as being harder to measure anyway). So low damping should be at an advantage for accurate determination of mechanical resonant frequency perhaps?


Anyway if you measure V, I and SPL over a slow frequency sweep you'll get the same information whatever the driving impedance.
 
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