Hi
I need to measure my speakers impedance. Is there a simple method?
I am thinking about sending a tone with 1 volt amplitude and mesure current. I could then calculate impedance for that frequency with Ohm Law. V=Z*I Then, i would repeat for many different frequencies.
Will that work? I have read method about constant voltage or constant current. But the method above is so simple, why do it differently.
Thanks
F
I need to measure my speakers impedance. Is there a simple method?
I am thinking about sending a tone with 1 volt amplitude and mesure current. I could then calculate impedance for that frequency with Ohm Law. V=Z*I Then, i would repeat for many different frequencies.
Will that work? I have read method about constant voltage or constant current. But the method above is so simple, why do it differently.
Thanks
F
The method you propose will work but it has two shortcomings- measuring the current, which can be problematic above a few milliamps, and doing all the calculating. Using a resistor network is far easier as you basically can just run a sinewave tone into the speaker at whatever frequency you wish and instantly read on your DMM the impedance.
I do all calculating with excel.
I tried yesterday with a 1K resistor in serie with the driver and reading the voltage drop in the driver. I had a few problems:
1- The voltage drop is small in the driver because much of the voltage drop is in the 1K resistor.
2- I had problems with unstable values. At first, the value was large and then, the level dropped slowly during 1 minute before setting to a certain value. Ex: at 1000Hz, At first, the voltage drop across driver was equivalent to 13 Ohm, slowly going down to set at 9.5 Ohm for example. The value of 9.5 seemed plausible but I am worry about the drop.
3- It is possible to calibrate the system using a 10 Ohm resistor, and adjusting gain to read a 0.100 volt drop across driver. But if the sound card has irregular FR, then the calibration is screwed?
F
I tried yesterday with a 1K resistor in serie with the driver and reading the voltage drop in the driver. I had a few problems:
1- The voltage drop is small in the driver because much of the voltage drop is in the 1K resistor.
2- I had problems with unstable values. At first, the value was large and then, the level dropped slowly during 1 minute before setting to a certain value. Ex: at 1000Hz, At first, the voltage drop across driver was equivalent to 13 Ohm, slowly going down to set at 9.5 Ohm for example. The value of 9.5 seemed plausible but I am worry about the drop.
3- It is possible to calibrate the system using a 10 Ohm resistor, and adjusting gain to read a 0.100 volt drop across driver. But if the sound card has irregular FR, then the calibration is screwed?
F
I use a 100 ohm series resistor and a 10 ohm calibrating resistor and calibrate to a 1 volt input. The downside here is that the higher the impedance the greater the deviation from a true reading, but a 100 ohm resistor is a lot more load friendly for the average amp, and high accuracy above 20 ohms matters little, it's where the impedance peaks are that count for cabinet tuning.
I'd never use a soundcard. I use a tone generator going through a 100 watt amp. I suspect your instability problems are related to using a soundcard rather than a power amplifier.
I'd never use a soundcard. I use a tone generator going through a 100 watt amp. I suspect your instability problems are related to using a soundcard rather than a power amplifier.
Hi,
my first measurements were with a 100 Ohms resistor and a 4 Ohm woofer (3.9 Ohms DC).
Using a simple 50W amplifier , sine-wave generator, frequency meter, and a multi-meter.
Then a table in Excel, with 2 voltage measurements for each frequency: 1 for the voltage over all (ouput of the amplifier) and 1 over the 100 Ohms resistor.
I measured 100 points from 20 Hz to 2 kHz (for a midwoofer).
Be sure to have a FULL battery in the multi-meter!
I thought my meter wasn't precise enough (deviation because of a high frequency), and that's the reason I also measured the voltage overall (to compensate that deviation).
A few month later I bought a "real" measurement-system, DLSA Pro, and when measuring the same woofer again, all measurements turned out to be within 3%, smaller than most capacitors and inductors.
So, when it's only for a few units, this simple (time consuming) approach, is surely good enough.
Grtz, Joris
my first measurements were with a 100 Ohms resistor and a 4 Ohm woofer (3.9 Ohms DC).
Using a simple 50W amplifier , sine-wave generator, frequency meter, and a multi-meter.
Then a table in Excel, with 2 voltage measurements for each frequency: 1 for the voltage over all (ouput of the amplifier) and 1 over the 100 Ohms resistor.
I measured 100 points from 20 Hz to 2 kHz (for a midwoofer).
Be sure to have a FULL battery in the multi-meter!
I thought my meter wasn't precise enough (deviation because of a high frequency), and that's the reason I also measured the voltage overall (to compensate that deviation).
A few month later I bought a "real" measurement-system, DLSA Pro, and when measuring the same woofer again, all measurements turned out to be within 3%, smaller than most capacitors and inductors.
So, when it's only for a few units, this simple (time consuming) approach, is surely good enough.
Grtz, Joris
Rather you than me. 
If that's what it takes to do it manually, then I would certainly build the jig if you intend to do it more than once. It's nice to build DIY tools to aid building your DIY speakers
The jig and SW combo can produce insanely accurate results:
http://www.vikash.info/audio/sw_jig/measurements.asp
gary f said:
The method you descibe is constant voltage, though your meters
AC current measurement capabilities come into question.
Constant current is just as easy with a ~ 1Kohm series resistor,
here your your meters AC voltage readings need to be up to scratch.
Note that for basic simulation only two readings are required, DC
resistance and impedance at say 1KHz to work out inductance.
To an extent effective acoustic phase angle is the real issue,
and this is worth a full frequency sweep analysis of a driver,
but note this is acoustic not driver terminal voltage / phase.
sreten.Vigier said:
If you still insist on doing it manually I would make 2 suggestions (which you may already be doing). Firstly, choose a resistor that is on the same order of magnitude as you expect from your load. So since you speaker is going to go from 4-ohms to 40-ohms, maybe a 20-ohm power resistor might work slightly better. 2ndly, when you choose your 100 frequencies, you might want to consider dividing them evenly along a logarithmic or exponential kind of scale. So your 20Hz - 20KHz range is either 10 octaves OR 3 decades. Split up your 100 points into sa 10 points per octave or 30 pts per decade.
@Vikash: well, when you have a rainy sunday, with nothing else to do....
@azira: the 100 ohms resistor was the only resistor that was more than 0,25W and of higher resistance than the unit itself. And to Excel it doesn't matter. The measured frequencies where indeed (rounded) logarithmic, to get a good curve (linear steps is pretty useless)
Fortunately, DLSA Pro does this now for me
For the people that are too lazy to measure a few hundred points, I really advise you to build some of the jigs that are on the internet and download Speaker Workshop. Please do one of both (hand-measure or jig), because you can make designs SO much better!
Grtz, joris
@azira: the 100 ohms resistor was the only resistor that was more than 0,25W and of higher resistance than the unit itself. And to Excel it doesn't matter. The measured frequencies where indeed (rounded) logarithmic, to get a good curve (linear steps is pretty useless)
Fortunately, DLSA Pro does this now for me
For the people that are too lazy
to measure a few hundred points, I really advise you to build some of the jigs that are on the internet and download Speaker Workshop. Please do one of both (hand-measure or jig), because you can make designs SO much better!Grtz, joris
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