Impedance matching…can we Just add resistors? - diyAudio
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Old 19th July 2008, 03:41 PM   #1
aim is offline aim  Wales
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Default Impedance matching…can we Just add resistors?

This might sound like a stupid question and I’ve a feeling it is, but I’m going to ask it anyhow,

Firstly my assumption is that a speaker driver is much like a resistor. Maybe it has other affects such as adjusting the phase and what not…feel free to elaborate.

So, if speakers need to be matched to the amplifiers preferred impedance for optimal output efficiency and parallel wiring of drivers uses the following equation to determine impedance: -

(R1 * R2) / (R1 + R2)

Then could we not simply place resistors (instead of additional drivers) to modify the impedance?

So for example if I have a speaker with 4 drivers, each with an impedance of 8 ohms, I could wire them in series to make a combined impedance of 32 ohms. Of course this isn’t very suitable for most amps as this is quite a load and power output would be seriously affected. Preferred impedance would be 4 or 8 ohms… so, could I not wire a 4 ohm resistor in parallel to make the impedance 8 ohms?

Speaker wired in series:-

8+8+8+8 = 32

Now take the speaker and wire a 4 ohm resistor in parallel should give me the following: -

(32*4) / (32 + 4) = 8 ohms yes?

This would suit the amplifier much better but I’m sure that there is a very good reason not to do this. Anyone like to explain what affect this would have?

Please try not to mock me for this question as I’m sure that many other people would be interested in the answer also.
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Old 19th July 2008, 03:59 PM   #2
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I'm not trying to pee on the parade but methinks you should get a book and read up on simple circuits and see how they can be applied to speakers. Your calculations are not correct.

If you have 4-8 ohm drivers and want 8 ohms total you would wire two of them in series (16 ohm). Wire the other two in series also (16 ohm). Then connect the two pairs in parallel to return to 8 ohms. You can also do it the other way round. Wire two in parallel and the other two in parallel and then series those two pairs. Both will give a total of 8 ohms.

If you are combining different impedances you invert the numbers, add them together and flip them back over so:

4 drivers wired for 32 ohms plus a 4 ohm resistor wired parallel is:
1 divided by 32 = 0.03125
And 1 divided by 4 = 0.25

together they add up to 0.28125. Inverted = 3.5555 ohms.

Hope that helps.
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Old 19th July 2008, 04:03 PM   #3
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The only reason I did that last calculation is to show how impedances work. That circuit is one you would never wire. Most all of the power is going to the resistor.
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