Axiom Home Page Forums General Discussion Technical Questions Speaker Resistance (Ohm) Wiring Question

I also want to add bass shakers to go with my 3 subs, anxious to read about this post.

Since I am traveling for work this week, I am dealing with a "limited" laptop. So I am just going to copy/paste some diagrams.

In the following picture, change everything that says 8 Ohms, to 4 Ohms and you have one "leg" of the 8 shaker setup that I am talking about.

The only problem with this is that the shaker is a transducer and, just like a regular speaker, has an impedance that varies with frequency. Putting a resistor in series will then alter the frequency response of shakers.

Indeed. Using resistors, when you want to feel the "WUB WUB WUB" of your dubstep beats, you might only get "wib wib wib". Definitely a let down.

I am sitting here at dinner at a Saltgrass Steakhouse near Denver, and I about spit my water out when I read "wib, wib, wib"

Good thing that I am here alone.

Yeah, Peter(and Alex); but Andrew is emphasizing that the added resistance not only loses some power(at a given volume control setting), but more importantly results in an uneven frequency response. It's the same principle that applies when a speaker wire is too thin and therefore has too high resistance for the run involved.

Resistances connected in series form a voltage divider. For example, if a 4 ohm resistor was connected in series with a nominally 4 ohm speaker, the resistor and speaker would divide the voltage equally at a frequency where the speaker resistance actually was 4 ohms. In practice, speakers can vary widely in impedance as the frequency varies. If the impedance was say 12 ohms at a certain frequency the speaker would take a larger "share" of the voltage at that frequency and would be relatively louder than at the 4 ohm frequency. So, the speaker response would go up and down as frequency varied instead of being relatively flat.

Speaker wires with too high resistance or amplifier designs(e.g., certain tube designs with output impedances of several ohms)with high output impedance are examples of practices that can lead to flawed sound.

Wouldn't it just shift the impedance graph up by 4ohms? Basically shifting the Y intercept, while not altering the slope; since a resistor is a "real" load that does not vary with AC, like an inductive or capacitive load?

Speakers are basically inductors, that are used to create a magnetic field, which drives the speaker cone... The same principle a bass shaker works off of, accept without a cone, correct?

I know were getting in to AC theory here a little... But, i dont know why the resistor option wouldn't just push the entire impedance line up as a whole?

I know it would waist some power, at this point i would consider my option an academic discussion.

No, Alex; the voltages are divided in proportion to the relative magnitudes of the two resistances. In the example I gave above the speaker would take a relatively larger proportion of the voltage at 12 ohms(the resistor remaining fixed at 4 ohms)and following Ohm's Law, as usual, would be relatively louder at the 12 ohm frequency than at the 4 ohm frequency even if the output voltage from the amplifier was the same at the two frequencies and they should be equal in loudness.

John, this is correct, a 4ohm+4ohm load in seres would be 8ohms, and would be the reference point. As the speaker's impedance varied up or down, it would vary the same amount, and the over all circuits resistance would vary in line with the speakers impedance.

Andrew, are you talking about the voltage loss across the resistor? So, the voltage loss across the resistor would alter the voltage seen at the driver, which would skew the impedance graph?

It's been a couple years since i have done KVL's or KCL's so please bear with me.. But this seems more like a KVL problem, at a static point, the more i think about it...

I should be working on applied regression... But no.... gr

John, ohms law is for an entire circuit. you have to use Kirchhoff's voltage/current law for components within a circuit.

KVL's are used to calculate the voltage seen at a component level from a voltage source. There is a voltage loss across each component in a series circuit and can be calculated by using a KVL...

http://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws

However, at this point we are using DC equations.. AC gets much more complicated, I chose the KVL for the ease of conceptual understanding.