Question about resistors

[adwilk]

Shopping for an assortment of Resistors to experiement with and i was wondering... what do the values mean exactly in terms of changing the audible properties of the driver. I just need a good read on what resistor does what..
Obviously you need to use non-inductive but after that, what audio characteristics are different from using a 1 or a 2.3

 

[Eric Apple]

That's easy... The bigger the number the quieter the speaker will be. No audible difference except the volume will be reduced for higher values. If you are interested, this site has an introduction to electronic theory that maybe helpful in your experiment:

http://www.play-hookey.com/dc_theory/

 

[ScottMayo]

That's easy... The bigger the number the quieter the speaker will be. No audible difference except the volume will be reduced for higher values. If you are interested, this site has an introduction to electronic theory that maybe helpful in your experiment:

http://www.play-hookey.com/dc_theory/

It might not be quite that simple - it depends on where the resistor goes and what sort of crossover and amp, if any, is involved. Upping a series resistance can change how the amp views the load; in some amps, that can change behaviour. But increasing the overall resistance is not likely to make an amp *less* stable.

On the other hand, if you have a volume control, you probably don't need to start adding resistors..? Unless the amp is rated for 8 ohms and the speaker is 2, or something.

 

[MacManNM]

Eric is completely wrong, and Scott is partially wrong. You have to balance the resistors. Matching a pair of resistors, in a series parallel configuration, with the impedance of the driver, does this. You can't use any resistor. You must use one that has a power rating equivalent to that of the speaker (series) and one that will take the shunt current (parallel). There are plenty of online calculators that can help you do this. I do not recommend a novice doing this; chances are they are going to make things worse than they were when they started.

 

[Eric Apple]

Eric is completely wrong, and Scott is partially wrong. You have to balance the resistors. Matching a pair of resistors, in a series parallel configuration, with the impedance of the driver, does this. You can't use any resistor. You must use one that has a power rating equivalent to that of the speaker (series) and one that will take the shunt current (parallel). There are plenty of online calculators that can help you do this. I do not recommend a novice doing this; chances are they are going to make things worse than they were when they started.

Hey I resemble that reply! I am not completely wrong -- in fact if they are fiddling with a crossover it still applies but only to one of the drivers... You assumed way more then I did - I assumed someone was new to the electronics field and was simply putting a resistor in series with a speaker. I still believe this poster would benefit from some basic electronics knowledge even if they are fiddling around with a crossover. I was in the same spot about 28 years ago and turned the curiosity into a living back then.

 

[WmAx]

Hey I resemble that reply! I am not completely wrong -- in fact if they are fiddling with a crossover it still applies but only to one of the drivers... You assumed way more then I did - I assumed someone was new to the electronics field and was simply putting a resistor in series with a speaker.

You stated: "No audible difference except the volume will be reduced for higher values. "

If you add simple series resistance between the driver and crossover, you will completely throw off the crossover since the crossover properties are totally dependant on the terminated load impedance. Also, if you add simple series resistance in circuit anywhere between the amplifier and speaker system, you increase the apparent source impedance of the circuit, therefor altering[ degree of which is dependant on the particular variables in a specific circumstance ] the frequency response at and around frequencies where impedance is not constant at the apparent load.

-Chris

 

[adwilk]

Ok, Excellent... i've done some reading and have researched all the posts... I get it.. i think. I understand what happens to the circuit when said resistor is applied. How can i predict how it will affect sound, imaging, blah blah? Will it be more of an assumption than prediction? I've built plenty of speakers but have always just used somebody elses schematics and i'm interested in knowing how to select a resistor.... Do i just need to stick to online "calculators".. can i F*** anything up by just playing with different ones???

 

[adwilk]

actually, a good answer might go ahead and include selections of capacitors, inductors, and the works.. i read all about those too. Maybe i just need a really in depth read about all the different types of crossovers and what makes them "sound" the way they do...

 

[Eric Apple]

Adwik - What you are talking about changing are the properties of the crossover and with that how each of the drivers are blended into total sound of the speaker.

There's not really anyway to say how just changing "a resistor" is going to change the sound. It really depends upon where it is located electronically in the crossover.

My original answer doesn't apply to what you are trying to do, the real answer is much more complicated as others in this thread have stated. A good starting point would be to read up first on crossover designs, and calculations. Then design a crossover meeting what you would guess would be a decent design for your speaker.

Then start altering the cross over frequencies by changing component values based upon calcuations for the crossover. Just replacing parts is really a shot in the dark as to what the final effect on sound will be.

PS.. It's unlikely you will damage your amp by correctly replacing crossover components will the same component types of different values. There however is a fair chance you could fry a tweeter or midrange. It wouldn't be impossible to damage an amp, but it would be pretty difficult and most likely the result of a wiring error or a solder bridge.