forgive me for asking just a dumb question, but im unsure on one part.. Ok, if the impedence of a capacitor is 1/jwc ... that means at dc (w=0), the impedence is infinity.. it acts as a open circuit... at hf (w---> infinity), the impedence goes to 0.. so by this means, a cap in series with a speaker makes a HPF, and a cap in parallel would make a LPF... i understand that--what i dont understand is why do people use inductors as well?? I understand that an inductor would work as well (since z=jwl), but they are more expensive and larger... can i use caps for hpf and lpf?? whats the advantage to using an inductor---phasing issues??? Thanks in advance
-chris
-chris
Good Question!!!
Even though I have been designing, building, and modding crossovers for years I have never thought of that. I just calculate using published formuli, and then alter components to tweek the sound till it works with listening and measurement
I proudly display my ignorance, and wait to be enlightened!
Even though I have been designing, building, and modding crossovers for years I have never thought of that. I just calculate using published formuli, and then alter components to tweek the sound till it works with listening and measurement
I proudly display my ignorance, and wait to be enlightened!
zx3chris, could you please post a drawing how i have to understand your propoed C - crossover?
Is it on C in parallel for bass and one C in series for high? this maybe first order and normally you need higher order crossovers?
Or you put one C in series and one in parallel? so the parallel will short everything to ground what the series one wants to let pass through. (unless it has higher resistance at higher frequency because its an not so ideal C)
this would be a crossover that burn most of the energy you bring to the speaker through the cable.
Anotherthing is : what imedance sees the amp if you use a C in parralel and no spool before it? and what happens to the current that should be provided to the tweeter ?
Is it on C in parallel for bass and one C in series for high? this maybe first order and normally you need higher order crossovers?
Or you put one C in series and one in parallel? so the parallel will short everything to ground what the series one wants to let pass through. (unless it has higher resistance at higher frequency because its an not so ideal C)
this would be a crossover that burn most of the energy you bring to the speaker through the cable.
Anotherthing is : what imedance sees the amp if you use a C in parralel and no spool before it? and what happens to the current that should be provided to the tweeter ?
An externally hosted image should be here but it was not working when we last tested it.
ok this is what i was talking about
for this arguments sake lets set f=1000hz
so w=2pi*f = 6280
C1=100uF
R is a speaker, 8ohm
as w increases, the impedence of C1 drops, and more current will flow in that branch rather than through the speaker. w will never hit 0, because even at 20khz
freq Z
100 15.9 ohm
1000 1.59ohm
5000 .318ohm
etc
obvisouly, as f increases, Z gets pretty low, but i chose 100uF arbitralily... in any event, would this not function as a filter??
Sorry to say no.
D'ont forget that it will short at high frequqency. And that an amp is a voltage source(in theory) and that the voltage of a voltage source does not vary with the load. Thus the cap would have no effect on what the speaker see(In theory). In practice it will just blow in a spectacular way(and maybe bringing the amp with him)
D'ont forget that it will short at high frequqency. And that an amp is a voltage source(in theory) and that the voltage of a voltage source does not vary with the load. Thus the cap would have no effect on what the speaker see(In theory). In practice it will just blow in a spectacular way(and maybe bringing the amp with him)
Using all capacitors works quite well in active filters where we have load isolation, or at least a controlled load for the input signal, and a near zero Ohm driving impedance for the output.
With passive crossovers, I don't think you can control the Q unless you can use both C's and L's. Remember, a speaker can be both inductive and capactivie, depending upon frequency. For an example, consider a woofer. At high f's, an inductor works well to isolate it. We could place a capacitor across the woofer's inputs to short out the high frequency, but the C could resonate with the L of the woofer at some frequency, and that would not be a good thing. By using an inductor, the woofer is effectively removed from the high frequency circuit. The same, but opposite, works for a tweeter.
By "Q", we are referring to what kind of filter response we have. That is, the classic Butterworth response results when the Q is .707. Using both L and C in a passive filter we can set the Q. If only C's are used, I think it would be near impossible to achieve a desired Q.
There is nothing wrong with using an active crossover where you can pretty much eliminate the inductors. This is often done, in part, to get rid of the relatively poor performance of inductors compared to capacitors.
With passive crossovers, I don't think you can control the Q unless you can use both C's and L's. Remember, a speaker can be both inductive and capactivie, depending upon frequency. For an example, consider a woofer. At high f's, an inductor works well to isolate it. We could place a capacitor across the woofer's inputs to short out the high frequency, but the C could resonate with the L of the woofer at some frequency, and that would not be a good thing. By using an inductor, the woofer is effectively removed from the high frequency circuit. The same, but opposite, works for a tweeter.
By "Q", we are referring to what kind of filter response we have. That is, the classic Butterworth response results when the Q is .707. Using both L and C in a passive filter we can set the Q. If only C's are used, I think it would be near impossible to achieve a desired Q.
There is nothing wrong with using an active crossover where you can pretty much eliminate the inductors. This is often done, in part, to get rid of the relatively poor performance of inductors compared to capacitors.
ok thanks alot for the help--its cleared up a bit.. i knew there was a reason for it.. main reason i thought of this was bc in my ee lab a few weeks ago, we were dealing with op amp filters, and we used just caps for hp and lpf... just curious why it woudlnt apply to speakers as well--thanks again for the help
halojoy gives one of his classes
As the impedance of L increases with frequency
and the impedance of C decreases.
The sum of those 2, that the amplifier sees
is somewhat constant.
The LC, they act in fact in the same way
that 2 resistors can act as a voltage divider.
But the point between them is variable
in an LC, depending on the freq.
That means that the Speaker
connected across one part of that variable resistance,
will have more or less voltage over the voice coil.
And that is how a crossover works.
If you have 2 speakers with identical LC-dividers.
And connect one across the L
and the other over the C,
you let one of them take over
the bigger voltage above a certain crossover freq.
And the other takes over below that frequency.
Crossover = Hand over the sound
LC-filter have 12 dB slope /octave (double the freq)
RC-filter have 6 dB slope /octave
The difference comes out of that
in the LC one get bigger
while the other get lower. A kind of push-pull.
In the RC, R is the same,
while only the C get bigger or smaller
in impedance (resistance)
As the impedance of L increases with frequency
and the impedance of C decreases.
The sum of those 2, that the amplifier sees
is somewhat constant.
The LC, they act in fact in the same way
that 2 resistors can act as a voltage divider.
But the point between them is variable
in an LC, depending on the freq.
That means that the Speaker
connected across one part of that variable resistance,
will have more or less voltage over the voice coil.
And that is how a crossover works.
If you have 2 speakers with identical LC-dividers.
And connect one across the L
and the other over the C,
you let one of them take over
the bigger voltage above a certain crossover freq.
And the other takes over below that frequency.
Crossover = Hand over the sound
LC-filter have 12 dB slope /octave (double the freq)
RC-filter have 6 dB slope /octave
The difference comes out of that
in the LC one get bigger
while the other get lower. A kind of push-pull.
In the RC, R is the same,
while only the C get bigger or smaller
in impedance (resistance)
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