Hello
Just a newbie trying to do the first build, so some of the questions here may be pretty basic:
I "inherited" some old and crappy Sony speakers which I am trying to re-vitalize by replacing the drivers. This is more of a learning project than a real build that will act as a stepping stone for a nicer build in the future.
The speakers were 3 way: a 8 ohm woofer, a 6 ohm "midrange" and a third speaker whose impedance is unknown. Sony had marketed these as "6 ohm" speakers. I plan to convert this into a 2 way arrangement.
Questions:
In a loudspeaker taken as a whole, arent the 3 speakers in parallel with each other typically? So how does a 8 ohm woofer in parallel with 6 ohm midrange result in a "6 ohm speaker"? (It is unlikely that the third speaker has high values)
I plan to pair a 4 ohm tweeter driver with a 8 ohm woofer driver when I replace. My understanding is that this will reduce the impedance of the loudspeaker as seen from the amplifier's point of view. Is this correct? (I assume so..)
I also need to compensate for sensitivity for the tweeter. This will be done using a L-Pad. Can I put a 2 ohm series resistor in series with the L-pad to compensate for the 4 ohm tweeter? So effectively it will look like:
High pass Filter circuit -> 2 ohm resistor -> L pad -> Tweeter
The high pass filter will be designed assuming a 6 ohm value for the tweeter rather than 4 ohms.
Is the above approach daft? What are the disadvantages or other ways to compensate for a 4 ohm tweeter ( I do not want my amplifier to see a 4 ohm tweeter).
I am aware impedance is a function of frequency, I am being simplistic here.
Just a newbie trying to do the first build, so some of the questions here may be pretty basic:
I "inherited" some old and crappy Sony speakers which I am trying to re-vitalize by replacing the drivers. This is more of a learning project than a real build that will act as a stepping stone for a nicer build in the future.
The speakers were 3 way: a 8 ohm woofer, a 6 ohm "midrange" and a third speaker whose impedance is unknown. Sony had marketed these as "6 ohm" speakers. I plan to convert this into a 2 way arrangement.
Questions:
In a loudspeaker taken as a whole, arent the 3 speakers in parallel with each other typically? So how does a 8 ohm woofer in parallel with 6 ohm midrange result in a "6 ohm speaker"? (It is unlikely that the third speaker has high values)
I plan to pair a 4 ohm tweeter driver with a 8 ohm woofer driver when I replace. My understanding is that this will reduce the impedance of the loudspeaker as seen from the amplifier's point of view. Is this correct? (I assume so..)
I also need to compensate for sensitivity for the tweeter. This will be done using a L-Pad. Can I put a 2 ohm series resistor in series with the L-pad to compensate for the 4 ohm tweeter? So effectively it will look like:
High pass Filter circuit -> 2 ohm resistor -> L pad -> Tweeter
The high pass filter will be designed assuming a 6 ohm value for the tweeter rather than 4 ohms.
Is the above approach daft? What are the disadvantages or other ways to compensate for a 4 ohm tweeter ( I do not want my amplifier to see a 4 ohm tweeter).
I am aware impedance is a function of frequency, I am being simplistic here.
Patnio
i don't know which model Sony's you've got but even the cheapest will have at least a cap or two which is why the overall impedance isn't a simple "paralleling" of three speakers.
the L pad will effect your tweeter sensitivity so the 2 ohm resistor may not be needed.
designing your high pass filter with a different value than what will actually be used may change your x-over frequency
i don't know which model Sony's you've got but even the cheapest will have at least a cap or two which is why the overall impedance isn't a simple "paralleling" of three speakers.
the L pad will effect your tweeter sensitivity so the 2 ohm resistor may not be needed.
designing your high pass filter with a different value than what will actually be used may change your x-over frequency
X-over is parallel not the speakers. With no x-over the speakers will be parallel, and resulting impedance would be very low. For three 8 ohm speakers the resulting impedance would be 2,667 ohms - not very healthy for the amplifier.
With a parallel x-over the resulting impedance would still be 8 ohm. IF the impedance was a straight line of the frequency. But impedance is a function of the frequency and varies heavily with the frequency. An 8 ohm speaker is just an average.
No easy task to understand, but with a little reading on the topic you will learn.
With a parallel x-over the resulting impedance would still be 8 ohm. IF the impedance was a straight line of the frequency. But impedance is a function of the frequency and varies heavily with the frequency. An 8 ohm speaker is just an average.
No easy task to understand, but with a little reading on the topic you will learn.
Thanks for the replies
I am not sure I understand:
The impedance seen by the amplifier will be the impedance of the crossover in parallel with the impedance of the speaker assuming a parallel crossover. Both impedances are a function of frequency. Now if I change a nominal 6 ohm speaker to a 4 ohm speaker, surely the overall impedance seen by the amplifier will reduce.
Do I not need to compensate for the changed speaker value? I understand that there is in reality no such thing as a purely resistive 4 ohm speaker: it is more of an equivalent RLC circuit. But surely when one changes from a 6 ohm nominal to a 4 ohm nominal, there is greater strain placed on the amplifier due to higher current draw? Do I need to compensate for that , by using a 2 ohm resistor mentioned above? If not, why not?
Please note I am not disputing what you are saying, just trying to understand this a bit better.
I am not sure I understand:
The impedance seen by the amplifier will be the impedance of the crossover in parallel with the impedance of the speaker assuming a parallel crossover. Both impedances are a function of frequency. Now if I change a nominal 6 ohm speaker to a 4 ohm speaker, surely the overall impedance seen by the amplifier will reduce.
Do I not need to compensate for the changed speaker value? I understand that there is in reality no such thing as a purely resistive 4 ohm speaker: it is more of an equivalent RLC circuit. But surely when one changes from a 6 ohm nominal to a 4 ohm nominal, there is greater strain placed on the amplifier due to higher current draw? Do I need to compensate for that , by using a 2 ohm resistor mentioned above? If not, why not?
Please note I am not disputing what you are saying, just trying to understand this a bit better.
patnio
in the case of your tweeter which would be after a capacitor the "nominal" impedance is no longer "seen" by the amp.
you've removed part of the frequency spectrum(the cap)and as you stated impedance is a function of frequency.
your "compensating" resistor would reduce tweeter level and serve to not deviate the x-over frequency but recalculating the x-over with the final (actual) value may be better
in the case of your tweeter which would be after a capacitor the "nominal" impedance is no longer "seen" by the amp.
you've removed part of the frequency spectrum(the cap)and as you stated impedance is a function of frequency.
your "compensating" resistor would reduce tweeter level and serve to not deviate the x-over frequency but recalculating the x-over with the final (actual) value may be better
Just so I am sure I get this correctly:
From an electrical circuit perspective, a speaker can be modeled as
http://sound.westhost.com/lrp-f36.gif
(Ignore the series RLC circuit at left, thats the impedance compensation)
When I replace the old tweeter with another one, all the parameters represented at far right by the resistors, inductors, capacitors change. Obviously the impedance of these are all a function of frequency. However, if the new tweeter is nominally "4 ohms", the equivalent impedance of the new tweeter has reduced. Exactly by how much is difficult to state since it is a function of frequency.
Adding a L-pad type of circuit + a 2 ohm resistor (or combining it with the series resistor of the L-pad) will somewhat compensate for the drop in the impedance, though this will not be exact.
The result of adding the 2 ohm resistor will necessitate a re-calculation of the crossover assuming a 6 ohm tweeter (rather than the actual value of a 4 ohm tweeter) to ensure crossover frequency remains correct.
Effect of this will be a reduction in tweeter sensitivity and possibly more power being dissipated in the series resistor of the L-pad.
Am I Correct?
ummm!
i had a look at the link and i think you may have that a little wrong.
i may be mistaken but i believe that the components on the left are intended to represent the equivalent of a loudspeaker and the components on the right are the compensation.
i'm a little unclear as to the L pad configuration your planning and where your putting the two ohm resistor so i'll leave that alone.
"Am I Correct?" close but no cigar! i'm not sure how to help you at this point not knowing your level of knowledge or background i'm at a loss as to how to help clarify this for you.
sorry.
i had a look at the link and i think you may have that a little wrong.
i may be mistaken but i believe that the components on the left are intended to represent the equivalent of a loudspeaker and the components on the right are the compensation.
i'm a little unclear as to the L pad configuration your planning and where your putting the two ohm resistor so i'll leave that alone.
"Am I Correct?" close but no cigar! i'm not sure how to help you at this point not knowing your level of knowledge or background i'm at a loss as to how to help clarify this for you.
sorry.
Patnio, You are fine.
I think your problem is the tweeter/amp. I don't have experience with amps but I don't know of a problem with that. Usually is with low impedance in the lower frequencies. (Check that first). In this case you can use the 4R tweeter. Calculate the crossover and attenuate. In case you want 8R go with a 6 or 8 Ohm tweeter. Look at real values of Re in spec sheets.
Not if you placed it between the crossover and amp.
I think your problem is the tweeter/amp. I don't have experience with amps but I don't know of a problem with that. Usually is with low impedance in the lower frequencies. (Check that first). In this case you can use the 4R tweeter. Calculate the crossover and attenuate. In case you want 8R go with a 6 or 8 Ohm tweeter. Look at real values of Re in spec sheets.
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No, I dont think so. Please take a look at the the complete article:
Passive Crossover Network Design
Figure 3.1 is the equivalent circuit of a woofer, Figure 3.3 is that of the same woofer with a Zobel network. Similarly Figure 3.6 is that of a tweeter. Perhaps it was wrong of me to just link the picture, but it was the only one that was closest to what I wanted to get across. The diagrams are of an actual equivalent circuit that mimics a loudspeaker. I suppose if one simply wanted to represent a loudspeaker, a simple R, L and C components would do.
Please take a look at Figure 6.1 in the same link above. Resistors 2R0 and 12R are the L-pad resistors. My two ohm resistor would be in series with 2R0, on the left side of it connecting to the capacitor and the inductor.i.e. after the filter.
My understanding is from an amp's perspective replacing a tweeter/woofer from one that is 8 ohms nominal to a 4 ohms nominal makes a difference since the amp now has to send more current due to reduced impedance. And this has to be compensated somehow. My solution is to use a series resistor as above.
IF someone says that it does not make a difference, I would like to know why.
patnio
you are indeed correct i was looking at the circuit bass ackwards. in the immortal words of Homer J. Simpson "dooh!"
as to the L pad adding your 2 ohm resistor would reduce your tweeter level but keep in mind this attenuator is after the filter.
the resultant new impedance is not going to put undue demands on your amps output it's not like your simply paralleling another driver.
me thinks your neglecting the effect your x-over has perhaps.
still gotta figure out how to properly use the quote thingy
you are indeed correct i was looking at the circuit bass ackwards. in the immortal words of Homer J. Simpson "dooh!"
as to the L pad adding your 2 ohm resistor would reduce your tweeter level but keep in mind this attenuator is after the filter.
the resultant new impedance is not going to put undue demands on your amps output it's not like your simply paralleling another driver.
me thinks your neglecting the effect your x-over has perhaps.
still gotta figure out how to properly use the quote thingy
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Hi, Patnio! I'm working now on a similar project for my girlfriend. The foam woofer surrounds on her old 3-ways disintegrated, and I knew at a glance that redesigning the speakers would be more satisfactory than repairing them. I'm converting them to 8" 2-ways with Vifa drivers and 2nd-order Linkwitz-Riley crossovers.
What you propose with your L-pad will certainly work; you can add any value you want to the series resistor. But I think you've got the cart before the horse somewhat if you're taking the rated impedances of your various drivers at face value. My 8" Vifa woofer, for example, measures 13 ohms at my crossover point, 9 an octave below, and 19 an octave above. With compensation, it measures 6.4 ohms plus or minus 0.4 from 150Hz to 20,000. So now it's a stable load, but still not 8 ohms. Don't even get me started on the tweeter.
The existing crossover in my gf's speakers was just 2 electrolytic capacitors, nothing more. If yours looks similar, I'd suggest you toss it and start from scratch. Regardless, you really don't know how your crossover will act until you know more about your driver's impedance than simply that it's rated at 4, 6, or 8 ohms.
Since you're working up to a better build, I recommend you equip yourself to measure the actual impedance of the drivers you're using at various frequencies. Then start with calculated component values for your compensation circuit, test it, and refine as necessary. The calculated circuit for my woofer was 6 ohms in series with 12uf, but I found the best results using 7.5 ohms in series with 15uf.
What you propose with your L-pad will certainly work; you can add any value you want to the series resistor. But I think you've got the cart before the horse somewhat if you're taking the rated impedances of your various drivers at face value. My 8" Vifa woofer, for example, measures 13 ohms at my crossover point, 9 an octave below, and 19 an octave above. With compensation, it measures 6.4 ohms plus or minus 0.4 from 150Hz to 20,000. So now it's a stable load, but still not 8 ohms. Don't even get me started on the tweeter.
The existing crossover in my gf's speakers was just 2 electrolytic capacitors, nothing more. If yours looks similar, I'd suggest you toss it and start from scratch. Regardless, you really don't know how your crossover will act until you know more about your driver's impedance than simply that it's rated at 4, 6, or 8 ohms.
Since you're working up to a better build, I recommend you equip yourself to measure the actual impedance of the drivers you're using at various frequencies. Then start with calculated component values for your compensation circuit, test it, and refine as necessary. The calculated circuit for my woofer was 6 ohms in series with 12uf, but I found the best results using 7.5 ohms in series with 15uf.
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