Question about series/parallel speaker setup

[ShadowTek]

At my disposal are three 4-ohm speakers and one 16-ohm speakers.
I'm considering wiring the three 4-ohm speakers in series, then
connecting that in parallel to an 16-ohm speaker, which should result
in in overall 8-ohm load.

4-4-4
|_____amp
|
16

Is there any danger with having 1 speaker on one side of the parallel,
while having 3 speakers on the other side?

[David Nebenzahl]

On 2/10/2010 2:26 PM ShadowTek spake thus:

> At my disposal are three 4-ohm speakers and one 16-ohm speakers.
> I'm considering wiring the three 4-ohm speakers in series, then
> connecting that in parallel to an 16-ohm speaker, which should result
> in in overall 8-ohm load.
>
> 4-4-4
> |_____amp
> |
> 16

Wrong.

The formula for parallel resistance is:

R1 R2
Rt = --------
R1 + R2

yielding ~6.8 ohms in this case. Close, but no cigar.

> Is there any danger with having 1 speaker on one side of the parallel,
> while having 3 speakers on the other side?

Nope, although don't expect equal power distribution among the various
speakers. But so far as the amp goes, it doesn't care who is where.

Oh, and be sure to get the speakers connected in phase. You can check
this by connecting a battery to the terminals and watching whether the
cone moves in or out. If they aren't already marked, label the speaker
terminals "+" and "-", then connect them in phase.


--
You were wrong, and I'm man enough to admit it.

- a Usenet "apology"

[ShadowTek]

On 2010-02-11, David Nebenzahl > wrote:
> On 2/10/2010 2:26 PM ShadowTek spake thus:
>
>> At my disposal are three 4-ohm speakers and one 16-ohm speakers.
>> I'm considering wiring the three 4-ohm speakers in series, then
>> connecting that in parallel to an 16-ohm speaker, which should result
>> in in overall 8-ohm load.
>>
>> 4-4-4
>> |_____amp
>> |
>> 16
>
> Wrong.
>
> The formula for parallel resistance is:
>
> R1 R2
> Rt = --------
> R1 + R2
>
> yielding ~6.8 ohms in this case. Close, but no cigar.

LMFAO I'd added 4+4+4 in my head and got 16!

I did it right, I just messed up. lol


>> Is there any danger with having 1 speaker on one side of the parallel,
>> while having 3 speakers on the other side?
>
> Nope, although don't expect equal power distribution among the various
> speakers. But so far as the amp goes, it doesn't care who is where.

Assuming that there would be 16 ohms on *both* sides of the parallel
(like I was originally thinking), the power distribution among the four
4-ohms and the one 16-ohm would be the same, right?

4+4+4+4
|______amp
|
16

[David Nebenzahl]

On 2/10/2010 6:26 PM ShadowTek spake thus:

> Assuming that there would be 16 ohms on *both* sides of the parallel
> (like I was originally thinking), the power distribution among the four
> 4-ohms and the one 16-ohm would be the same, right?
>
> 4+4+4+4
> |______amp
> |
> 16

Yes; half on each side. You basically have a parallel network, so
current divides evenly between the two sides. (Voltage is equal across
both sides.)


--
You were wrong, and I'm man enough to admit it.

- a Usenet "apology"

[Mr.T]

"David Nebenzahl" > wrote in message
.com...
> On 2/10/2010 6:26 PM ShadowTek spake thus:
> > Assuming that there would be 16 ohms on *both* sides of the parallel
> > (like I was originally thinking), the power distribution among the four
> > 4-ohms and the one 16-ohm would be the same, right?
> >
> > 4+4+4+4
> > |______amp
> > |
> > 16
>
> Yes; half on each side. You basically have a parallel network, so
> current divides evenly between the two sides. (Voltage is equal across
> both sides.)

Considering they are speakers with complex impedances rather than resistors,
and considering the 4 ohm and 16 ohm speakers are likely to be different in
frequency Vs impedance, once again it is not that simple. And who knows what
the efficiencies, power rating, and interactions between speaker radiation
patterns are, so the likely acoustic outcome is anyone's guess!
However at least the amp will probably be OK with such a load if that's all
that matters.

MrT.

[David Nebenzahl]

On 2/10/2010 11:13 PM Mr.T spake thus:

> "David Nebenzahl" > wrote in message
> .com...
>
>> On 2/10/2010 6:26 PM ShadowTek spake thus:
>>
>>> Assuming that there would be 16 ohms on *both* sides of the parallel
>>> (like I was originally thinking), the power distribution among the four
>>> 4-ohms and the one 16-ohm would be the same, right?
>>>
>>> 4+4+4+4
>>> |______amp
>>> |
>>> 16
>>
>> Yes; half on each side. You basically have a parallel network, so
>> current divides evenly between the two sides. (Voltage is equal across
>> both sides.)
>
> Considering they are speakers with complex impedances rather than resistors,
> and considering the 4 ohm and 16 ohm speakers are likely to be different in
> frequency Vs impedance, once again it is not that simple. And who knows what
> the efficiencies, power rating, and interactions between speaker radiation
> patterns are, so the likely acoustic outcome is anyone's guess!
> However at least the amp will probably be OK with such a load if that's all
> that matters.

Meaningless quibbles from you as usual.

There will be as close to half the power on each side as is matters in
this case. How the speakers handle that power is a complex subject, as
you point out. But there will be half on one side and half on the other.


--
You were wrong, and I'm man enough to admit it.

- a Usenet "apology"

[Mr.T]

"David Nebenzahl" > wrote in message
.com...
> Meaningless quibbles from you as usual.

To *you*, quite probably.

> There will be as close to half the power on each side as is matters in
> this case. How the speakers handle that power is a complex subject, as
> you point out.

I'm glad you agree on something then.

MrT.

[ShadowTek]

On 2010-02-11, David Nebenzahl > wrote:
>
> There will be as close to half the power on each side as is matters in
> this case. How the speakers handle that power is a complex subject, as
> you point out. But there will be half on one side and half on the other.

Thanks.

This is my first time attemping to wire speakers in both series/parallel, so I
wanted to be sure.

[Peter Larsen[_3_]]

ShadowTek wrote:

> On 2010-02-11, David Nebenzahl > wrote:

>> There will be as close to half the power on each side as is matters
>> in this case. How the speakers handle that power is a complex
>> subject, as you point out. But there will be half on one side and
>> half on the other.

> Thanks.

> This is my first time attemping to wire speakers in both
> series/parallel, so I wanted to be sure.

What problem are you trying to solve?

Kind regards

Peter Larsen

[David Nebenzahl]

On 2/11/2010 11:26 PM Peter Larsen spake thus:

> ShadowTek wrote:
>
>> On 2010-02-11, David Nebenzahl > wrote:
>
>>> There will be as close to half the power on each side as is matters
>>> in this case. How the speakers handle that power is a complex
>>> subject, as you point out. But there will be half on one side and
>>> half on the other.
>
>> Thanks.
>
>> This is my first time attemping to wire speakers in both
>> series/parallel, so I wanted to be sure.
>
> What problem are you trying to solve?

Didn't you see his original message? [Assuming gender here]

> At my disposal are three 4-ohm speakers and one 16-ohm speakers.
> I'm considering wiring the three 4-ohm speakers in series, then
> connecting that in parallel to an 16-ohm speaker, which should result
> in in overall 8-ohm load.

[I think he actually has *4* 4-ohm speakers, or maybe not.] So he's
trying to use what speakers he has at hand. Sounds reasonable to me.


--
You were wrong, and I'm man enough to admit it.

- a Usenet "apology"

[Peter Larsen[_3_]]

>> What problem are you trying to solve?

> Didn't you see his original message? [Assuming gender here]

>> At my disposal are three 4-ohm speakers and one 16-ohm speakers.
>> I'm considering wiring the three 4-ohm speakers in series, then
>> connecting that in parallel to an 16-ohm speaker, which should result
>> in in overall 8-ohm load.

And he doesn't say what problem he wants to solve, what sounds he wants to
amplify.

> [I think he actually has *4* 4-ohm speakers, or maybe not.] So he's
> trying to use what speakers he has at hand. Sounds reasonable to me.

No, as I see this it comes across as plain silly if it is to be used to
distribute sound.

Kind regards

Peter Larsen

[ShadowTek]

On 2010-02-12, Peter Larsen > wrote:
>[i]
>>> What problem are you trying to solve?
>
>> Didn't you see his original message? [Assuming gender here]
>
>>> At my disposal are three 4-ohm speakers and one 16-ohm speakers.
>>> I'm considering wiring the three 4-ohm speakers in series, then
>>> connecting that in parallel to an 16-ohm speaker, which should result
>>> in in overall 8-ohm load.
>
> And he doesn't say what problem he wants to solve, what sounds he wants to
> amplify.
>
>> So he's
>> trying to use what speakers he has at hand. Sounds reasonable to me.

First off, I currently have what I assume is a 4-ohm stable amp. It's a Cambridge
Soundworks FPS2000. The manual does not give the amp's specs, and I can
only assume that it is at least a 4-ohm stable amp since the four speakers
that came with it are 4-ohm.
The speakers that came with it are 7-watt 3-inch, and they produce
respectable highs, but they don't dip down into the mid-range worth a
crap. So, I found some 4-ohm cabinet speakers that I forgot I still had
and replaced the old 3-inch speakers with them.
These 4-ohm cabinet speakers have a satisfying mid-range, but they don't
go into the high-range very far.
So, I thought that if I could connect both the 3-inch speaker *and* the
cabinet speakers, I could get the wider frequency range that I desire.

Anyway, that way the original concern. I could have just stopped
there, wired them in a series pair, and been done with it, but I wanted to
play around, experiment with what was possible, and learn something
along the way.

I only actually have three 4-ohm speakers (per channel) BTW--two of the
3-inchers and the cabinet speaker. I just went ahead a said four since I
had previously discussed a 16/16 parallel, and I just wanted to stick by
that to avoid confution.

[David Nebenzahl]

On 2/12/2010 1:36 PM ShadowTek spake thus:

> On 2010-02-12, Peter Larsen > wrote:
>>
>>>> What problem are you trying to solve?
>>
>>> Didn't you see his original message? [Assuming gender here]
>>
>>>> At my disposal are three 4-ohm speakers and one 16-ohm speakers.
>>>> I'm considering wiring the three 4-ohm speakers in series, then
>>>> connecting that in parallel to an 16-ohm speaker, which should result
>>>> in in overall 8-ohm load.
>>
>> And he doesn't say what problem he wants to solve, what sounds he wants to
>> amplify.
>>
>>> [I think he actually has *4* 4-ohm speakers, or maybe not.] So he's
>>> trying to use what speakers he has at hand. Sounds reasonable to me.
>
> First off, I currently have what I assume is a 4-ohm stable amp. It's a Cambridge
> Soundworks FPS2000. The manual does not give the amp's specs, and I can
> only assume that it is at least a 4-ohm stable amp since the four speakers
> that came with it are 4-ohm.
> The speakers that came with it are 7-watt 3-inch, and they produce
> respectable highs, but they don't dip down into the mid-range worth a
> crap. So, I found some 4-ohm cabinet speakers that I forgot I still had
> and replaced the old 3-inch speakers with them.
> These 4-ohm cabinet speakers have a satisfying mid-range, but they don't
> go into the high-range very far.
> So, I thought that if I could connect both the 3-inch speaker *and* the
> cabinet speakers, I could get the wider frequency range that I desire.

So now you're into trying to assign different frequencies to different
speakers; you didn't explain that before.

Not to pop your balloon or anything, but you might not get the results
you're hoping for. When you write about certain speakers having or not
having a satisfactory midrange, etc., you seem to be forgetting
something important about speaker design. It's called a "crossover", and
it's a (usually, not always) simple circuit that actually separates
different frequencies to different speakers.

Without it, all bets are kinda off as to what kind of performance you're
going to get out of this lash-up of mongrel speakers. It sounds as if
you're going to need a 3-way crossover (separate outputs for woofer,
midrange and tweeter).

I'm explaining all this in detail assuming you don't already know about
it. If you do, my apologies. But otherwise, you might want to do some
reading about it.


--
You were wrong, and I'm man enough to admit it.

- a Usenet "apology"

[ShadowTek]

On 2010-02-12, David Nebenzahl > wrote:
>
> So now you're into trying to assign different frequencies to different
> speakers; you didn't explain that before.
>
> Not to pop your balloon or anything, but you might not get the results
> you're hoping for. When you write about certain speakers having or not
> having a satisfactory midrange, etc., you seem to be forgetting
> something important about speaker design. It's called a "crossover", and
> it's a (usually, not always) simple circuit that actually separates
> different frequencies to different speakers.

You mean built *into* the speaker? Unless that's what you mean (and
assuming that it would filter the signal that is being sent to the rest
of the speakers in the circuit), I don't see how the outcome would be any
different than I had originally expected.

I've used *external* crossovers before for subwoofers, so I *do* know
what their purpose is. I've never used them for mids and highs though,
so I don't have any experience with how noticable the difference is.


> Without it, all bets are kinda off as to what kind of performance you're
> going to get out of this lash-up of mongrel speakers. It sounds as if
> you're going to need a 3-way crossover (separate outputs for woofer,
> midrange and tweeter).
>
> I'm explaining all this in detail assuming you don't already know about
> it. If you do, my apologies. But otherwise, you might want to do some
> reading about it.

I'f I *were* to buy some inline mid and high crossovers for each channel, that would
mean that I would have the 4-ohm cabinet speaker and the 16-ohm speaker
on the mid crossover, and the two 4-ohm 3-inch speakers on the high
crossover, so the result would have to be a parallel of the first pair, and
a series of the second pair?

Would a series of the first pair (24-ohm) paralleled with a series of the
second pair (8-ohm) actually work out? (6-ohm for whole circuit)

[David Nebenzahl]

On 2/12/2010 4:39 PM ShadowTek spake thus:

> On 2010-02-12, David Nebenzahl > wrote:
>>
>> So now you're into trying to assign different frequencies to different
>> speakers; you didn't explain that before.
>>
>> Not to pop your balloon or anything, but you might not get the results
>> you're hoping for. When you write about certain speakers having or not
>> having a satisfactory midrange, etc., you seem to be forgetting
>> something important about speaker design. It's called a "crossover", and
>> it's a (usually, not always) simple circuit that actually separates
>> different frequencies to different speakers.
>
> You mean built *into* the speaker?

No, of course not: crossovers are external to the speakers.

> I've used *external* crossovers before for subwoofers, so I *do* know
> what their purpose is. I've never used them for mids and highs though,
> so I don't have any experience with how noticable the difference is.

Well, the crossover keeps high-frequency signals out of low-frequency
speakers and vice versa. A 3-way crossover divides the audio spectrum
into 3 bands: the woofer gets the bottom, the midrange the middle and
the tweeter the top.

The idea is to keep frequencies that the particular speaker (midrange,
tweeter, etc.) isn't capable of handling well out of it entirely.

You also probably don't want more than one crossover per set of
speakers, for a number of reasons, among them complexity.


--
You were wrong, and I'm man enough to admit it.

- a Usenet "apology"

[Peter Larsen[_3_]]

David Nebenzahl wrote:

>> You mean built *into* the speaker?

> No, of course not: crossovers are external to the speakers.

Warning: David, with speaker you mean loudspeaker unit, the OP means
loudspeaker unit(s) in box.

> You also probably don't want more than one crossover per set of
> speakers, for a number of reasons, among them complexity.

The solution to poor sounding loudspeakers is to get better loudspeakers. I
appreciate the intent to learn, and experiments combined with reading is a
great method for learning, but experimenting with how resistors combine is
best done with resistors.

Also there is a simple "order of magnitude" issue with the original concept
of combining 3 4 Ohm loudspeakers and one 16 Ohm loudspeaker on two
amplifier channels (assuming that was what the OP asked about), two 4 Ohm
loudspeakers in series on one channel is in a low quality context
acceptable, but there is no good way to combine the remaining 4 Ohm unit and
a 16 Ohm unit on the other amplifier channel, the solution is to omit the 16
Ohm unit based on the amplifier being spec'ed for a minimum load of 4 Ohms.

Kind regards

Peter Larsen

[Mr.T]

"David Nebenzahl" > wrote in message
s.com...
> > You mean built *into* the speaker?
>
> No, of course not: crossovers are external to the speakers.

He probably means "built into the speaker" *enclosure* though, and that is
where most crossovers are usually located.


> The idea is to keep frequencies that the particular speaker (midrange,
> tweeter, etc.) isn't capable of handling well out of it entirely.

And usually not have two sets of drivers covering the same frequency range
and perhaps creating a response peak.
And lets not even talk about frequency dependant phase cancellation
problems! :-)

MrT.