I have an old Yamaha receiver with two speaker outputs.
The B speakers run outside, the A inside. I would like to add another
set of speakers outside. Is this possible? If so, is it difficult.
Thanks

Steve,

Is this possible? If so, is it difficult.

The answer is a definite "maybe." Some issues are the receiver's amp's power
capability, how low an impedance it can drive, and the impedance of the
speakers you connect.

But don't let this scare you. Try connecting a second set of speakers to
either the A or B terminals - depending on which existing set you want them
to work with. Then turn the volume down low, play some music, and slowly
turn up the volume. If the sound distorts at low volume, then the amp is
probably not capable of driving the additional speakers. But if it sounds
good then it's probably okay.

--Ethan

"Ethan Winer" ethan at ethanwiner dot com wrote in message
...
Steve,

Is this possible? If so, is it difficult.

The answer is a definite "maybe." Some issues are the receiver's amp's
power
capability, how low an impedance it can drive, and the impedance of the
speakers you connect.

But don't let this scare you. Try connecting a second set of speakers to
either the A or B terminals - depending on which existing set you want
them
to work with. Then turn the volume down low, play some music, and slowly
turn up the volume. If the sound distorts at low volume, then the amp is
probably not capable of driving the additional speakers. But if it sounds
good then it's probably okay.

--Ethan

I don't agree with this advice. Adding a second pair of speakers will cut
the impendence in half . If both pairs are 4 ohms, this will cut the
impedance to 2 ohms and damage your receiver (few receivers can handle less
than 4 ohms). If both pairs are 8 ohms and your receiver can handle 4 ohms
(or have a switch to set at 4 ohms) you will be OK (but only if both pairs
are 8 ohms).

Mark,

Adding a second pair of speakers will cut the impendence in half . If both
pairs are 4 ohms, this will cut the impedance to 2 ohms and damage your
receiver (few receivers can handle less than 4 ohms).

Maybe, maybe not. First the wire itself adds a fair amount of resistance.
Especially if the set he's doubling up is the one outside, and presumably
connected by fairly long wire. Also, most receivers have built-in protection
to keep them from self-destructing if the output is accidentally sorted out.
That's why I suggested trying it at low volume to start, to get a sense of
how the receiver copes with the load. Finally, some power amps have A and B
speakers, and some add a C set of output terminals. In all the receivers
I've ever seen, all three terminals are fed from the same amp(s).

--Ethan

On 17 Sep 2003 12:08:11 -0700, (Steve)
wrote:

I have an old Yamaha receiver with two speaker outputs.
The B speakers run outside, the A inside. I would like to add another
set of speakers outside. Is this possible? If so, is it difficult.
Thanks

Hello,
This can be easily and safely accomplished by simply connecting the
additional speakers in series rather than parallel. Just connect the
wires as indicated below.

Amp Out B Left + Left Speaker #1 +
Left Speaker #1 - Left Speaker #2 +
Left Speaker #2 - Amp Out B Left -

Amp Out B Right + Right Speaker #1 +
Right Speaker #1 - Right Speaker #2 +
Right Speaker #2 - Amp Out B Right -

A series connection will cause the load impedance to equal the sum of
the speakers' impedances, rather than a parallel connection which will
cause the load impedance to be lower than that of either speaker
alone. Solid-State amplifiers, almost without exception, are entirely
happy with higher load impedances.

Formula for parallel connection

zt = 1 / ( (1/S1) + (1/S2) )

Where zt = total load impedance
S1 = Speaker #1 impedance
S2 = Speaker #2 impedance
All values in Ohms

In article ,
Ethan Winer ethan at ethanwiner dot com wrote:
Mark,

Adding a second pair of speakers will cut the impendence in half . If both
pairs are 4 ohms, this will cut the impedance to 2 ohms and damage your
receiver (few receivers can handle less than 4 ohms).

Maybe, maybe not. First the wire itself adds a fair amount of resistance.
Especially if the set he's doubling up is the one outside, and presumably
connected by fairly long wire.

Okay, so say he runs 50 feet of 18 guage wire. That'd add
altogether a whopping 0.6 ohms of DC resistance. Hardly a "fair
amount," as it would constitute only about 10% or the typical
impedance of most loudspeakers.

--
| Dick Pierce |
| Professional Audio Development |
| 1-781/826-4953 Voice and FAX |
| |

Dick,

Okay, so say he runs 50 feet of 18 guage wire. That'd add altogether a
whopping 0.6 ohms of DC resistance.

Don't forget up and back smile. But you're right, it's a small factor to
be sure.

--Ethan

In article ,
Ethan Winer ethan at ethanwiner dot com wrote:
Dick,

Okay, so say he runs 50 feet of 18 guage wire. That'd add altogether a
whopping 0.6 ohms of DC resistance.

Don't forget up and back smile. But you're right, it's a small factor to
be sure.

I didn't. 50 feet of speaker cord is 100 feet total run. 18
gauge annealed copper wire 18 gauge is about 6.3 ohms per 1000
feet, thus about 0.6 ohms.
--
| Dick Pierce |
| Professional Audio Development |
| 1-781/826-4953 Voice and FAX |
| |

Dick,

Yeah, I think/hope the main fallback is that any decent power amp will have
over-current protection.

So Steve, did you try it? What happened?

--Ethan

Okay, so say he runs 50 feet of 18 guage wire. That'd add
altogether a whopping 0.6 ohms of DC resistance. Hardly a "fair
amount," as it would constitute only about 10% or the typical
impedance of most loudspeakers.

--
| Dick Pierce |
| Professional Audio Development |
| 1-781/826-4953 Voice and FAX |
| |

Here we have more non-sensical tech garbage from
Dick Pierce, the most incoherent curt poster ever in this
group - he is just pure bull****, just ignore Dick Pierce.

"Richard D Pierce" wrote in message
...
In article ,
Ethan Winer ethan at ethanwiner dot com wrote:
Dick,

Okay, so say he runs 50 feet of 18 guage wire. That'd add altogether a
whopping 0.6 ohms of DC resistance.

Don't forget up and back smile. But you're right, it's a small factor to
be sure.

I didn't. 50 feet of speaker cord is 100 feet total run. 18
gauge annealed copper wire 18 gauge is about 6.3 ohms per 1000
feet, thus about 0.6 ohms.
--
| Dick Pierce |
| Professional Audio Development |
| 1-781/826-4953 Voice and FAX |
| |

Yawn,
Dick, you are a total ******, so just stop posting.
Most of us have you on kill file - so just be quiet.

"Ethan Winer" ethan at ethanwiner dot com wrote in message
...
Dick,

Okay, so say he runs 50 feet of 18 guage wire. That'd add altogether a
whopping 0.6 ohms of DC resistance.

Don't forget up and back smile. But you're right, it's a small factor to
be sure.

--Ethan

Ethan, why do you bother replying to cocksure ill-mannered Dickless Pierce?
He hasn't got a clue what he is talking about, Dick Pierce only posts
bull****
tech garbage, just put him on killfile as the rest of us do...............

"NonSenZe" wrote in message
...
Ethan, why do you bother replying to cocksure ill-mannered
Dickless Pierce?
He hasn't got a clue what he is talking about, Dick Pierce only
posts
bull****
tech garbage, just put him on killfile as the rest of us
do...............

Huh? What have you been smoking? Dick is one of the few people
who post accurate and relevant information.

If he was on killfile why did you read his post?

Ethan made an exaggeration and Dick called him on it. Ethan said
the wire "adds a fair amount of resistance". Dick posted that a
very long run of very small wire adds only small amout. For the
same power to the speaker the amp would need to source just a
little more power (107W vs 100W). Ethan's claim just didn't hold
water.

On Thu, 18 Sep 2003 22:17:51 -0500, "Rusty Boudreaux"
wrote:

"NonSenZe" wrote in message
...
Ethan, why do you bother replying to cocksure ill-mannered
Dickless Pierce?
He hasn't got a clue what he is talking about, Dick Pierce only
posts
bull****
tech garbage, just put him on killfile as the rest of us
do...............

Huh? What have you been smoking? Dick is one of the few people
who post accurate and relevant information.

If he was on killfile why did you read his post?

Ethan made an exaggeration and Dick called him on it. Ethan said
the wire "adds a fair amount of resistance". Dick posted that a
very long run of very small wire adds only small amout. For the
same power to the speaker the amp would need to source just a
little more power (107W vs 100W). Ethan's claim just didn't hold
water.

What, you didn't realise that 'NonSenZe' *is* Ethan Winer? :-)

Lots of people don't like Dick Pierce, because they just can't handle
the truth of the real physical world!
--

Stewart Pinkerton | Music is Art - Audio is Engineering

Stewart,

What, you didn't realise that 'NonSenZe' *is* Ethan Winer? :-)

I assure you I always post under my real name. There's probably a way to
verify an IP trace in the message headers but I can't be bothered.

I don't think Dick, Rusty, or you are incompetant. Unenlightened? Sure.
Rude? Absolutely! But totally incompetant? Probably not.

--Ethan

If I Remember correctly connecting speakers in series are only a good idea
if both sets of speakers are identical. The reason why eludes me at this
time


"Robert Rowton" wrote in message
...
On 17 Sep 2003 12:08:11 -0700, (Steve)
wrote:

I have an old Yamaha receiver with two speaker outputs.
The B speakers run outside, the A inside. I would like to add another
set of speakers outside. Is this possible? If so, is it difficult.
Thanks

Hello,
This can be easily and safely accomplished by simply connecting the
additional speakers in series rather than parallel. Just connect the
wires as indicated below.

Amp Out B Left + Left Speaker #1 +
Left Speaker #1 - Left Speaker #2 +
Left Speaker #2 - Amp Out B Left -

Amp Out B Right + Right Speaker #1 +
Right Speaker #1 - Right Speaker #2 +
Right Speaker #2 - Amp Out B Right -

A series connection will cause the load impedance to equal the sum of
the speakers' impedances, rather than a parallel connection which will
cause the load impedance to be lower than that of either speaker
alone. Solid-State amplifiers, almost without exception, are entirely
happy with higher load impedances.

Formula for parallel connection

zt = 1 / ( (1/S1) + (1/S2) )

Where zt = total load impedance
S1 = Speaker #1 impedance
S2 = Speaker #2 impedance
All values in Ohms

DS,

If I Remember correctly connecting speakers in series are only a good idea
if both sets of speakers are identical. The reason why eludes me at this
time

I'm not sure if it's a good idea even then. One problem that comes to mind
is the crossover components will be in series, which may affect their
operation.

--Ethan

D Shadford wrote:

If I Remember correctly connecting speakers in series are only a good idea
if both sets of speakers are identical. The reason why eludes me at this
time

Connecting in series won't hurt anything but the sound and volume. The resulting sound could be significantly inferior to just having one speaker on a channel.

Although the "nominal" impedance of a speaker is rated at (say) 8 ohms, it actually varys significantly over the bandwidth of the unit--That's why it's not "resistance" that is important. An 8 ohm speaker system could vary from 6 ohms to 8Kohms depending on frequency, and the impedance/frequency curve is quite complex. This means that given a constant voltage, the current that passes through a speaker will vary significantly with frequency. If (and only if) that speaker is put in series with a complex load that perfectly matches that speaker, the voltage drop across the speaker will always be half of what it is across the total series load.

If you do not match the speaker, then the frequency dependant voltages across the speaker can be all over the place. For example, if you had a speaker with a very low impedance bass and high impedance treble range in series with a load that had a flatter impedance curve (e.g., medium impedance bass and medium impedance treble) then you can see that the current hungry bass end would get starved and the treble end would be fed harder than necessary resulting in a very bright sound. If the load in series with that speaker was another different speaker, it would tend to be just the opposite on that speaker (although no where near acoustically balanced between the two :-)

In fact, simply putting a resistor/reostat in series with a speaker to control its volume will have this same type of effect, that's why L-pads are used on remote speaker controls instead of reostats.

If you ever get a chance to see an impedance plot for a speaker system, you will be amazed at how many peaks, dips, and inconsistencies there are. The impedance can vary by over 3 orders of magnitude! In fact it gets even more complex when you add the reactance plots. An amplifier needs to guarantee that a particular voltage point on the source shows up amplified a fixed amount at the speaker regardless of frequency and the varying loads that the speaker system places on the amp. Some very exotic speakers are very ill-behaved in that their impedance and reactance values are all over the place and they need a high quality amplifier to drive them. This is the reason that a pair of $5000 speakers might sound like absolute trash on a cheap amp where a pair of $80 speakers (usually much better behaved. i.e., less reactance) would sound great. Take both pairs and put them on a better amp and then the ill-behaved ones can sound much better than the cheaper ones were ever capabile of.

-Jeff

Jeff,

Great post. Thanks for clarifying.

--Ethan

In article ,
Ethan Winer ethan at ethanwiner dot com wrote:
DS,

If I Remember correctly connecting speakers in series are only a good idea
if both sets of speakers are identical. The reason why eludes me at this
time

I'm not sure if it's a good idea even then. One problem that comes to mind
is the crossover components will be in series, which may affect their
operation.

Can you show why that would be the case?

A proper analysis of the situation shows that this is not the
case. Consider a simple case, a first-order series pass element
constructed with non-ideal components. Yes, series losses will
be doubled, but so will the load impedance. By Kirchoff, you can
show that the net effect of putting two such networks and their
driver loads in series, even considering the non-ideal loss
components of the reactive elements, that when you double the
load, double the reactive values and double the loss component,
you end up with the same net complex transfer function.

By superposition, this can be shown to be the case in more
complex speaker systems and networks. It can also be shown to be
the case empirically as well.
--
| Dick Pierce |
| Professional Audio Development |
| 1-781/826-4953 Voice and FAX |
| |

Dick,

series losses will be doubled, but so will the load impedance.

Ah, good point. Thanks for elucidating.

--Ethan

Ethan Winer wrote:

Jeff,

Great post. Thanks for clarifying.

--Ethan


It's interesting but some older receivers (and even some new ones) that allowed you to hook up 2 or 3 speaker sets to them would internally hook them up in series when running two together due to their inability to handle lower impedances. The old Sansui 9090dB comes to mind.

This brings up a question that I have. Several Yamaha RX-xxx series receivers (I have an RX-V995) have a switch on the back for 4 or 8 ohm speaker selection. I don't have a schematic and I wonder what it is for. If the amp can handle 4 ohms, it can handle 8, right? I originally surmised that it might simply be that the amp really only handles 8 ohm units and the switch is some kind of limiter to handle overloading.

Now, based on the current discussion, I wonder if it is only to ensure that speakers are put in series when both A and B systems are enabled and they are using 4 ohm speakers (the default perhaps being a parallel arrangement).

I'd like to see a schematic.

- Jeff

Jeff,

some older receivers ... would internally hook them up in series

I'm more into pro audio than consumer gear, so I'm not up on all the models.
Heck, I'm not up on all the models of pro gear either! :-)

I'd like to see a schematic.

That's the only way to know for sure. (And I'm old enough to remember when
most audio gear came with a schematic...)

--Ethan

Dick,

Can you show why that would be the case?

Then again, the damping factor will go way up with the added woofer and
series inductor of the crossover.

--Ethan

In article ,
Ethan Winer ethan at ethanwiner dot com wrote:
Dick,

Can you show why that would be the case?

Then again, the damping factor will go way up with the added woofer and
series inductor of the crossover.

No, it will not.

First, despite it's widespread usage, "damping factor" is a
pretty meaningless and useless term. The more accurate parameter
is the total Q at resonance as a measure of how the system is
damped.

Be that as it may, the Q at resonance is determined primarily by
the combination of the emchancial and electrical Q, with the
electrical Q being by far the most dominat in most high-quality
speakers. The electrical Q is determined simply by:

Qe = 2 pi Fc Mm Rs / (Bl)^2

where Fc is the resonant frequency, Mm is the moving mass, Re is
the effective total loop series resistance, B is the magnetic
field density in and l is the length of the voice coil
wire immersed in the field B.

If we put two speakers in series, watch what happens: we end up
doubling the moving mass, we end up doubling the resistance, we
also end up doubling the length of the wire immersed in the
magnetic field, or:

Qe = 2 pi Fc 2*Mm 2*Rs / (B 2*l)^2

combining terms and simplifying:

Qe = 2 pi Fc 4*Mm Rs / 4(Bl)^2

since the new added coefficients of 4 in the numerator and 4 in
the denominator cancel ourt, we end up with:

Qe = 2 pi Fc Mm Rs / (Bl)^2

Identical to our original equation. The damping, therefore is
unchanged.

Even in the case of the useless "damping factor" specification,
we're still unchanged. If we consider the resistance of the
crossover components as smoehow "separate" from the system, and
given the definition of damping factor, i.e.,

DF = Znom / Rg

where Znom is the nominal impedance of the speaker, and Rg is
the effective source resistance, we are doubling BOTH the
nominal impedance and the source resistance due to the crossover
series losses, the ratio remains essentially the same.


--
| Dick Pierce |
| Professional Audio Development |
| 1-781/826-4953 Voice and FAX |
| |

Dick,

Wow, great explanation. Especially, "...we are doubling BOTH the nominal
impedance and the source resistance ... the ratio remains essentially the
same."

Of course.

Thanks.

--Ethan