I have spent two days trying to figure out the math for series and parallel wiring.
Can someone help me?? I know series is easier on the amp and parallel is harder on the amp.

The amp sees half of the speaker impedence right?? I need a breakdown of how to do the math. I had it down pat at one time...some 10 or 12 years ago. Along with some hair some of my memory has went also.

Thanks

Tim

I got all kinds of info if you'd like you could pm your e-mail and I can send you a word .doc with whatever electronics info you need reguarding DC/AC

Ed

or you could just tell him...

Series:
Total resistance = R1 + R2 +R3...
Total resistance = 4 + 4
Total resistance = 8 ohms for 2 series connected 4 ohm speakers.

Parallel:
Total parallel resistance = (R1*R2)/(R1+R2)
Total parallel resistance = (4*8)/(4+8)
Total parallel resistance = 32/12
Total parallel resistance = 2.67 ohms

Very true but I have all kinds of info from where I go to school where I'm a Computer & Electronics Engineering Technology major, I could pretty much send him a whole book on DC if he wanted; overkill I guess.

Ed

Series: you just add the ohm ratings together correct??

now after you get that figure does the amp see half that amount = 4+4 =8 then the amp sees 1/2 making it a 4 ohm load to the amp right??

Parallel: If R1 is 4 ohms, how do you get R2??

Thanks,

Tim

Series:
Total resistance = R1 + R2 +R3...
Total resistance = 4 + 4
Total resistance = 8 ohms for 2 series connected 4 ohm speakers.

Parallel:
Total parallel resistance = (R1*R2)/(R1+R2)
Total parallel resistance = (8 * 8_) / (8 + 8_)
Total parallel resistance = 64 / 16
Total parallel resistance = 4 ohms


--> Total parallel resistance = 2.67 ohms with smileys I didnt follow?
Were you working 1 8 ohm and 1 4 ohm together?
8 * 4 / 8 + 4 = 2.67
32 / 12 = 2.6666666


Series: you just add the ohm ratings together correct??

The above is correct...... Excellent!

How did you arrive at this?
now after you get that figure does the amp see half that amount = 4+4 =8 then the amp sees 1/2 making it a 4 ohm load to the amp right??

no simple addition of resistances the load at the amp is 8 ohm.... NOMINAL

remember math and reality differ with magnetic coils who'* resistance load
vary do to input so in a sense it'* the mean average. At times on the load
the resistance may be 6.785 or others 10.02 ohms it'* never exactly 8ohms

-->Parallel: If R1 is 4 ohms, how do you get R2??
----> Whats the rating of the next resistor (coil) the drivers rating?!
multiply the resistances together
add the resistances together
divide the top from the bottom....

As a cheat you can half the smallest resistance say 2 ohm...

Lets do the math and see ifin were close...

4 x 4 = 16
4 + 4 = 8
16 / 8 = 2

Mr Silent gets the gold star... Head of the class!

Mac 'man y'a need to wake up and get to class on time!

Remember the Half the smallest resistance in parallel trick
it works well for speaker pairs but dont cheat if your building
crossovers or real RC networks because circuit harmonics
need better precision to filter and assemble....

Class Dismissed! :P

Ok if were talkin' series you just add the ohm ratings together and that is the ohm load to the amp right?

Now for parallel:

for instance if you were to use say 3 JL W3'* it would work out like this
3x3x3 =27 then 3+3+3 = 9

27/9 =3 for a 3 ohm load.

If this is so then why do I remember a number of industry experts saying the amp sees half of the load.


Tim

:?: HowY Wept

James 3:13

2 Co 10:12

2Ti 2:7

Luke 2:47

Owners manual page ....

I know what is right and what isn't with the resistances, but, HowY, as correct as you are, you are kinda off the wall...

Thanks! I guess...

Really my references were to describe humility and the desire
to come to an understanding and a willingness to learn even more.

Here'* a little take on the Ohms issue that might better apply.
It'* a little lengthly but will draw some interesting applications.
In particular Failure conditions and Safety measures:

My Home Theatre system employs some bass transducers.
These are basically 4 ohm voice coils that use whatever
they are attached to as a diaphram as opposed to a "cone".

Here'* the rig....

A Monoblock amplifier that is 4ohm stable
(fed via lfe channel - as in a summed output)

4 transducers rated at 4ohms per unit.

Layout:
2 shakers are mounted to a sofa and 2 chairs each have 1 (one) mounted.

How can I wire them up to present a 4 ohm (or higher) load to the amp.

Lets do some math to see where we are:
-- Series (_ R1 + R2 _) -- Parallel (_ R1 * R2 _) / (_ R1 + R2 _)

Series : 4 + 4 + 4 + 4 = 16 ohm
Parallel: 256 / 16 = 16 ohm

Well thats way too much resistance - won't throw the protect circuit
but wont have any volume either so theres got to be another way.....

How about 2 parallel strings of 2 drivers - then wired in series to the amp?

Math:
4 * 4 / 4 + 4 = 16 / 8 = 2 ohms per string in parallel

2 ohm + 2 ohm = 4 ohm wired in series at the amp!

- Id argue most folk would wire a pair of dual coil subs in this fashion

BUT This is mounted to furnishing and I'd like to be able to disconnect
a string (pair of shakers) and still run... then what happens if a coil fails?

If one string or pair is removed from the circuit the amp is facing 2 ohm which
is now in protect mode (ifin' I'm lucky) remember the amp is 4 ohm stable.

Say I only loose one coil in one of the parallel circuits....

4ohm + 2 ohm = 6 ohm at the amp <Whew!> now at least I'm not in protect
mode, I'm above the rated load - But the coils are wired in parallel so
what really happens if I loose a coil - Fails closed I've lost the whole circuit!
- literally fried (both) coils. Now I'm presenting only 2 ohm to the amplifier!

But for argument lets say 1 coil fails open (infinity) and the other is uneffected
my parallel circuit is at 4 ohm and when series'd to the other parallel circuit
the amp see'* 6 ohm. Not the best but it is a workable solution.

Lets do pairs in series - wired in parallel to the amplifier (the solution)
math:
4 + 4 = 8 ohm per circuit
8 * 8 / 8 + 8 = 64 / 16 = 4 ohm - seen at the amplifier

Here now we have to consider the requirement of detaching a circuit pair....
If one of the series circuits is removed the amp see'* 8 ohm well within
performance range and won't trip the protection circuit.

What happens if I loose one of the coils in the series circuit.....
The entire circuit is disabled so the amp see'* 8 ohm - no protect
and there is a chance that I've not sacrificed the working coil to
the failure of the other.

This is the fun in working out how you should wire up your systems
I know we've got 1 ohm stable amplifiers now (wonder why?)
DIY'ers wiring their multi - dual coil subs in parallel and losing a coil?

Here'* another interesting twist - you can wire your coils to different
channels! Thats right you can use one coil for Left and the other for
Right. We'll talk about wiring the multi-driver IB sub for the HT at
another time -- should I parallel each channel or series? What
about failure conditions and what the amp will see then....
How about the channels negating each others movement... Hmmm....

Anyway I've seem to have struck many of you all the wrong way
and I'ts not by intent. It'* with much humility and my personal
willingness to accept when I'm entirely wrong and be taught
"more perfectly" the right way. I dont post to flame and only
post on material that I have some (modest) level of understanding.

Please consider fully how you wire up your systems and what
loads your amps are seeing in best case and catastrophic failure...

Hope this get'* you all thinkin'!