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Speaker Resistance (Ohm) Wiring Question
#374784 04/30/12 09:55 PM
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Lets say that I have 8 speakers that are 4 ohms each.
I would like to set them up in groups of 4 speakers, so two groups of 4...

How can I get it so that they don't tax a basic stereo receiver that is only rated for 8 ohms/channel?


OK. So these are bass shakers, not speakers, but you get the idea.

I have a pair of wires running to my front row of seats, and a pair going to my back row of seats. Each row has 4 seats, each seat has a bass shaker.

I can wire them, it seems, so that I can get 8 or 2 ohms per pair and then combining two pairs (thus 4 seats) I can get 16 ohms, 4 ohms, 10 ohms (8+2), or 5 ohms (8+2/2)...

The receiver I was looking at was the Sherwood RX-4105. It costs all of $100 and offers 100 watts/channel (perfect match for two sets (channels) of four 25 watt shakers)...

If I go with a plate amp, they seem to all be mono output (which is fine), but when combining 8 shakers, I get 8 ohms (or other "odd" ohm numbers like the 5 & 10 in my first example)...

So either I need some help with the wiring to get something that won't tax the little $100 Sherwood (which I do NOT own yet) ohms wise but still utilize two channels to get the right wattage, or I need to find a mono amp that is 200 watts, not expensive, and can do 8 watts.

Ideas?


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Re: Speaker Resistance (Ohm) Wiring Question
nickbuol #374787 04/30/12 09:59 PM
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Resistance is futile.



Did that help?


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Re: Speaker Resistance (Ohm) Wiring Question
nickbuol #374788 04/30/12 10:03 PM
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LOL. No.


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Re: Speaker Resistance (Ohm) Wiring Question
nickbuol #374789 04/30/12 10:08 PM
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I was just searching and saw these.

AudioSource AMP100 - $99
Which the specs state...
"Output power 50 watts RMS per channel @ 8 ohms, 60 watts RMS per channel into 4 ohms, 160 watts RMS @ 8 ohms bridged mono"

And
Dayton Audio APA150 - $128
Which the specs state...
"Measured power output: 2 x 75 watts RMS (into 4 ohms @ less than 0.01% THD)"

I know that 75 watts is not 100 watts, but the trick I learned years ago about bass shakers is that they don't need to be "cranked up" to be effective, and in fact, they are more impressive at lower settings...

Other thoughts about any of this?


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Re: Speaker Resistance (Ohm) Wiring Question
nickbuol #374791 04/30/12 10:19 PM
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You need to make sure that the effective impedance of each branch or group is identical. Otherwise you will get level imbalances as more current will flow through the lower impedance branches. In other words you can forget about the 10 or 5 ohm options. If you need to use both channels, with 4 speakers per channel you can have 4ohms or 16ohms. With a single channel you can get 8ohms, as you have pointed out. The Sherwood will likely be fine with either the 16 or 4 ohm configuration, but there is no way to know for sure without trying it. Personally I would look for a cheap plate amp designed for DIY subwoofers.

Re: Speaker Resistance (Ohm) Wiring Question
nickbuol #374796 04/30/12 11:31 PM
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Nick, have you already bought the bass shakers yet? i would try to find some 16ohm shakers.. That would allow you to get a nominal impedance of 8ohms.. In your 8 driver example.

Re: Speaker Resistance (Ohm) Wiring Question
nickbuol #374799 04/30/12 11:57 PM
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Thanks Andrew. So many "diagrams" online show all sorts of configurations and odd ohm levels, thus my confusion.

Dakkon, yup, I've owned these shakers for years.


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Re: Speaker Resistance (Ohm) Wiring Question
nickbuol #374801 05/01/12 12:13 AM
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Here is a $120 plate amp. 250w at 4 ohms, 156w at 8 ohms...

Dayton Audio SPA250 Plate Amp

I would need to build some sort of "box" for it, but that wouldn't be a problem I wouldn't think.

Thoughts on this one?

I would just wire up each set of 4 shakers to either 4 ohms or 16 ohms, and then take the two wire runs that go back to where the amp would be and wire them in series or parallel (depending on which way I went - 4 or 16 ohm each set) and get 8 ohms.

I just want to make sure that I am thinking right before ordering something.


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Re: Speaker Resistance (Ohm) Wiring Question
nickbuol #374802 05/01/12 12:19 AM
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Nick, please do draw a map or take a picture or something. I'm interested in adding bass shakers, but the wiring has always baffled me.


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Re: Speaker Resistance (Ohm) Wiring Question
nickbuol #374805 05/01/12 12:36 AM
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Nick, you could always put a resistor in the circuit to increase the overall resistance.... It would be a "waste" of power, but would reduce the current draw from the amp.

Re: Speaker Resistance (Ohm) Wiring Question
nickbuol #374807 05/01/12 12:47 AM
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I also want to add bass shakers to go with my 3 subs, anxious to read about this post.


M80s VP180 4xM22ow 4xM3ic EP600 2xEP350
AnthemAVM60 Outlaw7700 EmoA500 Epson5040UB FluanceRT85


Re: Speaker Resistance (Ohm) Wiring Question
nickbuol #374808 05/01/12 12:49 AM
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Since I am traveling for work this week, I am dealing with a "limited" laptop. So I am just going to copy/paste some diagrams.

In the following picture, change everything that says 8 Ohms, to 4 Ohms and you have one "leg" of the 8 shaker setup that I am talking about.




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Re: Speaker Resistance (Ohm) Wiring Question
dakkon #374812 05/01/12 01:56 AM
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Originally Posted By: dakkon
Nick, you could always put a resistor in the circuit to increase the overall resistance.... It would be a "waste" of power, but would reduce the current draw from the amp.


The only problem with this is that the shaker is a transducer and, just like a regular speaker, has an impedance that varies with frequency. Putting a resistor in series will then alter the frequency response of shakers.

Re: Speaker Resistance (Ohm) Wiring Question
Andrew #374814 05/01/12 02:23 AM
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Indeed. Using resistors, when you want to feel the "WUB WUB WUB" of your dubstep beats, you might only get "wib wib wib". Definitely a let down.

Re: Speaker Resistance (Ohm) Wiring Question
nickbuol #374816 05/01/12 02:34 AM
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I am sitting here at dinner at a Saltgrass Steakhouse near Denver, and I about spit my water out when I read "wib, wib, wib"

Good thing that I am here alone.


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Re: Speaker Resistance (Ohm) Wiring Question
pmbuko #374818 05/01/12 03:03 AM
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Yeah, Peter(and Alex); but Andrew is emphasizing that the added resistance not only loses some power(at a given volume control setting), but more importantly results in an uneven frequency response. It's the same principle that applies when a speaker wire is too thin and therefore has too high resistance for the run involved.

Resistances connected in series form a voltage divider. For example, if a 4 ohm resistor was connected in series with a nominally 4 ohm speaker, the resistor and speaker would divide the voltage equally at a frequency where the speaker resistance actually was 4 ohms. In practice, speakers can vary widely in impedance as the frequency varies. If the impedance was say 12 ohms at a certain frequency the speaker would take a larger "share" of the voltage at that frequency and would be relatively louder than at the 4 ohm frequency. So, the speaker response would go up and down as frequency varied instead of being relatively flat.

Speaker wires with too high resistance or amplifier designs(e.g., certain tube designs with output impedances of several ohms)with high output impedance are examples of practices that can lead to flawed sound.


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Re: Speaker Resistance (Ohm) Wiring Question
JohnK #374819 05/01/12 03:20 AM
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Originally Posted By: Andrew
Originally Posted By: dakkon
Nick, you could always put a resistor in the circuit to increase the overall resistance.... It would be a "waste" of power, but would reduce the current draw from the amp.


The only problem with this is that the shaker is a transducer and, just like a regular speaker, has an impedance that varies with frequency. Putting a resistor in series will then alter the frequency response of shakers.



Wouldn't it just shift the impedance graph up by 4ohms? Basically shifting the Y intercept, while not altering the slope; since a resistor is a "real" load that does not vary with AC, like an inductive or capacitive load?

Speakers are basically inductors, that are used to create a magnetic field, which drives the speaker cone... The same principle a bass shaker works off of, accept without a cone, correct?

I know were getting in to AC theory here a little... But, i dont know why the resistor option wouldn't just push the entire impedance line up as a whole?

I know it would waist some power, at this point i would consider my option an academic discussion.


Last edited by dakkon; 05/01/12 03:28 AM.
Re: Speaker Resistance (Ohm) Wiring Question
dakkon #374822 05/01/12 03:33 AM
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No, Alex; the voltages are divided in proportion to the relative magnitudes of the two resistances. In the example I gave above the speaker would take a relatively larger proportion of the voltage at 12 ohms(the resistor remaining fixed at 4 ohms)and following Ohm's Law, as usual, would be relatively louder at the 12 ohm frequency than at the 4 ohm frequency even if the output voltage from the amplifier was the same at the two frequencies and they should be equal in loudness.


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Re: Speaker Resistance (Ohm) Wiring Question
JohnK #374823 05/01/12 03:45 AM
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Originally Posted By: JohnK
For example, if a 4 ohm resistor was connected in series with a nominally 4 ohm speaker, the resistor and speaker would divide the voltage equally at a frequency where the speaker resistance actually was 4 ohms. In practice, speakers can vary widely in impedance as the frequency varies.


John, this is correct, a 4ohm+4ohm load in seres would be 8ohms, and would be the reference point. As the speaker's impedance varied up or down, it would vary the same amount, and the over all circuits resistance would vary in line with the speakers impedance.

Andrew, are you talking about the voltage loss across the resistor? So, the voltage loss across the resistor would alter the voltage seen at the driver, which would skew the impedance graph?

It's been a couple years since i have done KVL's or KCL's so please bear with me.. But this seems more like a KVL problem, at a static point, the more i think about it...

I should be working on applied regression... But no.... gr wink

Re: Speaker Resistance (Ohm) Wiring Question
JohnK #374825 05/01/12 03:53 AM
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Originally Posted By: JohnK
No, Alex; the voltages are divided in proportion to the relative magnitudes of the two resistances. In the example I gave above the speaker would take a relatively larger proportion of the voltage at 12 ohms(the resistor remaining fixed at 4 ohms)and following Ohm's Law, as usual, would be relatively louder at the 12 ohm frequency than at the 4 ohm frequency even if the output voltage from the amplifier was the same at the two frequencies and they should be equal in loudness.


John, ohms law is for an entire circuit. you have to use Kirchhoff's voltage/current law for components within a circuit.

KVL's are used to calculate the voltage seen at a component level from a voltage source. There is a voltage loss across each component in a series circuit and can be calculated by using a KVL...

http://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws

However, at this point we are using DC equations.. AC gets much more complicated, I chose the KVL for the ease of conceptual understanding.

Last edited by dakkon; 05/01/12 04:12 AM.
Re: Speaker Resistance (Ohm) Wiring Question
dakkon #374831 05/01/12 04:20 AM
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A basic discussion of voltage dividers is shown here . The voltage drop across each of the fixed resistors is proportional to its relationship to the total resistance in the circuit. Likewise, the voltage drop across the speaker voice coil is proportional to its resistance relative to the total resistance including the fixed resistor. At frequencies where the impedance is higher, the voltage drop across the speaker voice coil is higher and results in a relative increase in loudness at that frequency.


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Re: Speaker Resistance (Ohm) Wiring Question
nickbuol #374867 05/01/12 03:13 PM
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Here is a partial copy/paste of what I posted over at AVS and haven't gotten a response back on yet...
-------------------------------------------------------
The bass shakers I have are the non-pro version (4 ohms and 25 watts each). I know that "ideally" I would have 200 watts total, but my previous experience with these shakers is that they are most effective in adding to the experience (without being just a stupid level of shaking) that they don't require that much power. I mean, they won't be run at max anyway.

So, I can wire up each row with a combination of series/parallel and get either 4 ohms or 16 ohms per set of four shakers.

So far so good.

So if I go with that, I would need something with 2 channels that can handle 4 ohms or 16 ohms per channel and output up to about 100 watts (any more is overkill).

OR

I can tie both rows together and get a single 8 ohm connection at about 200 watts.

So can I assume that any of the option options would work?
#1) Dayton Audio SPA250 - 156 Watt at 8 ohms plate amp (connect both rows together)
#2) Dayton Audio APA150 - 2x75 watts in 4 ohm mode, 150 watts in bridged 8 ohm mode
#3) Dayton Audio SA70 - 70 watts at 4 ohms - Buy 2 of them, one for each row.

#1 is priced between the others, and would just require a simple box to house it in.
#2 is the most expensive, but still within reason, it is a finished product not needing a box to be constructed.
#3 is the cheapest, requires a box to house both amps, but adds control for both rows independently.

First up, is my math correct on the Ohms?
Second, opinions/options for powering this setup? While I would have loved something like a receiver with remote capabilities to dial in the level of shake from my seat, I can deal without. Otherwise I would need a 2 channel receiver that could handle 4 ohms at 100 watts per channel...

Thanks!


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Re: Speaker Resistance (Ohm) Wiring Question
nickbuol #374877 05/01/12 07:10 PM
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This all seems unnecessarily complicated. Why not just wire them in series? (Rcv+ to spkr1+, spkr1- to spkr2+, spkr2- to rcvr-; keep extending until you've added as many speakers as you want.) You'll wind up with a 16 ohm resistance with 4, but that won't hurt any receiver, just makes in a little less loud.

Re: Speaker Resistance (Ohm) Wiring Question
nickbuol #374879 05/01/12 07:55 PM
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So if I picked up the originally named Sherwood receiver which has 100 watts at 8 ohms per channel, what could I expect if I had each row of 4 shakers (16 ohms each 'load') for wattage? Half of the rated 100? If it would work from the ohms perspective, I could have the advantage of each row being separate and have remote control....


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Re: Speaker Resistance (Ohm) Wiring Question
nickbuol #374892 05/02/12 01:52 AM
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Nick, although I've commented on the interesting(to me, at least)technical aspects of drivers connected in series, I should also reply to your basic question. I tend to agree with Cork, and you may be over-thinking this shaker setup. Wiring four in series for each row and using that Sherwood receiver(theoretically about 50 watts per channel into 16 ohms, which is also just a theoretical number)would be relatively simple and should be plenty to get the folks all shook up.

Any concerns about frequency fluctuations would be minimal with identical drivers with identical impedance graphs. Also, any possible irregularities caused by series connection would be less bothersome in shaking as compared to listening, although ardent "audiophiles" might claim that their butts were just as golden as their ears.


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Re: Speaker Resistance (Ohm) Wiring Question
nickbuol #374893 05/02/12 02:04 AM
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Wasn't it Mark that had an avatar years ago with a picture of himself with giant golden ears? I don't think that I want to see his picture of his golden butt...

Almighty, so if memory serves me correctly from the last time I had shakers booked up, I don't even think that I came close to 50% power, so maybe I will just go the route of the Sherwood and be done with it.


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Re: Speaker Resistance (Ohm) Wiring Question
nickbuol #374902 05/02/12 11:01 AM
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Yes, it was me.

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Re: Speaker Resistance (Ohm) Wiring Question
MarkSJohnson #374910 05/02/12 01:35 PM
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Originally Posted By: MarkSJohnson
Yes, it was me.

The Golden Butt photos aren't free.


But I've found that they do hold their value.


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