# 4ohm DVC running each coil off separate channel questions



## DonutHands (Jan 27, 2006)

I was previously running dual subs. Both were 4ohm DVC wired to present 4ohm load to amp together. Amp is 4 ohm stable bridged, not stable 2ohm.

Im going down to one sub to save space, same sub 4ohm DVC.

Instead of wiring for 2ohm load which the amp cant handle while bridged, I am going to run each coil off of each channel. Gain for both channels are on one knob, so no mis match gain issues.

Real question is about RCA's. Sub out is going to be stereo. So sub out will wired like this?

PRE Out Stereo RCA >Y-adatper to mono > Y-adapter back to Stereo RCA at amp. 

Will this sum the stereo source to mono, then break out to dual mono at the amp? Don't want to fubar anything by having the coils playing different source material.


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## ca90ss (Jul 2, 2005)

You'll get the same power bridging it at 8 ohms as running each coil on a separate channel.


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## DonutHands (Jan 27, 2006)

ca90ss said:


> You'll get the same power bridging it at 8 ohms as running each coil on a separate channel.


Really? I thought power would be reduced as ohm rises.

Amp is Alpine PDX 4.150, just dealing with channels 3/4 for sub duty.
Channels 3/4 - 150x2 @ 4 ohm
Channels 3/4 - 300x1 @ 4 ohm

I was under the assumption at 8 ohm amp would output
Channels 3/4 - 75x2 @ 8 ohm
Channels 3/4 - 150x1 @ 8 ohm


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## ca90ss (Jul 2, 2005)

Bridgeable Amplifiers


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## Holmz (Jul 12, 2017)

DonutHands said:


> ...
> 
> Real question is about RCA's. Sub out is going to be stereo. So sub out will wired like this?
> 
> ...


That will work.
As the frequency gets lower there is no difference in L/R channels, but it will not hurt to wire it that way.
This assumes that the channel you are using is separate from other channels... if not, then making them mono could make the signal mono back towards the HU.

I assume that you would be using rear channels for this and the front channels for higher freqs?

If your sub is crossed over low and steep, then I think running separate L/R channels would be OK.

Driving each 4-ohm coil separately will double the power compared to driving just 1 coil, or driving both coils in series.

I would probably just skip the stereo->mono->stereo.


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## DonutHands (Jan 27, 2006)

ca90ss said:


> Bridgeable Amplifiers


i don't believe that link validates what you are recommending, could be wrong, but doesn't seem like it.


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## DonutHands (Jan 27, 2006)

Holmz said:


> That will work.
> As the frequency gets lower there is no difference in L/R channels, but it will not hurt to wire it that way.


probably in studio mixed tracks. but live recordings could have different L/R low frequency data. better safe than sorry IMO


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## ca90ss (Jul 2, 2005)

DonutHands said:


> i don't believe that link validates what you are recommending, could be wrong, but doesn't seem like it.


Sure it does


> 2 Ohm Stereo vs 4 Ohm Mono Loads
> 
> There seems to be some confusion as to why a 4 ohm mono and a 2 ohm stereo load are the same, as far as the amplifier is concerned. When two 4 ohm speakers are connected to each channel of a 2 channel amplifier, the amplifier is capable of driving the speakers with half of the total power supply voltage. If the amplifier has a power supply which produces plus or minus 20 volts, it will not be able to drive the speakers on a single channel with any more than 20 volts at any point in time. If we have a 2 ohm load on each channel, at the highest point on the waveform the amplifier will apply 20 volts to the speaker load. Remember that we are only considering a single point in time for this example. If we go back to ohms law...
> 
> ...


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## Holmz (Jul 12, 2017)

He is saying bridge the amp and the coils in series.


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## Justin Zazzi (May 28, 2012)

The post above is good but could be more complete. I had to think a while to figure it out too since it's not immediately obvious.

Power is voltage squared divided by resistance (or impedance).
P = N * V^2 / R

_where P is power in watts, V is voltage, and R is impedance
and N is number of channels (2 for a pair of normal channels, or 1 for a bridged channel)_

If you go from a pair of normal channels to a bridged channel, you double the voltage which quadruples the power but you also half the number of channels in the N term which halves the power (so you're left with a net gain of 2x power). If you also double the impedance at the same time by going from a pair of 4ohm coils to a single 8 ohm coil, you cancel out half of the power gain again, so now you are left with the same amount of power as you began with.

So if you bridge a pair of channels into an 8ohm load, you'll get the same power as playing a pair of channels into a 4ohm load.
But if you bridge the channels, you can eliminate the stereo-to-mono-to-stereo problem which is what I think you're concerned with, right?


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## DonutHands (Jan 27, 2006)

Ok, where am i blowing it?? I can admit im wrong, but id like to know what im not understanding.

Lets say V=10 to make things simple

4ohm 2 channel ---- 2*100/4 = 50
4ohm bridged ---- 1*400/4 = 100

8ohm 2 channel ---- 2*100/8 = 25
8ohm bridged ---- 1*400/8 = 50


The above makes it a no brainer to run the coils independently ay 4ohm each so the sub gets 100 watts. If i wire sub for a single 8ohm load and bridge the amp i get 50 watts. Again, unless im missing something which im open to admitting.


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## Justin Zazzi (May 28, 2012)

DonutHands said:


> Ok, where am i blowing it?? I can admit im wrong, but id like to know what im not understanding.
> 
> Lets say V=10 to make things simple
> 
> ...


This is a tricky thing, so don't worry!

Your math looks good, and I also theorized a 10v amplifier when I was thinking about this earlier too.

Your first equation would be two channels to two 4ohm coils giving you 50 watts total.
Your fourth equation would be the same two channels bridged, to the same two voice coils but wired in series at 8ohms, also giving you 50 watts.

The second equation resulting in 100 watts would be the same amplifier bridged, but into a 4 ohm load. To do that, you would have to connect a single voice coil on your subwoofer and leave the other coil unconnected. You can do that if the one coil can handle the power, otherwise you cannot really compare the second equation to the others. I think this is the source of the confusion?

Or maybe you're thinking you can double the first equation to get 100 watts from it. You can't do that because the N=2 term is already doubling it for you.


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## DonutHands (Jan 27, 2006)

Justin Zazzi said:


> This is a tricky thing, so don't worry!
> 
> Your math looks good, and I also theorized a 10v amplifier when I was thinking about this earlier too.
> 
> ...


This does not make sense in my head, are are saying that if the amp does 2x25 @ 4 ohm, total 50watts. When bridged, it makes 1x100 @ 4 ohm, total 100 watts.

How does the amp create an extra 50 watts when bridged at the same ohm load? I understand this when the ohm load drops, but when it stays the same??? Seems like magic to me.


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## ParDeus (May 10, 2014)

DonutHands said:


> This does not make sense in my head, are are saying that if the amp does 2x25 @ 4 ohm, total 50watts. When bridged, it makes 1x100 @ 4 ohm, total 100 watts.
> 
> How does the amp create an extra 50 watts when bridged at the same ohm load? I understand this when the ohm load drops, but when it stays the same??? Seems like magic to me.


Becuase bridged, it's at 2ohm per channel.


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## DonutHands (Jan 27, 2006)

ParDeus said:


> Becuase bridged, it's at 2ohm per channel.


Bridged does not mean 2 ohms per channel if your resistance stays at 4 ohms right? Just because the amp is bridged, does not mean the speakers resistance has changed.


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## Bayboy (Dec 29, 2010)

This is getting too complex. Plain and simple the power is merely combined from two channels. Running an 8 ohm load bridged does not halve the power of 4 ohm bridge. It will still sum the power of 2 channels in 4 ohm stereo. 4 ohm bridged sums the power of 2 channels in 2 ohm stereo. 

The problem comes in when you're thinking that 8 ohm bridged is half of 4 ohm bridged. That's not how bridged amps work. You're still combining two channels


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## SPLEclipse (Aug 17, 2012)

DonutHands said:


> Bridged does not mean 2 ohms per channel if your resistance stays at 4 ohms right? Just because the amp is bridged, does not mean the speakers resistance has changed.


The technical explanation is above explaining the increase in output voltage, but if you doubt this go look at any amplifiers ratings. You will always see something like (for example) "50w x 2 @ 4ohm, 200w x 1 @ 4ohm bridged". You can think of a mono load impedance as being divided among each channel equally with half the impedance going to each channel.

BTW you can not merge left and right signals together without some kind of active or passive protection. 

Why Not Wye?

The advantage of bridging the amp as opposed to using a discreet chanel per coil is that the signal is automatically summed to mono in the amp, so you don't have to worry about summing the signal before it gets to the amp.


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## ParDeus (May 10, 2014)

DonutHands said:


> Bridged does not mean 2 ohms per channel if your resistance stays at 4 ohms right? Just because the amp is bridged, does not mean the speakers resistance has changed.


When you present a 4ohm load to 2 bridged amplifier channels, it will be like a 2ohm per channel nominal load. As to how the amplifier does that with the 4ohm overall load, it doubles the voltage while keeping current the same. So in effect, it's the same as a 2ohm load per channel. It basically inverts the signal of one channel, so know you have one channel driving to the positive rail, and one to the negative. 

So instead of say, your 4ohm speaker seeing 0v to 10v at the top of the waveform, it now sees 10v to -10v, effectively doubling the voltage.

That's why in this scenario, running with the speaker wired in series @8ohm, each channel is seeing a 4ohm load, so it's the same power as running the channels individually at 4ohm per channel.

I'm sure I didn't explain that perfectly, but I think that's the gist of it. I ain't no expert.


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## Holmz (Jul 12, 2017)

ParDeus said:


> When you present a 4ohm load to 2 bridged amplifier channels, it will be like a 2ohm per channel nominal load. As to how the amplifier does that with the 4ohm overall load, it doubles the voltage while keeping current the same. So in effect, it's the same as a 2ohm load per channel. It basically inverts the signal of one channel, so know you have one channel driving to the positive rail, and one to the negative.
> 
> So instead of say, your 4ohm speaker seeing 0v to 10v at the top of the waveform, it now sees 10v to -10v, effectively doubling the voltage.
> 
> ...


What power?
It would be better to talk voltage until one has the speaker impedance to work out the current and power.

The amp does not provide power, just the capability to provide power.


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## DonutHands (Jan 27, 2006)

Bayboy said:


> This is getting too complex. Plain and simple the power is merely combined from two channels. Running an 8 ohm load bridged does not halve the power of 4 ohm bridge. It will still sum the power of 2 channels in 4 ohm stereo. 4 ohm bridged sums the power of 2 channels in 2 ohm stereo.
> 
> The problem comes in when you're thinking that 8 ohm bridged is half of 4 ohm bridged. That's not how bridged amps work. You're still combining two channels
> 
> ...


I understand that, @ 4 ohm you combine the power of 2 channels and that is bridged at 4 ohm. 

So here is where im confused, if the amp makes half the power at 8 ohm, then its half the power. People are saying its not, wtf?


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## minbari (Mar 3, 2011)

I have a question that I dont thinkn anyone has asked and will have an impact on this discussion.

your DVC, are EACH coil 8ohms or 2 ohm? how are you getting 4 ohms out of it.

because it is 8ohm coils each, this will greatly effect how this ends 

if it is 2 ohms each then you only have one choice, 2ch, not bridged.

--------------------------------------------------------------------------------

I think where alot of the confusion is how you determine ohm load when bridged.

1) you first measure your DC ohm load. 
2) divide by 2

that is it. whatever your ohm load is, the amplifier will see half that as the load if it is bridged. so if it you have a 4ohm load, as measured by a meter. then bridged the amplifer sees 2 ohms. in your case, not a good idea.

if it is an 8 ohm load, as measured. then the amplifier sees 4 ohms.


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## Justin Zazzi (May 28, 2012)

DonutHands said:


> This does not make sense in my head, are are saying that if the amp does 2x25 @ 4 ohm, total 50watts. When bridged, it makes 1x100 @ 4 ohm, total 100 watts.
> 
> How does the amp create an extra 50 watts when bridged at the same ohm load? I understand this when the ohm load drops, but when it stays the same??? Seems like magic to me.


Look at the equations you wrote out in your earlier post. You had the math all correct. I think you're missing the part where voltage is *squared*, and that bridging channels *doubles* the voltage that the speaker receives. Try writing it out on paper, or using another method to visualize it more if that helps.

Using your example, if you double the voltage by bridging two channels, you get 4x the power because voltage is a squared term. But to do that, you consume two channels and turn them into one channel (bridged), so you loose half the power output because of the N term. The result is twice the power output (half of 4x), and this is where the "extra 50 watts" comes from in your example above.

Just keep in mind this is if the impedance stays the same, which is not your original question.


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## minbari (Mar 3, 2011)

Bayboy said:


> This is getting too complex. Plain and simple the power is merely combined from two channels.* Running an 8 ohm load bridged does not halve the power of 4 ohm bridge*. It will still sum the power of 2 channels in 4 ohm stereo. 4 ohm bridged sums the power of 2 channels in 2 ohm stereo.
> 
> *The problem comes in when you're thinking that 8 ohm bridged is half of 4 ohm bridged. * That's not how bridged amps work. You're still combining two channels
> 
> ...


that is exactly how it works! if you get 200watts @ 4ohms bridged, then you will get 100 watts @ 8 ohms.. simple ohms law and watts law. that amplifier can only put out a set amount of voltage per channel. when you bridge the channels you playing one channel inverted from the other, so you double the voltage, but into a single channel. if you double the load, you halve the wattage.


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## DonutHands (Jan 27, 2006)

minbari said:


> I have a question that I dont thinkn anyone has asked and will have an impact on this discussion.
> 
> your DVC, are EACH coil 8ohms or 2 ohm? how are you getting 4 ohms out of it.
> 
> ...


Its in the first post. 4ohm DVC (each coil is 4 ohm)


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## minbari (Mar 3, 2011)

DonutHands said:


> Its in the first post. 4ohm DVC (each coil is 4 ohm)


ok, I must have just missed it.

in that case, makes little difference.

4ohm stereo or 8ohm bridged will give you the same power. I would lean towards running it stereo just to avoid running the coils in series. (doubling the series inductance.)


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## Justin Zazzi (May 28, 2012)

minbari said:


> ok, I must have just missed it.
> 
> in that case, makes little difference.
> 
> 4ohm stereo or 8ohm bridged will give you the same power. I would lean towards running it stereo just to avoid running the coils in series. (doubling the series inductance.)


Inductance worries on a subwoofer? Why would you be concerned about that?


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## minbari (Mar 3, 2011)

Justin Zazzi said:


> Inductance worries on a subwoofer? Why would you be concerned about that?


depends on how high it is. I have seen subs that in series wouldnt play above 60hz.

QTS also changes and you get 2x the BL.


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## ParDeus (May 10, 2014)

Holmz said:


> What power?
> It would be better to talk voltage until one has the speaker impedance to work out the current and power.
> 
> The amp does not provide power, just the capability to provide power.


Noted.


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## DonutHands (Jan 27, 2006)

Ok, i guess ill just accept it that this Alpine PDX will make the exact same 300 watts bridged at 4ohm or at 8ohm. 

I thought that resistance changed things, but i guess everyone says it does not.


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## minbari (Mar 3, 2011)

DonutHands said:


> Ok, i guess ill just accept it that this Alpine PDX will make the exact same 300 watts bridged at 4ohm or at 8ohm.
> 
> I thought that resistance changed things, but i guess everyone says it does not.


The main reason for the PDX and other amplifiers like it (JL rips) is because it actually changes the rail voltage based on the load. So ohms law still applies, BUT it is massaging it a bit 

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## Jscoyne2 (Oct 29, 2014)

Hm. Maybe I'll get one of those for my 8ohm 10s in the kicks. How's the SQ?

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## minbari (Mar 3, 2011)

Jscoyne2 said:


> Hm. Maybe I'll get one of those for my 8ohm 10s in the kicks. How's the SQ?
> 
> Sent from my XT1710-02 using Tapatalk


It will only adjust up to 4 ohm stereo and 8 ohm bridged

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