# Cabinet impedance question

Discussion in 'Amps and Cabs [BG]' started by skadet, May 2, 2006.

May 2, 2006
Hello,

I have an SVT-410HLF (4x10, 4 ohm speakers, 4 ohm cabnet)

and an SVT15 (1x15, 8 ohm speaker)

driven by a B2-R (4 ohm stable)

Is it safe to run these two cabinets together for a final impedance of 2.7 ohms? I'm assuming not, so what would you do? Re-wire the cabinet?

2. ### joelb79

Mar 22, 2006
Lansing, Michigan
No. Dont do it.

You cant rewire the single 15, so the question is if you can rewire the 410 to make it work.

Heres the options:

All Speakers in Paralell = 2ohms
All speakers in Series = 16ohm

So... basicaly you would be decreasing the wattage in the 410 to 1/4 of our 4ohm power (150 watts), or blowing your head due to too low impedance.

Your best bet is to use the 410 by itself, its a killer cab. And then use the 15 where you dont need as much output and a lighter load to carry.

either that, or get yourself something 2 ohm stable.

3. ### joelb79

Mar 22, 2006
Lansing, Michigan
or you could wire the cabinets themselves in series.

the 15 would get 33% power, the 410 would get 66% power, however your total load would be 6 ohms, which would get you somewhere around 300watts, where as the 410 by itself gets 450 watts.

without doing some math which requires me to know the xmax of all the drivers, i cant tell you that you the 6ohm option would be louder.

May 2, 2006

5. ### ulrich

Parallel would give the 6 ohms etc. that you describe. Series would give 12 ohms.

6. ### Richard Lindsey

Mar 25, 2000
SF Bay Area
It's not a good idea. I say don't do it.

You can't rewire the 4-10 to anything that would really help you. If the drivers are 4 ohm drivers wired series-parallel, then your only options are to wire them all paralle, which gives you 1 ohm (!), or all series, which gives you 16 ohms.

I'd say use the 4-10 alone. Alternatively, you might consider getting a head that can handle a 2.67 ohm load reliably.

7. ### joelb79

Mar 22, 2006
Lansing, Michigan

no. google search ohms law. you are wrong.

paralell always drops the impedance. ALWAYS.
series will always take the two and add them together. but in this case, it will take the 4 ohm load higher, since they are not equal impedances.

8. ### Herman

Dec 25, 2005
Lynchburg, VA
I'm not sure whay you're trying to say. You're right about the first part "series will always take the two and add them together" but I don't understand what you think happens to the 4 ohm cab when it's in series with the 8 ohm cab - it's impedance goes up?

I think your original post had it backwards as well. In the series case with an 8 and 4 ohm cab, the 8 ohm cab will dissipate 66% of the power.

The 8 and 4 ohm cab in series would be 12 ohms total load for the amp.

9. ### ulrich

Fascinating. All the years I've missunderstood ohms. Now that I think about it, it makes perfect sense.

10. ### BassIan

Apr 27, 2003
Cupertino, California
FAQ?

Connecting the two cabs in parallel leaves 2.67 ohms as you know. Connecting them in series leaves them as a 12 ohm load, with 1/3 of power disippated by the 410, and 2/3 of power dissipated by the 15 (this is not the same as with the parallel connection!).

Run one or the other, not both, or see if you can find a head to handle a 2 ohm load if you must run them both.

Either that or connect it anyway, play it, and don't push it. If, as with most new amplifiers, the B2R has thermal protection, it will shut down the amp before anything else goes wrong if it overheats from excessive current.

11. ### fdeckSupporting MemberCommercial User

Mar 20, 2004
HPF Technology LLC
Connecting two _identical_ cabs in series gives you twice the impedance.

Connecting two _different_ cabs in series results in a complete mess, because impedance changes with frequency, and the two cabs are likely to have their impedance peaks at different frequencies.

12. ### joelb79

Mar 22, 2006
Lansing, Michigan
i dont know how an 8 ohm load gets 66% of the power when there is a 4 ohm load with it.

the lower the load the more the power/voltage, so therefor the 4 ohm cabinet gets the greater share of the power.

Series: ZT = Z1 + Z2
Parallel: ZT = (Z1 * Z2)/(Z1 + Z2)

SO yes.. in series they are 12 ohms. My bad:scowl:

but, ohms law states that a higher resistance allows a small amount of current to flow. A lower resistance allows a large amount of current to flow. Resistance is measured in ohms.

So the 4 ohm cabinet gets more power, the 8 ohm cabinet gets less power. More ohms = more resistance to power. the 4 ohm cabinet is resisting the power less than the 8 ohm cabinet.

13. ### Herman

Dec 25, 2005
Lynchburg, VA
It doesn't - it gets 66.7% of the power when in series with a 4 ohm cab.

When in parallel, yes. When in series, the higher impedance dissipates the greater power. Go look up Ohm's Law and you'll see.

14. ### joelb79

Mar 22, 2006
Lansing, Michigan
note i edited my post, but it was replied to before i could copy past all the ohms law into it..

[sarcasm]
And to think, my college professor was wrong. All the things i learn on talkbass.
[/sarcasm]

ohms law of resistance still applies to resistors in series. But i see where you are getting wrong. yess, the higher impedance dissipates the greatest power. But were not creating a heater element which would be dissipating. its.. not... dissipating, rather RESISTING power. therefore, the higher impedance (8ohm) RESISTS more power, meaning it gets 33% of the share of 12 ohms to move its cone.

15. ### Herman

Dec 25, 2005
Lynchburg, VA

Yes, it dissipates power because it resists it. But, in a series circuit, the current flow through both cabs would be equal and determined by the voltage across the two cabs divided by their series impedance (ohm's law). The power that the 8 ohm cab - "sees", "dissipates", "resists", "feels", whatever - is twice that of the power in the 4 ohm cab. Power = Current^2 X Impedance - so for two cabs with the same current going through them, the one with the greater impedance would dissipate (or "see", or "get" or whatever) more power.

The opposite is true when you have an 8 and 4 ohm cab in parallel. With an equal voltage appplied to both, the current through each is now different and determined by Current = Voltage/Impedance. If you had an amp putting 8 volts out to both cabs, the 8 ohm cab would be passing 1 amp while the 4 ohm cab would pass 2 amps. Using the same Power = Current^2 X Impedance you get 8 watts dissipated by the 8 ohm cab, 16 watts by the 4 ohm cab. The dissipation % are reversed from the series case with the lower impedance getting 2X the power.

16. ### joelb79

Mar 22, 2006
Lansing, Michigan
wow.. dude get it right.. and get rid of that word dissipation. your math, correct. but your theory is wrong. resistance/dissipation resists the circuit/power.

your adding resistance, plain and simple. more resistance = less power. regardless of series or parallel, the higher resistance cabs get less power than the lower ones.

how about plain english. in NO WAY will an 8 ohm cabinet get more power than a 4 ohm, regardless of how its wired up.

science experiment: lets take 5,000,000 volts and .5 amps, and i'll let you choose what resistor you want wired up with it.. a 1 ohm or a 1,000,000 ohm resistor, and then i'll attempt to shock you. which one will hurt less. (note, you are 0 ohms, since your grounded and in a bucket of water)

17. ### Herman

Dec 25, 2005
Lynchburg, VA
And this from a guy who, if even for just the briefest moment, thought that 8 plus 4 equals 6.

In my last post, I explained how there's a difference in the power dissipated by each cab depending on whether they're wired series or parallel. What do you think it means when someone says "my amp can put 300W into my 4 ohm cab"? The 300W is the power dissipated by the cab and calculated by either Voltage^2/Impedance OR Current^2 X Impedance. Your logic says to use these equations only when cabs are in parallel and throw them away if the cabs are in series.

Yes, it will when in series. The series circuit creates a voltage divider where 2/3 of the voltage is across the 8 ohm and only 1/3 is across the 4 ohm cab - this is how you have more power across the 8 than the 4.

First, explain to me how you're providing 5 million volts at 1/2 an amp when I get to choose the resistor?

18. ### Mark Reccord

Herman is correct. I don't think it's him that needs to review Ohm's law.

Little example: Say we have a 4 Ohm load and an 8 Ohm one wired in series and an amp is putting out 12V. In series, both loads have the same current passing through them, no? 1A in this case. Power=I^2 * R. So the 8 Ohm load will dissipate (because that's what every load does) 8W and the 4 Ohm load will dissipate 4W.

Or....

The voltage drop over the 8 Ohm load is equal to IR, which in this case is 8V ,so 4V drops across the 4 Ohm load. Power is also V^2/R so we have 64/8=8W and 16/4=4W. It's also VI, which is 8*1=8W and 4x1=4W.

All loads dissipate power. A typical speaker dissipates about 95% of the power put into it as heat and the other 5 or so % as acoustic energy.

19. ### Herman

Dec 25, 2005
Lynchburg, VA
Thanks, Mark - I was beginning to wonder if I'd fallen through the looking glass.

20. ### Mark Reccord

Nope, you've got it dead on.