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Ohms FAQ

Dang, great stuff! But after reading all of that I can STILL manage a question. I will have (this week) a Mesa 400+ 2,4,8 ohm outputs. Connect this to a 410 8o, a 15" 8o, and an 18" 4o.

Daisy chain the 2 8o's out of the 4o output and the 4o out of the 4? OR daisychain them all out of the 2o? Hmmm? Someone knows the answer.
 
Intune said:
Dang, great stuff! But after reading all of that I can STILL manage a question. I will have (this week) a Mesa 400+ 2,4,8 ohm outputs. Connect this to a 410 8o, a 15" 8o, and an 18" 4o.

Daisy chain the 2 8o's out of the 4o output and the 4o out of the 4? OR daisychain them all out of the 2o? Hmmm? Someone knows the answer.

If you start daisy chaining offa different taps on that amp, I'm sure your new 400+ will become a doorstop right quick.

Pick one of the sets and work from that.

When I was looking at an uneaven load 2.67 ohms, I got different advice about either using the 2 or 4 ohm outs...it's a different beast than a solid state amp. Anyway, I believe it was Jorg who told me that he didn't think the 400+ was strong enough to run at 2 ohms reliably...and he's a guy who would know. I'm currently running 1 or 2 8ohm cabs...and it's plenty loud. You could run all those cabs off the 2 ohm taps, but I can't imagine needing that much.
 
I have an Ampeg svt 610 cab. I am looking for a head now. The cab is 4 ohm @1200 watts. I have been looking at peavey amps (I'm poor). Lets say a peavey t max, wich is rated for 500@2 ohms, and 350@4 ohms, I would have to have another cab to get to two ohms, right? There are sevral speaker inputs on the back of the cab. What is the speakon input for? Whould the t max or a firebass be a good fit? I had a mesa 400+ head and didnt really care for the sound. Thanks Joe
 
You would need to add another 4 ohm cabinet to get 2 ohms, but I wouldn’t get too hung up over trying to get maximum power out of the amp. You can barely hear the difference between 350W and 500W (less than a 3dB difference). Adding another cab makes no sense unless it’s efficiency is close to the Ampeg’s.

Speakon's are higher current connectors than the usual 1/4" jack, and are preferred for higher power rigs.
 
however, adding a second cab that had equal efficiency to the ampeg would give you a theoretical maximum of a 6 db increase in volume. i wouldn't go mental about it. see how it sounds all by its lonesome, but see if you can do that before you buy!!!!!
 
Theoretically, yes, but going from 350 to 500 watts and doubling cab efficiency (by adding a second, equal efficiency cab) only gives a 4.4dB boost. You only get the full 6dB boost if the amp can truly deliver twice the power with the load impedance halved. So there’s really three things to look at here.

1. What’s the cost of adding a second cab? Don’t forget the cost of hauling around the extra bulk to every gig. Mike Dimin once told me that he doesn’t get paid to play, he gets paid for cartage.

2. What’s the cost of buying more power? 970W into 4 ohms also gets a 4.43dB boost, but the higher power amp is much easier to move than a second large cab.

3. Do you really need 3dB more volume capability?
 
IvanMike said:
Also, if you hook up a 4 ohm cabinet and an 8 ohm cabinet to an amp that can power a 2 ohm load, the 4 ohm cabinet will get twice the wattage as the 8 ohm one. Depending of the sensitivity and size of the cabinets, you may notice a significant difference in volume between the 2, or none at all.

Mike, I just wanted to expand on this a bit, as I have seen
this question asked over and over, and have seen people
erroneously post that the speakers in such a circuit
get the same power.

The scenario is as follows: Poster X asks if he can
use a 8 ohm and a 4 ohm cab in the same circuit of his
head. AnswermanMike posts the above answer same as yours. Poster Y hops in and tells him the speakers get the same power. We sigh ...

We than have to trot out Ohm's law, which hasn't yet
been repealed since we last checked.

===========


V = I x R
or Voltage in a circuit is defined as
Voltage = Amps x Resistance.
Example:
40 Volts = 10 Amps x 4 ohms

or for example
R = V/I 4 ohms = 40volts/10amps
and 8ohms = 40volts/5 amps

Now Power is defined as:
P = I x V
Power = Amperage x Voltage

Since v = I x R then we substitute that into the equation.

V = I^2 x R

Substituting our example values above in your circuit
with a fixed voltage input and two different impedance values, the value of each branch calculates out to be:

P = 10^2Amps x 4ohms = 400 watts and
P = 5^2 Amp x 8 ohms = 200 watts.

Thus the 4 ohm section of the circuit gets twice the power
in the same circuit with a fixed voltage such as the output
circuit of your amp.

Q.E.D

Of course in the real world there are lots more variables,
but I've seen this basic question asked time and again.
 
But if I understand the example correctly, in the post that occasioned your response, the 4 ohm cab had *two* 8 ohm drivers in it, whereas the 8 ohm cab had only one. If the 4 ohm *cab* gets twice the power, but that cab has two drivers rather than one, this would suggest that each *driver* gets the same power, which I think was the original assertion.
 
44me said:
What’s the cost of adding a second cab? Don’t forget the cost of hauling around the extra bulk to every gig. Mike Dimin once told me that he doesn’t get paid to play, he gets paid for cartage.


I've never shown up to a gig. brought my amp in, showed it to the owner of the venue, shoved it back in my car, and been paid. They always seemed to want me to make music too. Go figure. ;)
 
44me said:
1. What’s the cost of adding a second cab? Don’t forget the cost of hauling around the extra bulk to every gig. Mike Dimin once told me that he doesn’t get paid to play, he gets paid for cartage.

I agree with him.

I play for fun...the club pays me for:

-the drive
-loading gear
-the wait before and after playing



...that's the real work.
 
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Sure it’s a slight exaggeration, but there’s a lot of wisdom there. I’ve never met anyone that said they’re in this because they love to haul heavy equipment, but I’ve met plenty of people that will show up to a jam and play for free through the house rig. I look at every gear choice I make as a tradeoff between how it’s going to alter my sound vs. the effort to haul it night after night. After a bunch of gigs, even the purchase price becomes secondary.

- John
 
Richard Lindsey said:
But if I understand the example correctly, in the post that occasioned your response, the 4 ohm cab had *two* 8 ohm drivers in it, whereas the 8 ohm cab had only one. If the 4 ohm *cab* gets twice the power, but that cab has two drivers rather than one, this would suggest that each *driver* gets the same power, which I think was the original assertion.

I was quoting the original post, last paragraph of Post #1.
It only mentions the cabinet, not the driver configuration.

It is true that in any circuit, you can analyze and compute
the discrete Voltage, Amperage and Resistance of each
branch of the circuit. And the Power of each branch as well,
all on a theoretical basis.

Assuming that that case you mention was the case, that
would be true. But I was referring to the cabs, not the
individual components and drivers. Sorry if I was not clear
enough about that.

The paragraph in question did not specifiy that, and I did not
intend to muddy the issue by analyzing individual potential
driver configurations within each cab.

The overall resistance of the circuit still comes out to be
2.67 ohms, and twice as much Power is manifested in the
4 ohm branch (or cabinet). That overall 2.67 Ohm figure is a
problem for Solid State Amps that require a minimum
4 ohm load.

I just wanted to restate this simply for the benefit of those
who wonder over and over again why this is so, as is seems
to come up constantly.

BTW, Richard, I have seen some of your posts from time to time, and always found them thoughtful and incisive ...
 
Thor said:
I was quoting the original post, last paragraph of Post #1.
It only mentions the cabinet, not the driver configuration.

It is true that in any circuit, you can analyze and compute
the discrete Voltage, Amperage and Resistance of each
branch of the circuit. And the Power of each branch as well,
all on a theoretical basis.

Assuming that that case you mention was the case, that
would be true. But I was referring to the cabs, not the
individual components and drivers. Sorry if I was not clear
enough about that.

The paragraph in question did not specifiy that, and I did not
intend to muddy the issue by analyzing individual potential
driver configurations within each cab.

The overall resistance of the circuit still comes out to be
2.67 ohms, and twice as much Power is manifested in the
4 ohm branch (or cabinet). That overall 2.67 Ohm figure is a
problem for Solid State Amps that require a minimum
4 ohm load.

I just wanted to restate this simply for the benefit of those
who wonder over and over again why this is so, as is seems
to come up constantly.

BTW, Richard, I have seen some of your posts from time to time, and always found them thoughtful and incisive ...

You know what, it was my error in this case--I responded to the wrong thread. I'd meant to post this response to this thread:

http://www.talkbass.com/forum/showthread.php?p=2167668#post2167668

where MuzikMan refers to having two drivers in the 4 ohm cab and one in the 8 ohm cab. I think I looked at the two threads in close succession and just got mixed up. Sorry about that!

Thanks for the kind words, BTW.
 
Richard Lindsey said:
You know what, it was my error in this case--I responded to the wrong thread. I'd meant to post this response to this thread:

http://www.talkbass.com/forum/showthread.php?p=2167668#post2167668

where MuzikMan refers to having two drivers in the 4 ohm cab and one in the 8 ohm cab. I think I looked at the two threads in close succession and just got mixed up. Sorry about that!

Thanks for the kind words, BTW.

No problem. That kind of brain fart happens to me all the time.
:eek: And you're welcome, I just never a chance to mention
it, but you earned it, imho. ;)
 
I am pretty sure this is alright but I just want to double check with you bass gurus to be sure I am not going to fry anything:

Here are my stewart specs:

FTC Power Rating
<0.1% THD
20Hz-20kHz
both channels driven 200W x 2 @ 8 Ohms
350W x 2 @ 4 Ohms

Can I run one 4ohm cab on one channel and one 8ohm cab on the other channel?
 
TORIN said:
I am pretty sure this is alright but I just want to double check with you bass gurus to be sure I am not going to fry anything:

Here are my stewart specs:

FTC Power Rating
<0.1% THD
20Hz-20kHz
both channels driven 200W x 2 @ 8 Ohms
350W x 2 @ 4 Ohms

Can I run one 4ohm cab on one channel and one 8ohm cab on the other channel?

Yes, if the amp performs up to its specs, which it should.
 
First off, great thread. Very informative. But as usual I am still left with a question:

I have a 300PRO Fender all tube head and a matching 2x15 cab (8ohms, 96dB sensitivity). I wanted to hook this up with my ampeg 8x10 (4ohms, 100dB sensitivity). The two cabs would give me a total impedance of 2.67 ohms. The fender amp has a switch for 2,4, and 8 ohms so I would set it to 2ohms for this application, correct?

Also, the fender amp has a primary speaker cabinet output, and a secondary cabinet ouput. Would using both outputs be the same as "daisy-chaining" the two cabinets directly?

Thanks!
Justin
 
StingrayKid21 said:
First off, great thread. Very informative. But as usual I am still left with a question:

I have a 300PRO Fender all tube head and a matching 2x15 cab (8ohms, 96dB sensitivity). I wanted to hook this up with my ampeg 8x10 (4ohms, 100dB sensitivity). The two cabs would give me a total impedance of 2.67 ohms. The fender amp has a switch for 2,4, and 8 ohms so I would set it to 2ohms for this application, correct?

Also, the fender amp has a primary speaker cabinet output, and a secondary cabinet ouput. Would using both outputs be the same as "daisy-chaining" the two cabinets directly?

Thanks!
Justin


I'd talk to fender to get their input. tube amps are a different animal alltogether. many times they do ok with a 100% impedance mismatch either way, but this is still non-optimal. In many cases operating at an impedance higher than what is selected on the amp raises the plate voltage, which can do bad things if it gets high enough.
 

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