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Umm...Ohms? Ehh..help

P

Pawn

Aug 5, 2005
26
0
#1
Hello..im new to Guitar amps. Im going to get a used Peavey 115 amp. I understand that it is 200 watts in to 2 ohms, and 150 watts in to 4 ohms, what the hell does this mean? Does this mean i could hook up 2 4 ohm cabs and get an extra 200 watts? or 2 8 ohm cabs and get an extra 150 watts? Please help because i have no clue what this means. :eyebrow:

Thanks :help:
 
#3
Hiya,

I taken it you're new to Talkbass.com as well as amplifiers.

What Bassline Delux is trying to say is that this topic is covered quite a lot on here so check out the sticky at the top of the page about using the search function.

Type in Ohms and cabs and some previous threads will probably come up with your answer. If you're still in trouble PM and i'll see if i can help

Cheers and Beers

Andy
 
#5
its a combo amp, you wont usually use more than one external cab, if you do you would have to daisy chain them

the 115 is 150 watts by itself, the speaker it has is 4 ohms, adding an external 4ohm speaker cabinet will lower the impedance to 2 ohms (4 ohms + 4 ohms in parallel = 2 ohms) and hence mean the amp will put out 200 watts
 
#7
OK, I'll take a wag at it....

Of course, in order to get sound from your amp, you must connect it to a speaker. All speakers have a certain impedance, which, oversimplified, is a sort of electrical "resistance". If the impedance is too low (below 2 ohms), your amp will likely burn up.

If you hook up an 4-ohm speaker (or a connection of multiple speakers which combine to give an overall impedance of 4 ohms), your amp will put out 150 watts of power.

Connect to a 2-ohm speaker (which are actually very rare; more likely, you'll connect to an array of multiple speakers which combine to give 2 ohms total impedance), and it'll pump out 200 watts.

Does that help?
 
#9
I've seen far too much confusion over this. Perhaps this will help.

Ohms is the measure of resistance in a circuit. Resistance calculations can be made for either series or parallel circuits using very simple math, as defined by Thevenin (physics really is fun!) In series circuits, the resistance of a circuit is the sum of the resistors in the circuit. As an example two 10 Ohm resistors in series = 10+10=20 Ohms.

In a parallel circuit, which is what is almost always the case when adding speaker cabinets, the resistance is determined by 1/total resistance = 1/resistance 1 + 1/ resistance 2. As an example, lets consider two 16 Ohm cabinets: 1/16 + 1/16 = 2/16 or 1/8 as such the result is 8 Ohms.

Applied to the resistances typically found in production cabinets, here we go...

Two 4 Ohm cabs 1/4 + 1/4 = 2/4 which reduces to 1/2 as such the solution is 2 Ohms.

Two 8 Ohm cabs 1/8 + 1/8 = 2/8 which reduces to 1/4 as such the solution is 4 Ohms.

One 4 Ohm plus one 8 Ohm 2/8 + 1/8 = 3/8. Huh? To reduce this to 1/x, the solution is found by dividing by top number, which is three in this case, so, 8/3 = 2.67.

Sorry, I'm convinced at some point, we just got to face the math and get over it.
 
#10
Guess what, there is more of this math / physics garbage....

Power amplifiers output levels in solid state amps are primarily determined by the voltage of the amplifiers power supply and the load resistance. A "perfect" amplifier would exactly follow the model of Volts (Squared)/resistance = Watts. It isn't really that simple in practice, due to internal resistances, power supply design, etc. When we break it down once again in pretty simple math, the amplifier's voltage remains a constant, with resistance being the variable. As an example, an amplifier with a 50 Volt power suppy into an 8 Ohm load:

(50*50)/8 = 2500/8 = 312.5 Watts.

Now, let's try that amp (in a perfect world) at 4 Ohms

(50*50)/4 = 2500/4 = 625 Watts

I hope this helps
 
#11
Interceptor said:
I've seen far too much confusion over this. Perhaps this will help.

Ohms is the measure of resistance in a circuit. Resistance calculations can be made for either series or parallel circuits using very simple math, as defined by Thevenin (physics really is fun!) In series circuits, the resistance of a circuit is the sum of the resistors in the circuit. As an example two 10 Ohm resistors in series = 10+10=20 Ohms.

In a parallel circuit, which is what is almost always the case when adding speaker cabinets, the resistance is determined by 1/total resistance = 1/resistance 1 + 1/ resistance 2. As an example, lets consider two 16 Ohm cabinets: 1/16 + 1/16 = 2/16 or 1/8 as such the result is 8 Ohms.

Applied to the resistances typically found in production cabinets, here we go...

Two 4 Ohm cabs 1/4 + 1/4 = 2/4 which reduces to 1/2 as such the solution is 2 Ohms.

Two 8 Ohm cabs 1/8 + 1/8 = 2/8 which reduces to 1/4 as such the solution is 4 Ohms.

One 4 Ohm plus one 8 Ohm 2/8 + 1/8 = 3/8. Huh? To reduce this to 1/x, the solution is found by dividing by top number, which is three in this case, so, 8/3 = 2.67.

Sorry, I'm convinced at some point, we just got to face the math and get over it.
Click to expand...

When you're dealing with two cabs (and it's the rare bass rig that uses more than two cabs), I've always found the multiply/sum calculation method much easier:

Two 4-ohm cabs: 4X4/4+4 = 16/8 = 2 ohms
Two 8-ohm cabs: 8X8/8+8 = 64/16 = 4 ohms
One 8-ohm, one 4-ohm: 8X4/8+4 = 32/12 = 2.67 ohms

Nothing wrong with your method at all; this is just a little more intuitive, and easier to remember (at least for me)....
 
#12
so if i wanted with the peavey i could add

1) 2 4 ohm cabs @ 2 ohms to get a total of 200 watts
2) 4 8 ohm cabs @ 2 ohms to get 200 a total of 200 watts
(if i was crazy)
3) 1 4 ohm cab @ 4 ohms to get a total of 150 watts
4) 2 8 ohm cabs @ 4 ohms to get a total of 150 watts
5) 8!!! 16 ohm cabs @ 2 ohms to get a total of 150 watts
(if i was evean more crazy!)
6) 4 16 ohm cabs @ 4 ohms to get a total of 150 watts

right? and its not really # of cabs but number of speakers right?

Thanks for all the help bigbeefdog,Interceptor,i_got_a_mohawk :hyper:
 
#13
the thing is, its a combo amp (i take its a TNT if those are the wattages you were told)

so you more than likely can only add ONE cab in parallel (because the speaker in the combo is always there, so think of it as an amp that always has a 4ohm cab attatched), as there will only be one spare output

so, you could only really add another 4ohm cab (to give 200watt) or an 8ohm cab (2.67 ohms so say about 175 watts)


And the speakers have thier own impedance (ohm rating), but because of the way they are wired its the cab's impedance you look at, a 4ohm cab coould have 4x10" speakers, each speaker at 16 ohms, but because they are wired in parallel it means the cab (all the speakers in the box together) are 4ohms, because thats what the amp would "see"
 
#15
i_got_a_mohawk said:
of course, you could daisy chain cabs, but thats another story, thats putting the cabs in series instead of parallel, so two 4 ohm cabs in series would give 8ohms, but, i dont want to confuse you :p
Click to expand...

Not always.

In many cases, daisy-chaining cabs actually connects them in parallel.... depends on the particular cab, and it's internal wiring......
 
#16
Pawn said:
so if i wanted with the peavey i could add

1) 2 4 ohm cabs @ 2 ohms to get a total of 200 watts
2) 4 8 ohm cabs @ 2 ohms to get 200 a total of 200 watts
(if i was crazy)
3) 1 4 ohm cab @ 4 ohms to get a total of 150 watts
4) 2 8 ohm cabs @ 4 ohms to get a total of 150 watts
5) 8!!! 16 ohm cabs @ 2 ohms to get a total of 150 watts
(if i was evean more crazy!)
6) 4 16 ohm cabs @ 4 ohms to get a total of 150 watts

right? and its not really # of cabs but number of speakers right?

Thanks for all the help bigbeefdog,Interceptor,i_got_a_mohawk :hyper:
Click to expand...

You have it now. :)

But no, it's not the number of speakers - it is the number of cabs, or "loads". For example, you might have an 8x10 cab that has an overall (cab) load of 4 ohms. Or you might have a 1x12 that is also 4 ohms.

They both factor in the same in the calculation - as a 4-ohm cab, and as a single 4-ohm load when you calculate total connected load.
 
#17
Pawn said:
Im going to get a used Peavey 115 amp. I understand that it is 200 watts in to 2 ohms, and 150 watts in to 4 ohms
Click to expand...

It's a great amp to start with. I wouldn't get too worried about connecting more cabs- I think you'll find that you spend a lot to only get a little louder; 200w just doesn't go that far. It's a good enough amp for practice, and if it isn't loud enough for shows, I'd look at another amp completely and keep this one for practice (saves toting junk around).
 
#20
Pawn said:
so if i wanted with the peavey i could add

1) 2 4 ohm cabs @ 2 ohms to get a total of 200 watts
2) 4 8 ohm cabs @ 2 ohms to get 200 a total of 200 watts
(if i was crazy)
3) 1 4 ohm cab @ 4 ohms to get a total of 150 watts
4) 2 8 ohm cabs @ 4 ohms to get a total of 150 watts
5) 8!!! 16 ohm cabs @ 2 ohms to get a total of 150 watts
(if i was evean more crazy!)
6) 4 16 ohm cabs @ 4 ohms to get a total of 150 watts

right? and its not really # of cabs but number of speakers right?

Thanks for all the help bigbeefdog,Interceptor,i_got_a_mohawk :hyper:
Click to expand...

No, you can't "add" a 2 ohm load *if you're already driving the 4 ohm load represented by your internal speaker*. If you drive a 2 ohm load in addition to a 4 ohm load, you get a total lower than 2 ohms, which is bad. All the speakers hooked up to the amp, *including the internal one* (if you're using that one too)* must add up to a load no lower than 2 ohms.

Thus, if you use the 4 ohm internal speaker, any other cabs you hook up, whether a single cab or more than one, must add up to a total of *no less than 4 ohms*. This means, for example, one 4 ohm cab, two 8s, an 8 plus two 16s, or four 16s.
 

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