NevadaPete, I appreciate reading about this. Please explain the "electronics theory" you're referring to. I wanna have a setup that will be appropriate, balanced, and also undamaged. This is why I am asking all of you guys this info - I'm humble enough to say I don't know what I'm talking about.
So, please, without judging or preaching, can you please help out? I'd like to make sure I'm not buying a cab that I'll f*** up at the first note I play through it. Thank you.
Okay. Yeah sure. With regard to electronic theory, I think the best technician level on-line tutorial on electronics is, “All About Circuits”. The lessons are straight forward and explained in a way that makes them easy to understand. It’s a marvelous effort by author Tony R. Kuphaldt. If you decide to go through the entire tutorial, I think you will find “All About Circuits” to be fascinating, and very enjoyable.
Invalid Link Removed
So I’ll go over some of this with you as it relates to your original post, and your subsequent message to me with the hope that my comments may be helpful. The chapter in All About Circuits that pertains to what you are doing mixing and matching bass cabs would be Chapter 5, “Series and Parallel Circuits”. This is because as you know, when you connect two bass cabs together to drive with one bass head, your almost always going to be connecting those cabs together in what is known as a
parallel circuit.
In the subchapter “Simple Parallel Circuits”, we use “The Equation for Resistance in Parallel Circuits” to figure out what’s called the
load on your bass head / bass amplifier. [Note: There’s a caviet on this at the end of the diatribe]. Some solid state power amplifiers / bass heads will be damaged if you attempt to drive
anything less than a 4 Ohm load with them, so figuring out the load on an amplifier / bass head is a very important calculation.
Equation for Resistance in Parallel Circuits:
R total=
___
1____
1 +
1 +
1
R1 R2 R3
So the load for two 8 Ohm cabs in parallel would be:
(1 ÷ 8) + (1 ÷ 8) = .25
1 ÷ .25 =
4 ohms
The load for two 4 Ohm cabs in parallel would be:
(1 ÷ 4) + (1 ÷ 4) = 0.5
1 ÷ 0.5 =
2 ohms
[Note: Not all bass heads will safely operate with a 2 Ohm load. You must check the manufacturer’s specifications].
What if you want to connect an 8 Ohm and a 4 Ohm cab together? As per the above formula, that would be:
(1 ÷ 8) + (1 ÷ 4) = .375
1 ÷ .375 =
2.66 Ohms
[Note: Not all bass heads will safely operate with a 2.66 Ohm load. You must check the manufacturer’s specifications].
Connecting an 8 Ohm and a 4 Ohm cab together in parallel will turn out to be very important later.
Now that you know what the overall load on your bass head is going to be for two 8 Ohm cabs, for two 4 Ohm cabs or for one 8 Ohm and one 4 Ohm cab, your going to want to know how much power is going to be distributed to the various cab combinations respectively. Obviously, to get this information you can just check the owners manual for your bass head. But a more in depth explanation is useful in this case.
To calculate the power delivered to the various cab combinations, you need to know how much voltage and how much current is being delivered to each cab, and that will depend in part on your bass head. I don’t have the specs available for your SVT III (Non-Pro), so I’m going to guess it’s output is aproximately 35 Volts which will yield aproximately 300 Watts into a 4 Ohm load.
In the subchapter “Simple Parallel Circuits”, under “Ohm’s Law Applications for Simple Parallel Circuits” it shows how to calculate how much
current an 8 Ohm and a 4 Ohm cab will receive, respectively, assuming your solid state bass head will produce 35 volts output. The Ohm’s law application we’ll use to calculate
current will be:
E = I
R
or, voltage divided by resistance equals current.
For an 8 Ohm load or cab at 35 volts:
E = I
R
35 ÷ 8 =
4.375 Amperes
For a 4 Ohm load or cab at 35 volts:
E = I
R
35 ÷ 4 =
8.75 Amperes
The point here is that with the 4 Ohm load or cab, 8.75 Amperes is
twice the current compared to the 8 ohm cab (i.e., 4.375 Amperes) because the load on the amplifier / bass head has been cut in half.
In the subchapter “Power Calculations”, we’ll use one of the Power Equations to analyze the power available to the various loads (cabs) using our imaginary 35 volt output amplifer, which should be pretty close to your SVT III (Non Pro). The power equation we’ll use is I x E = P, or current times voltage equals power, and we’ll obtain the value for the current from the calculations we made just above.
For an 8 Ohm load or cab:
I x E = P
4.375 X 35 =
153 Watts
For a 4 Ohm load or cab:
I x E = P
8.75 X 35 =
306 Watts
The point here is that power has doubled with the 4 Ohm load compared to the 8 ohm load because the current has doubled in the 4 Ohm load, compared to the 8 Ohm load.
Now, let’s use the numbers from our calculations above for our imaginary amplifier / bass head to analyze what will happen when you you combine an 8 Ohm 4 X 10 bass cab with an 8 Ohm 2 X 10” bass cab, as proposed in your original message.
First, how much power were we originally delivering to the single 8 Ohm 4 X 10 cab. The power delivered to the 4 X 10 cab was 153 Watts.
Next, what is the overall load going to be combining the two 8 Ohm cabs? The overall load for the two 8 Ohm cabs (in parallel) will be 4 Ohms.
Now, how much power is going to be delivered to this 4 Ohm (overall) load? The power will be 306 Watts overall into 4 ohms.
So under the above circumstances, with two 8 Ohm cabs in parallel, how much power will be delivered to the 4 X 10 cab in such an array? The power will still be 153 Watts.
How much power will be delivered to the additional 2 X 10 cab we have added to the system. The power delivered to that cab will also be 153 Watts because both cabs are rated at 8 Ohms.
If the power delivered to the 2 X 10 cab is 153 Watts, how much power will be delivered to each 10” loudspeaker in the 2 X 10 cab? The power will be 153 ÷ 2 =
76.5 Watts per loudspeaker.
If the power delivered to the 4 X 10 cab is also 153 Watts, how much power will be delivered to each 10” loudspeader in the 4 X 10 cab? The power will be 153 ÷ 4 =
38.25 Watts per loudspeaker.
What is wrong with this picture?
What’s wrong is that as long as 2 cabs are both 8 ohms, power will be distributed evenly between them. Yet, one cab has 4 ten inch loudspeakers while the other has only 2 ten inch loudspeakers. If you bring the 4 X 10 up to volume (assuming you had an amplifier with the power to do that), you would be pounding the snot out of the 2 X 10 because each loudspeaker in the 2 X 10 cab will always be receiving precisely twice the power as each loudspeaker in the 4 X 10 cab. Thus, at any volume level, the entire loudspeaker array is always going to be totally out of balance.
Ideally, with a loudspeaker array consisting of six 10 inch loudspeakers, you want all the loudspeaker cones to move (produce sound) in unison. In this case, however, because the loudspeakers in the 2 X 10 cab are receiving twice the power as the loudspeakers in the 4 X 10 cab, the loudspeakers in the two cabs will not be moving in unison as they should be, and therefore, the loudspeakers in the array will not couple together properly. This will cause the sound coming from such an array to not be as “solid”, or, in other words, to not have as much “depth” as it would at the same volume if all six loudspeakers were moving in unison. This detrimental effect will also vary depending on the musical note playing through the array and will no doubt prove to be unperdictable and impossible to control.
Can you play out with a system like that? Sure. Will it ruin the show? No, I doubt it. Will it sound as good as it would if the two cabs were properly matched regarding how much power they are each receiving? No, an out of balance set up like that will never sound as solid as a properly configured setup, and the louder it is played, the worse it will get. Some will say we’re talking about a subtle difference here. That is a matter of opinion. There will be a difference.
So, how do you properly configure a 2 X 10 cab with a 4 X 10 cab for them to be able to play well together? You must ensure that
each loudspeaker in each of the two cabs receives the same power. To do that, obviously, the array must be set up so that the 2 X 10 cab receives one half the power that the 4 X 10 cab receives. This may be accomplished using only one amplifier by selecting a 4 X 10 cab with a rated impedance of
4 Ohms, and paring it with an
8 Ohm 2 X 10 cab (i.e., a cab with the same or very similar loudspeakers as the 4 X 10). In that case, the impedance of the entire array will be
2.66 Ohms as calculated above, and an amplifier will have to be selected that can safely drive such a load.
A Mesa D800 Series bass head or something like it would be an ideal bass head to drive a 2.66 Ohm setup as described just above. Unfortunately, you would have to acquire a 4 Ohm 4 X 10 Powerhouse cab to be able to safely use it with the 8 Ohm 2 X 10 Powerhouse cab you already have. A Mesa D800 Series head would be ideal to use with either your 2 X 10 Powerhouse or with your 4 X 10 Powerhouse (nice cabs!), but I don’t personally think it would be a good idea to use a D800 Series head with both of them together for the reasons mentioned above. I am not associated with Mesa in any way.
Attempting to use an Ampeg SVT III (Non Pro) bass head is going to be problematic because such a bass head cannot hope to drive either your Powerhouse 4 X 10 (rated at 600 Watts) or your Powerhouse 2 X 10 (rated at 400 Watts) to anywhere near their potential full volume output. The SVT III (Non Pro) is, however, known as being a particularly sweet sounding bass head and there’s no reason not to use it with one or the other of your cabs, as long as it plays loud enough.
All I’ve done here is reiterate what agedhorse has already said, but in more detail. Tried to, anyway.
Caviet:
Everything above the sentence that says, “What is wrong with this picture?” is incorrect. Technically incorrect, anyway. The formulas cited above are for DC current and are ordinarily used, for example, to calculate current, power, resistances, etc., in a DC circuit. The signals delivered to a musical instrument amplifer, and the signals such an amplifier delivers to a musical instrument loudspeaker are AC, of course, and AC circuits have impedance. Not resistance. Resistance is a component of impedance so it is similar to impedance in some ways, but it is not identical by any means. In AC circuits, impedance varies with frequency on account of something called reactance. Therefore, the maths required to calculate factors related to impedance are complex, and the results ordinarily also have to be plotted out and graphed to be useful. For daisy chaining musical instrument loudspeaker cabinets, PA arrays, Hi-Fi’s, etc., the math cited above works plenty well enough for what we’re using it for, and that’s why it’s universally adapted for these purposes. Just know there’s more to the story.
To see for yourself, use an Ohm meter and measure a loudspeaker rated at 8 Ohms impedance. It it really 8 Ohms? Try measuring a loudspeaker rated at 4 Ohms. Is it really 4 Ohms?
