# Series/Parellel Cab Hookup

Discussion in 'Amps and Cabs [BG]' started by Larzito, Oct 20, 2004.

1. ### Larzito

Aug 1, 2000
Dallas, Texas
Scenario 1
Speakon out of the head into cab1, speakon out of cab1 to cab2. This hookup is in series with both cabs 8ohms, so I have a 4ohm load, right?

Scenario 2
Speakon out of head into cab1, speakon out of head into cab2. This hookup is also series with both cabs 8ohms, for a total load of 4ohms, right?

I get a little foggy on the whole impedence thing from time to time. Anyone care to explain how the loads work?

2. ### robb.

you got everything right except for terminology. only in rare cases will cabinets be connected in series. in both of the scenarios you described, the connection is in parallel. 8 // 8 = 4 (that's 8 Ohms in parallel with 8 Ohms makes a 4 Ohm equivalent load).

robb.

3. ### buzzbassShoo Shoo Retarded Flu !Supporting Member

Apr 23, 2003
NJ
no, the parallel hookup would yield a 2 ohm load, be careful, most amps don't like 2 ohm loads very much. Speakers are like big resistors. when you put them in parallel, the resistance is halfed (sp?). Speakers in series are added, so 2-8 ohm speakers in series would be a 16 ohm load.

4. ### natrab

Dec 9, 2003
Bay Area, CA
Both of those scenarios would yield 4 ohm loads, however they would be parallel connections. Notice most dual jacks in the back of amps or cabs are "parallel" just for this purpose. So you're almost always halving the Ohms when you put two equal ohmed cabs together into an amp.

5. ### Christopher

Apr 28, 2000
New York, NY
Scenario 1 is a parallel connection, even though it might seem like a series connection. Whether a cab runs in parallel or series depends on the internal wiring of the cab. Merely running the signal out of one cab and into another does not change a parallel-wired cab into a series-wired one.

Scenario 2 is also a parallel connection.

The way to figure out total impedance for parallel connections is: (1/R1) + (1/R2) + (1/R3) + ... = (1/Rt)

In this case (1/8 ohm) + (1/8 ohm) = (2/8 ohm) = (1/4 ohm). 2 8 ohm cabs in parallel presents your amp with a 4 ohm load.

The series calculation is simpler: R1 + R2 + R3 + ... = Rt. If you have a series connection (which is rare, as rcz notes), 2 8 ohm cabs will present your amp with a 16 ohm load.

6. ### Larzito

Aug 1, 2000
Dallas, Texas
Thanks for the formula...I knew I had seen it before.

Now, I have heard that with it is possible to hook two 4ohm cabs together in series to present the amp with an 8ohm load. Edens with binding posts for example. How would one accomplish this?