The most commonly asked question in guitar amp world is, in our experience this, “Hey dude, what impedance do I set my tube head to when I’m using two cabs?”
Before we can answer that question, we need to discuss a little thing called Ohms Law. Well, actually we really don’t need to discuss it as such; all we really need do is remember is that there are two formulas that enable us to find out the combined impedance of speakers (or cabinets) when joined together in either Series or Parallel.
The following equations help you match the impedance of speakers to amplifiers for optimized performance (avoiding overloads and other issues). Impedance (Z) is how much a device resists the flow of an AC signal, such as audio. Impedance is similar to resistance, which is how much a device resists the flow of a DC signal. Both impedance and resistance are measured in Ohms
For ease of understanding, we’ll start with series calculations:
R = resistance (the ohm rating of your loudspeaker)
T = total
THE FORMULA: RT = R1 + R2 + R 3 etc….
If we have 4 speakers, each with a 4 Ohm rating, using the above equation for our example gives: 4+4+4+4 = 16, RT = 16 Ohms
So, in this case 16 Ohms of resistance is presented to the amp, or in other words, the output current of the amp would meet with 16 Ohms of load resistance.
THE FORMULA: 1/Rt = 1/R1 + 1/R2 + 1/R3 + etc.
To keep life as simple as possible, most people put enclosures of the same impedance in a parallel circuit (wiring separate enclosures in series is kind of a pain). If you do this it’s all just a matter of dividing that impedance by the number of speakers. If you connect speakers of different impedances, the power output will be greater to some, less to others, which means some will be louder than others. (In higher tech circles, we commonly refer to this condition as “very not good.”) Example; Two 16-ohm loads in parallel = 16 / 2 = 8 ohms (Similarly, two 8-ohm loads in parallel = 8 / 2 = 4 ohms).
Let’s look at the Parallel law in action. This is the one you need to be most familiar with because when it comes to hooking-up cabinets to combos or heads it is almost always done in parallel.
Getting back to our original question, “Hey dude, what impedance do I set my tube head to when I’m using two cabs?” Well, let’s take a look shall we…
Let’s suppose that the dude in question has a Marshall DSL100 head and a Marshall 1960A and 1960B.
1. Make sure both cabinets are switched to Mono operation.
2. Make sure you have two SPEAKER cables to do the hook-up with (NEVER use guitar cables).
3. Plug one end of one lead into the 16 Ohm, MONO input of the 1960A (NOT the 4 Ohm input…remember, there are two due to the stereo option).
4. Plug one end of the other cable into the16 Ohm, MONO input of the 1960B.
5. Since we’re hooking two 16 Ohm cabinets to our DSL100 head in parallel, let’s plug R1 (16) and R2 (also 16) into our trusty formula:
1/RT = 1/16 + 1/16, RT = 8 ohms
6. Set the 4 Ohm/8 Ohm switch below the two parallel speaker outputs on the back panel of our DSL100, plug the speaker cable ends into said outputs….
7. Switch the amp on, plug in your axe, dial in a tone and rock!!
And there you have it. In fact, a simplified offshoot of the Parallel law we’ve just learned is this:
Whenever you’re connecting two cabinets of the same impedance (and let’s call that impedance “X”) in parallel, the resulting impedance is X/2.
So, a 16 & 16 = 8; 8 & 8 = 4; 4 & 4 = 2, etc…