You should always use the recommended load for
your amplifier. However, if you must use a load other than the recommended load, here is
a short guide.

**Solid-state Amps:** Most solid-state amplifiers would rather look at
an open circuit (no speaker at all) than a load. Therefore, you can usually use a load
that is higher than the recommended load. Do not use a lower impedance load, as this
could cause serious damage to your SS amp.

**Tube Amps:** On the other hand, most tube amplifiers need a load to
avoid transformer or tube socket damage. If a mismatch can not be avoided on a tube amp,
it is better to go towards a lower impedance rather than too high of a speaker impedance.
This may stress the tubes some, but, tubes are cheaper than transformers.

The impedances **add** in a series connection. If speaker A equals 8ohms
and speaker B equals 8ohms, the total load is A plus B, or 16ohms. If you string three 8
ohm speakers in series the net impedance will be (3X8) 24 ohms.

You can use speakers with unlike impedances and as before, you simply add to get the total load. When you use unlike speaker impedances in a series connection, the larger impedance will have to dissipate more watts than the smaller impedance, and if the sensitivity ratings are the same, the speaker with the larger impedance will also be louder. This is not a problem if you assure that each speaker is properly rated and you don't mind that one is louder than the other.

Example:

Speaker A and B are each rated at 25 watts. Speaker A=8ohms and
speaker B=4ohms. A total of 50 watts is delivered to the circuit. In this example the
8ohms speaker will be asked to dissipate about 33 watts, while the 4ohm speaker will need
to dissipate only 16 watts or so. As you can see, this could be a real problem for
speaker A. The roles are reversed in a parallel circuit.

If all of the speakers have a common impedance, you can simply divide the
common impedance by the total number of speakers to get the net impedance.

Examples:

Two 8 ohm speakers wired in parallel will have a net impedance of (8 divided by 2) 4
ohms.

If you have three speakers, each at 8ohms, divide 8 by three to get a total impedance of
2.66ohms.

If the speakers do not have a common impedance, you can use formula 1 for two speakers or formula 2 for more than two speakers.

Un-like Impedances Formula 1:

Speaker A=8ohms and Speaker B=4ohms. (8*4)/(8+4)=2.66ohms.

Un-like Impedances Formula 2:

Speaker A=8 ohms, Speaker B=4 ohms, Speaker C=16 ohms.

We'll solve the bottom part of the equation first.

1divided by 8= .125, 1 divided by 4= .25, 1 divided by 16= .0625

.125 + .25 + .0625= .4375

Now the top part of the equation.

1 divided by .4375= 2.2857 ohms.

We'll solve the bottom part of the equation first.

1divided by 8= .125, 1 divided by 4= .25, 1 divided by 16= .0625

.125 + .25 + .0625= .4375

Now the top part of the equation.

1 divided by .4375= 2.2857 ohms.

You must be sure that each speaker is rated for the demand that it will be expected to dissipate.

Example:

We'll use the same example as before, but this time the speakers
are wired in parallel. Speaker A and B are each rated at 25 watts. Speaker A=8ohms and
speaker B=4ohms. A total of 50 watts is delivered to the circuit. The 8ohms speaker will
be asked to dissipate about 16 watts, while the 4ohm speaker will need to dissipate only
33 watts or so. In this case speaker B is being ask to dissipate more watts than it is
rated to do. Also, speaker B will sound louder than speaker A, assuming the sensitivity
ratings are the same for both speakers (and assuming that speaker B is not yet blown
[grin]). Note, this is backwards from the series connection.

Consider the S/P 1 connections. Again we'll assume that each speaker's impedance is 8ohms. Speaker A is connected in parallel to speaker B and together they make up a network equal to 4ohms. Speakers C and D are also connected together in parallel and also make a network equal to 4 ohms. Now, the two networks are connected in series to give us an 8ohm output.

Now consider the example labeled S/P2. Assume that each speaker is 8ohms.
Speakers A and B are wired in series to make up a network that equals 16ohms. Likewise,
speakers C and D make up a series network equal to 16ohms. Now network A/B is connected
in parallel to network C/D. The result is a net impedance of 8ohms out.