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Impedance Fender Amp Sees w/Ext. Spkr.

Posted: 28 Feb 2010 11:36 am
by Jeff Watson
What impedance is a Fender Twin seeing if your adding an 8 ohm extension speaker to the existing 4 ohm load? 6 ohm? Thanks

Posted: 28 Feb 2010 11:43 am
by Dave Mudgett
This is equivalent to 3 8-ohm speakers in parallel, so the equivalent nominal impedance is 1/({1/8}+{1/8}+{1/8}) = 8/3 or 2-2/3 Ohms.

When you add a load in parallel, it always lowers the impedance.

Posted: 28 Feb 2010 2:11 pm
by Paul Graupp
Dave; Correct me if I'm wrong but doesn't the twin have a switched jack for external speaker that adds a different tap on the output transformer secondary ?? I only have a partial recollection of this and no schematics but it is a sticking point to my mind.

As I said correct me if I'm wrong and I'm on a streak lately...

Regards, Paul

Posted: 1 Mar 2010 9:27 am
by Gary Jones
Here is a link to a web site that has a speaker impedance calculator, if you need such a thing. http://www.about-guitar-amps.com/speake ... lator.html
Dave is right, the total impedance the amp would see is 2.67 Ohms.
Also, keep in mind when you have speakers in parallel that the speaker with the lowest impedance will get the most power. That is to say that if you have a 4 Ohm, and an 8 Ohm speaker connected together, the 4 Ohm speaker will get twice as much power, so you may see a volume difference between the 2 cabinets

Gary Jones

Posted: 1 Mar 2010 11:18 am
by Dave Mudgett
Paul - some of the later Twin Reverbs like the Twin Reverb II may have had a multi-tap output transformer, but the classic Twin Reverb design definitely didn't. Of course, that doesn't change the load impedance presented, but simply how it matches with the output impedance of the amp.

The total impedance Z[sub]T[/sub] of N series impedances is Z[sub]T[/sub] = Z[sub]1[/sub] + Z[sub]2[/sub] + ... + Z[sub]N[/sub] - i.e., the impedances simply add to give the total impedance.

The total impedance Z[sub]T[/sub] of N parallel impedances is Z[sub]T[/sub] = 1/{(1/Z[sub]1[/sub]) + (1/Z[sub]2[/sub]) + ... + (1/Z[sub]N[/sub])} - i.e., the reciprocals of the impedances add to give the reciprocal of the total impedance.

The output impedances of both a tube amp and speaker are complex, having both resistive and reactive components. The rule about electrical power transfer is that maximum power transfer is achieved when the load impedance is equal to the complex conjugate of the output impedance of the amp - http://en.wikipedia.org/wiki/Impedance_matching. For purely resistive loads, this means that the load resistances are the same, but this is more complex for a reactive load. The power transferred is given by this expression

P = {(1/2)*|V|[sup]2[/sup]*R[sub]L[/sub]}/{{R[sub]S[/sub] + R[sub]L[/sub])[sup]2[/sup] + (X[sub]S[/sub] + X[sub]L[/sub])[sup]2[/sup]}

where V is the applied voltage, X is reactance (S=source, L=load) and R is resistance (S=source, L=load). So maximum power is transferred when

(1) X[sub]S[/sub] = -X[sub]L[/sub]
and
(2) R[sub]S[/sub] = R[sub]L[/sub]

Remember that purely capacitive reactance and purely inductive reactance have different signs, so conjugate reactance matching is possible in principle, but not generally achieved in practice. But regardless of whether or not reactance matching is achieved, for a given source and load reactance, power transfer is generally achieved when source and load resistance are matched. There is a possible second-order effect that changing the load resistance may change its reactance in a favorable or unfavorable way, but it is generally a good idea to match source and load resistances.
Also, keep in mind when you have speakers in parallel that the speaker with the lowest impedance will get the most power. That is to say that if you have a 4 Ohm, and an 8 Ohm speaker connected together, the 4 Ohm speaker will get twice as much power, so you may see a volume difference between the 2 cabinets.
This can get somewhat complicated by the complex load impedances, but to the extent that the resistive impedances dominate, this is correct.

Posted: 1 Mar 2010 2:23 pm
by Paul Graupp
Thanks Dave: Exactly what I was trying to recall.

Regards, Paul