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Out-put jacks on a HD 130 reverb

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MikeST
Posted on Friday, September 10, 2004 - 01:20 pm:   

I just got an HD 130 reverb head in great shape.
Still building my speaker cabinet.
Why are there two out-put jacks on the back of the head. Panel says " Output jacks are wired in series" or something very similar.
What does this mean?
I expected to find one output jack.
Forgive me if this is a dumb question
Mike / Chicago
Doug Elick
Posted on Friday, September 10, 2004 - 02:40 pm:   

There are two output jacks so you can run two cabs (that do not have "through" jacks themselves)at the same time.

Since the jacks are series wired, if you plug a 4 ohm cab into each jack (total of two cabs), the total impedance will be 8 ohms. Select the "8 Ohms" postion of the switch on the rear of your amp to correctly match the speaker load to it's output transformer.

Use the "main" jack unless you have two cabs to run.

Doug
Steve Kennedy (admin)
Posted on Tuesday, November 09, 2004 - 01:48 pm:   

All Music Man heads and 2-speaker combos (theoretically) have their speaker jacks wired in series. If you use 2 8-ohm cabinets, each one plugged into a jack, they will be wired in series which will exhibit a 16 ohm load to the amp.

You could probably get away with running the amp in the 8-ohm position, but it would be better if the cabinets themselves had parallel "pass-through" jacks (as many do). You could plug one 8-ohm cabinet into the other 8-ohm cabinet for a total impedance of 4-ohms, then put the amp's impedance switch to the 4-ohm position.

There isn't any magic here, it is simple math (known as "Ohm's Law"). Speakers in series, add the impedances together:

8-ohms + 8-ohms = 16 ohms

If running 2 speakers (same impedance) in parallel makes a total impedance of 1/2 that of a single speaker, so:

8-ohms in parallel with 8-ohms = 4-ohms total

When paralleling 2 speakers of unequal impedance, the following formula must be used:

R1 x R2 / R1 + R2 (Known as "Product over the Sum"). In the case of a 4-ohm and an 8-ohm in parallel:

4 x 8 = 32 / 4 + 8 = 12
So, 32/12 (32 divided by 12) = 2.66 ohms total.

In parallel connections, the total impedance is ALWAYS less than the value of the smallest impedance. In series connections, the total is ALWAYS higher than any of the impedances and is always the sum of all of them.

There are two links in the "Speaker Info" topic on this site that should help:

http://www.abrown.com/spkrwire.htm

4-Speaker Cabinet Wiring


Steve



(Message edited by admin on November 09, 2004)
Fabrizio
Posted on Wednesday, January 18, 2006 - 12:47 am:   

Hi
i got a hd130 reverb (head).
My question is :
can i plug my head into a cab in stereo mode (for example a Marshall 1960 4x12
[16ohm mono-8ohm stereo])wiring them with two cables?
-Assuming that i switch the output impedance of the head into 8 ohm -does the output impedance of my MM turn into 4ohm or would remain the same(8ohm)?
thanks
and regards from Italy!
mike kaus
Posted on Wednesday, January 18, 2006 - 04:21 am:   

No, if I read you right, if you use the cab in stereo mode, you basically have two 8 ohm cabs and plugged in in tha back of the head in series it will net 16 ohms. If you REALLY want to use it, I would open the cabinet up, and wire all fout 16 ohm speakers in parallel giving you 4 ohms. Mike.