A Stupid Question.

[Dolebludger]

I should know this, but I don’t!

OK. I have this tube amp head. It has two speaker output jacks. I intend to plug two 8 OHM speakers into the head, one speaker for each output jack. The amp has an OHM switch for 4, 8, or 16 OHMS. How should I set that switch?

 

[BDW60]

Doesn’t it depend on if the speakers are wired in series or parallel? Hell, I’m probably stupid too.

 

[Dolebludger]

The speakers are in separate 1X12 cabs, both wired the same — if that matters!

 

[truckermde]

4 ohm

Two 8 ohm speakers run in parallel will put a 4 ohm load on the amp.

 

[sonar1]

There are no stupid questions.

Just piles and piles of inop gear owned by guys that didn’t bother to ask.
Who’s stupid now?

 

[machine17]

seperate 8ohm cabs, set it on 8 ohms

 

[Pop1655]

It is a stupid question. Just not that complicated. I've read the answer a hundred times. Still, it never sinks in. The answer always confuses me and scares me to the point that my end result is to chicken out and never run two cabs.

 

[hbucker]

When it comes to ohms and resistance I always have to check and double check.

For some reason this topic is a lot like the, I before E rule...

 

[efstop]

There should be a manual for whatever amp it is, unless it's a custom built.
If the amp has one speaker jack, the amp usually tells you the minimum load. For two jacks, it should tell you the load for one or both.
For two jacks and a switch, I have no idea unless you know how the jacks are wired.

 

[bluesoul]

"A stupid question". How does one not take a look! Click bait..
Hope ya sort it out!

 

[Roxy13]

My DSL40c is marked on the back. The speaker that comes in the combo is 16 ohm. Then if you want to run that plus an extra 16 ohm cab you use the two 8 ohm jacks.

So based on that I believe when you run 2 they are in parallel, which would mean 4 ohm for your amp, but if nothing is marked I don't want to say that for certain. If you don't have the owner's manual for it maybe you can find it online in PDF format.

 

[dc007]

2 8 ohm cabs will present a 4 ohm load to the amp

 

[Steven]

Actually, it's a very good question for those of us that are technically inept, but have many amps and face the same sitiuation time to time.

 

[Dolebludger]

Thanks guys. I didn’t title this post as click bait. I just wanted to call myself stupid and save you all the trouble!

 

[tzd]

Actually it depends on how that selector switch is implemented, and whether those two jacks are connected to the amp in series or parallel.

The usual way of doing it in the amp would be two parallel jacks. And the switch will set the overall impedance. The best example of such an amp is the EVH 5150 III amp, where the two jacks are clearly labeled as "Parallel Speaker Outputs".

The manual also provides a very clear table of settings:

 

[cybermgk]

2 8 ohm cabs will present a 4 ohm load to the amp

Not necessarily. That is only the case if those two jacks are wired parralel to the Output Transformer. If they are in Series, then it's 16 ohms. AND, they can be on separate taps to the OT, so each present 8 ohms to their respective tap.

@Dolebludger it depends on the amp.

What amp is it?

 

[Dolebludger]

The amp is a Genz-Benz El Dorado 60 Tribal Series (whew!).

 

[ehb]

Thanks guys. I didn’t title this post as click bait. I just wanted to call myself stupid and save you all the trouble!

Not stupid. Few have a clue as far as impedance....

Redneck logic:

If you add another pipe in parallel and said pipe is the same as primary, then twice as much water can be delivered so the OPPOSITION TO WATER FLOW HAD TO HAVE DECREASED BY SAME FACTOR.

If you fill up a 55 gal drum with water and let it sit. Flow is zero and opposition to flow is infinite.
With a sharpened stretch of rebar, jab a hole. flow went from zero to a flow rate determined by hole dimensions and opposition to flow decreased from infinite (no hole) to opposition dictated by hole dimensions.
Jab another hole of same size, TWICE as much water is coming out, so opposition to the water leaking out was just decreased by the same factor, HALF the previous opposition.

Water is the current. Pressure is the same across all parallel paths. Opposition is the relationship between rate of flow and pressure.


Plug one 8Ω cab in an amp and you will have 8Ω total opposition to current (load).
Plug in another 8Ω cab and you create another identical path for current. So, since pressure is the same across all parallel outputs, then the load current must DOUBLE. If current doubled, pressure (voltage) stayed the same, then total opposition had to have been affected by the same factor but inverse.... Sooo, total output impedance is 4Ω...

The source (amp) sees 4Ω. It neither knows nor cares what that 4Ω consists of.


Every outlet in you house is in parallel. Plug one lamp in. Voltage stayed the same, current INCREASED soooooo the total impedance (opposition to current flow) had to have DECREASED....

Plug another identical lamp/bulb in and Voltage is STILL 120v....BUT another identical path for current has been created so current doubled. Sooooo, total impedance had to have been halved...



Lots of folks have blown up amps from NOT understanding load (current). The lower the impedance, the more current the source must supply.... If an amp is rated at 4Ω, if you go below 4Ω, you are forcing the source ( amp ) to supply more current than it was designed to handle......and sooner or later something is gonna give....


Remember: Everything can be a smoke machine if you operate it wrong enough...

edro.

 

[ehb]

I've never had an amp with series outputs.

 

[dc007]

@Dolebludger we'll have this figured out in a few more pages