oooohm

[T-bone]

Because I didn't major in electrical engineering, I find myself stumbling over what should be a very simple question. So I turn to you, the movers and shakers, the Kings and Queens (I'm thinking Psychicpet here), the absolute Guhru's of the Bass Word.

My little 2x10 combo has an external speaker out jack which says 4 ohm. Now, am I able to connect an 8 ohm cab?

And while you're pondering this rather deep question, any warnings on the SWR workingman's 1x15? A friend is offering a killer deal.

Thanks

Tbone

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and if you can't fly you'll have to move in with the rhythm section - Steely Dan

 

[maddog]

Yes.

 

[prickly_pete]

Unless you have a tube power amp, you should be fine. That 4 ohm listing is probably a minimum impedance load. The power amp won't have to work as hard with an 8 ohm load, and it will not put out as much power.

 

[phatduckk]

ya you can plug it in ...

i dunno much about the ohm thing either but i do know a few rudimentary things (someone correct me if im on crack) ...

1) best thing todo is to match the ohms of the head and cab(s)
2) example: if the head can take a 4ohm load dont connect cabs rated at "less" (so no 4 ohm cab to an 8ohm head). that's bad
3) when connecting multiple cabs the math for the ohm stuff sounds "backwards": 2 x 8ohm cabs = 4 ohm load on the head, 2 x 4 ohm cabs = 2 ohm load on the head

i dunno all the technical jargin but that's my understanding.

dunno about the workingman 15 but i had a workingman head (4004) and didnt like it at all.

 

[DBD-Bass]

From what I understand of impedance, you need to pay attention to matches. Phatduckk is correct about mixing cabs and changing the impedance. Solid state power amps are more forgiving than tube power amps. I have also heard that if you turn on an all tube amp with no speakers/cab plugged in, it could damage the power amp because there is no load at all. So rule of thumb. Learn the impedance of your amp and match your cabs accordingly. I have an Ampeg SVT3pro head... it can handle a 8ohm or 4ohm load. Nothing more, nothing less. My cab is a Ampeg SVT610hlf, rated at 4ohms. I cannot add another cab to it unless I got a seperate power amp and slaved it off of the pre-amp.

Hope that rambling made sense.

 

[SteveB]

Unless you have a tube power amp, you should be fine. That 4 ohm listing is probably a minimum impedance load. The power amp won't have to work as hard with an 8 ohm load, and it will not put out as much power.

+1

Of course, matched loads are preferable, but you should be fine.

It may be possible to re-wire the speakers in your cabinet to make the cabinet a 4 ohm load. Depends on the impedance for each speaker (listed on the speaker magnet, in Ohms), and the number of speakers.

If you have for example, two 8-ohm speakers in the cab, you can wire those speakers in parallel and get a 4-ohm load.

If you have four 16-ohm speakers, you could wire them parallel and get 4 ohms.

Other combinations are also possible using series and parallel wiring.

 

[T-bone]

2) example: if the head can take a 4ohm load dont connect cabs rated at "less" (so no 4 ohm cab to an 8ohm head). that's bad
3) when connecting multiple cabs the math for the ohm stuff sounds "backwards": 2 x 8ohm cabs = 4 ohm load on the head, 2 x 4 ohm cabs = 2 ohm load on the head

phatduckk - you're making my head hurt

So, if I follow everyone's logic, I can add an 8 ohm cab and end up with 1/2 the watts? So my little 200 watt combo then becomes a 100 watt beast? Do I gain enough bottom to justify the loss of power?

Don't hurt me now........tbone

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me and you will listen to a little bit of what made the preacher dance - Steely Dan

 

[maddog]

phatduckk - you're making my head hurt

Time to make it hurt worse :

If you have a stiff power supply (can produce lots of current), as you halve the impedance you double the power.

Ohm's Law: Voltage = Current*Impedance

and

Power = Voltage*Current

Solve Ohm's Law for Current:

Current = Voltage/Impedance.

Substitute current into the formula for Power,

Power = Voltage^2/Impedance

So setting up amp1 and amp2 to sit at the same voltage (or gain) but amp2 is hooked up to a cab that is half the impedance of the cab hooked up to amp1 results in twice the power being drawn from amp2.

You can play the impedance game to get more power until either the power supply sags (can't hold the voltage cuz the current is too much) or the output transistors go into thermal runaway and melt.

I used to be big into audio as a hobby. Built my own speakers and amps. Fun stuff. Been bitten a few times by 1kV or more.

On using your combo output, if it is solid state, the 4 ohms marking is a minimum impedance allowed. If it is tube, best to match to 4 ohms due to transmission theory and proper transformer loading, etc.

So you will actually be gaining power but power isn't everything. Always good to check out the sensitivity of the cabinet to actually make sure it will be loud enough with the power given.

 

[SteveB]

Great post, maddog.

 

[Dr. Nick]

phatduckk - you're making my head hurt

So, if I follow everyone's logic, I can add an 8 ohm cab and end up with 1/2 the watts? So my little 200 watt combo then becomes a 100 watt beast? Do I gain enough bottom to justify the loss of power?

Don't hurt me now........tbone

Okay, I'm no techie, but was has been said above makes sense to me....

Sounds like the same situation I had with my previous rig: Peavey 210 combo. 240 watts or so with internal speakers only. External jack said "4 ohms or greater". When I plugged in a 4 ohm, cab, the wattage supposedly jumped to 340 w (why not 480? I dunno...) But I COULD use an 8 ohm cab, which I figure increased wattage to 300 watts or so. I can testify that adding a 1x15 cab, 4 or 8 ohm, greatly increased the low end capability of the amp and made it sound a heck of a lot better...

 

[maddog]

When I plugged in a 4 ohm, cab, the wattage supposedly jumped to 340 w (why not 480? I dunno...)

Most likely due to power supply limitations.

 

[shamus63]

I've always wondered about mixing two cabs with different ohm ratings (one @ 4; one @ 8), and what the end result would be...hopefully not negative for the bass head.

Reason I'm asking is that I'm picking up an SWR 8x8 rated @ 4ohms, and want to pair it up with a Mesa-Boogie RR 1x15, which is rated @ 8ohms (need the bottom-end filled out). In fact, most of the 1x15 cabs I've seen (or at least want) are rated @ 8hms.

Any electricians want to jump in on this one?

 

[maddog]

for series hookups, resistance (or impedance) adds.

Rt = R1 + R2

for parallel hookups

Rt = R1*R2/(R1+R2)

So for you problem

Rt = 4*8/(4+8) = 2.67.

So your bass head needs to be rated for atleast 2 ohms.

 

[AnthonyD]

Maddog - you are my hero!

 

[SteveB]

I'd like to add that some amp heads have multiple speaker output jacks, and these are usually wired in parallel.

Therefore, if you have such a setup and connect an 8-ohm load to each jack, you are in fact placing a 4-ohm load on the amp.

If you were to connect a 4-ohm cabinet to each jack, you'd be running a 2-ohm load which is probably not within the tolerance of your amp.

 

[Golem]

Okay, I'm no techie, but was has been said above makes sense to me....

Sounds like the same situation I had with my previous rig: Peavey 210 combo. 240 watts or so with internal speakers only. External jack said "4 ohms or greater". When I plugged in a 4 ohm, cab, the wattage supposedly jumped to 340 w (why not 480? I dunno...) But I COULD use an 8 ohm cab, which I figure increased wattage to 300 watts or so. I can testify that adding a 1x15 cab, 4 or 8 ohm, greatly increased the low end capability of the amp and made it sound a heck of a lot better...

Unlike combining cabs, where two eights make a four, the effects of impedence on your power potential is less direct. Assuming no other limitations [like a wimpy internal power supply], you don't get double the wattage if you drop from 8 ohms down to 4 ohms. You'd have to drop from 8 ohms down to 2 ohms to accomplish a doubling of the wattage.

When a pair [two] of 8 ohms cabs produce a 4 ohms result, that's direct, flat out arithmetic. You get two of something and the result is half [like in 'two']. But the power deal doesn't follow a plain 'two'. It follows 'the square root of two'. If you hate roots and powers, just know one fact: square root of 'two' = 1.4, and you can then forget higher math forever. Note that the wattage changes resulting from changing up or down between 8 ohms and 4 ohms are by a factor of 1.4 in the example above:

1.4 x 240 = 340,

and 340 / 1.4 = 240.

Printed specs on the rear of the amp are rounded off.

If the published specs for an amp don't follow this pattern of wattage changes in responces to speaker load changes, then that amp is restricted by a limited internal power supply unit.

 

[SteveB]

Golem,

That was great info. I knew the wattage thing wasn't as easy a computation as the impedance, but I never knew just how to figure it out.

And I do hate square roots, so your 1.4 multiplier will stick with me.

 

[shamus63]

for series hookups, resistance (or impedance) adds.

Rt = R1 + R2

for parallel hookups

Rt = R1*R2/(R1+R2)

So for you problem

Rt = 4*8/(4+8) = 2.67.

So your bass head needs to be rated for atleast 2 ohms.

Good enough. Then I'll be shopping for the Big Block 750 (750w @ 2ohms).

Thanks 'Dog!

 

[Dr. Nick]

Unlike combining cabs, where two eights make a four, the effects of impedence on your power potential is less direct. Assuming no other limitations [like a wimpy internal power supply], you don't get double the wattage if you drop from 8 ohms down to 4 ohms. You'd have to drop from 8 ohms down to 2 ohms to accomplish a doubling of the wattage.

When a pair [two] of 8 ohms cabs produce a 4 ohms result, that's direct, flat out arithmetic. You get two of something and the result is half [like in 'two']. But the power deal doesn't follow a plain 'two'. It follows 'the square root of two'. If you hate roots and powers, just know one fact: square root of 'two' = 1.4, and you can then forget higher math forever. Note that the wattage changes resulting from changing up or down between 8 ohms and 4 ohms are by a factor of 1.4 in the example above:

1.4 x 240 = 340,

and 340 / 1.4 = 240.

Printed specs on the rear of the amp are rounded off.

If the published specs for an amp don't follow this pattern of wattage changes in responces to speaker load changes, then that amp is restricted by a limited internal power supply unit.

See? I knew you guys would know this stuff! Thanks Golem!

The "limited internal power supply" theory is probably dead-on. At least I don't have to worry about this stuff anymore! My new head can handle 2x4 ohm, 2x 8 ohm, 8 ohm bridged, 4 ohm bridged...whatever. (Yes, I realize that the power ratings are all different depending on the configuration, but is always loud! ) I purchased an 8 ohm cab to start with figuring it will be easier to add cabs starting at 8 ohms and working down (ohm-wise, that is) than starting at 4.

(Does that make any sense at all?)

All I know is "Hit string here. Big Noise come out there."

 

[T-bone]

Yet another reason why we ask these questions to the group. Thanks gentlemen (and ladies), I appreciate all the great responses and math lessons. Now how about that SWR 1x15 cab? Anybody out there that would not spend $150 on one?

Ciao Ciao Ciao

tbone
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I'm so excited I can barely cope
I'm sizzling like an isotope
I'm on fire - Steely Dan