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Amp/Cab ohms help


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Ideally, ohms should match. However, I've been told that if they don't match, it is much better for the amp if the amp ohms are lower than the cab ohms, so 4ohms amp into 8ohms cab is fine, but not ideal to do over a long period of time. Putting 8ohms amp into a 4ohms cab is regarded as a bit more damaging as far as I'm aware.

Ohms not matching will affect the volume a little, but won't really be noticeable. If the ohms on the amp are lower than the cab, then it can often sound a bit different. Different sometimes meaning bad. Depending on the amp and cab

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Damn. i'm going into a studio to do some recording, and i'm hoping to borrow a Park JTM45, which is 16 ohms, and the only cab i has is 4/8 ohms.....

is there any sort of.... magic box you could plug the head into, and then the cab into, that drops the ohm down?

What kinda cab is it you have and how many speakers?

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you would get a cab that has an 8 ohm impedance

whether that would be a 2x12, 4x10 or 1x12

just make sure that if it does end up to be a single speaker cab that the speaker is capable of taking 30watts, impedance and wattage should be on the back of the speaker

my amp is 120 watts but my speakers are only 75watts but there is 4 of them

to work out series just adding up the individual impedance of the speakers

4 speakers all 4 ohms = 16ohms

wiring speakers in parallel is different

first multiply the amount of speakers by impedance say a 4 x 10 at 4ohms each speaker

4 x 4 = 16

then add the the same numbers together 4+4 = 8

then divide the first result with the second

16/8

which is 2 ohms

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I have a similar question. I've got a Vox CH-30 which has an 8 Ohm/30 watt output for an external speaker/cab. Does that mean i have to get a cab which is only 30 watts or a cab (e.g 4 X 12) which has the same Ohms impotence as my amp?

Your Vox will output 30 watts. An 8ohm speaker rated below 30 watts wouldn't be a good idea but 30watts or over will be fine, although the closer to 30 the better.

I believe you'd be fine running a cab with a higher impedance than your amp, although the power output of your amp would be reduced. I think it would be about half if you had a 16 ohm cab.

If you use a cab with a lower impedance than the amp, you will probably damage your amp. So don't.

Please feel free to correct me if I'm wrong folks. It's been a while since I researched speaker impotence.

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I've always been under the impression that if in doubt use 16ohms. Ohms is resistance, so if you don't want to blow the speaker than you want to use 16 to prevent putting to much through.

A decent rule of thumb. Obviously if you know the signal impedance, then it's better to match it to the speaker to get maximum power. I think some people reckon it can affect tone too. But I'd contest that.

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As far as I understand, the impedance rating on most amps relates to the minimum impedance cab you can safely use.

Eg, if an amp is rated at 8 ohms, then it will deliver it's maximum output at this impedance. If you attach a cab rated at 16 ohms then all that's likely to happen is the amp won't be able to deliver it's maximum output. If you attached a cab rated at 4 ohms then the amp would get damaged, I believe most commonly it would be the output stage that would suffer.

If in doubt, match your amp to your cab or use a cab with a higher impedance.

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you would get a cab that has an 8 ohm impedance

whether that would be a 2x12, 4x10 or 1x12

just make sure that if it does end up to be a single speaker cab that the speaker is capable of taking 30watts, impedance and wattage should be on the back of the speaker

my amp is 120 watts but my speakers are only 75watts but there is 4 of them

to work out series just adding up the individual impedance of the speakers

4 speakers all 4 ohms = 16ohms

wiring speakers in parallel is different

first multiply the amount of speakers by impedance say a 4 x 10 at 4ohms each speaker

4 x 4 = 16

then add the the same numbers together 4+4 = 8

then divide the first result with the second

16/8

which is 2 ohms

My recollection of school physics is that for parallel circuits:

1/Rtotal = 1/R1+ 1/R2 + 1/R3 + 1/R4 etc.....

....so if you've got 4 x 4 ohm cabinets, you get the following:

1/4 + 1/4 + 1/4 + 1/4 = 1/Rtotal....so Rtotal = 1 ohm

Wouldn't 2 x 4 ohm cabinets go to 2 ohm, rather than 4 cabs?

ie. 1/4 + 1/4 = 1/Rtotal....so Rtotal = 2

Dunno if I'm right, just interested is all.

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My recollection of school physics is that for parallel circuits:

1/Rtotal = 1/R1+ 1/R2 + 1/R3 + 1/R4 etc.....

....so if you've got 4 x 4 ohm cabinets, you get the following:

1/4 + 1/4 + 1/4 + 1/4 = 1/Rtotal....so Rtotal = 1 ohm

Wouldn't 2 x 4 ohm cabinets go to 2 ohm, rather than 4 cabs?

ie. 1/4 + 1/4 = 1/Rtotal....so Rtotal = 2

Dunno if I'm right, just interested is all.

That's the proper way to work out impedance in a parallel circuit. I think the afore mentioned one works too though. I've seen it before anyway.

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My recollection of school physics is that for parallel circuits:

1/Rtotal = 1/R1+ 1/R2 + 1/R3 + 1/R4 etc.....

....so if you've got 4 x 4 ohm cabinets, you get the following:

1/4 + 1/4 + 1/4 + 1/4 = 1/Rtotal....so Rtotal = 1 ohm

Wouldn't 2 x 4 ohm cabinets go to 2 ohm, rather than 4 cabs?

ie. 1/4 + 1/4 = 1/Rtotal....so Rtotal = 2

Dunno if I'm right, just interested is all.

you are correct with your formula, but the formula you chose is for working out the RESISTANCE of a circuit based on voltage and current not impedance

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Well they can't both work cos they give different answers!

I totally misread your original post pointing out the two methods were different.

As for resistance and impedance. Good point; my bad. Although, the ohm rating in this case will still be the same, as nominal impedance should pretty much be the same as if it were purely resistive, would it not? I'm a bit rusty with reactance and inductance and sll that shite at the moment.

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I totally misread your original post pointing out the two methods were different.

As for resistance and impedance. Good point; my bad. Although, the ohm rating in this case will still be the same, as nominal impedance should pretty much be the same as if it were purely resistive, would it not? I'm a bit rusty with reactance and inductance and sll that shite at the moment.

using the formula to work out resistance for impedance wont give you the same results as impedance formula because they are different

example

image33.gif

To calculate the load impedance for the series-wired channel in the picture , add up the impedances of each speaker in the chain. The math involves a simple equation in which Zt stands for the cab as a whole, Za and Zb represent the impedances of Speakers A and B, respectively:

Speakers in Series

Zt = Za + Zb

image34.gif

Calculating the load impedance for the parallel-wired channel, multiply the impedances of each speaker and then divide the result by the sum of the speakers' impedances, whol;e cab is Zt. Za and Zb represent the impedances of speakers A and B

Equation 4: Speakers in Parallel

Zt = (Za x Zb) / (Za + Zb)

substituting the resistance with ohms will not end up in the same result

its like working out accelaration using the time, distance and speed triangle you wont get the right answer if you use the wrong formula

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using the formula to work out resistance for impedance wont give you the same results as impedance formula because they are different

example

Zt = (Za x Zb) / (Za + Zb)

substituting the resistance with ohms will not end up in the same result

its like working out accelaration using the time, distance and speed triangle you wont get the right answer if you use the wrong formula

1/Zt=1/Za+1/Zb

Multiply 1/Za by Zb/Zb(=1) and 1/Zb by Za/Za(=1)

1/Zt= Zb/ZaZb + Za/ZaZb

1/Zt= (Za+Zb)/ZaZb

Zt=ZaZa/(Za+Zb)

Same thing innit.

A common way of wiring is parellelled series

With a 4 speaker cab this results in the same cab impedance as speaker impedance.

2 *2 8 ohm speakers in series=2 16 ohm pairs. wire them in parallel and you get 8 ohms total.

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