It may be a bit much to ask . . .

[Bostonrock]

OK,
I love my Desert Gold Sterling to death, and I need to match up an amp
that'll get the most out of her.
Which brings a few questions:

If I have a 350 Watt at 4ohm/200 watt at 8 Ohm head with two speaker outs on the back, does that split the wattage/ohms in half on each side?
Could I attach two 300 Watt 8 Ohm speaker cabs and get a good sound,
or should these numbers match up differently?

How much "headroom" is good in an amp/cab combination?

Can someone explain the Ohms/Watts relationship?

Thanks folks!

 

[maddog]

Ohm's Law

Ohm's Law sez:

Voltage = Current * Resistance (or impedance if AC)

and then you have the power relationship:

Power = Voltage * Current = Voltage^2 / Resistance = Current^2 * Resistance

Series connections of resistance add, Rnew = R1 + R2

And parallel connections add reciprocally, Rnew = 1/(1/R1 + 1/R2)

Armed with all this knowledge, if your amp is rated for 4ohms then two 8ohm cabs in parallel is OK since Rnew = 1(1/8 + 1/8) = 4. You shouldn't be too concerned with the actual wattage number unless the cabinets aren't loud enough before clipping. For that you need efficiency numbers of the cabs (spl@1W/1m).

Hope this helps,
MD

 

[johans]

Ohm's Law sez:

Voltage = Current * Resistance (or impedance if AC)

and then you have the power relationship:

Power = Voltage * Current = Voltage^2 / Resistance = Current^2 * Resistance

Series connections of resistance add, Rnew = R1 + R2

And parallel connections add reciprocally, Rnew = 1/(1/R1 + 1/R2)

Armed with all this knowledge, if your amp is rated for 4ohms then two 8ohm cabs in parallel is OK since Rnew = 1(1/8 + 1/8) = 4. You shouldn't be too concerned with the actual wattage number unless the cabinets aren't loud enough before clipping. For that you need efficiency numbers of the cabs (spl@1W/1m).

Hope this helps,
MD

please tell me you're not a physics major will you =)

lol

 

[Mobay45]

Ohm's Law sez:

Voltage = Current * Resistance (or impedance if AC)

and then you have the power relationship:

Power = Voltage * Current = Voltage^2 / Resistance = Current^2 * Resistance

Series connections of resistance add, Rnew = R1 + R2

And parallel connections add reciprocally, Rnew = 1/(1/R1 + 1/R2)

Armed with all this knowledge, if your amp is rated for 4ohms then two 8ohm cabs in parallel is OK since Rnew = 1(1/8 + 1/8) = 4. You shouldn't be too concerned with the actual wattage number unless the cabinets aren't loud enough before clipping. For that you need efficiency numbers of the cabs (spl@1W/1m).

Hope this helps,
MD

OK, from now on I will just post all the numbers and you can come up with the answer for me. I weren't too very gooden with rithmetic but, my englisch scills is very pretty good. Hyuk, hyuk.

 

[jubjub721]

i would say yes

 

[Aussie Mark]

If I have a 350 Watt at 4ohm/200 watt at 8 Ohm head with two speaker outs on the back, does that split the wattage/ohms in half on each side? Could I attach two 300 Watt 8 Ohm speaker cabs and get a good sound,
or should these numbers match up differently? How much "headroom" is good in an amp/cab combination?

The two speaker outs on your (mono / single channel) head are in "parallel", which means that the amp output (power/wattage) is split equally to both speaker outs (assuming you have a speaker cabinet connected to both speaker outs). If you connect one speaker cabinet to the head, using either speaker out, that single speaker cabinet will get the entire output of the head.

Two 300 watt 8 ohm cabs connected to that head will result in a 4 ohm load, so the amp will put out 350 watts, split between each cab. So, each cab will get 175 watts each.

This combination will "work" and if its a quality amp and cabinets will probably sound good, however if you run the amp at full volume you could send a clipped signal to the speakers and fry them. It's unlikely, but possible, especially if you're trying to compete with two Marshall stacks and a John Bonham type stick man.

The general rule of thumb for matching amps to speakers is to ensure that you have more headroom than you need. For example, using your 2 x 300 watt @ 8 ohm cabinet example, the 300 watt rating is likely an RMS ("average") rating, which means those cabinets can actually handle 600 watts each, so theoretically you could drive both cabinets with a amp that puts out 1200 watts @ 4 ohms. That type of headroom will ensure that you will never be in a situation where you are forced to run the amp at maximum gain, so the risk of frying your speakers is non existent (stupidity aside).

Hope this helps.

 

[maddog]

Not a physics major...

even worse, a math major who thinks he understands electronics.

What aussie mark said sums it up better than what I said.

MD

 

[johans]

even worse, a math major who thinks he understands electronics.

What aussie mark said sums it up better than what I said.

MD

LOL

 

[cgworkman]

...stupidity aside.

s**t!!!!