I read it that he has the idea of one day buying some stuff. The TC amp was just a cheapest best option and the cab he was looking at in Jan is out of stock.
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4 ohm amp into 8 ohm cab- will I blow the speaker?
- Thread starter Jordy4strings
- Start date
4 ohm 500-600w amp is meant for a 8 ohm 250-400w 2 cab, but if you use only one i think it will be ok ,too only on half amp volume max.
If you mean stage worthy bass amplifier and single driver cab then yeah but even then it's not hard to find exceptions with 100w tube amps thrashing away with impunity. I hope you aren't meaning the mythical square waved DC killing kind of little amp overdriving speakers to death thing that has to be debunked every month, it's only the 5th today.
The never ending story Lol...but it's great, we're learning!(Unless someones using some speaker combination that's less than the "4 ohms".)
OP, if your amp is 500 watts @ 4 ohms, you should have zero issues running an 8 Ohm 300 watt cabinet.
Now, you haven't stated which cab it is, so I will say (as always) use caution and your ears. If you find that you have to crank your volume way up to be heard, then you should get a 2nd "matching" 8 ohm cabinet to achieve more volume.
As was stated, its easier for the gang to help if you post exactly what your equipment is that you are asking for advice on.
T$
Sure looks like some right answers in there.
Maybe he asked twice for a best of 3?
But yeah, no your SS head is not putting out 500W at 8-ohms. A lot of amp makers will nestle the 8-ohm values in the spec sheet because the 4-ohm ones look more impressive. Yes you CAN blow a 300W cab even if the head's rated at 200W RMS at 8-ohms because the rated value doesn't account for exceptional signal boost via onboard preamp settings, pedal settings, or even extreme EQ settings on the head. Its actual RMS could end up way higher if you're diming it.
If you're not getting it, don't make another thread. Ask for clarification. It will be provided.
WRT musical instrument amplifiers this double impedance = half power trope will never die and you’ll not see it stated by any electrical engineer, me for example, unless it’s a specific case like a DC circuit or we’re talking in really broad terms.
Matching source and load impedance in a circuit yields the maximum transfer of power between the amp and speaker system. That is elementary electrical network theory. A load (speaker system) that is greater or lesser than the amp’s design spec reduces the power transfer but, since we’re talking an AC signal with a wealth of frequencies and harmonics, it’s only roughly proportional to the impedance mismatch. Any impedance mismatch reduces the power transfer but not simplistically. IOW halving the load (say, from 8 ohm to 4 ohm) won’t double output power and doubling the load (say, from 8 to 16) won’t halve it.
In the OP case that amp will not deliver 500 watts to that speaker system. Nor will it likely deliver 250 watts though 250 is closer. It may work fine forever or it may cause speaker failure followed soon by amplifier failure. There are too many (ignored) variables to hazard a guess about the likelihood of success with the OP’s rig. It’s been said here often: it’s possible to blow out a 300 watt cab with a 50 watt amp. When you “crank the amp” you choose to experiment with your gear.
My advice is to consider carefully your desire to use that combination. The more you understand the more likely you’ll tend to your economic welfare.
I'm going to need an explanation of why this is true, especially the last half of the statement. Are you saying that an amp that produces, let's say 250 watts into an 8-ohm load doesn't produce exactly half that (125 watts) when the load is 16 ohms? I'll need that math explained to me. Are you saying that because it's an AC signal, that it is not valid to use the formula Watts=Voltage squared/Impedance?
Long story short, the impedance ratings of amps and speakers are what we call “nominal.” They are a reasonable approximation of the actual value through the working range of the amp or speaker. It’s a simplification that's useful if not taken too far.
The power equation you gave there is correct for DC networks and also correct but incomplete for AC networks. In a DC network the voltage and current are locked together so changing one proportionally changes the other. DC network math is clean and easy.
You’ve doubtless read here on TB that using a multimeter to check the coil of an 8-ohm speaker gives a reading of something around 5.5 ohms. The reason for the difference is called reactance, an effect of capacitance and inductance. AC network theory includes reactance that’s not a factor in DC. Sound, audio, is an AC phenomenon.
AC, like DC, is affected by resistance but, unlike DC, it’s also affected by reactance (e.g. the speaker voice coil is an inductor) thus the applied voltage and current are not locked together a.k.a. “in phase.” How far the phase shift goes and how it affects what we hear varies by a variety of factors such as the amp’s output stage design, speaker voice coil size/shape/wire and position in the magnet structure, design of the magnet structure and the frequencies and waveforms we apply. As you’re playing the effect of each one is small but not negligible and it’s the sum of all of them that determines how the speaker behaves as a load for the amp. These and a few other factors gang up to make AC math messy vs. the DC side of things.
There’s a tendency to latch onto the ideas that are easy to understand such as P=E^2/I but it’s valuable to understand when it’s incomplete or even wrong. I hope I haven’t left you too far out in the weeds with this.
The power equation you gave there is correct for DC networks and also correct but incomplete for AC networks. In a DC network the voltage and current are locked together so changing one proportionally changes the other. DC network math is clean and easy.
You’ve doubtless read here on TB that using a multimeter to check the coil of an 8-ohm speaker gives a reading of something around 5.5 ohms. The reason for the difference is called reactance, an effect of capacitance and inductance. AC network theory includes reactance that’s not a factor in DC. Sound, audio, is an AC phenomenon.
AC, like DC, is affected by resistance but, unlike DC, it’s also affected by reactance (e.g. the speaker voice coil is an inductor) thus the applied voltage and current are not locked together a.k.a. “in phase.” How far the phase shift goes and how it affects what we hear varies by a variety of factors such as the amp’s output stage design, speaker voice coil size/shape/wire and position in the magnet structure, design of the magnet structure and the frequencies and waveforms we apply. As you’re playing the effect of each one is small but not negligible and it’s the sum of all of them that determines how the speaker behaves as a load for the amp. These and a few other factors gang up to make AC math messy vs. the DC side of things.
There’s a tendency to latch onto the ideas that are easy to understand such as P=E^2/I but it’s valuable to understand when it’s incomplete or even wrong. I hope I haven’t left you too far out in the weeds with this.
Half with mosfets used to be ''close enough for jazz''. Then class D came along and it's more like 2/3 to 3/4. Which is still half close enough to half for jazz.
If your cab is half low in wats to your amp it will react normally till the half max of your amp, if you add another cab you will get the full amp power.
