4/8 Ohm Impedance Explanation

Good technical thread on amps that I feel like bumping!

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Actually you do need to worry about this if you wish to expand your combo amp to get more wattage, and sound out of the limits put on the combo amp.

Like the case @Gio stated in his first post, needing another cabinet to get the full rated wattage from his 600w head.

I know there has been a more accurate and technical explanation posted in this thread, but I was in a situation where this mattered, and lead me ultimately to my decision when I purchased my SWR Combo. It is 160w at 4ohms. It CLEARLY STATES. To expand DO NOT USE A CABINET LESS THEN 8OHMS.

My SWR WorkingMan 15 Combo rated at 180 RMS is 200 @4ohms with with addition of an 8ohm Cab, but if I were to add a 4ohm cab, I risk burning up the internal amp head.

The reason it mattered is that I have a Hartke 8x10 cabinet already, and it has not sold yet. When I was researching Combo’s I was hoping to find something that I could use with that Cab nstead of getting, a cab with the combo amp now or in the future, however the Hartke is 4ohms, and almost every combo amp I looked specified expansion cabs must be no less then 8ohms.
I think a lot of expansion cabs are 8ohms. I think the reason the Hartke cab is 4ohms is because it is 8x10, a full stack, and may typically not be used along with other cabs on the same head. And / or because it’s 800 watts, and any head that is going to put out enough power for two of them will likely be rated down to 2ohm output.

Just thoughts, and posting in a dead thread for conversation

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Parallel resistance equations… this brings me back to my Navy days when I learned all this stuff. Glad there are already some real solid explanations in this thread already so I don’t need to jump into it. Great info in here. Howard’s two hoses one pump analogy is the simplest way to wrap your head around it without more complex electrical theory.

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@Jazzbass19 thanks for posting that video Joe. That answers many of my questions.

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So, after watching the video that @Jazzbass19 posted, it’s my understanding that whatever watts the amp head is, the cabinet has to be that many watts or greater.
Also, whatever ohms the head is rated, the cabinet must be the same ohms.

Are those two assumptions correct?

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“should” be the same for optimal performance is my understanding. Cab has to have at least the same rating as the head. The cab can have a higher ohm rating than the amp head but won’t be as loud then.

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TLDW; always match your ohms. ESPECIALLY if you have a tube/valve amplifier.

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That is a method that stops a bass player from accidentally turning the amp up so loud that it ruins the speaker. It’s an attempt to simplify a complex topic for one goal. It’s a good answer and it’s perfectly fine to stop there and call it done.

If one wants to explore the topic further, there’s no such thing as a one number power rating for a speaker. It has to be based against time. A speaker can take so much power on average long term before it overheats. It has a higher power rating for what it can take for 1 second and it has an even higher power rating for what it can take for 1 millisecond. All these aren’t on the spec sheet. It’s OK to put more power into a speaker than it’s rated for a short period of time, like plucking or high string attack transients.

The chances of most of us turning up an amp high enough to damage a 410 cabinet are quite low. If you have the master gain at max and you are wishing you had more, the speaker may be in danger.

Engineers like to think about operating margin, the difference between what you are using and what you have. It’s more comfortable to have more power in the engines than what you need, even if max power could damage the plane under certain conditions.

If we leave bass land for a moment, it’s more commonly the case where people blow up speakers in PA reinforcement systems because the amps are too low power and get pushed too hard, which overheats the speakers.

This is one of those cases where all answers are correct depending on how you want to see it.

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Here’s an example from a manufacturer where they explain in the first paragraph that they are feeding a cabinet rated at 450 Watts with an 1800 Watt amp generating 1800 watt peaks. It’s no problem. Every manufacturer uses this test method.

https://barefacedbass.com/technical-information/understanding-power-handling.htm

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Thanks @DaveT for explaining this stuff over and over :wink:

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@DaveT I thank you for the over and over :slight_smile:
It’s slowly sinking in :rofl:
Jamie

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It’s a ridiculously complicated topic. Even Fender chooses to say to match continuous power of the cabinet to amp power. It’s too much to fully explain.

https://www.fender.com/articles/tech-talk/what-you-need-to-know-about-bass-amp-power-ratings

My point is that a 1000 Watt head isn’t nearly as big as it sounds in audio terms and that feature alone shouldn’t stop someone from getting one if it otherwise has everything they want. On that Peavey head they have fewer tone options on the lower power model.

I have a larger amp than my cabinet is average power rated for and I know I’ll never turn it up that high. It’s not like there’s a stick of dynamite on the end table, just 3dB at the most.

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If you don’t mind, do the math on this so we can see the relationship in abc’ s and 123’s.
800 watt head at 4 ohm with 2ea 115 cabs rated for 300 watts at 8 ohms in parallel. Music or smoke? How about in series?

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I know this depends a great deal on how far up you crank it so assume a deaf and overzealous bassist.

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CASE 1
Head: solid state (not tube), rated 800 Watts average (RMS) at 4 Ohms
Cabinet 1: Single 15", 8 ohm impedance, 300 Watt average (RMS) power rating, sensitivity 99 dB/1 Watt at 1 meter
Cabinet 2: Same
Wiring: Cabinets in parallel
Impedance of cabinets shown to amplifier = 1 / (1/R1 + 1/R2) = 1 / (1/8 + 1/8) = 4 ohms
Wiring the cabinet in parallel produces a 4 ohm load which is OK for this amplifier rating, so far so good. The 800 Watts the amplifier is capable of will be distributed equally to both cabinets, each will get 400 Watts. I’ll leave the math off of this since it’s sufficient to say that the cabinets must have the same voltage across them when wired in parallel and we know the power delivered is V-squared / R. Since they have the same R, they must have the same power delivered to each. The amp will not deliver more than it has, so that means each gets 400 Watts RMS, adding up to the total capability of 800 Watts.

The sound pressure level produced 10 meters away by each cabinet will be dB-SPL = Sensitivity + 10log(Power) - 20log(Distance in Meters) = 99 + 10log(400) - 20log(10) = 105 dB-SPL. Normally in PA systems when you stack two speakers you get a 3dB increase for them adding together (Trust me, I’m leaving off the math on that one). However, the bass frequencies are very large wavelengths compared to the size of the cabinet, so they will add together more efficiently, up to as much as 6dB rather than 3dB. Soooo, the total SPL from both cabinets working together will be 105 + 3 or 6, likely 110dB, maximum RMS level, peaks could be higher.

If the player is standing 1 meter away from their cabinet they will be exposed to 99 + 10log(400) - 20log(1) = 125 dB from one driver. The level could be as high as 131 dB for this person.

What about the idea that the cabinets are rated at 300 Watts and we have the capability to deliver 400 Watts to them? It’s probably inconsequential. How many more dB that is is given by 10log(P1/P2) = 10log(400/300) = 1.25 dB additional power in the amp. The cabinet also has a peak power rating of at least 2x the average power rating (for bass, 4x for normal PA) and we won’t be over that number.

Result: No smoke or destruction, possible hearing damage.

CASE 2
Wiring: Cabinets in series
Impedance of cabinets shown to amplifier = R1 + R2 = 16 ohms
Wiring the cabinet in series produces a 16 ohm load which is OK with the penalty of not being able to draw the full power from the amp. Our amplifier can only make its maximum voltage on the output and power is given by V-squared / R. If we change R from 4 to 16, we have reduced the power that can come out of the amp by a factor of 4. Therefore, we now have a Watts * R-Rated / R-Actual = 800 * 4 / 16 = 200 Watts amp. The 200 Watts the amplifier is capable of will be distributed equally to both cabinets, each will get 100 Watts. In this case both cabinets will have the same current and power is given by I-squared*R, so each will receive the same power. Another way to look at it is that the voltage across the series circuit will divide proportional to their load. Since they have the same R, they must have the same power delivered to each. The amp will not deliver more than it has, so that means each gets 100 Watts RMS, adding up to the total capability of 200 Watts for a total 16 Ohm load.

The sound pressure level produced 10 meters away by each cabinet will be dB-SPL = Sensitivity + 10log(Power) - 20log(Distance in Meters) = 99 + 10log(100) - 20log(10) = 99 dB-SPL. Normally in PA systems when you stack two speakers you get a 3dB increase for them adding together (Trust me, I’m leaving off the math on that one). However, the bass frequencies are very large wavelengths compared to the size of the cabinet, so they will add together more efficiently, up to as much as 6dB rather than 3dB. Soooo, the total SPL from both cabinets working together will be 99 + 3 or 6, likely 104 dB, maximum RMS level, peaks could be higher.

If the player is standing 1 meter away from their cabinet they will be exposed to 99 + 10log(100) - 20log(1) = 119 dB from one driver. The level could be as high as 125 dB for this person.

Result: No smoke or destruction, possible hearing damage.

Is that what you wanted to see presented @Gorch?

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IDK if that is what he wanted or not, but it was fascinating, thanks for the lesson.

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That. Is. Crazy! And completely unexpected.

I really thought you would get 4 ohms either way. Extrapolating this out…

If you have four 8 ohm cabinets
with two in parallel
and a second one hanging of of each of them in series
You end up with each of the four cabinets getting 100 watts.

Based on this you would be using only 400 of the 800 watts. Is this correct? It feels like I’m way off base on this.

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What you describe is equivalent to having two 16 ohm loads in parallel. So total impedance at the amp should be 8 ohms, which would produce 400 watts split evenly between the four cabs, at 100W per cab.

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This rig would produce sound pressure levels 3-6 dB higher than my Case #2. You are delivering 100 Watts to twice as many cabinets.

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@DaveT Would there be any advantage to running the four cabinets at 400 watts over just running two cabinets in parallel at 800 watts?

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