Intentionally pairing 8 and 4 ohm cabinets

[Old Garage-Bander]

This was also my next concern after impedance and power sharing.

It is impossible to determine ahead of time, how the two cabs will sound together out in the wild.
And every venue can be different in that respect.

But since you already have the cabs, you'll begin to know these answers once you have an amp that can drive them both. (Sounds like a solid reason to get a new amp IMO. )

Best of luck.

 

[aprod]

When you run them that way each speaker cab gets equal power. What I found doing this is the 8 ohm is not as loud as the 4 ohm. And it is noticeable. It was very apparent when I ran 2 4 ohm cabs.

 

[Wasnex]

Why does that amp need an impedance switch?

When current flows, it produces heat. The reason for the switch is so the amp can operate at a lower impedance without incurring damage.

Solid state amps try to produce the same voltage regardless of impedance. As the impedance goes down the current and power increase.

Here is the AC Ohm's Law Formula Wheel.

Assume we have an amp that makes 400W at 8 ohms, 800W at 4 ohms, and 800W at 2 ohms.

Solving for V = sq rt of PZ

At 8 ohms at 400W

V = sq rt of (400 x 8) = ~56.57V
​

At 4 ohms at 800W

V = sq rt of (800 x 4) = ~56.57V
​

These answers are rounded to the 100s space, which is why I used "~".

​

Notice when the impedance is cut in half, the voltage stays the same but the power doubles.

Now lets see what happens with the current by solving for I = sq rt (P/Z)
At 8 ohms and 400W

I = sq rt (400/8) = ~7.07A
​

At 4 ohms and 800W

I = sq rt (800/4) = ~14.14A​

You get the same answer by using I = V/Z

At 8 ohms and 400W

I = ~56.57/8 = ~7.07A​

At 4 ohms and 800W

I = ~56.57/4 = ~14.14A
​

So as you see, the current also doubles when the impedance is cut in half.

Let's calculate the Voltage and Current for 800W at 2 ohms

V = sq rt (800 x 2) = 40V

I = sq rt (800/2) = 20A
​

Notice that even though the voltage was reduced from ~56.57V to 40V that the current still increased from ~14.14A to 20A

P = E x I

40 x 20 = 800W​

With out the switch, the amp would try to maintain the voltage at ~56.57V

I = ~56.57/2 = ~28.29A
P = ~56.57 x ~28.29 = ~1600W.
​

Unfortunately the amp won't work reliably at this power and current level, so it has to be derated.

Some amps reconfigure the power supply to lower the voltage in the power supply. Some amps adjust a limiter so the drive sent to the output section is kept within a safe range.

FYI, The switch is not new or specific to class-D. This image is from a 70's era Traynor PM-300 power amp:

 

[Phaidrus]

There are different ways of accomplishing this, but whatever the method, it’s necessary on these models that have the switch to select the correct nominal impedance mode for the speakers being driven.

How bad is it to connect two 4 Ohm cabs (=2 Ohm) to an amp that is only rated for 8 and 4 Ohm load? Will it definitely blow? Will it work safely at lower volumes?

 

[CochinoBass]

When current flows, it produces heat. The reason for the switch is so the amp can operate at a lower impedance without incurring damage.View attachment 4969934

I think the question was "why would an solid state amp designed to handle a 2 ohm load need a 2.67ohm switch?". Your answer is correct, but it does not address why an amp designed for a 2 ohm load would need a switch to run more than 2 ohm loads, IE 2.67 ohms. It shouldn't need one any more than an amp designed for a 4 ohm minimum load needs one to run an 8 ohm load.

 

[vvvmmm]

How bad is it to connect two 4 Ohm cabs (=2 Ohm) to an amp that is only rated for 8 and 4 Ohm load? Will it definitely blow? Will it work safely at lower volumes?

Well, if you connect 'em in series ...

 

[Wasnex]

I think the question was "why would an solid state amp designed to handle a 2 ohm load need a 2.67ohm switch?". Your answer is correct, but it does not address why an amp designed for a 2 ohm load would need a switch to run more than 2 ohm loads, IE 2.67 ohms. It shouldn't need one any more than an amp designed for a 4 ohm minimum load needs one to run an 8 ohm load.

The way the manual is written is amazingly confusing. Based on your response, I assume it's a 3-way impedance switch, instead of the normal 2-way switch you find in most amps. My guess is having an additional setting would further maximize performance. It's the same concept taken one step further.

Here are the Specs...perhaps if you calculate the voltage and current that occurs at these power and impedance levels, you will get a better understanding.

Legacy 500:

190W @ 1% THD+N, 8Ω
300W @ 1% THD+N, 4Ω
450W @ 1% THD+N, 2.7Ω
300W @ 1% THD+N, 2Ω​

Legacy 800:

400W @ 1% THD+N, 8Ω
800W @ 1% THD+N, 4Ω
800W @ 1% THD+N, 2.7Ω
800W @ 1% THD+N, 2Ω​

 

[HolmeBass]

No problems re: impedance, and let’s assume power is good also (your power computation is correct).

The issue I see is one of sensitivity paired with the unequal power distribution.

You will be putting twice the power into a speaker cab that probably also has a higher sensitivity rating, the smaller cabinet my just be superfluous.

But there is no harm in trying.

 

[lowtom]

I run a 4 ohm and 8 ohm cab with my set up and it works fantastic. Mine is also matching 15's so each of the speakers is getting the same amount of power though.

View attachment 4969630

Same, but with the GRBass 1400 Amp

 

[agedhorse]

How bad is it to connect two 4 Ohm cabs (=2 Ohm) to an amp that is only rated for 8 and 4 Ohm load? Will it definitely blow? Will it work safely at lower volumes?

Not a good idea. Some amps will protect themselves, some amps will fail.

 

[Wasnex]

No problems re: impedance, and let’s assume power is good also (your power computation is correct).

The issue I see is one of sensitivity paired with the unequal power distribution.

You will be putting twice the power into a speaker cab that probably also has a higher sensitivity rating, the smaller cabinet my just be superfluous.

But there is no harm in trying.

Using cabs with drastically different sensitivity ratings can produce ironic results...two cabs not as loud as one.

However you might want to consider how power is shared and the sensitivity of the individual drivers, rather than the sensitivity of the cab.

Hopefully this will illustrate my point. Assume we have an 8 ohm 110 and a 4 ohm 210. Both cabs use the same driver and are tuned the same way.

The 210 will have a 3dB higher sensitivity rating than the 110, but all three drivers will actually share the power equally and produce the same SPL.

AFAIK when you double the drivers the sensitivity rating goes up about 3dB. The SPL coming from each driver actually decreases by 3dB because each is receiving 1/2 Watt, but you get +6dB from mutual summing. So the net is +3dB.

If you triple the drivers the sensitivity goes up by about 4.8dB. If you quadruple the drivers, the sensitivity goes up by about 6dB. Know we are talking about a 410. Double the drivers one final time (octuple ) and you get an 810 and the sensitivity of the system has gone up a total of 9dB.

 

[HolmeBass]

Using cabs with drastically different sensitivity ratings can produce ironic results...two cabs not as loud as one.

However you might want to consider how power is shared and the sensitivity of the individual drivers, rather than the sensitivity of the cab.

Hopefully this will illustrate my point. Assume we have an 8 ohm 110 and a 4 ohm 210. Both cabs use the same driver and are tuned the same way.

The 210 will have a 3dB higher sensitivity rating than the 110, but all three drivers will actually share the power equally and produce the same SPL.

AFAIK when you double the drivers the sensitivity rating goes up about 3dB. The SPL coming from each driver actually decreases by 3dB because each is receiving 1/2 Watt, but you get +6dB from mutual summing. So the net is +3dB.

If you triple the drivers the sensitivity goes up by about 4.8dB. If you quadruple the drivers, the sensitivity goes up by about 6dB. Know we are talking about a 410. Double the drivers one final time (octuple ) and you get an 810 and the sensitivity of the system has gone up a total of 9dB.

I’ve been in the situation where adding an inefficient cabinet to an efficient one made me quieter twice.

I have also used two cabinets of different impedance and sensitivity that made them blend really well.

Adding more of the same driver in a
box tuned the same is not much of a case, that really is apples to apples.

 

[Barticus]

I run
SL112 8ohm
SL212 4ohm
With my TH700 @2.67ohm
Sounds awesome

 

[Boneless]

Not a good idea. Some amps will protect themselves, some amps will fail.

If keeping volumes low and/or using a limiter to keep levels low, would it fail regardless?

 

[Boneless]

When current flows, it produces heat. The reason for the switch is so the amp can operate at a lower impedance without incurring damage.

Solid state amps try to produce the same voltage regardless of impedance. As the impedance goes down the current and power increase.

Here is the AC Ohm's Law Formula Wheel.

Assume we have an amp that makes 400W at 8 ohms, 800W at 4 ohms, and 800W at 2 ohms.

Solving for V = sq rt of PZ

At 8 ohms at 400W

V = sq rt of (400 x 8) = ~56.57V
​

At 4 ohms at 800W

V = sq rt of (800 x 4) = ~56.57V
​

These answers are rounded to the 100s space, which is why I used "~".

​

Notice when the impedance is cut in half, the voltage stays the same but the power doubles.

Now lets see what happens with the current by solving for I = sq rt (P/Z)
At 8 ohms and 400W

I = sq rt (400/8) = ~7.07A
​

At 4 ohms and 800W

I = sq rt (800/4) = ~14.14A​

You get the same answer by using I = V/Z

At 8 ohms and 400W

I = ~56.57/8 = ~7.07A​

At 4 ohms and 800W

I = ~56.57/4 = ~14.14A
​

So as you see, the current also doubles when the impedance is cut in half.

Let's calculate the Voltage and Current for 800W at 2 ohms

V = sq rt (800 x 2) = 40V

I = sq rt (800/2) = 20A
​

Notice that even though the voltage was reduced from ~56.57V to 40V that the current still increased from ~14.14A to 20A

P = E x I

40 x 20 = 800W​

With out the switch, the amp would try to maintain the voltage at ~56.57V

I = ~56.57/2 = ~28.29A
P = ~56.57 x ~28.29 = ~1600W.
​

Unfortunately the amp won't work reliably at this power and current level, so it has to be derated.

Some amps reconfigure the power supply to lower the voltage in the power supply. Some amps adjust a limiter so the drive sent to the output section is kept within a safe range.

FYI, The switch is not new or specific to class-D. This image is from a 70's era Traynor PM-300 power amp:
View attachment 4969934

Yes, I got that; I assume the reason most solid state amps do not include an impedance switch is that, in the absence of robust protection circuits, setting the switch to the wrong position can potentially be catastrophic, whereas with tube amps a mismatched impedance will mostly result in lower performance and greater tear of easily replaceable parts (the output tubes). I also assume that these GK amps, for instance, have been thoroughly tested to ensure that an incorrect setting will engage the protection circuits and prevent damage, in order to protect against user error. I realise that damage could still be caused in the absence of such a switch, just by connecting the wrong combination of cabinets, but I guess that a sternly worded warning not to connect a load lower than 4 ohms is more effective in preventing damage compared to a switch that enables you to go lower than 4 ohms, which you can forget to set. Users who aren't scared will often do what they're not supposed to.

 

[agedhorse]

If keeping volumes low and/or using a limiter to keep levels low, would it fail regardless?

It depends. Maybe, maybe not.

 

[Auspuff]

In regards to questions, regarding one of my previous posts, here is the back of the Gallien-Krueger Legacy 800. Note the three-way Impedance Switch, on the right. Comes in real handy.