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Author Topic: Crossover, L-Pad, Impedence, Resistance question  (Read 10248 times)

Tim McCulloch

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Re: Crossover, L-Pad, Impedence, Resistance question
« Reply #60 on: November 22, 2017, 02:23:37 pm »

Very true. It's a reason I like these fora.

However, what I'm not too keen on is you saying that my post is "bullshit" without any presentation of evidence to the contrary.

So, lets try again.

Bridging an amplifier will double the voltage across the terminals. It does this by driving one side positive, and the other side negative. When that happens, each channel "sees" half the load impedance. Imagine a centre-tap on the driver and this becomes obvious.

So, the bridged power output of the amplifier is equal to the total power output from each channel if it were driving a load of half of the impedance of the load connected.

We can see this from a variety of spec sheets. The Crown MA5000VZ, for example, is as follows:

Stereo:
8ohm - 1300w/ch
4ohm - 2000w/ch
2ohm - 2500w/ch

Bridge mono:
16ohm - 2600w
8ohm - 4000w
4ohm - 5000w

We can see that there's a little voltage sag going from 8ohm to 4ohm (stereo), so the amp won't quite manage 4x the power going from 8ohm on one channel to 8ohm bridged. It will come pretty close, though.

I could go through a thousand spec sheets and consistently draw the same conclusion. So, Tim, which bit is bullshit?

Chris
The part you're not looking at correctly....

Flip this the other way - in 2 channel mode, where did the power disappear to?

Have we forgotten the law of conservation of energy?
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Chris Grimshaw

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Re: Crossover, L-Pad, Impedence, Resistance question
« Reply #61 on: November 22, 2017, 02:29:17 pm »

The part you're not looking at correctly....

Flip this the other way - in 2 channel mode, where did the power disappear to?

Have we forgotten the law of conservation of energy?

When using a 2-channel amplifier in stereo mode to drive a single load, one of the channels will be disconnected.

Conservation of energy works just fine. I happen to have a Physics degree from a decent university, so CoE and the like is still just about second-nature.

Chris
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Art Welter

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Re: Crossover, L-Pad, Impedence, Resistance question
« Reply #62 on: November 22, 2017, 03:13:29 pm »

So, anyhoo, regarding my latest drawing, Anybody know where I can get a 6-pole double throw for less than the $500 Mouser wants for one?

Also, still off topic, but still wondering which is better - two channels into two speakers, 8 ohms each, (from an XTi-6k this would be 1,200W each) or bridged into both paralleled, at 4 ohms (Resulting in a theoretical 6,000W from the amp)?  The Speakers are rated at 3.6KW program power each, and 1.8KW AES.  Power compression is not an issue...
Jeff,
1) Just wire your passive crossover in to a separate box and use separate high and low outputs. This will make it much easier to change components when you realize that you don't like the passive crossover ;^).
2) More headroom is "better", 3000 watts is a bit more than 3dB more than 1200 watts. With your bass guitar as a source, power compression should not be an issue unless you use a lot of compression in your signal chain.

To elaborate on Ivan's explanation of AES power, the difference in "real world" compared to the AES rating comes down to the impedance. An "8 ohm" speaker in a band limited open air test will be much higher impedance over most of the pass band than it's nominal rating, so only "draws" a fraction of it's rated power, as the voltage used would develop the rated power at 8 ohms. When in a bass reflex cabinet, at Fb, the impedance is usually right around the driver's DC resistance, consequently the driver can generally only handle about half the AES rating.

Art

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Stephen Swaffer

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Re: Crossover, L-Pad, Impedence, Resistance question
« Reply #63 on: November 22, 2017, 04:28:45 pm »

Very true. It's a reason I like these fora.

However, what I'm not too keen on is you saying that my post is "bullshit" without any presentation of evidence to the contrary.

So, lets try again.

Bridging an amplifier will double the voltage across the terminals. It does this by driving one side positive, and the other side negative. When that happens, each channel "sees" half the load impedance. Imagine a centre-tap on the driver and this becomes obvious.

So, the bridged power output of the amplifier is equal to the total power output from each channel if it were driving a load of half of the impedance of the load connected.

We can see this from a variety of spec sheets. The Crown MA5000VZ, for example, is as follows:

Stereo:
8ohm - 1300w/ch
4ohm - 2000w/ch
2ohm - 2500w/ch

Bridge mono:
16ohm - 2600w
8ohm - 4000w
4ohm - 5000w

We can see that there's a little voltage sag going from 8ohm to 4ohm (stereo), so the amp won't quite manage 4x the power going from 8ohm on one channel to 8ohm bridged. It will come pretty close, though.

I could go through a thousand spec sheets and consistently draw the same conclusion. So, Tim, which bit is bullshit?

Chris

Ohms law and power equation always apply-it's simply physics.  If you double the voltage with the same current, you double the power.  If it is a pure resistive load, doubling the voltage will cause the current to double, and in that case the power will be four times.  However, this assumes an unlimited power supply with zero internal impedance.

Power amps have an internal impedance-how well this internal impedance matches the load determines how much power is transferred.  Hence the need for spec sheets.  You could test everything and do the math yourself, but doesn't it make more sense to simply match your loads to the amps impedance for each scenario?

Intersting pro's and con's for each method-might be easier if there weren't so many "right ways" to do things!
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Steve Swaffer

Ivan Beaver

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Re: Crossover, L-Pad, Impedence, Resistance question
« Reply #64 on: November 22, 2017, 05:12:32 pm »

Ohms law and power equation always apply-it's simply physics.  If you double the voltage with the same current, you double the power.  If it is a pure resistive load, doubling the voltage will cause the current to double, and in that case the power will be four times.  However, this assumes an unlimited power supply with zero internal impedance.

Power amps have an internal impedance-how well this internal impedance matches the load determines how much power is transferred.  Hence the need for spec sheets.  You could test everything and do the math yourself, but doesn't it make more sense to simply match your loads to the amps impedance for each scenario?

Intersting pro's and con's for each method-might be easier if there weren't so many "right ways" to do things!
There have been a couple of "perfect" amplifiers. 

By that, I mean one that doubles the power when the load is halved.

And then there are others that can drive any load, but the power is not what you would expect.

The Peavey CS800 is a good example.  It can drive any load-in fact the test after a repair is to drive a dead short into full clip until it thermals out.

The specs are 8 ohms 300 watts, 4 ohms 400 watts, 2 ohms 250 watts.

So once you get below 4 ohms, the power capacity goes down, but it will keep on working.

Even some of the modern big amps drop their power at 2 ohms.  Some are optimum at 2.67 ohms (3x8 ohm loads)

Reading and understanding specs IS an important aspect of understanding performance
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Jeff Schoonover1

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Re: Crossover, L-Pad, Impedence, Resistance question
« Reply #65 on: November 22, 2017, 09:04:22 pm »

1) Just wire your passive crossover in to a separate box and use separate high and low outputs. This will make it much easier to change components when you realize that you don't like the passive crossover ;^).
I like this idea.  Simple, except there's another thing to remember/carry.  But not like it's huge.
Why don't you simply use the 4 pole drawing presented earlier?  It will work perfectly fine, be less complicated to wire, a cheaper switch etc.
I thought we had decided that drawing won't work, as it leaves the positives connected to the X-Over outputs.  The last drawing I put up has two switches to lift those.  It's really just a 4 pole and a 2 pole switch.  I just need to be sure to throw both up, or down.  What I don't like is having two inputs.  More likely it insert the wrong cable or have the amp set incorrectly...
2) More headroom is "better", 3000 watts is a bit more than 3dB more than 1200 watts. With your bass guitar as a source, power compression should not be an issue unless you use a lot of compression in your signal chain.
More headroom = better was always my thinking too.  But reading the other guys reasons for using Bi-Amp, made a lot of sense too and got me thinking.  there's damping factor, and THD to think of too.  Also, it's easier on the amplifier, right?  I won't ever be running my low box at near its full-out capabilities, and you're right with bass guitar also - the continuous load will fluctuate considerably.  It's mostly to get low - cleanly.  I suppose the only way to really know is to try both.  I feel like 3dB is trivial when compared to less distortion, better damping, and easier loading.  Hmm...
To elaborate on Ivan's explanation of AES power, the difference in "real world" compared to the AES rating comes down to the impedance. An "8 ohm" speaker in a band limited open air test will be much higher impedance over most of the pass band than it's nominal rating, so only "draws" a fraction of it's rated power, as the voltage used would develop the rated power at 8 ohms. When in a bass reflex cabinet, at Fb, the impedance is usually right around the driver's DC resistance, consequently the driver can generally only handle about half the AES rating.
  Interesting.  So, would this also be more support for Bi-Amping?
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Ivan Beaver

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Re: Crossover, L-Pad, Impedence, Resistance question
« Reply #66 on: November 22, 2017, 09:24:39 pm »

  But not like it's huge. I thought we had decided that drawing won't work, as it leaves the positives connected to the X-Over outputs.
PLEASE look at the drawing in question again.  I don't know where "we" decided that it would not work-because the Hots are still connected.

Connected to what?  Nothing that I can see.  I still say it will work just fine. If it won't PLEASE show me where or why it won't work.

The crossover is completely disconnected from the drivers and there is no path for any signal on the common.  So therefore no path for any signal.

That is the same thing I have said on SEVERAL posts in this thread.  I still stand by my comments.

BTW, you have quotes associated with the wrong people who said them.
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Jeff Schoonover1

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Re: Crossover, L-Pad, Impedence, Resistance question
« Reply #67 on: November 22, 2017, 11:35:21 pm »

PLEASE look at the drawing in question again.  I don't know where "we" decided that it would not work-because the Hots are still connected.  Connected to what?  Nothing that I can see.  I still say it will work just fine. If it won't PLEASE show me where or why it won't work.  The crossover is completely disconnected from the drivers and there is no path for any signal on the common.  So therefore no path for any signal.  That is the same thing I have said on SEVERAL posts in this thread.  I still stand by my comments.
Let's figure it out... Let's start with my original drawing in post 24.  Can you see that the positive of the woofer is always connected to the positive low output of the crossover?  The same is true with the positive on the tweeter.  It is always connected to the positive of the high output on the x-over.
BTW, you have quotes associated with the wrong people who said them.
Apologies to all concerned.
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Paul G. OBrien

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Re: Crossover, L-Pad, Impedence, Resistance question
« Reply #68 on: November 22, 2017, 11:52:53 pm »

Let's figure it out... Let's start with my original drawing in post 24.  Can you see that the positive of the woofer is always connected to the positive low output of the crossover?
The same is true with the positive on the tweeter.  It is always connected to the positive of the high output on the x-over.

No they aren't. The diagram represents fullrange operation using the crossover, when the switches are thrown to bi-amp mode the only thing left connected to the crossover is the woofer common(-) which is fine because electrically it just goes straight through anyway.
« Last Edit: November 22, 2017, 11:55:04 pm by Paul G. OBrien »
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Jeff Schoonover1

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Re: Crossover, L-Pad, Impedence, Resistance question
« Reply #69 on: November 23, 2017, 12:03:48 am »

The diagram represents fullrange operation using the crossover
No, as pictured, the input to the crossover is switched out.  It shown in Biamp mode. 

The problem is that even though the x-over input is switched out, the positive poles of the drivers remain connected to the positive poles on the output of the x-over.
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Re: Crossover, L-Pad, Impedence, Resistance question
« Reply #69 on: November 23, 2017, 12:03:48 am »


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