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Old 27th December 2012, 10:06 AM   #21
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Hi Tony....crossed again during an edit! Think we might be on the same page now.
Ta, J
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Old 27th December 2012, 10:18 AM   #22
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I have a suspicion that all that happens is that the amp delivers more power, but the same amount still goes through the tweeter Just more goes off to earth through the shunt... The impedance is constant but nothing changes other than the amp delivering more power (the extra going through the shunt) at that frequency.

I Just did a sim and absolutely no difference between the LCR and No LCR as far as spl is concerned.

I have a feeling I've been through this before

Tony.
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Old 27th December 2012, 10:58 AM   #23
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Sorry guys. What determines the excursion is the voltage across the terminals of the tweeter. PERIOD. The only function of the LRC is to flatten the driver's Z. The only way that affects the voltage across the tweeter's terminals is by the way it affects the crossover's transfer function. If the same voltage transfer function can be obtained with and without the LRC (by changing the crossover component values) then the excursion is the same with and without the LRC. Low impedance in the return path yielding greater damping is not accurate. As I stated before, if the same acoustic output is obtained (which means the same voltage applied across the tweeter terminals) everything is the same, excursion, damping, everything.

Remember, this is a minimum phase response of a linear, time invariant system (for the most part). It's frequency and impulse response are related by the FFT. So if you have the same frequency response you have the same impulse response and two linear systems with the same impulse response have the same transient response and damping.

The only unresolved question is whether or not the exact same transfer function can be obtained with and without the LRC. In the passive case it is possible that you may not be able to make them exact. But if you go with an active crossover it should be apparent that the voltage across the tweeter's terminals is not a function of the LRC.

So here we are again. The LRCs effect on excursion is only a result of how it affects the Z load seen by the rest of a passive crossover, and how that difference in Z affects the crossover transfer function. But the LRC has no direct effect on excursion by itself.

I think part of the confusion is due to the common belief that a resistance in series with a driver increases the driver's Q. That is not correct. The driver's Q does not change. What changes is the system Q. The system is the driver + series resistance. This idea leads to the belief that a high return path impedance leads to poor damping of a tweeter (or other driver). That is not correct either because, again, you have to look at the complete system. Certainly it would appear that a high return path Z would reduce the electrical damping of the tweeter. Taken alone, that is true. But what is ignored by that simple observation is that there is energy stored in the crossover that must also be dissipated. Some of that energy is dumped in the amplifier's source resistance. Some is dumped in the driver voice coil. The result is that the damping of the system is that determined by the transfer function of the voltage across the driver's terminals.

Let's look at an active system driven by a true voltage source, for example. With the driver connected directly to the amp there is no argument about what the return path Z is. But now, let's place a high Q boost circuit at the input of the amplifier. Certainly nothing has changed between the amp and driver as they are still directly connected. But now the response of the driver to an impulse rings like a bell at the center frequency of the Q boost. The driver's damping hasn't changed here, but the damping of the system composed of the Q boost + amp + driver has changed drastically at the Q boost center frequency. Add the LRC and nothing changes other than the Z load seen by the amplifier. The point is that you can not look only at the driver; it is the system response that matters and the system response is observed by looking at the output.

It's the same with the LRC in a passive crossover. It is part of a system. It either has an effect on the system response of it doesn't. In cases where it has an effect that effect willbe observed in the output and may alter excursion. In cases where it has no effect (like with the active crossover), there is no change to observe in the system output and there is no chnage in driver excursion. The LRC alone has no effect on excursion.
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Old 27th December 2012, 11:28 AM   #24
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John I don't know how to lift and "quote" sections of previous posts as I've seen some people do so can I just go back to your 5th paragraph beginning; "I think....etc" where you discuss the stored energy from the x-over and the paths it takes. Granted as far as you go but you omit the ONE element under discussion i.e. the LCR shunt. If that is a path then surely it increases power dissipation as Vifa suggest.

I just feel you are introducing too many distractions and putting up "straw men" to knock down. No one has raised active systems (which I am fully conversant with) and no one is suggested we look at frequency response with 1st order networks etc or series resistance on driver Q.

I simply sought an explanation as to why Vifa ask that the shunt network they propose be inserted across their tweeter "To ensure maximum electrical damping and consequently minimum excursion" and as they go on to say; "This is essential for high power handling." Their words....

Why does this network achieve these goals?

Cheers, Jonathan.
Btw turning in soon so will not see any further post for about 8 hrs......
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Old 27th December 2012, 11:47 AM   #25
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Hi Tony. Bits are falling into place. Re; you simulation in #22. I once read an explanation that showed that speaker SPL was voltage dependent and so your work is right. The voltage across the tweeter terminals (above the x-over frequency) will not alter but current drawn down (or shunted) by the LCR network may well be relevant (at resonance that is)......

Jonathan
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Old 27th December 2012, 11:55 AM   #26
sreten is offline sreten  United Kingdom
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Hmmm........

There seems little point saying the LCR does not reduce excursion.

The datasheet is from the days when its never going to explain
the effects of different filters and the resulting acoustic response.

Two things are clear : 1) if the tweeter is driven by a low impedance
source the LCR has no effect, 2) if its driven by a high impedance
source the LCR reduces the excursion and peaking around Fs due
to the drivers impedance peak, the impedance is flattened.

Quote:
"[a] The load provided by the cross over should be as low as possible at the
tweeter resonance frequency. [b] To ensure maximum electrical damping and
consequently minimum excursion, it is also recommended to apply a parallel
compensation circuit. [c] This is essential for high power output."
Is the sort of confusing statement many engineers make, and its poor.
(Specifically for ideal [a], active drive, [b] is pointless, [c] wrong.)

[b]Only follows [a] if you've not managed to create low impedance drive,
[b] doesn't automatically follow [a], and [c] is simply not always true.

You can say [c] is true for [b] if [a] is not true, for a typical x/o point,
and [a] is unlikely to be true for a typical design with tweeter L-pads.

JohnK is of course correct in that its the target acoustic response that
determines the tweeter excursion, and nothing else. The datasheet is
written for someone who might think they can just stick any electrical
x/o of their choosing in front of the tweeter, it is not that simple.

rgds, sreten.
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Last edited by sreten; 27th December 2012 at 11:58 AM.
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Old 27th December 2012, 12:15 PM   #27
DrNick is offline DrNick  United Kingdom
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If I was at home I'd sim the driver excursion and post the graphs, but I just have my phone on me.

The excursion of the driver goes up as frequency goes down, for the same power. If you have a large hump in the frequency response at the high end, it will not limit the power handling but a hump at the low end if the working range will mean the driver will reach Xmax at a lower power.

If you measure the power handling of a speaker using a signal comprising a spread of frequencies, then removing the hump at the lower end of the tweeter response will allow higher power before the tweeter reaches Xmax. This is very similar to saying that if you cross higher you can get higher power handling.
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Old 27th December 2012, 01:57 PM   #28
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Well it seems pointless to argue the point further except to say that like all the other components in a passive crossover, when an LRC shunt is present it is nothing more than additional elements in the crossover. Like any passive crossover those LRC elements serve to shape the voltage transfer function across the tweeter. That is all that matters. And like any other passive crossover, you may or may not need specific elements. The LRC does noting but alter the impedance seen by the upstream components. It does so by storing and releasing energy. It does not reduce excursion simple by being there. The change in impedance alters the transfer function the upstream components generate. That alters excursion (and radiated SPL) where ever the transfer function differers.

The reason I brought up active circuits was to make the point that when driven by a voltage source the presence of an LRC shunt has ZERO effect on the driver's behavior.

Let me put it another way. Suppose you have a driver connected to a passive crossover. The passive crossover is potted in resin so there is no way to tell what is inside. You then measure the voltage transfer function across the driver terminals. Is there anything you don't know about the drivers response at that point? Does it make any difference if there is an LRC shunt in that potted x-o? If I reverse engineer the crossover and come up with a circuit which, when connected to the driver, yields the same transfer function will the diver not behave in the exact same way? Does it make any difference is that circuit is different than the potted circuit? There are no straw men to knock down in my arguments. There is only physics.

Certainly the LRC shunt creates another current path, but the L and C are also energy storage elements which have there own effect.

I'll say it one more time. You can place any passive circuit you want between an amplifier and a driver. The result will be some transfer function across the driver's terminals. It makes no difference what that circuit is. If another circuit yields the same transfer function the driver will be have exactly the same. The only question left to answer is if the same transfer function can be achieved with and without an LRC present.
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Old 27th December 2012, 09:18 PM   #29
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John I admit to still being a little confused but may I ask you to further clarify the portion of post #23

("This idea leads to the belief that a high return path impedance leads to poor damping of a tweeter (or other driver). That is not correct either because, again, you have to look at the")

Do I take this as an answer to my question in post#14 about the DCR of the shunt coil??

By the way thanx for your patience in trying to make the explanations understandable by those of us who do not have an electro-mechanical background
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Old 27th December 2012, 11:11 PM   #30
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I really don't know how to address it further, but I'll try. You have an amplifier connected to a driver through some passive network. That is a system. The voltage output of that system can be measured at the terminals of the driver. That defines the transfer function. Since it's fundamentally a linear, time invariant system, that transfer function defines the system completely, both in the frequency and time domain. So it defines the impulse response as well. It defines what the system damping is as well. So, if there are two different systems that produce the same impulse response then they must have the same damping. That's just the physics of linear systems. The details of how that damping comes about are irrelevant.

Now, if you have system A without an LRC and just add the LRC to create system B, system B will not behave the same. It may have more or less damping as will be reflected in its impulse response. It may be possible to change the values of other components of the system to make system C which then behaves identically to system A. Then system A and C will have the same impulse and damping, etc,\.

Perhaps a simpler example might help. Suppose we have a 2nd order HP network connected to a purely resistive 8 ohm load. It would be a series C with a shunt L. It has some specific impulse. Now, change the resistance to 4 ohms. The system behaves differently. But double the values of C and half the value of L and now the system behaves exactly as the 8 ohm system. This is in spite of the fact that the dissipative part of the circuit is less, because the energy storage elements (L and C) have been adjusted to compensate.

Adding an LRC shunt across the driver is similar to changing the load from 8 to 4 ohms except that it only alters the impedance around the driver peak. It changes the dissipative character of the system, but if the other elements can be adjusted so that the transfer function across the driver remains the same then no change in damping occurs.

Let me give you another example of some of the myths that circulate around the net. You may have read that 2nd order electrical high pass networks are superior because the shunt inductor creates a low impedance path across the driver and helps damping. The problem is that if you actually look at the impedance the driver sees when that network is connected to an amplifier with Rs output impedance what you find is that at high frequency the driver see Rs, at low frequency the driver sees the DC R of the inductor, and in between there is an LC resonance peak in the impedance which approaches infinity as the DC R of the inductor goes to zero. This LC resonance peak occurs at the corner frequency of the filter. So the idea that the L shunt element yields a low impedance path to ground and therefore improves damping is obviously incorrect. So what is the damping of the system? It can be determined by looking at the impulse response or transfer function measured at the terminals of the driver.
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